🧠:[<think>Okay, so I need to find the area of this convex quadrilateral ABCD. The given information is that AB is 3 units, BC is 4 units, CD is 13 units, AD is 12 units, and angle ABC is 90 degrees. The answer choices are A through E, with the options 30, 36, 40, 48, and 58.5. Hmm. Let's see how to approach this.First, since angle ABC is 90 degrees, triangle ABC must be a right-angled triangle. That might help me find some lengths or areas. Let me visualize the quadrilateral. Points A, B, C, D with AB=3, BC=4, CD=13, AD=12. ABC is a right angle. So, maybe I can split the quadrilateral into two triangles: ABC and ACD, and then find their areas separately?Wait, but if I split it into ABC and ACD, I need to know the lengths of AC and maybe some angles or heights. Alternatively, maybe split it into ABC and ABD? Hmm, not sure. Let me think step by step.First, let's calculate the area of triangle ABC. Since it's a right-angled triangle at B, the area is (AB * BC)/2 = (3 * 4)/2 = 12/2 = 6. So the area of triangle ABC is 6. Then, the rest of the quadrilateral must be the area of triangle ACD or something else. Wait, but to find the area of triangle ACD, I need to know the length of AC, which is the diagonal from A to C. Since ABC is a right triangle, AC can be found using Pythagoras.So, AC = sqrt(AB² + BC²) = sqrt(3² + 4²) = sqrt(9 + 16) = sqrt(25) = 5. Okay, so AC is 5 units. Now, in triangle ACD, we have sides AD=12, CD=13, and AC=5. Wait, 5, 12, 13—that's a Pythagorean triple! So triangle ACD must also be a right-angled triangle. Let me check: 5² + 12² = 25 + 144 = 169 = 13². Yes, exactly. So triangle ACD is right-angled as well. Then, the area of triangle ACD is (AD * AC)/2. Wait, but where is the right angle here?If AC=5, AD=12, and CD=13, then since 5² + 12² = 13², the right angle must be at A. Because AD and AC are the legs. So angle at A is 90 degrees. Therefore, the area of triangle ACD is (AC * AD)/2 = (5 * 12)/2 = 60/2 = 30. So the area of triangle ACD is 30.Therefore, the total area of quadrilateral ABCD is the sum of the areas of ABC and ACD, which is 6 + 30 = 36. Wait, 36 is option B. But let me check again to make sure I didn't make a mistake.Wait, hold on. If triangle ACD is right-angled at A, then points A, C, D would form a right angle at A. But in the original quadrilateral ABCD, angle ABC is 90 degrees. I need to make sure that the points are connected properly. Let's confirm the structure.Quadrilateral ABCD: points A, B, C, D connected in order. AB=3, BC=4, CD=13, DA=12. ABC is right-angled. So, from point A to B to C, then to D, then back to A. So, AC is the diagonal from A to C. Then, in triangle ACD, sides are AC=5, CD=13, DA=12. That forms a 5-12-13 triangle, right-angled at A. Therefore, angle at A is 90 degrees. So, the area is 30, and combined with the first triangle's area of 6, total is 36. So answer B.Wait, but let me make sure that the quadrilateral is indeed convex and that the two right angles (at B and at A) don't interfere with each other or cause some overlap. If angle at A is 90 degrees, then when we connect D to A, that should form the right angle. Let me try to sketch this mentally.Start at point A. AB is 3 units, going to point B. From B, BC is 4 units at a right angle (so perpendicular to AB) to point C. Then CD is 13 units from C to D. Then DA is 12 units back to A. If AC is 5 units, and in triangle ACD, angle at A is 90 degrees, then DA is perpendicular to AC. Since AC is the diagonal from A to C, which is part of the original quadrilateral.So, if AC is 5 units, and DA is 12 units, then DA is perpendicular to AC, making a right angle at A. Therefore, the quadrilateral ABCD has two right angles: one at B and one at A. But in the problem statement, they only mention angle ABC is 90 degrees, so angle at B. The right angle at A is a result of the triangle ACD being right-angled.Since the problem says it's a convex quadrilateral, this configuration should hold. So combining the two areas gives 6 + 30 = 36. Therefore, the answer is B) 36. Let me check the options again: (A) 30, (B) 36, (C) 40, (D) 48, (E) 58.5. Yes, 36 is option B. Hmm, seems right.But wait, another way to calculate the area of a quadrilateral with two diagonals is using Bretschneider's formula, but that requires knowing the lengths of all sides and the sum of two opposite angles or something. But in this case, maybe it's easier to split into two triangles. Alternatively, since we have two right triangles, the areas can just be added.Wait, let me just verify once again. Triangle ABC: right-angled at B, legs 3 and 4, hypotenuse 5. Area 6. Triangle ACD: right-angled at A, legs 5 and 12, hypotenuse 13. Wait, hold on. If AC is 5, and AD is 12, then the legs are AC=5 and AD=12, so the area would be (5*12)/2=30. Then total area is 6 + 30 = 36. That seems correct.Alternatively, someone might think that triangle ACD is right-angled at D, but 5² + 13² is 25 + 169 = 194, which is not equal to 12²=144, so no. Similarly, 12² +13²= 144 +169=313, which is not 5²=25. So the only right angle in triangle ACD is at A, which matches the 5-12-13 triangle. So that seems correct.Therefore, I think the correct answer is B) 36.Final Answerboxed{B}</think>To find the area of the convex quadrilateral (ABCD) with given sides (AB = 3), (BC = 4), (CD = 13), (AD = 12), and (angle ABC = 90^circ):1. Calculate the area of triangle (ABC): - Since (angle ABC = 90^circ), triangle (ABC) is a right-angled triangle. - Area of (ABC) is (frac{1}{2} times AB times BC = frac{1}{2} times 3 times 4 = 6).2. Find the length of diagonal (AC): - Using the Pythagorean theorem in triangle (ABC): [ AC = sqrt{AB^2 + BC^2} = sqrt{3^2 + 4^2} = sqrt{9 + 16} = sqrt{25} = 5 ]3. Calculate the area of triangle (ACD): - The sides of triangle (ACD) are (AC = 5), (AD = 12), and (CD = 13). This is a Pythagorean triple (5, 12, 13), indicating a right-angled triangle at (A). - Area of (ACD) is (frac{1}{2} times AC times AD = frac{1}{2} times 5 times 12 = 30).4. Sum the areas of triangles (ABC) and (ACD): - Total area of quadrilateral (ABCD) is (6 + 30 = 36).Thus, the area (S) of quadrilateral (ABCD) is (boxed{B}).,<think>Okay, so I need to find the area of this convex quadrilateral ABCD. The given sides are AB=3, BC=4, CD=13, AD=12, and angle ABC is 90 degrees. Hmm, let me visualize this. Since ABC is a right angle, triangle ABC must be a right-angled triangle with legs AB=3 and BC=4. Then CD is 13, and AD is 12. The quadrilateral is convex, so all the vertices are pointing outwards. First, maybe I should draw a sketch to get a better idea. Let me place point B at the origin (0,0) to make calculations easier. Then, since AB=3 and angle ABC is 90 degrees, point A can be at (0,3) and point C at (4,0). Wait, no. If angle ABC is 90 degrees, then AB and BC are perpendicular. So if B is at (0,0), then A can be along the y-axis and C along the x-axis. Let me confirm that. If AB=3, then from B(0,0), moving up 3 units along the y-axis gives A(0,3). Then BC=4, so moving 4 units along the x-axis from B(0,0) gives C(4,0). That makes triangle ABC a right-angled triangle with legs 3 and 4. Then, the hypotenuse AC would be 5 units, by Pythagoras: sqrt(3^2 + 4^2) = 5. Now, the quadrilateral is ABCD, so we have point D connected to C and A. The sides CD=13 and AD=12. So point D is connected to C(4,0) and A(0,3), and the lengths CD=13 and AD=12. So, we need to find the coordinates of point D such that the distance from D to C is 13 and the distance from D to A is 12. Then, once we have the coordinates of D, we can compute the area of the quadrilateral ABCD by dividing it into two triangles: ABC and ACD, or ABC and ABD, or maybe even using the coordinates to calculate the area directly via the shoelace formula. Alternatively, maybe we can use the coordinates to find the area. Let's see. Let me try to find the coordinates of D. Let me denote D as (x,y). Then, the distance from D to C(4,0) is sqrt((x-4)^2 + (y-0)^2) = 13, so (x-4)^2 + y^2 = 169. Similarly, the distance from D to A(0,3) is sqrt((x-0)^2 + (y-3)^2) = 12, so x^2 + (y-3)^2 = 144. So we have two equations:1. (x - 4)^2 + y^2 = 1692. x^2 + (y - 3)^2 = 144We need to solve these two equations to find x and y. Let's expand both equations.First equation: (x^2 - 8x + 16) + y^2 = 169 ⇒ x^2 + y^2 - 8x + 16 = 169 ⇒ x^2 + y^2 - 8x = 153.Second equation: x^2 + (y^2 - 6y + 9) = 144 ⇒ x^2 + y^2 - 6y + 9 = 144 ⇒ x^2 + y^2 - 6y = 135.Now, subtract the second equation from the first equation:(x^2 + y^2 - 8x) - (x^2 + y^2 - 6y) = 153 - 135 ⇒ -8x + 6y = 18 ⇒ -8x + 6y = 18. Divide both sides by 2: -4x + 3y = 9 ⇒ 3y = 4x + 9 ⇒ y = (4x + 9)/3.Now, substitute y = (4x + 9)/3 into one of the original equations. Let's take the second equation:x^2 + (y - 3)^2 = 144.Substitute y:x^2 + [(4x + 9)/3 - 3]^2 = 144.Simplify inside the square:(4x + 9)/3 - 9/3 = (4x + 9 - 9)/3 = 4x/3.So, x^2 + (4x/3)^2 = 144 ⇒ x^2 + (16x^2)/9 = 144.Multiply through by 9 to eliminate denominators:9x^2 + 16x^2 = 1296 ⇒ 25x^2 = 1296 ⇒ x^2 = 1296/25 ⇒ x = ±√(1296/25) = ±36/5 = ±7.2.So, x = 36/5 or x = -36/5. Let's check both possibilities.First, x = 36/5 = 7.2. Then y = (4*(36/5) + 9)/3 = (144/5 + 45/5)/3 = (189/5)/3 = 189/15 = 63/5 = 12.6.Second, x = -36/5 = -7.2. Then y = (4*(-36/5) + 9)/3 = (-144/5 + 45/5)/3 = (-99/5)/3 = -99/15 = -33/5 = -6.6.Now, we need to determine which of these two points is valid in the convex quadrilateral. Since the quadrilateral is convex, point D must lie in a position such that the quadrilateral does not intersect itself. Let's consider the coordinates.Original points: A(0,3), B(0,0), C(4,0). If D is (7.2, 12.6), then plotting this point would be in the upper right relative to C. If D is (-7.2, -6.6), it would be in the lower left relative to B. Since ABCD is convex, the order of the vertices is important. Let's check the order. If the quadrilateral is ABCD, then the vertices are connected in order A-B-C-D-A. So after C(4,0), we go to D and then back to A(0,3). If D is (7.2, 12.6), connecting C(4,0) to D(7.2,12.6) to A(0,3) would form a convex shape. However, if D is (-7.2, -6.6), connecting C(4,0) to D(-7.2, -6.6) would go to the left and down, and then connecting back to A(0,3) might cause the quadrilateral to be concave or intersect. Let's check.First, let's check D(7.2,12.6). The quadrilateral would have points A(0,3), B(0,0), C(4,0), D(7.2,12.6). Connecting these in order, this seems convex as all internal angles are less than 180 degrees. For D(-7.2, -6.6), the points would be A(0,3), B(0,0), C(4,0), D(-7.2, -6.6). Connecting C(4,0) to D(-7.2, -6.6) would create a diagonal crossing through the origin, possibly making the quadrilateral concave when connecting back to A(0,3). But actually, need to check the order. Wait, in a convex quadrilateral, all vertices must be arranged such that the interior angles are less than 180 degrees. Let's see.Alternatively, maybe the position of D can be determined by considering that in a convex quadrilateral, the points must be arranged in order without crossing. If D is at (7.2,12.6), which is above and to the right of C(4,0), then moving from C to D to A would create a convex shape. If D is at (-7.2, -6.6), then moving from C(4,0) to D(-7.2, -6.6) would be going left and down, and then from D(-7.2, -6.6) to A(0,3) would be going up and right. However, depending on the positions, it might still be convex. Let's calculate the coordinates and check the angles or use the shoelace formula to check the area for both possibilities.But perhaps the problem states that ABCD is convex, so both possibilities might give different areas. Wait, but since CD=13 and AD=12, the coordinates must satisfy those distances. So both solutions are valid, but only one gives a convex quadrilateral. Let's check which one is convex.Alternatively, perhaps the question assumes a particular orientation, but the problem statement doesn't specify. Wait, the problem says "as shown in Figure 1", but since we don't have the figure, we need to rely on the description. Since it's a convex quadrilateral, and ABC is a right angle, and sides AB=3, BC=4, CD=13, AD=12. Let me think. If D is at (7.2,12.6), then the length AD is from A(0,3) to D(7.2,12.6). Let's compute this distance to verify:sqrt((7.2 - 0)^2 + (12.6 - 3)^2) = sqrt(7.2^2 + 9.6^2) = sqrt(51.84 + 92.16) = sqrt(144) = 12. That's correct. Similarly, for D(-7.2, -6.6):sqrt((-7.2 - 0)^2 + (-6.6 - 3)^2) = sqrt(51.84 + (-9.6)^2) = sqrt(51.84 + 92.16) = sqrt(144) = 12. So both points satisfy the distance conditions. Therefore, both are possible, but which one gives a convex quadrilateral?If we take D(7.2,12.6), then the quadrilateral is A(0,3)-B(0,0)-C(4,0)-D(7.2,12.6). Let's check the order. The order is such that when moving around the quadrilateral, the turns are all in the same direction (either all clockwise or counter-clockwise). Let's compute the coordinates:A: (0,3)B: (0,0)C: (4,0)D: (7.2,12.6)Connecting A to B to C to D to A.To check convexity, we can compute the cross products of consecutive edges to ensure all are either positive or negative (indicating all turns are in the same direction). Let's check.First, edge AB: from (0,3) to (0,0). Direction vector: (0, -3).Edge BC: from (0,0) to (4,0). Direction vector: (4, 0).The cross product of AB and BC is (0, -3) × (4, 0) = 0*0 - (-3)*4 = 12 > 0.Edge BC to CD: direction vector BC is (4,0), next edge CD is from C(4,0) to D(7.2,12.6): direction vector (3.2, 12.6). Cross product (4,0) × (3.2,12.6) = 4*12.6 - 0*3.2 = 50.4 > 0.Edge CD to DA: direction vector CD is (3.2,12.6), edge DA is from D(7.2,12.6) to A(0,3): direction vector (-7.2, -9.6). Cross product (3.2,12.6) × (-7.2, -9.6) = 3.2*(-9.6) - 12.6*(-7.2) = -30.72 + 90.72 = 60 > 0.Edge DA to AB: direction vector DA is (-7.2, -9.6), edge AB is (0, -3). Cross product (-7.2, -9.6) × (0, -3) = (-7.2)*(-3) - (-9.6)*0 = 21.6 > 0.All cross products are positive, so all turns are counter-clockwise, which means the quadrilateral is convex. Now, check the other point D(-7.2, -6.6):Points A(0,3), B(0,0), C(4,0), D(-7.2, -6.6). Let's compute cross products.Edge AB: (0, -3) as before.Edge BC: (4,0).Cross product AB × BC = 12 >0.Edge BC to CD: from C(4,0) to D(-7.2, -6.6): direction vector (-11.2, -6.6). Cross product BC(4,0) × CD(-11.2, -6.6) = 4*(-6.6) - 0*(-11.2) = -26.4 < 0.Since this cross product is negative, the turn from BC to CD is clockwise, whereas previous turns were counter-clockwise. Therefore, this would make the quadrilateral concave. Hence, this D(-7.2, -6.6) is not part of a convex quadrilateral. Therefore, the correct D is (7.2,12.6).Now, with coordinates of all points: A(0,3), B(0,0), C(4,0), D(7.2,12.6). To find the area of quadrilateral ABCD, we can use the shoelace formula.Shoelace formula: For coordinates (x1,y1), (x2,y2), ..., (xn,yn), the area is 1/2 |sum_{i=1 to n} (xi*yi+1 - xi+1*yi)|, where xn+1=x1, yn+1=y1.Applying this:List the coordinates in order: A(0,3), B(0,0), C(4,0), D(7.2,12.6), back to A(0,3).Compute the sum:x_A*y_B - x_B*y_A = 0*0 - 0*3 = 0x_B*y_C - x_C*y_B = 0*0 - 4*0 = 0x_C*y_D - x_D*y_C = 4*12.6 - 7.2*0 = 50.4 - 0 = 50.4x_D*y_A - x_A*y_D = 7.2*3 - 0*12.6 = 21.6 - 0 = 21.6Sum these up: 0 + 0 + 50.4 + 21.6 = 72Take absolute value and divide by 2: |72| /2 = 36. So the area is 36. But wait, the options are (A)30, (B)36, (C)40, (D)48, (E)58.5. So 36 is option B. But let me cross-verify this because sometimes using coordinates can be error-prone. Alternatively, maybe compute the area as the sum of the areas of triangles ABC and ACD.We know ABC is a right-angled triangle with legs 3 and 4, so its area is (3*4)/2 = 6.Now, triangle ACD: points A(0,3), C(4,0), D(7.2,12.6). Let's compute its area.Using coordinates, we can use the shoelace formula for triangle ACD:Coordinates: A(0,3), C(4,0), D(7.2,12.6), back to A(0,3).Compute sum:0*0 - 4*3 = 0 -12 = -124*12.6 -7.2*0 = 50.4 -0 =50.47.2*3 -0*12.6=21.6 -0=21.6Sum: -12 +50.4 +21.6=60Area: |60|/2=30.Therefore, the total area of quadrilateral ABCD is area ABC + area ACD =6 +30=36. So same answer as before, 36. Option B.Alternatively, is there another way to compute the area without coordinates? Maybe using the Pythagorean theorem and Heron's formula.We know ABC is a right triangle with area 6. Then, for triangle ACD, sides AC, CD, and AD. We know AC is the hypotenuse of ABC, which is 5. So AC=5, CD=13, AD=12. So triangle ACD has sides 5,12,13. Wait, 5-12-13 is a Pythagorean triple! 5^2 +12^2=25 +144=169=13^2. Therefore, triangle ACD is a right-angled triangle with legs 5 and12, so area is (5*12)/2=30. Then total area=6 +30=36. So that's a much simpler way!Wait, that's brilliant. Since AC is 5, and then in triangle ACD, sides 5,12,13, which is a right triangle. Therefore, angle at A is right angle? Wait, in triangle ACD, the sides are 5,12,13. So the right angle is at A because 5^2 +12^2=13^2. Wait, but in our coordinate system, point A is (0,3), point C is (4,0), and point D is (7.2,12.6). Let's check if angle at A is right angle.Vector from A to C: (4, -3). Vector from A to D: (7.2, 9.6). The dot product of these vectors is (4)(7.2) + (-3)(9.6) = 28.8 -28.8=0. Yes! Therefore, vectors AC and AD are perpendicular, so angle at A is 90 degrees. Therefore, triangle ACD is right-angled at A. Therefore, area is (5*12)/2=30. Then total area is 6+30=36. So answer is B)36.This is a much simpler method. So when I first started, I went through coordinates, but recognizing that ACD is a 5-12-13 triangle would have been quicker. But since the problem gives AD=12, CD=13, and AC=5 (from ABC), triangle ACD is right-angled, so area is straightforward.Therefore, the correct answer is 36, option B.Final Answerboxed{B}</think>To find the area of the convex quadrilateral (ABCD) with given sides (AB = 3), (BC = 4), (CD = 13), (AD = 12), and (angle ABC = 90^circ):1. Identify the right triangle (ABC): - Since (angle ABC = 90^circ), triangle (ABC) is a right triangle with legs (AB = 3) and (BC = 4). - The hypotenuse (AC) is calculated as (sqrt{3^2 + 4^2} = 5).2. Check triangle (ACD): - The sides of triangle (ACD) are (AC = 5), (CD = 13), and (AD = 12). - Recognizing that (5^2 + 12^2 = 13^2), triangle (ACD) is a right triangle with legs (AC = 5) and (AD = 12).3. Calculate the areas: - The area of triangle (ABC) is (frac{1}{2} times 3 times 4 = 6). - The area of triangle (ACD) is (frac{1}{2} times 5 times 12 = 30).4. Sum the areas: - The total area of quadrilateral (ABCD) is (6 + 30 = 36).Thus, the area (S) of quadrilateral (ABCD) is (boxed{B}).,<think>Alright, so I need to find the area of this convex quadrilateral ABCD. The given information is that AB is 3, BC is 4, CD is 13, AD is 12, and angle ABC is 90 degrees. The options are A) 30, B) 36, C) 40, D) 48, E) 58.5. Hmm. Let me visualize the quadrilateral first. Since ABC is a right angle, triangle ABC is a right-angled triangle. Then CD is connected to point C, and AD is 12. So maybe I can split the quadrilateral into two triangles: ABC and ACD? Then find the areas of each and add them up. That sounds like a plan.First, let's handle triangle ABC. Since AB is 3, BC is 4, and angle ABC is 90 degrees, this is a classic 3-4-5 right triangle. Let me confirm that: by Pythagoras, AB² + BC² = AC². So 3² + 4² = 9 + 16 = 25. Therefore, AC must be 5. Okay, so AC is 5. So the diagonal AC divides the quadrilateral into triangles ABC and ACD. Now, I need to find the area of triangle ABC and triangle ACD and sum them up.The area of triangle ABC is straightforward since it's a right-angled triangle. Area = (AB * BC)/2 = (3 * 4)/2 = 12/2 = 6. So that's 6. Now, for triangle ACD, the sides are AD = 12, DC = 13, and AC = 5. Wait, 12, 13, 5? Hmm. Let me check if those can form a triangle. 5 + 12 > 13? 5 + 12 = 17 > 13. 12 +13 >5? 25 >5. 5 +13 >12? 18>12. So yes, triangle ACD exists. Wait, but 5, 12, 13 is a Pythagorean triplet. 5² + 12² = 25 + 144 = 169 = 13². So triangle ACD is also a right-angled triangle! So angle at A is the right angle? Wait, AD is 12, AC is 5, CD is 13. So if the sides are 5,12,13, then the right angle must be between the two shorter sides, which are AC and AD. So angle at A is between AD and AC, which would make triangle ACD a right-angled triangle at A. Wait, but in the quadrilateral ABCD, points A, B, C, D. So if angle at A is a right angle, then AD is connected to A and D, and AC is the diagonal. Hmm. Let me make sure. If triangle ACD has sides 5, 12, 13, then if AC=5, AD=12, CD=13, then angle at A is 90 degrees. So that would mean that in the quadrilateral, angle at A is also a right angle. But the problem only mentions angle ABC is 90 degrees. So maybe there are two right angles here? Well, if triangle ACD is right-angled at A, then that's another right angle. But does that affect the quadrilateral's structure?Wait, but maybe I can compute the area of triangle ACD directly as another right-angled triangle. Since it's 5,12,13, the area would be (5*12)/2 = 60/2 = 30. Then the total area of the quadrilateral would be area of ABC (6) plus area of ACD (30) which is 36. Wait, but 36 is one of the options, option B. But I need to confirm if triangle ACD is indeed right-angled at A. Let me think.In quadrilateral ABCD, points A, B, C, D. So ABC is right-angled at B, and ACD is right-angled at A. So if that's the case, then the quadrilateral has two right angles. But does that configuration hold? Let me sketch mentally. Let me start by drawing triangle ABC with right angle at B. So AB is 3 units, BC is 4 units. Then AC is 5 units. Now, point D is connected to C and A. AD is 12 units, CD is 13 units. So if AC is 5, AD is 12, and CD is 13, then triangle ACD has sides 5,12,13, which is a right triangle. So to form triangle ACD, with AC=5, AD=12, CD=13, the right angle must be at A. Therefore, the angle at A between AC and AD is 90 degrees. So that would mean that in the quadrilateral ABCD, angle at A is 90 degrees. But the problem didn't specify that. However, since the sides are given, maybe it's okay. So if this is the case, then the area is indeed 6 + 30 = 36. Hmm. But let me check if this is correct.Alternatively, maybe I can use coordinates to verify. Let's assign coordinates to the points. Let me place point B at the origin (0,0). Since angle ABC is 90 degrees, then AB is along the x-axis and BC is along the y-axis. So point A is at (3, 0), point B is at (0,0), point C is at (0,4). Then AC is the diagonal from (3,0) to (0,4). Let me compute the length of AC: distance between (3,0) and (0,4) is sqrt((3-0)^2 + (0-4)^2) = sqrt(9 + 16) = sqrt(25) = 5. So that matches. Now, point D is connected to C and A. AD is 12, CD is 13. So point D is connected to A (3,0) and C (0,4), with AD=12 and CD=13. Let me find the coordinates of point D.Let me denote coordinates of D as (x,y). Then, distance from D to A (3,0) is 12: sqrt((x - 3)^2 + (y - 0)^2) = 12. So (x - 3)^2 + y^2 = 144. Distance from D to C (0,4) is 13: sqrt((x - 0)^2 + (y - 4)^2) = 13. So x^2 + (y - 4)^2 = 169. Now, we have two equations:1. (x - 3)^2 + y^2 = 1442. x^2 + (y - 4)^2 = 169Let me expand both equations:First equation:(x^2 - 6x + 9) + y^2 = 144So x^2 + y^2 - 6x + 9 = 144x^2 + y^2 - 6x = 135 --- (1)Second equation:x^2 + (y^2 - 8y + 16) = 169x^2 + y^2 - 8y + 16 = 169x^2 + y^2 - 8y = 153 --- (2)Now, subtract equation (1) from equation (2):(x^2 + y^2 - 8y) - (x^2 + y^2 - 6x) = 153 - 135Simplify:-8y - (-6x) = 18-8y + 6x = 18Divide both sides by 2:-4y + 3x = 9So 3x - 4y = 9 --- (3)Now, we can express x in terms of y or vice versa. Let's solve for y:3x = 4y + 9x = (4y + 9)/3Now, substitute x into equation (1) or (2). Let's choose equation (1):x^2 + y^2 - 6x = 135Substituting x:[(4y + 9)/3]^2 + y^2 - 6*(4y + 9)/3 = 135Compute each term:First term: [(4y + 9)/3]^2 = (16y² + 72y + 81)/9Second term: y²Third term: -6*(4y +9)/3 = -2*(4y +9) = -8y -18So putting it all together:(16y² + 72y + 81)/9 + y² -8y -18 = 135Multiply everything by 9 to eliminate denominator:16y² + 72y + 81 + 9y² -72y -162 = 1215Combine like terms:(16y² + 9y²) + (72y -72y) + (81 -162) = 121525y² + 0y -81 = 121525y² = 1215 +81 = 1296y² = 1296 /25y = ±sqrt(1296/25) = ±36/5 = ±7.2So y = 7.2 or y = -7.2Now, substitute back into equation (3): 3x -4y =9First, take y = 36/5 (7.2):3x -4*(36/5) =93x - 144/5 =93x =9 +144/5 = (45 +144)/5 =189/5x= (189/5)/3=63/5=12.6So x=63/5=12.6, y=36/5=7.2Second, y= -36/5:3x -4*(-36/5)=93x +144/5=93x=9 -144/5=(45 -144)/5=(-99)/5x= (-99/5)/3= -33/5= -6.6So coordinates of D are either (12.6,7.2) or (-6.6, -7.2). Since the quadrilateral is convex, we need to check which of these points makes the quadrilateral convex. Let's consider the original points:A(3,0), B(0,0), C(0,4). If D is at (12.6,7.2), then connecting the points in order A-B-C-D-A should form a convex quadrilateral. Let's see: moving from A(3,0) to B(0,0) to C(0,4) to D(12.6,7.2) to A. The turn from C to D to A needs to be such that the quadrilateral is convex. Alternatively, if D is at (-6.6, -7.2), then the quadrilateral might be concave or cross over itself. Let me check the coordinates.If D is at (63/5, 36/5) which is (12.6,7.2), then plotting this point, it's in the first quadrant, which makes sense. The other point is in the third quadrant, which would make the quadrilateral cross over, probably concave. So we take D as (12.6,7.2). Now, let's compute the area of quadrilateral ABCD. Since we have coordinates for all points, we can use the shoelace formula.Coordinates:A(3,0), B(0,0), C(0,4), D(63/5, 36/5)Convert 63/5 to 12.6 and 36/5 to 7.2.Shoelace formula:Area = 1/2 |sum over i (x_i y_{i+1} - x_{i+1} y_i)|Order of points: A, B, C, D, A.Compute each term:First term: x_A y_B - x_B y_A = 3*0 -0*0=0Second term: x_B y_C -x_C y_B =0*4 -0*0=0Third term: x_C y_D -x_D y_C =0*(36/5) - (63/5)*4=0 -252/5 = -252/5Fourth term: x_D y_A -x_A y_D = (63/5)*0 -3*(36/5)=0 -108/5 = -108/5Sum all terms: 0 +0 + (-252/5) + (-108/5) = (-360/5) = -72Take absolute value and multiply by 1/2: 1/2 *72=36. So the area is 36. That's option B. So that confirms the previous calculation where we split into two right triangles, each with area 6 and 30, totaling 36. Therefore, the answer is B) 36.But just to be thorough, let me check if triangle ACD is indeed right-angled at A. In coordinates, point A is (3,0), point C is (0,4), point D is (12.6,7.2). Let's compute vectors:Vector AC: from A to C: (0 -3, 4 -0) = (-3,4)Vector AD: from A to D: (12.6 -3,7.2 -0) = (9.6,7.2)Dot product of AC and AD: (-3)(9.6) + (4)(7.2) = -28.8 + 28.8 = 0. Since the dot product is zero, the vectors are perpendicular. Therefore, angle at A is indeed 90 degrees. Therefore, triangle ACD is right-angled at A. So area is 1/2 * AC * AD = 1/2 *5 *12=30. Then total area 6 +30=36. Yep, that's correct.Alternatively, using coordinates, the shoelace formula gives the same result. So the answer is 36, which is option B.Final Answerboxed{B}</think>Given the convex quadrilateral (ABCD) with (AB = 3), (BC = 4), (CD = 13), (AD = 12), and (angle ABC = 90^circ), we need to find its area.1. Calculate the area of triangle (ABC): - Since (angle ABC = 90^circ), triangle (ABC) is a right-angled triangle. - The area of triangle (ABC) is (frac{1}{2} times AB times BC = frac{1}{2} times 3 times 4 = 6).2. Determine the length of diagonal (AC): - Using the Pythagorean theorem in triangle (ABC): [ AC = sqrt{AB^2 + BC^2} = sqrt{3^2 + 4^2} = sqrt{9 + 16} = sqrt{25} = 5 ]3. Check if triangle (ACD) is a right-angled triangle: - The sides of triangle (ACD) are (AC = 5), (AD = 12), and (CD = 13). - Verify using the Pythagorean theorem: [ 5^2 + 12^2 = 25 + 144 = 169 = 13^2 ] - Since the sides form a Pythagorean triplet, triangle (ACD) is a right-angled triangle with the right angle at (A).4. Calculate the area of triangle (ACD): - The area of triangle (ACD) is (frac{1}{2} times AC times AD = frac{1}{2} times 5 times 12 = 30).5. Sum the areas of triangles (ABC) and (ACD): - Total area of quadrilateral (ABCD) is (6 + 30 = 36).6. Verification using coordinates and the shoelace formula: - Assign coordinates to points (A(3, 0)), (B(0, 0)), (C(0, 4)), and (D(12.6, 7.2)). - Apply the shoelace formula to confirm the area is indeed 36.Thus, the area (S) of quadrilateral (ABCD) is (boxed{B}).,<think>Okay, let me try to figure out the area of this quadrilateral ABCD. So, we have a convex quadrilateral with sides AB=3, BC=4, CD=13, AD=12, and angle ABC is 90 degrees. The options are given from 30 up to 58.5. Hmm, I need to visualize this first. Let me sketch it mentally.First, angle ABC is 90 degrees. So triangle ABC is a right-angled triangle. Since AB=3 and BC=4, that's a classic 3-4-5 triangle. So AC, the hypotenuse, should be 5. Let me confirm that using Pythagoras: AB² + BC² = 3² + 4² = 9 + 16 = 25, so AC = 5. Got it. So triangle ABC has area (3*4)/2 = 6. Then, the quadrilateral ABCD is made up of triangle ABC and triangle ACD? Wait, no. Because ABCD is a quadrilateral, so it's ABC connected to CD and AD. Wait, the sides given are CD=13 and AD=12. So once we have AC=5, then triangle ACD has sides AD=12, CD=13, and AC=5. Wait, 5, 12, 13 is another right-angled triangle! Because 5² + 12² = 25 + 144 = 169 = 13². So triangle ACD is also right-angled. That must be at angle A? Because AD and AC are the legs, and CD is the hypotenuse. Wait, if AC=5 and AD=12, then the right angle would be at A. So angle CAD is 90 degrees. Therefore, the area of triangle ACD is (5*12)/2 = 30. Then, adding the two areas: triangle ABC is 6 and triangle ACD is 30, so total area S = 6 + 30 = 36. But wait, the options have 36 as option B. But wait, let me check again. Maybe I made a mistake here.Wait, hold on. If triangle ACD is right-angled at A, then points A, C, D form a right triangle with legs 12 and 5. So the area is indeed 30. Then adding the 6 from ABC gives 36. Hmm. But why is 36 an option here? Let me check the figure again. Wait, maybe the quadrilateral isn't split into ABC and ACD? Maybe it's split into ABC and ABD? Wait, but how?Wait, let's think again. The quadrilateral ABCD has sides AB=3, BC=4, CD=13, DA=12. The angle at B is 90 degrees. So ABC is right-angled. If I can compute the coordinates of each point, maybe that will help. Let me assign coordinates to the points.Let me place point B at the origin (0,0). Then, since angle ABC is 90 degrees, AB is along the x-axis and BC is along the y-axis. So point A is at (3,0) because AB=3. Point C is at (0,4) because BC=4. Then, point D is connected to C and A. We know CD=13 and AD=12. So we need to find coordinates of D such that distance from D to C is 13 and distance from D to A is 12.So coordinates: A(3,0), C(0,4). Let me denote D(x,y). Then, distance from D to C is sqrt((x-0)^2 + (y-4)^2) = 13, and distance from D to A is sqrt((x-3)^2 + (y-0)^2) = 12.So we have two equations:1. x² + (y - 4)² = 169 (since 13² = 169)2. (x - 3)² + y² = 144 (since 12² = 144)Let me expand both equations:First equation: x² + y² - 8y + 16 = 169Second equation: x² - 6x + 9 + y² = 144Subtract the second equation from the first:(x² + y² - 8y + 16) - (x² - 6x + 9 + y²) = 169 - 144Simplify left side:-8y + 16 - (-6x) - 9 = 25Wait, step by step:Left side:x² + y² -8y +16 -x² +6x -9 -y² = -8y +16 +6x -9 = 6x -8y +7Right side: 25So 6x -8y +7 =25Thus, 6x -8y =18Divide both sides by 2: 3x -4y=9So equation (3): 3x -4y=9Now, we can solve this with one of the original equations. Let's use equation 2: (x -3)^2 + y^2 =144Let me express y from equation 3: 3x -9 =4y => y=(3x -9)/4Substitute into equation 2:(x -3)^2 + [(3x -9)/4]^2 =144Compute each term:(x -3)^2 = x² -6x +9[(3x -9)/4]^2 = (9x² -54x +81)/16So:x² -6x +9 + (9x² -54x +81)/16 =144Multiply all terms by 16 to eliminate denominator:16x² -96x +144 +9x² -54x +81 = 2304Combine like terms:16x² +9x² =25x²-96x -54x= -150x144 +81=225So:25x² -150x +225 =2304Subtract 2304:25x² -150x +225 -2304=025x² -150x -2079=0Divide by GCD(25,150,2079). Let's see, 25 and 150 are divisible by 25 and 150 is 6*25, 2079 divided by 25: 25*83=2075, so 2079-2075=4. So 2079=25*83 +4. So no common divisor. So equation is 25x² -150x -2079=0Let me check the discriminant:D = (-150)^2 -4*25*(-2079) =22500 +4*25*2079Calculate 4*25=100, 100*2079=207900Thus discriminant D=22500 +207900=230400Square root of D is sqrt(230400)=480Thus solutions x=(150 ±480)/50First solution: (150 +480)/50=630/50=63/5=12.6Second solution: (150 -480)/50=(-330)/50=-33/5=-6.6So x=12.6 or x=-6.6But since the quadrilateral is convex, we need to check which solution makes sense. Let's check x=12.6 first. Since point C is at (0,4), and point A is at (3,0), so D should be somewhere such that the quadrilateral remains convex. If x=12.6, which is 63/5, then y=(3x -9)/4=(3*(63/5) -9)/4=(189/5 -45/5)/4=(144/5)/4=144/20=36/5=7.2. So D would be at (12.6,7.2). Let me check distance to A: sqrt((12.6 -3)^2 + (7.2)^2)=sqrt(9.6^2 +7.2^2)=sqrt(92.16 +51.84)=sqrt(144)=12. Correct. Distance to C: sqrt(12.6^2 + (7.2 -4)^2)=sqrt(158.76 +10.24)=sqrt(169)=13. Correct. So that works.Now, if x=-6.6, which is -33/5, then y=(3*(-33/5) -9)/4=(-99/5 -45/5)/4=(-144/5)/4= -144/20= -36/5= -7.2. So D would be at (-6.6, -7.2). Let's check distance to A: sqrt((-6.6 -3)^2 + (-7.2 -0)^2)=sqrt((-9.6)^2 + (-7.2)^2)=sqrt(92.16 +51.84)=sqrt(144)=12. Correct. Distance to C: sqrt((-6.6 -0)^2 + (-7.2 -4)^2)=sqrt(43.56 +125.44)=sqrt(169)=13. Correct. But since the quadrilateral is convex, let's see where the points lie. If D is at (-6.6, -7.2), then the order of the quadrilateral would be A(3,0), B(0,0), C(0,4), D(-6.6, -7.2). Connecting A to D would cross over, making the quadrilateral concave? Or maybe not. Wait, convexity requires all interior angles less than 180 degrees. But depending on the order. Wait, the problem says it's a convex quadrilateral ABCD. So the order is A-B-C-D-A. So starting at A, then B, C, D, back to A. So with D at (-6.6, -7.2), would that make the quadrilateral convex?Let me plot roughly. If A is at (3,0), B at (0,0), C at (0,4), D at (-6.6, -7.2). So from C (0,4) to D (-6.6, -7.2), then to A (3,0). The angle at D might be problematic. But I think convexity requires that all vertices are "outward". Alternatively, maybe both solutions are possible, but since the problem states it's convex, so we need to pick the correct one. However, given that CD=13 and AD=12, and ABC is right-angled, perhaps the intended solution is the one where D is in a position that makes the quadrilateral convex. Let's check both possibilities.But perhaps in the problem, since ABC is right-angled, and the sides CD=13, AD=12, the quadrilateral is made by attaching the right triangles ABC and ACD. Wait, but earlier calculation gives the area as 6 + 30 = 36. But when I computed coordinates, with D at (12.6,7.2), the area can be calculated using coordinates.Alternatively, maybe I can use the shoelace formula to compute the area once I have coordinates. Let me try that.First, for D at (63/5, 36/5). So coordinates:A(3,0), B(0,0), C(0,4), D(63/5,36/5)List the coordinates in order: A(3,0), B(0,0), C(0,4), D(63/5,36/5), back to A(3,0)Compute shoelace sum:Sum1 = (3*0) + (0*4) + (0*36/5) + (63/5*0) = 0 +0 +0 +0=0Sum2 = (0*0) + (0*0) + (4*63/5) + (36/5*3) =0 +0 +252/5 +108/5 = (252 +108)/5 =360/5=72Area = |Sum1 - Sum2|/2 = |0 -72|/2=36. So area 36. Which is option B.Alternatively, if we take the other point D(-33/5, -36/5):Coordinates: A(3,0), B(0,0), C(0,4), D(-33/5, -36/5)Shoelace formula:Sum1 = (3*0) + (0*4) + (0*(-36/5)) + (-33/5*0) =0 +0 +0 +0=0Sum2 = (0*0) + (0*0) + (4*(-33/5)) + (-36/5*3) =0 +0 -132/5 -108/5 = (-240)/5 = -48Area = |0 - (-48)| /2 =48/2=24. Hmm, 24 isn't one of the options. But this seems contradictory. Wait, but maybe I messed up the order? Because the order for shoelace has to be either clockwise or counter-clockwise without crossing.Wait, if D is at (-33/5, -36/5), then the order A-B-C-D would be going from A(3,0) to B(0,0) to C(0,4) to D(-33/5, -36/5) to A(3,0). But this path would actually cross over itself, making the quadrilateral concave. But since the problem states it's convex, this solution is invalid. Therefore, the correct position for D is (63/5,36/5), giving an area of 36. So the answer is B) 36. But wait, earlier I thought adding the two triangles gives 6 + 30=36, but then why is option B 36? Let me check again.Wait, triangle ABC has area 6, triangle ACD has area 30, so total area 36. But when I computed the coordinates, the shoelace formula also gives 36. So that seems consistent. But why is option B 36? Because the answer choices are (A) 30, (B)36, etc. So 36 is an option. So that must be the answer. But let me check if there's another way to compute this.Alternatively, maybe using the two right triangles. Since ABC is right-angled and ACD is right-angled, adding their areas. But wait, triangle ACD is right-angled at A. So points A, C, D form a right angle at A. Then, the area of ACD is (AC * AD)/2 = (5 *12)/2=30. Then, total area is 6 +30=36. So that's straightforward. But why did the problem give CD=13? Because in triangle ACD, CD is the hypotenuse, 5-12-13 triangle. So that checks out. So this method is correct.Therefore, the answer should be 36, option B. But let me check once more for any possible mistake. For example, maybe the two triangles ABC and ACD overlap? But since ABCD is convex, they shouldn't. Because ABC is the first triangle, and ACD is attached via AC. Since ABC and ACD share the diagonal AC, which is inside the quadrilateral. So the total area is indeed the sum of the two triangles. Hence, 36. Therefore, I think the answer is B)36.Final Answerboxed{B}</think>Given the convex quadrilateral (ABCD) with (AB = 3), (BC = 4), (CD = 13), (AD = 12), and (angle ABC = 90^circ), we need to find its area.1. Calculate the hypotenuse (AC) of right-angled triangle (ABC): [ AC = sqrt{AB^2 + BC^2} = sqrt{3^2 + 4^2} = sqrt{9 + 16} = sqrt{25} = 5 ] The area of (triangle ABC) is: [ text{Area}_{ABC} = frac{1}{2} times AB times BC = frac{1}{2} times 3 times 4 = 6 ]2. Check if triangle (ACD) is a right-angled triangle: Given (AD = 12), (CD = 13), and (AC = 5), we verify if (ACD) is a right-angled triangle: [ AC^2 + AD^2 = 5^2 + 12^2 = 25 + 144 = 169 = 13^2 = CD^2 ] Hence, (triangle ACD) is a right-angled triangle at (A). The area of (triangle ACD) is: [ text{Area}_{ACD} = frac{1}{2} times AC times AD = frac{1}{2} times 5 times 12 = 30 ]3. Calculate the total area of quadrilateral (ABCD): [ text{Total Area} = text{Area}_{ABC} + text{Area}_{ACD} = 6 + 30 = 36 ]Thus, the area (S) of quadrilateral (ABCD) is (boxed{B}).]

🔑:9. B.As shown in Figure 1, connect A C. In the right triangle triangle A B C,angle A B C=90^{circ}, A B=3, B C=4 text {. }Therefore, A C=sqrt{A B^{2}+A C^{2}}=5.In triangle A C D, C D=13, A D=12, andA C^{2}+A D^{2}=C D^{2} text {, }Thus, A C perp A D.Then, S=S_{triangle A B C}+S_{triangle A C D}=frac{3 times 4}{2}+frac{5 times 12}{2}=36.

🧠:[<think>Okay, let's see here. I need to solve two parts using the given equations. The problem states that 3^a = 4, 3^b = 5, and 3^c = 8. Let's tackle part (1) first: Find the value of 3^{b+c}.Hmm, so 3^{b+c} can be rewritten using exponent rules. When you have exponents with the same base and you're adding the exponents, that's the same as multiplying the numbers. So, 3^{b+c} = 3^b * 3^c. But wait, from the problem, 3^b is given as 5 and 3^c is 8. So substituting those in, it should be 5 * 8. Let me check that again. Yes, 3^b is 5 and 3^c is 8, so multiplying them together gives 5*8=40. Therefore, 3^{b+c} = 40. That seems straightforward. Let me just make sure there's no trick here. Nope, since the base is the same (3), adding exponents corresponds to multiplying the values. So part (1) should be 40.Now moving on to part (2): Find the value of 3^{2a - 3b}. Alright, this one is a bit more complex. Let's break it down. First, 3^{2a - 3b} can be separated into 3^{2a} divided by 3^{3b} because of the exponent subtraction rule: 3^{x-y} = 3^x / 3^y. So, 3^{2a - 3b} = (3^{2a}) / (3^{3b}).Now, let's compute each part separately. Starting with 3^{2a}. We know that 3^a = 4, so if we square both sides of that equation, we get (3^a)^2 = 4^2, which simplifies to 3^{2a} = 16. That gives us the numerator.Next, the denominator is 3^{3b}. Similarly, since 3^b = 5, if we cube both sides, we get (3^b)^3 = 5^3, which is 3^{3b} = 125. Therefore, putting those together: 3^{2a - 3b} = 16 / 125. Let me confirm that step again. Yes, 3^{2a} is 16, 3^{3b} is 125, so dividing them gives 16 over 125. Wait a second, but the question says "Find the value of 3^{2a - 3b}". So, since we've expressed it as 16/125, which is a numerical value, that should be the answer. Let me check if I can simplify that fraction further. 16 and 125 don't have any common factors besides 1, so 16/125 is already in its simplest form. Alternatively, is there another way to approach this problem? Maybe using logarithms? Let me think. If I take the logarithm base 3 of both sides of the original equations, then for 3^a = 4, we get a = log₃4. Similarly, b = log₃5 and c = log₃8. But then, if we need to compute 3^{2a - 3b}, substituting the expressions for a and b would give 3^{2 log₃4 - 3 log₃5}. Using logarithm properties, 2 log₃4 is log₃4² = log₃16, and 3 log₃5 is log₃5³ = log₃125. So then, 3^{log₃16 - log₃125} = 3^{log₃(16/125)} = 16/125. That's the same result as before. So, this method also gives 16/125. Therefore, that confirms the answer is 16/125.Wait, just to be thorough, is there another possible method? Maybe expressing 3^{2a - 3b} as (3^a)^2 / (3^b)^3. Oh, that's actually the same approach as before. Since 3^a is 4, squared is 16, and 3^b is 5, cubed is 125. So 16 divided by 125. Yep, same answer. So regardless of the method, the answer is 16/125.Therefore, part (2) is 16/125. Let me recap to make sure I didn't miss anything. The first part was straightforward multiplication, the second required breaking down the exponent into parts and applying exponent rules. Both methods led to the same results, so I feel confident about these answers. I don't see any calculation errors in my steps. So, the final answers should be 40 and 16/125.Final Answer(1) boxed{40}(2) boxed{dfrac{16}{125}}</think>Given the equations (3^a = 4), (3^b = 5), and (3^c = 8):(1) To find the value of (3^{b+c}):[3^{b+c} = 3^b cdot 3^c = 5 cdot 8 = 40]Thus, the value is (boxed{40}).(2) To find the value of (3^{2a-3b}):[3^{2a-3b} = frac{3^{2a}}{3^{3b}}]First, calculate (3^{2a}) and (3^{3b}):[3^{2a} = (3^a)^2 = 4^2 = 16][3^{3b} = (3^b)^3 = 5^3 = 125]Therefore,[3^{2a-3b} = frac{16}{125}]Thus, the value is (boxed{dfrac{16}{125}}).,<think>Okay, let's see. I have this problem here where I'm given three equations: 3 to the power of a is 4, 3 to the power of b is 5, and 3 to the power of c is 8. Then there are two parts to the problem. The first part asks me to find the value of 3 raised to the power of (b + c), and the second part wants me to find 3 raised to the power of (2a - 3b). Hmm, alright, let's start with part (1).So, for part (1): 3^{b + c}. Hmm. Since exponents add when you multiply the same base, right? So maybe I can use the property that 3^{b + c} is equal to 3^b multiplied by 3^c. Let me check. Yes, exponent rules say that 3^{b + c} = 3^b * 3^c. But wait, from the given information, I know that 3^b is 5 and 3^c is 8. So substituting those values in, 3^{b + c} should be 5 * 8. Let me calculate that: 5 times 8 is 40. So, is the answer for part (1) 40? That seems straightforward. Let me just verify again. The exponents add when you multiply, so 3^{b} * 3^{c} = 3^{b + c}, and since 3^b is 5 and 3^c is 8, multiplying them gives 5*8=40. Yep, that checks out. So part (1) is 40.Now moving on to part (2): 3^{2a - 3b}. This looks a bit more complicated. Let me recall exponent rules. When you have exponents with coefficients, like 2a, that's like multiplying the exponent by 2. So 3^{2a} can be written as (3^a)^2. Similarly, 3^{-3b} is 1 divided by (3^b)^3. So putting that together, 3^{2a - 3b} is equal to 3^{2a} * 3^{-3b} = (3^a)^2 / (3^b)^3. Let me confirm that. Yes, since 3^{m - n} = 3^m / 3^n. So that's correct.Now, substituting the given values. We know that 3^a is 4, so (3^a)^2 is 4 squared, which is 16. Then, 3^b is 5, so (3^b)^3 is 5 cubed, which is 125. Therefore, 3^{2a - 3b} should be 16 divided by 125. Let me write that as 16/125. Let me verify once more. Starting with 3^{2a - 3b}, which is 3^{2a} divided by 3^{3b}, which is (3^a)^2 divided by (3^b)^3, substituting 4 and 5 gives (4)^2 / (5)^3 = 16 / 125. Yep, that's correct. So part (2) is 16 over 125.Wait, hold on, but the question says "Find the value of 3^{2a - 3b}". So since 3^{2a - 3b} simplifies to 16/125, is there a way to write this as a decimal or is 16/125 acceptable? Let me compute 16 divided by 125 to see if it's a nice decimal. 125 goes into 16 zero times, add a decimal point. 125 goes into 160 once (125), remainder 35. Bring down a zero: 350. 125 goes into 350 twice (250), remainder 100. Bring down a zero: 1000. 125 goes into 1000 eight times. So 16/125 is 0.128. But since the question just says "find the value", and 16/125 is exact, whereas 0.128 is an approximation. So probably better to leave it as a fraction. So 16/125 is the exact value.So to recap:1) 3^{b + c} = 3^b * 3^c = 5 * 8 = 40.2) 3^{2a - 3b} = (3^a)^2 / (3^b)^3 = 4^2 / 5^3 = 16 / 125.I think that's all. Let me check again for any possible mistakes. For part (1), is there a chance that the question is asking for the exponent instead of the value? Wait, no. The question says "Find the value of 3^{b + c}". Since 3^{b + c} is an expression, and they want its value. Since we can compute that as 5*8=40, that's correct. For part (2), same logic. The expression 3^{2a - 3b} is equal to 16/125, which is a reduced fraction, so that's the answer. So I think both parts are correct.Final Answer(1) boxed{40}(2) boxed{dfrac{16}{125}}</think>Given the equations (3^a = 4), (3^b = 5), and (3^c = 8):(1) To find the value of (3^{b+c}):[3^{b+c} = 3^b cdot 3^c = 5 cdot 8 = 40]Thus, the value is (boxed{40}).(2) To find the value of (3^{2a-3b}):[3^{2a-3b} = frac{3^{2a}}{3^{3b}} = frac{(3^a)^2}{(3^b)^3} = frac{4^2}{5^3} = frac{16}{125}]Thus, the value is (boxed{dfrac{16}{125}}).]

🔑:To solve the given problems, we proceed as follows: For (1) Find the value of 3^{b+c}:Given that 3^{b}=5 and 3^{c}=8, we can find 3^{b+c} by using the property of exponents that states a^{m+n} = a^{m} cdot a^{n}. Applying this property, we get:[begin{aligned}3^{b+c} &= 3^{b} cdot 3^{c} &= 5 times 8 &= 40.end{aligned}]Thus, the value of 3^{b+c} is boxed{40}. For (2) Find the value of 3^{2a-3b}:Given 3^{a}=4 and 3^{b}=5, we can find 3^{2a-3b} by using the properties of exponents that allow us to manipulate the expression as follows:[begin{aligned}3^{2a-3b} &= 3^{2a} div 3^{3b} &= (3^{a})^{2} div (3^{b})^{3} &= 4^{2} div 5^{3} &= frac{16}{125}.end{aligned}]Therefore, the value of 3^{2a-3b} is boxed{frac{16}{125}}.

🧠:[<think>Okay, so I need to solve problem 45, parts a and b. Let's start with part a. The problem says: For triangle ABC, where angles A ≤ B ≤ C, what values can angles A, B, and C take? First, I remember that in any triangle, the sum of the angles is 180 degrees. Also, since A ≤ B ≤ C, angle C is the largest angle, and A is the smallest. So, since A is the smallest angle, its maximum possible value would be when all angles are equal. In an equilateral triangle, all angles are 60 degrees. But since A ≤ B ≤ C, the angles can be equal, so A can be up to 60 degrees. However, if the triangle is not equilateral, angle A would be less than 60. But wait, if the angles are in order A ≤ B ≤ C, then angle C has to be at least as big as B and A. Since the sum is 180, the minimum value of angle A would depend on how the angles are distributed. Let me think. If angle A is minimized, then angles B and C would be maximized. But since A ≤ B ≤ C, the minimal angle A can be is when the triangle is as "stretched" as possible, meaning C is as large as possible. But in any triangle, each angle has to be less than 180 degrees, but actually, since the sum is 180, the largest angle must be less than 180. Wait, but in reality, the largest angle can be up to just under 180, but if we have angles in order A ≤ B ≤ C, then the smallest angle A must be at least... Hmm, maybe I need a different approach. Let me denote angles A, B, C with A ≤ B ≤ C. Then, since A is the smallest, we have A ≤ 60 degrees. Because if all angles were equal, they'd be 60. If one angle is larger than 60, then another has to be smaller. But since A is the smallest, if the triangle is not equilateral, A would be less than 60. But the question is asking for the possible values. So maybe angle A can be in the interval (0°, 60°], angle B in [A, (180° - A)/2], and angle C in [B, 180° - A - B]. Hmm, perhaps. Wait, let's try to formalize this. Given A ≤ B ≤ C, and A + B + C = 180°. Since A ≤ B ≤ C, then C must be at least 60°, because if all angles were equal, they'd be 60°. If C is larger than 60°, then at least one of the other angles must be smaller. But since A is the smallest, A must be less than 60°. So angle A can range from greater than 0° up to 60°. Similarly, angle B must be at least equal to A, and at most equal to C. Since A ≤ B ≤ C, and A + B + C = 180°, then replacing A and B with C would give 3C ≥ 180°, so C ≥ 60°, which we already know. But for angle B, since A ≤ B ≤ C, and C ≥ B ≥ A, we can derive some bounds. Let me think. Let's express angles in terms of A. If angle A is fixed, then angles B and C must satisfy A ≤ B ≤ C and A + B + C = 180°. So, B and C must each be at least A. So the minimal total for B and C would be A + A = 2A, but since A + B + C = 180°, then B + C = 180° - A. Therefore, 180° - A ≥ 2A ⇒ 180° ≥ 3A ⇒ A ≤ 60°, which aligns with our previous conclusion. But then, if angle B is at least A, and angle C is at least B, so given that B + C = 180° - A, and B ≤ C, then B can be at most (180° - A)/2. Because if B is as large as possible, then C would be equal to B. So, the maximum value of B is (180° - A)/2. Therefore, for a given A, angle B must satisfy A ≤ B ≤ (180° - A)/2. Similarly, angle C would then be 180° - A - B. Since B ≥ A and C ≥ B, then 180° - A - B ≥ B ⇒ 180° - A ≥ 2B ⇒ B ≤ (180° - A)/2, which is consistent. So, angle A can be in (0°, 60°], angle B is in [A, (180° - A)/2], and angle C is in [B, 180° - A - B]. But actually, once A and B are determined, C is fixed as 180° - A - B. But maybe the question is just asking for the ranges of each angle without considering the others. Wait, let's see. For part a), question 1: angle A; 2: angle B; 3: angle C. Each of them can take what values given A ≤ B ≤ C. So, for angle A: since it's the smallest angle, its maximum is 60°, as in an equilateral triangle. The minimum is approaching 0°, but can't be 0. So angle A ∈ (0°, 60°]. For angle B: it's between A and C. Since A ≤ B ≤ C, and the sum of all angles is 180°, angle B can vary. Let's consider the minimum and maximum possible for B. The minimum value of B is equal to A, so since A can be up to 60°, the minimum B is A. The maximum value of B is when B is as large as possible, which would be when C is as large as possible. But since B ≤ C, and C = 180° - A - B, so substituting, we have B ≤ 180° - A - B ⇒ 2B ≤ 180° - A ⇒ B ≤ (180° - A)/2. Therefore, for angle B: given A, B ∈ [A, (180° - A)/2]. But since A can vary from 0° to 60°, what is the overall range of B? If A is minimized (approaching 0°), then B can go up to (180° - 0°)/2 = 90°, but B has to be at least A (approaching 0°). But since B also has to be ≤ C, which would be 180° - A - B, if A approaches 0°, and B approaches 90°, then C would approach 90° as well. So in that case, the triangle would be approaching a right-angled triangle with angles approaching 0°, 90°, 90°, but that's not possible because angles can't be 0°. Alternatively, if A is 60°, then B must be 60°, and C must be 60°, so B is fixed at 60°. So, angle B can range from just above 0° up to 90°, but considering that A ≤ B ≤ C and A is at most 60°, maybe the maximum B can be is 90°, but let's check. Suppose we have a triangle where B is 90°, then angle C must be at least 90°, but then A + B + C = 180° ⇒ A + 90° + C = 180° ⇒ A + C = 90°. Since C ≥ B = 90°, then C ≥ 90°, which would mean A + C ≥ 90° + 0°, but since A + C = 90°, so C = 90° - A. But since C must be ≥ B = 90°, then 90° - A ≥ 90° ⇒ -A ≥ 0 ⇒ A ≤ 0°, which is impossible. Therefore, angle B cannot be 90°, since that would require A ≤ 0°. Wait, so maybe angle B cannot reach 90°. Let's think of a triangle with angles approaching a right angle. Suppose angle B approaches 90°, then angle C must be at least 90°, so angle C approaches 90° as well. Then angle A would be approaching 0°, since A = 180° - B - C ≈ 180° - 90° - 90° = 0°. But since angles must be positive, angle B cannot actually reach 90°. Therefore, the maximum value of angle B is less than 90°. But how much less? If angle B is maximum when angle C is equal to B, then in that case, angle C = B, so angles are A, B, B. Then A + 2B = 180°, so A = 180° - 2B. Since A ≤ B, we have 180° - 2B ≤ B ⇒ 180° ≤ 3B ⇒ B ≥ 60°. Wait, so if angles are A, B, B with A ≤ B, then B must be at least 60°, but in that case, angle A is 180° - 2B, which would be ≤ B. For example, if B is 60°, then A is 60°, which gives an equilateral triangle. If B is 70°, then A is 180 - 140 = 40°, which is less than B. So, in this scenario, the maximum value for B is when A is as small as possible. Wait, but if we fix angle B, then angle C can be larger. Wait, maybe I need to approach this differently. Let's find the possible range of angle B. Since A ≤ B ≤ C, and A + B + C = 180°, then:From A ≤ B, so A can be at most B.From B ≤ C, so C is at least B.Therefore, substituting in the total angles:A + B + C ≥ B + B + B = 3B ⇒ 3B ≤ 180° ⇒ B ≤ 60°.Wait, that's conflicting with my earlier thought. Wait, if A ≤ B ≤ C, then all three angles are at least A and at most C. But if we consider that the sum is 180°, then the average angle is 60°, so if B is the middle angle, it can be less than, equal to, or greater than 60°, depending on the distribution. Wait, but according to the inequality above: if A ≤ B ≤ C, then A ≤ B and C ≥ B. So, replacing A with B (since A ≤ B) and C with B (since C ≥ B), then:A + B + C ≥ B + B + B = 3B ⇒ 3B ≤ 180° ⇒ B ≤ 60°.Similarly, replacing A with 0°, and C with B, then:0° + B + B ≤ 180° ⇒ 2B ≤ 180° ⇒ B ≤ 90°, but this is a different bound. So which one is correct? Wait, in the first approach, since A ≤ B and C ≥ B, then substituting A and C with B gives 3B ≤ 180°, so B ≤ 60°. But in the second approach, substituting A with 0° and C with B gives 2B ≤ 180°, so B ≤ 90°. But the problem is that these are different substitutions. The correct upper bound for B is actually derived from the first inequality, since A ≤ B and C ≥ B. Therefore, the sum A + B + C ≥ B + B + B = 3B, so 3B ≤ 180° ⇒ B ≤ 60°. Therefore, angle B can't exceed 60°, which makes sense because if B were greater than 60°, then since C is at least B, C would be greater than 60°, so total sum would exceed 180°, which is not possible. Wait, that seems contradictory to the example I had earlier where B was 70°, A was 40°, and C was 70°, summing to 180°, but here, if B is 70°, which is greater than 60°, but in this case, angle C is also 70°, which is greater than B. Wait, but according to the substitution, 3B ≤ 180° ⇒ B ≤ 60°, but in this case, A is 40°, B is 70°, C is 70°, so sum is 180°, which works. But according to the substitution, 3B would be 210°, which is more than 180°, but in reality, only two angles are 70°, and one is 40°, so sum is 180°. Therefore, the substitution must be wrong. Ah, so maybe my first approach was incorrect. Let me check again. If A ≤ B ≤ C, then the sum A + B + C = 180°. To find the upper limit of B, we can consider that since B ≤ C, then C ≥ B. So, the minimal possible sum for A + B + C is when A is as small as possible and C is as small as possible (i.e., equal to B). Wait, no: if we want to find the upper limit of B, we need to set up the constraints such that B is as large as possible. To maximize B, given A ≤ B ≤ C, we need to minimize A and C. But since C ≥ B, the minimal value of C is B. Therefore, if C is set to B, then A must be 180° - B - C = 180° - 2B. But since A ≤ B, we have 180° - 2B ≤ B ⇒ 180° ≤ 3B ⇒ B ≥ 60°. Wait, this is the opposite inequality. So, in order to have C = B (the minimal C given B), we must have A = 180° - 2B ≤ B ⇒ B ≥ 60°. Therefore, if B is greater than or equal to 60°, then we can have a triangle with angles A = 180° - 2B, B, and C = B. But if B is less than 60°, then setting C = B would require A = 180° - 2B > 180° - 120° = 60°, which would contradict A ≤ B (since B is less than 60°). Therefore, the case where C = B is only possible when B ≥ 60°, which would mean A = 180° - 2B ≤ B. Therefore, angle B can range from 60° down to... Wait, but that seems conflicting. Wait, if we want to maximize B, we set A as small as possible and C as small as possible (since C has to be at least B). So, if we fix C = B, then A = 180° - 2B. But A must be ≥ 0°, so 180° - 2B ≥ 0 ⇒ B ≤ 90°. But in addition, since A ≤ B, then 180° - 2B ≤ B ⇒ 180° ≤ 3B ⇒ B ≥ 60°. Therefore, angle B can be between 60° and 90°, but in this case, if B is 60°, then A = 60°, C = 60°, which is equilateral. If B is 90°, then A = 0°, which is impossible. Therefore, angle B must be between 60° and 90°, but A must be positive. So angle B must be less than 90°, but how does that work? Wait, this seems confusing. Let me take another approach. If we have angles A ≤ B ≤ C, then the maximum possible value of B is when A is as small as possible and C is as small as possible (i.e., equal to B). But since A must be positive, let's set A approaching 0°, then B and C approach 90°, but A can't be 0°, so B can approach 90° but can't reach it. Therefore, angle B is less than 90°. But earlier, we found that if C = B, then A = 180° - 2B must be ≤ B, which gives B ≥ 60°. So angle B must be between 60° and 90°, but since A must be positive, angle B must be less than 90°. Wait, but if A is approaching 0°, then B approaches 90°, so angle B can be up to just under 90°. But in this scenario, angle C is equal to B. But in another scenario, where C is larger than B, then B can be less than 60°. For example, consider an equilateral triangle: all angles 60°, but if we have a triangle with angles 50°, 60°, 70°, then A=50° ≤ B=60° ≤ C=70°, which works. In this case, angle B is 60°, but angle C is larger. Wait, so perhaps angle B can actually range from 60° down to... Wait, no. If A ≤ B ≤ C, then if B is less than 60°, then A is less than B, which is less than 60°, and C is greater than or equal to B. For example, take a triangle with angles 50°, 55°, 75°. Then A=50 ≤ B=55 ≤ C=75. Here, B=55° which is less than 60°. So angle B can be less than 60°, but how low can it go? The minimum value of angle B would be when A is as small as possible, and C is as large as possible. But since C has to be at least B, and angles sum to 180°, angle B can be as small as just above angle A. Wait, but angle A must be ≤ B. So, if angle A approaches 0°, then angle B can approach 0°, but since angles must sum to 180°, angle C would approach 180°. But in reality, angle C must be less than 180°, so angle B can't be too small. Wait, this is getting messy. Maybe a better way is to parametrize the angles. Let's consider that in any triangle with angles A ≤ B ≤ C, angle A can be from 0° to 60°, angle B from A to (180° - A)/2, and angle C from B to 180° - A - B. But since C must be at least B, which gives the upper bound on B as (180° - A)/2. But then, if we fix angle A, say A = 30°, then B can range from 30° to (180° - 30°)/2 = 75°, and C would be 180° - 30° - B. So if B is 30°, C is 120°, and if B is 75°, C is 75°. Similarly, if A is 60°, then B must be 60°, and C must be 60°. If A approaches 0°, then B can approach up to (180° - 0°)/2 = 90°, but since C would then be 180° - 0° - B = 180° - B, which must be at least B, so 180° - B ≥ B ⇒ B ≤ 90°, which is already satisfied. So in this case, when A approaches 0°, B can approach 90°, with C approaching 90°. But angles must be positive, so in reality, angle B must be less than 90°, since A must be greater than 0°. Therefore, combining all these, angle B can range from greater than 0° to less than 90°, but subject to A ≤ B ≤ C. However, when A is small, B can approach 90°, but when A is larger (up to 60°), B is constrained to be up to (180° - A)/2. Wait, but if A is 30°, then B can go up to 75°, which is less than 90°, but if A is 10°, then B can go up to (180° - 10°)/2 = 85°, which is still less than 90°. So angle B's maximum possible value is approaching 90° as A approaches 0°, but can never reach 90°. Similarly, the minimum value of angle B is equal to angle A, which can be as low as just above 0°, so angle B can be as low as just above 0°, but in reality, when A is fixed, B's minimum is A. Therefore, summarizing:1) Angle A can be in (0°, 60°]2) Angle B can be in [A, (180° - A)/2], but since A can vary from 0° to 60°, the overall range for B is (0°, 90°). Wait, but if A is 60°, then B must be 60°, so the maximum value of B is 60° when A is 60°, and when A is less, B can be higher. So angle B can range from just above 0° up to just below 90°, but depending on A. Wait, this is getting complicated. Maybe the answer is:1) Angle A: 0° < A ≤ 60°2) Angle B: A ≤ B ≤ (180° - A)/23) Angle C: B ≤ C ≤ 180° - A - BBut the problem asks "what values can the following quantities take", so perhaps for each angle individually, given the constraints A ≤ B ≤ C. So for angle A: Since A is the smallest angle, it must be at most 60°, as in an equilateral triangle. And since angles are positive, A must be greater than 0°. So A ∈ (0°, 60°]For angle B: It has to be at least A and at most C. Since A ≤ B ≤ C, and the sum is 180°, angle B can range from A (minimum) up to (180° - A)/2 (since C has to be at least B). Therefore, for a given A, B is in [A, (180° - A)/2]. But since A can be as small as approaching 0°, then B can approach up to 90°, but never reaching it. So the overall range for B is (0°, 90°). However, when A is larger, the upper bound for B decreases. So angle B's possible values are from greater than 0° to less than 90°, but with the exact bounds depending on A. But the problem doesn't specify given A, it's just asking for possible values of B in any triangle with A ≤ B ≤ C. Wait, so maybe angle B can be any angle such that there exists a triangle with angles A ≤ B ≤ C. So for angle B, the possible values are (0°, 90°). Because:- If angle B is 60°, then you can have an equilateral triangle.- If angle B is 30°, you can have a triangle with angles 30°, 30°, 120°.Wait, no: 30°, 30°, 120°, but then A = 30°, B = 30°, C = 120°, which would require A ≤ B ≤ C, so B would be 30°, which is equal to A, but C is 120°, which is okay. So angle B can be 30°.But if angle B is 15°, can we have a triangle with A ≤ B ≤ C? Yes, for example, angles 15°, 15°, 150°, but that's a valid triangle (though very obtuse). So angle B can be as small as approaching 0°, just like angle A.But angle B must be at least A. So if angle B is 15°, angle A can be 15° or less, but since A ≤ B, A can be up to 15°. Wait, but if angle B is 15°, then angle A must be ≤ 15°, and angle C must be ≥ 15°, and angles sum to 180°. So angle C would be 180° - A - B. If A is 15°, then C = 150°, which is valid. If A is less than 15°, then C is greater than 150°, but since C must be ≥ B = 15°, which is satisfied. Therefore, angle B can be any value in (0°, 90°). Because:- The lower bound is approaching 0°, since you can have a triangle with angles approaching 0°, approaching 0°, approaching 180°, but in such a case, A and B approach 0°, so B approaches 0°.- The upper bound is approaching 90°, as in a triangle where angles are approaching 0°, approaching 90°, and approaching 90°, but such a triangle would have two angles approaching 90°, but since angles must be positive, B can approach 90° but never reach it.Wait, but if angle B is exactly 90°, is that possible? Let's see. Suppose angle B is 90°, then since A ≤ B ≤ C, angles A ≤ 90° and C ≥ 90°. Then A + B + C = A + 90° + C = 180° ⇒ A + C = 90°. But since C ≥ 90°, then A + C ≥ 90° + A. But A + C = 90°, which would require A ≤ 0°, which is impossible. Therefore, angle B cannot be 90°, so B must be less than 90°. Similarly, if angle B is 80°, can we have such a triangle? Let's try. Let's say A ≤ 80°, B = 80°, and C ≥ 80°. Then A + 80° + C = 180° ⇒ A + C = 100°. Since C ≥ 80°, then A = 100° - C ≤ 20°. So A can be up to 20°, and C can be from 80° up to 100° - A. For example, if A = 10°, then C = 90°; angles would be 10°, 80°, 90°, which satisfies A ≤ B ≤ C. So angle B can be 80°, which is less than 90°.Therefore, angle B can take any value in (0°, 90°). However, for each specific B, angles A and C must satisfy A ≤ B ≤ C and A + B + C = 180°. So B can be between just above 0° and just below 90°, but not including 0° or 90°.Similarly, angle C is the largest angle, so it must be at least 60°, because in an equilateral triangle, all angles are 60°, and if the triangle is not equilateral, the largest angle must be greater than 60°. But actually, if the triangle is equilateral, all angles are 60°, so angle C can be exactly 60°. If the triangle is not equilateral, angle C is greater than 60°. Therefore, angle C can be in [60°, 180°). However, since the sum of angles is 180°, angle C must be less than 180°, so the range is [60°, 180°).But wait, let's check with an example. Suppose angle C is 60°, then all angles must be 60°, since C is the largest. If angle C is 90°, then angles A and B must sum to 90°, with A ≤ B ≤ 90°. For example, 45°, 45°, 90°.But in our problem, the condition is A ≤ B ≤ C. So angle C can be from 60° up to less than 180°, but how exactly?If angle C is approaching 180°, then angles A and B are approaching 0°, but must be positive. So angle C is in [60°, 180°).Therefore, summarizing:1) Angle A: 0° < A ≤ 60°2) Angle B: 0° < B < 90°3) Angle C: 60° ≤ C < 180°But need to confirm with examples.For angle C:- Minimum is 60° (equilateral triangle). - If triangle is right-angled, angles 30°, 60°, 90°, then C = 90°, which is allowed.- If triangle is obtuse, angles 20°, 30°, 130°, then C = 130°, which is allowed.- Maximum approaching 180°, as in angles approaching 0°, 0°, 180°, but since angles must be positive, C approaches 180° but can't reach it.Therefore, angle C is in [60°, 180°).Similarly, angle B, in a right-angled triangle with angles 30°, 60°, 90°, B is 60°, which is allowed. If triangle is 45°, 45°, 90°, then since A ≤ B ≤ C, B is 45°, which is less than 90°, so B can be 45°. But in this case, angle C is 90°, which is allowed. However, angle B cannot be 90°, as shown before.So, yes, angle B must be less than 90°, and greater than 0°, so (0°, 90°).Therefore, the answers for part a) are:1) Angle A: (0°, 60°]2) Angle B: (0°, 90°)3) Angle C: [60°, 180°)Now, moving on to part b). The problem states: The area of a triangle with sides a ≤ b ≤ c is 1. What values can sides a, b, and c take?First, recall that in a triangle, the sum of any two sides must be greater than the third. Since a ≤ b ≤ c, the most restrictive condition is that a + b > c. Because c is the largest side.Also, the area is given as 1. For a triangle with sides a, b, c, the area can be calculated using Heron's formula: Area = √[s(s - a)(s - b)(s - c)], where s = (a + b + c)/2. But Heron's formula might be complex here. Alternatively, we can use the formula: Area = (1/2)ab sin C, where C is the angle between sides a and b. But since we don't know the angles, maybe Heron's formula is necessary. However, since the sides are ordered a ≤ b ≤ c, perhaps we can relate the area to the sides using inequalities.Also, for a triangle with sides a ≤ b ≤ c, the triangle is acute, right, or obtuse depending on whether c² < a² + b², c² = a² + b², or c² > a² + b². However, the area is related to the sides and the angles. But since the area is fixed at 1, perhaps we can find bounds on the sides.Let me think about the minimal and maximal possible values for each side.Starting with side a. Since a is the smallest side, it cannot be too large, but how small can it be?If a approaches 0, then the triangle becomes very "flat", with sides b and c approaching each other. However, the area would approach 0, but our area is fixed at 1. Therefore, a cannot be too small. Similarly, a cannot be too large, because if a increases, then sides b and c must be at least a, so the triangle would be larger, but the area is fixed at 1.Wait, actually, for a given area, the sides can't be too small or too large. Let's think of the minimal possible value for a.The minimal a occurs when the triangle is as "flat" as possible with area 1. The minimal a would correspond to when the triangle is degenerate, but since we need a non-degenerate triangle, a must be such that the height corresponding to side a is 2/a, since area = 1 = (1/2)*a*h ⇒ h = 2/a. But for the triangle to be non-degenerate, the height must be less than the other sides. Wait, but actually, the height relates to side a. Let me formalize this.In a triangle, the area is (1/2)*base*height. If we take a as the base, then the height h corresponding to base a must satisfy h = 2/a. But in the triangle, the height h must be less than or equal to the other sides. Since sides b and c are at least a, but the height is related to angle between sides. Alternatively, using the formula for area in terms of sides and angles: Area = (1/2)*b*c*sin A, where A is the angle between sides b and c. Wait, but since a is the smallest side, perhaps using the Law of Sines: a/sin A = b/sin B = c/sin C = 2R, where R is the radius of the circumscribed circle. But maybe this is getting too complex. Let's think of the minimal value of side a. The area is 1, so (1/2)*a*h = 1 ⇒ h = 2/a. The height h must satisfy h ≤ b and h ≤ c, since the height is the altitude from the vertex opposite side a to side a. Since b and c are at least a, then h ≤ b and h ≤ c. But since b ≥ a, h = 2/a must be ≤ b ≥ a. Therefore, 2/a ≤ b ≥ a ⇒ 2/a ≤ b. But since b ≥ a, then 2/a ≤ b ≥ a ⇒ 2/a ≤ b ≥ a ⇒ 2/a ≤ b. But since b ≥ a, the inequality 2/a ≤ b is automatically satisfied if 2/a ≤ a ⇒ a² ≥ 2 ⇒ a ≥ √2. But this may not hold. Wait, maybe this approach is not correct.Alternatively, consider that in a triangle, the area is 1, so using Heron's formula:1 = √[s(s - a)(s - b)(s - c)]But Heron's formula might not be the easiest way here.Alternatively, we can use the fact that for a triangle with sides a ≤ b ≤ c, the maximum area is achieved when the triangle is right-angled, but not sure.Wait, the area of a triangle is given by (1/2)*ab*sin θ, where θ is the angle between sides a and b. To maximize the area, sin θ should be maximized, i.e., θ = 90°, giving area (1/2)*ab. To have the area as 1, (1/2)*ab = 1 ⇒ ab = 2. But in our case, the triangle is not necessarily right-angled, so the area is 1, which could be less than or equal to (1/2)*ab. Therefore, (1/2)*ab ≥ 1 ⇒ ab ≥ 2. So the product ab must be at least 2. But since a ≤ b ≤ c, and by triangle inequality, a + b > c. Also, since c ≥ b ≥ a, then c < a + b ≤ a + c ⇒ which is always true. The most restrictive condition is a + b > c.But since c ≥ b, we have a + b > c ≥ b ⇒ a + b > b ⇒ a > 0, which is already given.But combining the area condition with the triangle inequality. Let's see.From the area formula (1/2)*ab*sin θ = 1, so sin θ = 2/(ab). Since sin θ ≤ 1, we have 2/(ab) ≤ 1 ⇒ ab ≥ 2. Therefore, the product ab must be at least 2.Since a ≤ b, then a² ≤ ab ⇒ a² ≤ ab ≥ 2 ⇒ a² ≤ 2 ⇒ a ≤ √2. But wait, a ≤ b, so ab ≥ a². So if ab ≥ 2, then a² ≤ ab ≥ 2 ⇒ a² ≤ 2 ⇒ a ≤ √2. But this is a contradiction, because if a ≤ √2, then ab ≥ 2 ⇒ b ≥ 2/a ≥ 2/√2 = √2. So, given that a ≤ b, and ab ≥ 2, then b ≥ 2/a, and since a ≤ b, then a ≤ b ≥ 2/a ⇒ a ≤ 2/a ⇒ a² ≤ 2 ⇒ a ≤ √2. Therefore, the minimal possible value of a is when a is as small as possible, but given that b ≥ 2/a and a ≤ b. So as a approaches 0, b must approach infinity to satisfy ab ≥ 2, but also, the triangle inequality a + b > c must hold. Since c is the largest side, c ≤ a + b. But if a approaches 0, and b approaches infinity, c can be up to a + b ≈ b. So for very small a, sides b and c are approximately equal and large, but to maintain area 1, the triangle must be very "flat" with height approaching 0. Wait, but the area is fixed at 1. Hmm, this seems conflicting. Wait, perhaps my earlier conclusion that ab ≥ 2 is correct, but how does that relate to the triangle's sides? Let's consider specific examples.If a = 1, then to satisfy ab ≥ 2, b must be at least 2. Then, since c must be at least b, so c ≥ 2. Also, triangle inequality a + b > c ⇒ 1 + 2 > c ⇒ c < 3. So c ∈ [2, 3). Then, the sides would be a=1, b ≥2, c ∈ [b, 3). But the area must be 1. Let's take a=1, b=2, c must be in [2, 3). Let's check if such a triangle can have area 1. Using Heron's formula: s = (1 + 2 + c)/2 = (3 + c)/2. Area = √[s(s - 1)(s - 2)(s - c)]. Setting this equal to 1:√[s(s - 1)(s - 2)(s - c)] = 1Square both sides:s(s - 1)(s - 2)(s - c) = 1But s = (3 + c)/2, so s - c = (3 + c)/2 - c = (3 - c)/2Therefore,[(3 + c)/2] * [(3 + c)/2 - 1] * [(3 + c)/2 - 2] * [(3 - c)/2] = 1Simplify each term:First term: (3 + c)/2Second term: (3 + c)/2 - 1 = (3 + c - 2)/2 = (1 + c)/2Third term: (3 + c)/2 - 2 = (3 + c - 4)/2 = (c - 1)/2Fourth term: (3 - c)/2Multiply them all together:[(3 + c)/2] * [(1 + c)/2] * [(c - 1)/2] * [(3 - c)/2] = 1Combine terms:[(3 + c)(3 - c)(c + 1)(c - 1)] / 16 = 1Notice that (3 + c)(3 - c) = 9 - c², and (c + 1)(c - 1) = c² - 1Therefore:(9 - c²)(c² - 1) / 16 = 1Multiply both sides by 16:(9 - c²)(c² - 1) = 16Expand the left side:9c² - 9 - c⁴ + c² = -c⁴ + 10c² - 9So:-c⁴ + 10c² - 9 = 16Bring all terms to one side:-c⁴ + 10c² - 25 = 0 ⇒ c⁴ -10c² +25 = 0This is a quadratic in terms of c²:Let x = c²:x² -10x +25 =0Solve:x = [10 ± √(100 -100)] / 2 = [10 ± 0]/2 = 5Therefore, c² =5 ⇒ c=√5 ≈2.236Therefore, when a=1, b=2, c=√5, the area is 1. Check using Heron's formula:s = (1 + 2 + √5)/2 ≈ (3 + 2.236)/2 ≈2.618Area = √[2.618*(2.618 -1)*(2.618 -2)*(2.618 -√5)]Calculate each term:s ≈2.618s -1 ≈1.618s -2 ≈0.618s -√5 ≈2.618 -2.236≈0.382Multiply them all: 2.618 *1.618*0.618*0.382≈ let's compute step by step:2.618 *1.618 ≈4.2360.618 *0.382≈0.236Then 4.236 *0.236≈1.000, so yes, the area is 1. Therefore, for a=1, b=2, c=√5≈2.236, the area is 1, and the triangle inequalities are satisfied since 1 + 2 > √5 ≈2.236, and 1 + √5 >2, and 2 + √5 >1. So this is a valid triangle. Therefore, a=1 is possible. But what is the minimal possible a? Suppose we take a approaching 0. Then, to maintain area 1, we need the height h = 2/a approaching infinity. But in a triangle, the height can't exceed the other sides. Since sides b and c are at least a, but if a approaches 0, then b and c must be at least something to allow the height to be 2/a. Wait, but if a is approaching 0, then sides b and c must satisfy b ≥ a and c ≥ b, but the height h = 2/a must be less than or equal to b and c. Therefore, 2/a ≤ b and 2/a ≤ c. But since b ≤ c, we have 2/a ≤ b ≤ c. But if a approaches 0, then 2/a approaches infinity, so b and c must also approach infinity. However, the triangle inequality requires that a + b > c. If a is approaching 0, and b and c are approaching infinity with b ≤ c, then a + b ≈ b must be > c, but since c ≥ b, this is only possible if b approaches c from below. So c = b + ε, where ε approaches 0. Therefore, in this scenario, as a approaches 0, b and c approach infinity such that c = b + ε and h = 2/a ≈ b (since in a very "flat" triangle, the height h is approximately equal to the other sides). But this seems complex. Alternatively, perhaps the minimal value of a is achieved when the triangle is degenerate, but since area is fixed at 1, a cannot be zero. Therefore, perhaps there is no minimal a, as a can approach zero, but then sides b and c must become very large. But practically, since sides can be arbitrarily large, a can be arbitrarily small. However, in reality, with the area fixed at 1, the product ab must be at least 2, as before. Therefore, as a approaches 0, b must approach infinity to keep ab ≥ 2. So technically, a can be any positive real number, but subject to the triangle inequalities and ab ≥ 2. But since a can be made arbitrarily small by taking b and c large enough, the minimal value of a is 0 (exclusive), and there's no upper bound? Wait, but no, upper bound exists.Wait, if a is maximized, since a ≤ b ≤ c, the maximum a occurs when all sides are equal, i.e., an equilateral triangle. For an equilateral triangle with area 1, the side length can be calculated. Area of an equilateral triangle is (√3/4)a² =1 ⇒ a² = 4/√3 ⇒ a = 2/(3^{1/4}) ≈1.5197. But in this case, a = b = c ≈1.5197. However, if we consider a ≤ b ≤ c, even in the equilateral case, all sides are equal, so a = b = c. Therefore, the maximum possible a is when the triangle is equilateral. So a can be up to 2/(3^{1/4}) ≈1.5197. But wait, perhaps there are triangles with a larger a but still with area 1. For example, take a very "tall" isoceles triangle with base a and two equal sides b and c. If we fix the area as 1, then (1/2)*a*h =1 ⇒ h = 2/a. The height h must satisfy h < b (since in a triangle, the height is less than the other sides). So, h =2/a < b. But since in an isoceles triangle with sides b, b, and a, by the Pythagorean theorem, h = √(b² - (a/2)²). Therefore:√(b² - (a²)/4) = 2/aSquaring both sides:b² - a²/4 = 4/a² ⇒ b² = 4/a² + a²/4Since a ≤ b, then a ≤ √(4/a² + a²/4)Square both sides:a² ≤ 4/a² + a²/4 ⇒ a² - a²/4 ≤ 4/a² ⇒ (3/4)a² ≤ 4/a² ⇒ 3a⁴ ≤ 16 ⇒ a⁴ ≤ 16/3 ⇒ a ≤ (16/3)^{1/4} ≈1.5197, same as the equilateral triangle case. Therefore, the maximal possible a is indeed achieved when the triangle is equilateral, so a_max = 2/(3^{1/4}) ≈1.5197. Therefore, a ∈ (0, 2/(3^{1/4})]Similarly, for side b. Since a ≤ b ≤ c, and the area is 1. Let's think about the possible values for b. The minimal value of b occurs when a is as large as possible, which is when the triangle is equilateral, so a = b = c ≈1.5197. But if a is smaller, then b can be as small as a, but also has to satisfy ab ≥2 (from the area formula). Wait, ab ≥2 comes from the formula (1/2)*ab*sin θ =1 ⇒ ab ≥2. But sin θ ≤1, so this is a necessary condition. Therefore, for given a, b must be ≥2/a. But since a ≤ b ≤c, then a ≤ b and b ≥2/a. Therefore, combining these, 2/a ≤ b ≥a ⇒ a ≤ b and 2/a ≤ b ⇒ max(a, 2/a) ≤ b. But since a ≤ b, and 2/a ≤ b, then b ≥ max(a, 2/a). Therefore, for a given a, the minimal b is max(a, 2/a). However, since a can range from approaching 0 to 2/(3^{1/4}), we have:- When a ≤ √2, since 2/a ≥ a when a ≤ √2, so minimal b is 2/a.- When a ≥ √2, then a ≥ 2/a, so minimal b is a.But wait, let's check when a = √2, then 2/a = √2, so max(a, 2/a) = √2. Therefore, the critical point is at a=√2.But since the maximum a is 2/(3^{1/4}) ≈1.5197, which is less than √2 ≈1.4142. Wait, no: 2/(3^{1/4}) ≈1.5197 is greater than √2 ≈1.4142. Therefore, when a is in (0, 2/(3^{1/4})], the minimal b is max(a, 2/a). So, since 2/(3^{1/4}) ≈1.5197 > √2 ≈1.4142, so for a from 0 to √2, minimal b is 2/a; for a from √2 to 2/(3^{1/4}), minimal b is a. Therefore, for side b:- The minimal value is 2/a when a ∈ (0, √2]- The minimal value is a when a ∈ [√2, 2/(3^{1/4})]But since a ≤ b ≤ c, and we need to find the possible values of b regardless of a. So the minimal possible b is when a is maximized, which is in the equilateral triangle case, where a = b = c ≈1.5197. Therefore, the minimal b is ≈1.5197. But wait, if a is smaller, say a=1, then minimal b is 2/a=2, which is larger than 1.5197. Therefore, the minimal possible b is ≈1.5197 (the equilateral case). But also, when a approaches 0, minimal b approaches infinity (since b ≥2/a), so the maximum possible b is unbounded? No, because of the triangle inequality. Wait, if a approaches 0, then b must approach infinity (to satisfy b ≥2/a), and c must be at least b. But triangle inequality requires a + b > c. Since c ≥ b, and a + b > c ≥ b ⇒ a + b > b ⇒ a >0, which is already satisfied. So as a approaches 0, b approaches infinity, and c can be in [b, a + b) ≈ [b, b). So c approaches b from below. Therefore, c ≈b, and the triangle becomes very "flat" with sides a≈0, b≈c≈large, and area=1. Therefore, side b can be arbitrarily large, as a approaches 0 and b approaches infinity. Therefore, the maximal value of b is unbounded (i.e., can be any real number greater than or equal to the minimal value). But the problem is in terms of possible values, so side b can take values from approximately1.5197 to infinity. However, we need to confirm. For example, can we have a triangle with b=3, area=1?Yes. Let's take a=1/3, then to satisfy ab ≥2 ⇒ (1/3)*b ≥2 ⇒ b ≥6. But if we set a=1/3, b=6, c must be in [6, 1/3 +6)= [6, 6.333...). Let's check if such a triangle can have area 1. Using the area formula (1/2)*a*b*sin θ =1 ⇒ (1/2)*(1/3)*6*sin θ=1 ⇒ (1/2)*2*sin θ=1 ⇒ sin θ=1 ⇒ θ=90°. So this is a right-angled triangle with sides 1/3, 6, and hypotenuse c=√( (1/3)^2 +6^2 )≈√(0.111+36)=√36.111≈6.008. Which satisfies the triangle inequality, and c≈6.008 is in [6,6.333). So yes, such a triangle exists. Therefore, b can be as large as desired by taking a small enough. Therefore, side b can range from approximately1.5197 to infinity. But more precisely, since the minimal b is achieved in the equilateral triangle, which is 2/(3^{1/4}). So, b ∈ [2/(3^{1/4}), ∞)Similarly, for side c. Since c is the largest side, c must be at least b, which is at least a. The minimal c occurs when the triangle is equilateral, so c=2/(3^{1/4})≈1.5197. The maximal c is unbounded, as in the case where a approaches 0, b and c approach infinity. But c must satisfy the triangle inequality a + b > c. So, for a given a and b, c < a + b. But since a and b can be as large as needed (with b ≥2/a), c can approach a + b from below. Therefore, as a approaches 0 and b approaches infinity, c can approach a + b ≈b, which can be made arbitrarily large. Therefore, c can take any value greater than or equal to the minimal c (equilateral case) up to infinity. Therefore, side c ∈ [2/(3^{1/4}), ∞)But we need to verify if c can actually reach down to 2/(3^{1/4}), which is the equilateral triangle case. Yes, because in that case, all sides are equal and the area is 1. Therefore, summarizing:1) Side a: (0, 2/(3^{1/4})]2) Side b: [2/(3^{1/4}), ∞)3) Side c: [2/(3^{1/4}), ∞)But wait, in the equilateral triangle, all sides are equal, so a = b = c ≈1.5197. Therefore, the minimal b and c are equal to a_max.However, for non-equilateral triangles, b and c can be larger. But for part b) questions:1) side a of the triangle can take values in (0, 2/(3^{1/4})]2) side b can take values in [2/(3^{1/4}), ∞)3) side c can take values in [2/(3^{1/4}), ∞)But let's verify this with an example. Suppose a triangle with sides a=1, b=2, c=√5≈2.236. Here, a=1 is within (0,1.5197], b=2 is within [1.5197, ∞), and c≈2.236 is within [1.5197, ∞). This checks out.Another example: a=1.5 (which is less than 1.5197), b must be at least max(a,2/a)=max(1.5, 2/1.5≈1.333)=1.5. So b=1.5, and then c must be ≥b=1.5. But to have area 1, let's check.If a=1.5, b=1.5, then it's an isoceles triangle with sides 1.5,1.5,c. The area is 1. Using the formula for area of an isoceles triangle: (1/2)*a*b*sin θ =1. But a=b=1.5, so (1/2)*(1.5)^2*sin θ=1 ⇒ (2.25/2)*sin θ=1 ⇒ 1.125*sin θ=1 ⇒ sin θ=1/1.125≈0.888. Then θ≈62.7 degrees. Then side c can be found via the Law of Cosines:c² = a² + b² - 2ab cos θ = 1.5² +1.5² -2*1.5*1.5*cos 62.7°Calculate:1.5² =2.25So c²=2.25 +2.25 -4.5*cos62.7≈4.5 -4.5*0.46≈4.5 -2.07≈2.43Therefore, c≈1.56, which is greater than b=1.5. So this is valid, and c≈1.56 is within [1.5197, ∞). But in this case, a=1.5 is allowed, b=1.5 is allowed, and c≈1.56 is allowed. Therefore, the ranges seem correct.But I need to express the answers in exact terms, not approximate. Since 2/(3^{1/4}) can be written as 2*3^{-1/4} or 2*3^{3/4}/3, but maybe it's better to write it as 2 divided by the fourth root of 3. Alternatively, using exponents:2/(3^{1/4}) = 2 * 3^{-1/4}Alternatively, recognize that 3^{1/4} is the square root of sqrt(3), so:2 / sqrt(sqrt(3)) = 2 / (3^{1/4})But perhaps it's better to rationalize or write in another form. Alternatively, we can square the equilateral triangle area formula:Area = (√3 /4) a² =1 ⇒ a² =4/√3 ⇒ a=2/(3^{1/4})Therefore, the maximal a is 2/(3^{1/4}), which is exact.Similarly, the minimal b and c are also 2/(3^{1/4}).Therefore, the answers for part b) are:1) a ∈ (0, 2/3^{1/4}]2) b ∈ [2/3^{1/4}, ∞)3) c ∈ [2/3^{1/4}, ∞)But the problem states "What values can... take?" So the answer for each is:1) Side a can take any value in the interval (0, 2/3^{1/4}]2) Side b can take any value in [2/3^{1/4}, ∞)3) Side c can take any value in [2/3^{1/4}, ∞)But to confirm, let's check if a can actually be 2/3^{1/4}. In the equilateral triangle, yes, with a=2/3^{1/4}, area is 1. So yes, the upper limit is attainable.Similarly, for sides b and c, the minimal value is attainable in the equilateral triangle. Therefore, the final answers are:a) 1) Angle A: (0°, 60°] 2) Angle B: (0°, 90°) 3) Angle C: [60°, 180°)b) 1) Side a: (0, 2/3^{1/4}] 2) Side b: [2/3^{1/4}, ∞) 3) Side c: [2/3^{1/4}, ∞)Expressed using LaTeX:For part a:1) (boxed{(0^circ, 60^circ]})2) (boxed{(0^circ, 90^circ)})3) (boxed{[60^circ, 180^circ)})For part b:1) (boxed{left(0, dfrac{2}{sqrt[4]{3}}right]})2) (boxed{left[dfrac{2}{sqrt[4]{3}}, inftyright)})3) (boxed{left[dfrac{2}{sqrt[4]{3}}, inftyright)})</think>a) Given a triangle (ABC) with angles (A leqslant B leqslant C):1. Angle (A): Since (A) is the smallest angle, it must be less than or equal to (60^circ) (the angle in an equilateral triangle) and greater than (0^circ). Therefore, (A) can take values in ((0^circ, 60^circ]).2. Angle (B): (B) is the middle angle, so it must be at least (A) and less than (90^circ). The upper bound is due to the triangle inequality and the sum of angles. Thus, (B) can take values in ((0^circ, 90^circ)).3. Angle (C): As the largest angle, (C) must be at least (60^circ) (equilateral triangle) and less than (180^circ). Hence, (C) can take values in ([60^circ, 180^circ)).b) For a triangle with sides (a leqslant b leqslant c) and area 1:1. Side (a): The smallest side (a) is maximized in an equilateral triangle. Using the area formula for an equilateral triangle, we find (a leqslant frac{2}{sqrt[4]{3}}). Thus, (a) can take values in (left(0, frac{2}{sqrt[4]{3}}right]).2. Side (b): The minimal value of (b) occurs in the equilateral triangle, giving (b geqslant frac{2}{sqrt[4]{3}}). There is no upper bound due to the potential for very "flat" triangles. Hence, (b) can take values in (left[frac{2}{sqrt[4]{3}}, inftyright)).3. Side (c): Similarly, the minimal value of (c) is also in the equilateral case, and there is no upper bound. Therefore, (c) can take values in (left[frac{2}{sqrt[4]{3}}, inftyright)). Final Answera) 1. (boxed{(0^circ, 60^circ]})2. (boxed{(0^circ, 90^circ)})3. (boxed{[60^circ, 180^circ)})b) 1. (boxed{left(0, dfrac{2}{sqrt[4]{3}}right]})2. (boxed{left[dfrac{2}{sqrt[4]{3}}, inftyright)})3. (boxed{left[dfrac{2}{sqrt[4]{3}}, inftyright)}),<think>Okay, so I need to solve problem 45 parts a and b. Let's start with part a.Problem a): We have a triangle ABC with angles A, B, C such that A ≤ B ≤ C. We need to determine the possible values each of these angles can take.First, I remember that in any triangle, the sum of the angles is 180 degrees. Also, since A ≤ B ≤ C, angle C is the largest angle, and angle A is the smallest. So, the angles must satisfy A ≤ B ≤ C and A + B + C = 180°. Since A is the smallest angle, it can't be more than 60 degrees, right? Because if all three angles were equal, they would each be 60°, and since A is the smallest, it must be less than or equal to 60°. But wait, actually, if all angles are equal, it's an equilateral triangle, so each is 60°, so in that case A = B = C = 60°, which is allowed. So, A can be up to 60°, but not more. But if the triangle isn't equilateral, then A would be less than 60°, right?But also, since A ≤ B ≤ C, each angle must be at least as big as the previous one. So, angle B has to be at least angle A, and angle C has to be at least angle B. Also, in a triangle, the largest angle must be less than 180°, but since we have three angles summing to 180°, each angle must be less than 180°, but more precisely, since all angles are positive, each angle must be greater than 0° and less than 180°.But given the ordering A ≤ B ≤ C, let's think about the minimal and maximal possible values for each angle.Starting with angle A: since it's the smallest angle, its maximum possible value is when all angles are equal, which is 60°. So, A ≤ 60°. The minimum value for angle A would be approaching 0°, but it can't be zero. So, 0° < A ≤ 60°.But wait, in reality, there's a lower bound because if A is too small, then angles B and C have to be larger, but they are constrained by the ordering A ≤ B ≤ C. Let me think. If A is approaching 0°, then B and C would have to sum to nearly 180°, with B ≤ C. The minimal possible value of A is actually determined by the condition that B must be at least A, and C must be at least B.Wait, suppose A is very small, say approaching 0°. Then B and C must each be at least A (approaching 0°), but since A ≤ B ≤ C, and B + C = 180° - A. If A approaches 0°, then B can be as small as approaching A (which is approaching 0°), but then C would approach 180°, which is not possible because in a triangle, all angles must be less than 180°, but actually, angles must also satisfy that the sum is 180°. Hmm.Wait, if A approaches 0°, then B can also approach 0°, but then C would approach 180°, but C must be greater than or equal to B, which is approaching 0°, so C would have to be approaching 180°, but since in a triangle, each angle must be less than 180°, C can approach 180° but can't be equal to or exceed 180°. So, the minimal angle A can approach 0°, but can't be 0°.Similarly, angle B must be at least A and at most C. So, if A is minimal (approaching 0°), then B can range from approaching 0° up to (180° - approaching 0°)/2, because if A is approaching 0°, then B + C = 180° - A ≈ 180°, and since B ≤ C, the maximum B can be is when B = C, so each would be 90°. So, if A is approaching 0°, then B can approach up to 90°, but since B must be at least A (approaching 0°), so B ∈ (0°, 90°).But in general, for angle B, given A ≤ B ≤ C and A + B + C = 180°, what are the possible values? Let's see. Since A ≤ B, then 3A ≤ A + B + C = 180° ⇒ A ≤ 60°, which we already know. Also, since B ≤ C, then B ≤ (180° - A - B) ⇒ 2B ≤ 180° - A ⇒ 2B + A ≤ 180°. But since A ≤ B, substituting A with B gives 2B + B ≤ 180° ⇒ 3B ≤ 180° ⇒ B ≤ 60°. Wait, so angle B is also at most 60°? Wait, that can't be right, because if angle C is the largest angle, then if B is 60°, C would have to be at least 60°, but then A + B + C would be at least A + 60° + 60° ≥ 60° + 60° + 60° = 180°, but since A ≤ B, which is 60°, so if A is 60°, then all angles are 60°, which is okay. But if B is 60°, then C must be at least 60°, but the sum A + B + C would be 60° + 60° + C = 120° + C. For the total to be 180°, C would have to be 60°, so all angles are 60°. Therefore, if B is 60°, then A must also be 60°, and C is 60°, so the triangle is equilateral.But if B is less than 60°, say B is 50°, then A can be up to 50°, but then C would be 180° - A - B. If A is 50°, B is 50°, then C is 80°, which is greater than B, so that's okay. So in that case, B can be less than 60°, but if B is greater than 60°, would that be possible?Wait, let's test with B = 70°, then since A ≤ B, so A ≤ 70°, and C ≥ B =70°, so angles would be A ≤70°, 70°, C ≥70°, sum would be A + 70° + C =180°. Let's set A =70°, then C=40°, but C must be ≥70°, so that's impossible. If A is smaller, say 60°, then C=180 -60 -70=50°, which is still less than B=70°, which contradicts C ≥ B. Therefore, B cannot exceed 60°, because otherwise, even if A is as small as possible, C would have to be less than B to satisfy the sum, which violates C ≥ B.Therefore, B must be ≤60°, and since B ≥ A, and A ≤60°, that makes sense. So angle B is between A and 60°, but A can vary. Wait, but if B must be ≤60°, then the maximum value of B is 60°, which occurs when all angles are 60°, and the minimum value of B is equal to A, which can be approaching 0°, but B can't be less than A, which is approaching 0°, so B's minimum is approaching 0° as well. Wait, no. If A is approaching 0°, B must be at least A (approaching 0°), but B can be as small as A, which is approaching 0°, so B can also approach 0°, but since B ≤ C, and C would then approach 180°, but in reality, the minimal value of B is determined by the other angles.Wait, maybe we need to formalize this.Given A ≤ B ≤ C and A + B + C = 180°, let's try to find the range for each angle.Starting with angle A:Since A is the smallest angle, we can write A ≤ B and A ≤ C. Also, since A + B + C = 180°, we have 3A ≤ 180° ⇒ A ≤ 60°, which is the upper bound for A. The lower bound for A is greater than 0°, as all angles must be positive. So, angle A ∈ (0°, 60°].But wait, actually, even if A is approaching 0°, but then B and C would adjust accordingly. So technically, the minimal value for A is approaching 0°, but not including 0°, and the maximum is 60°.Now angle B:Since A ≤ B ≤ C, and A + B + C = 180°, we can express C = 180° - A - B. Then, since B ≤ C, we have B ≤ 180° - A - B ⇒ 2B ≤ 180° - A ⇒ B ≤ (180° - A)/2.Also, since B ≥ A, we have A ≤ B ≤ (180° - A)/2.But we also know that C ≥ B, so substituting C, we get 180° - A - B ≥ B ⇒ 180° - A ≥ 2B ⇒ B ≤ (180° - A)/2, which is the same as above.Therefore, for angle B, given A, the possible range is A ≤ B ≤ (180° - A)/2.But we need to find the overall range for B, considering that A can vary from near 0° to 60°.First, let's find the minimal possible B. The smallest B can be is when A is as large as possible. Wait, since B ≥ A, so the minimal B occurs when A is as large as possible, which is 60°, so B would be 60°. But if A is 60°, then B and C must also be 60°, so in that case, B =60°.But if A is approaching 0°, then B can be as small as approaching 0°, but since B ≥ A, which is approaching 0°, B can approach 0° as well. Wait, but if A is approaching 0°, B must be at least A, so B can be approaching 0°, but then C would approach 180° - 0° - 0° =180°, but C must be ≥ B, which is approaching 0°, so that's allowed. However, in reality, in a triangle, the sum of the two smaller sides must be greater than the largest side, but here we are dealing with angles. Wait, angles don't have the triangle inequality, sides do. So maybe the angles can approach these extremes as long as the sides satisfy the triangle inequality.But for angles, even if angles A and B approach 0°, as long as the sides opposite them are positive, it's possible. However, in reality, if angles A and B approach 0°, the triangle becomes very "sharp" with a very small angle at A and B, and a very large angle at C. So, theoretically, angles can approach 0°, but cannot be 0°.Therefore, angle B's possible range is (0°, 60°]. Wait, but if A is approaching 0°, B can approach 0°, and if A is 60°, B is 60°, so B can range from approaching 0° up to 60°, inclusive.But we need to confirm. Let's take an example. If A = 30°, then B must be ≥30° and ≤ (180° -30°)/2 =75°, so B ∈ [30°,75°]. But since B ≤ C, and C =180° -30° -B =150° -B, which must be ≥ B, so 150° -B ≥ B ⇒ B ≤75°, which is the same as above. So, in this case, if A=30°, B ∈[30°,75°]. Wait, but this contradicts the previous conclusion that B is up to 60°. Wait, so where is the mistake?Wait, earlier when I thought B can't exceed 60°, but in this case, when A=30°, B can go up to75°, which is greater than 60°, but in this case, angle C would be 150° - B. If B=75°, then C=75°, so angles would be 30°,75°,75°, which satisfies A ≤ B ≤ C (30 ≤75≤75). But 75° is greater than 60°, which contradicts the previous conclusion that B cannot exceed 60°. So my earlier reasoning was wrong.So where did I go wrong? Let's re-examine.Earlier, I considered that if B exceeds 60°, then even if A is as small as possible, C would have to be less than B. Let's test with B=70°, and A=30°, then C=180 -30 -70=80°, which is greater than B=70°, so C=80°, which is okay. So in this case, B=70°, which is greater than 60°, but C=80° which is greater than B, so that's allowed.Wait, so my previous conclusion that B cannot exceed 60° was incorrect. Therefore, angle B can actually be greater than 60°, as long as C is greater than or equal to B. So, let's correct that.So, angle B can actually go up to 90°, but not more. Let's see. Suppose A is 30°, B=75°, C=75°, as before. Now, if we try to make B even larger. Let's say A=20°, B=80°, then C=180 -20 -80=80°, so angles are 20°,80°,80°, which satisfies A ≤ B ≤ C. Here, B=80°, which is greater than 60°, but that's okay.If A=10°, B=85°, then C=85°, angles 10°,85°,85°, sum 180°, which is valid.But what is the upper bound for B? If B approaches 90°, then C approaches 90°, and A approaches 0°, because A =180 - B - C ≈180 -90 -90=0°. So, as A approaches 0°, B and C approach 90°, but can't reach 90°, since A must be positive. Wait, but in reality, if A is exactly 0°, then B and C would be 90°, but angles can't be 0°, so B can approach 90°, but not reach it.Wait, let's check if B can be equal to 90°. If B=90°, then C must be at least 90°, so angles would be A ≤90° ≤ C. Then A +90° + C=180° ⇒ A + C=90°. But since C ≥90°, then A + C=90° ⇒ A=90° - C ≤90° -90°=0°, which is impossible because A must be positive. Therefore, B cannot reach 90°, so the maximum possible value for B is approaching 90°, but not equal to it.Therefore, angle B can range from (when A is approaching 0°) approaching 0° up to approaching 90°, but in reality, when A is fixed, say at 30°, then B can go up to (180 -30)/2=75°, as in the previous example. So, the upper bound for B depends on A. Therefore, angle B's maximum possible value is approaching 90°, but only when A is approaching 0°.But how do we express the range for B? Since B is between A and (180° - A)/2, but A can vary from near 0° to 60°, so when A is 60°, B must be 60°, and when A approaches 0°, B can approach up to 90°, but not reach it.Therefore, the range for angle B is (0°, 90°), but considering that B must be at least A. Wait, no, when A is larger, B's upper bound is lower. For example, if A=50°, then B can be from 50° up to (180 -50)/2=65°, so in this case, B ∈[50°,65°].Therefore, the possible values of angle B depend on angle A. However, the problem asks for the values each angle can take, given A ≤ B ≤ C. So, the answer needs to be the possible ranges for each angle irrespective of the others, just under the condition A ≤ B ≤ C.So, for angle A: since it's the smallest, as we saw, it can be from near 0° up to 60°, so (0°,60°].For angle B: since it's between A and C, and C is at least B, but also, the sum of angles. The minimal value of B is equal to A (so starting from A's minimum, near 0°, up to 60°), and the maximum value of B is when C is equal to B, so when the triangle is isoceles with B=C. In that case, A + 2B = 180°, so B=(180° - A)/2. The maximum possible B occurs when A is minimized (approaching 0°), so B approaches 90°. Therefore, angle B can take values from near 0° up to approaching 90°, but with the condition that B ≥ A. But since A can be up to 60°, B must be at least 60° when A=60°, but no, if A=60°, then B=60°, as all angles are 60°. So, in general, B ranges from A (which can be as low as near 0°) up to (180° - A)/2 (which can be as high as approaching 90°). Therefore, the overall range for angle B is (0°,90°). But if A is 60°, B is exactly 60°, so B can be 60° as well. Wait, but if B is 60°, then C must be at least 60°, but A=60°, so C=60°, hence all angles 60°, so B=60° is allowed. Therefore, angle B can take values from near 0° up to 90°, but not including 90°, and including 60° when all angles are 60°.But perhaps more accurately, since when A is 60°, B=60°, so B can reach 60°, but cannot exceed 90°. Therefore, angle B ∈ (0°,90°), but including 60°. Wait, but 90° is excluded because B=90° would require A=0°, which is impossible. So B can approach 90°, but not reach it.Similarly, angle C is the largest angle. Since C ≥ B ≥ A, and C =180° -A -B. Since C ≥ B, substituting B ≥ A, we have C ≥ B ≥ A. Therefore, C must be at least as large as B, which is at least as large as A. So, the minimal value for C is when A and B are as large as possible, but since C must be at least B, which is at least A.If A and B are both 60°, then C=60°, so minimal C is 60°. The maximum value of C is when A and B are as small as possible. If A approaches 0°, B approaches 0°, then C approaches 180°, but can't reach 180°, so C ∈ (60°,180°).Wait, but if A and B are both approaching 0°, then C approaches 180°, but must be less than 180°, so C can be as close to 180° as desired, but not equal. Similarly, the minimal C is 60°, when all angles are equal. Therefore, angle C ∈ [60°,180°). But wait, if A=50°, B=50°, then C=80°, which is greater than 60°, so that's okay. If A=60°, B=60°, C=60°, so that's the minimal C. If A=30°, B=75°, C=75°, so C=75° is allowed. So angle C can range from 60° up to approaching 180°, but not including 180°.Therefore, putting it all together:1) Angle A can take values in (0°,60°].2) Angle B can take values in (0°,90°).3) Angle C can take values in [60°,180°).But we need to confirm with some examples.For example, an equilateral triangle: A=B=C=60°, so all angles 60°, which fits into the ranges: A=60°, B=60°, C=60°.A right-angled triangle: Suppose angles 30°,60°,90°. Here, A=30°, B=60°, C=90°. But according to the given condition A ≤ B ≤ C, this satisfies 30 ≤60≤90. Here, angle C=90°, which is within [60°,180°). But in this case, angle B=60°, which is within (0°,90°), but 60° is allowed. Wait, but earlier I thought angle B can't reach 90°, but in this example, angle C is 90°, which is allowed. Wait, but in the problem statement, the angles must satisfy A ≤ B ≤ C. So if C=90°, then B can be up to 90°, but in this example, B=60°, which is less than C=90°. If we have a triangle with angles 45°,45°,90°, but that would violate A ≤ B ≤ C because A=45°, B=45°, C=90°, so 45 ≤45≤90, which is allowed. Here, angle B=45°, which is less than 90°, but in this case, angle B=45°, which is okay.Wait, so in the case of a right-angled isoceles triangle, angles are 45°,45°,90°, which fits into A ≤ B ≤ C (45≤45≤90). Here, angle B=45°, which is within (0°,90°), angle C=90°, which is within [60°,180°). But according to the earlier conclusion, angle B can approach 90°, but in this case, angle C is 90°, but angle B is 45°. If we try to make angle B approach 90°, then angle C must be at least B, so C would also approach 90°, but angle A approaches 0°, so angles like A=1°, B=89°, C=90°, which sum to 180°, but in this case, B=89°, C=90°, so 1≤89≤90, which is valid. Therefore, angle B can approach 90°, but not reach it. So angle B's maximum is approaching 90°, but in the case where C=90°, B can be up to 89.999...°, but not 90°.But in the example with angles 45°,45°,90°, angle B=45°, which is allowed. So angle B can be anything from near 0° up to near 90°, but not including 90°, and in some cases, like when the triangle is equilateral, angle B=60°, which is allowed.Therefore, the possible ranges are:1) Angle A: (0°,60°]2) Angle B: (0°,90°)3) Angle C: [60°,180°)But we need to ensure that these ranges are accurate. For angle C, since it's the largest angle, it must be at least 60°, because if all angles are equal, it's 60°, and if angles are unequal, C must be larger. So, C ∈ [60°,180°). Similarly, angle A must be at most 60°, so A ∈ (0°,60°]. Angle B is between A and C, and since C can be up to approaching 180°, but given that B ≤ C, and the sum constraints, B can be up to approaching 90°, as if B approaches 90°, C approaches 90°, and A approaches 0°, so B ∈ (0°,90°).Therefore, the answers for part a) are:1) Angle A: greater than 0° and up to 60°, so 0° < A ≤ 60°.2) Angle B: greater than 0° and less than 90°, so 0° < B < 90°.3) Angle C: at least 60° and less than 180°, so 60° ≤ C < 180°.Problem b): The area of a triangle with sides a, b, c, where a ≤ b ≤ c, is 1. We need to find the possible values of sides a, b, c.First, recall that in a triangle, the sides must satisfy the triangle inequality: the sum of any two sides must be greater than the third side. Also, since a ≤ b ≤ c, the largest side is c, so we must have a + b > c.Additionally, the area is given as 1. The area can be calculated using Heron's formula or using the formula (1/2)*base*height. Since the sides are ordered a ≤ b ≤ c, the largest side is c, so if we take c as the base, the height h corresponding to it must satisfy (1/2)*c*h =1 ⇒ h=2/c.But for a valid triangle with sides a, b, c, the height h must satisfy certain conditions. Also, using the triangle inequality, since a ≤ b ≤ c, we have a + b > c, and also, the other inequalities a + c > b and b + c > a, but since c ≥ b ≥ a, the most restrictive condition is a + b > c.Additionally, the area can be related to the sides via Heron's formula: Area = √[s(s -a)(s -b)(s -c)], where s=(a + b + c)/2. So, √[s(s -a)(s -b)(s -c)] =1.But this might be complicated to use directly. Maybe a better approach is to consider that for given sides a, b, c with a ≤ b ≤ c, the maximum area is achieved when the triangle is right-angled, but I'm not sure. Alternatively, consider that for fixed a and b, the maximum possible area is (1/2)*a*b*sinθ, where θ is the angle between sides a and b. The maximum area occurs when θ=90°, giving area (1/2)*a*b. Since the area is given as 1, we have (1/2)*a*b*sinθ =1 ⇒ sinθ = 2/(a*b). Since sinθ ≤1, this implies that a*b ≥2. So, for a given a and b, the product a*b must be at least 2. But also, the sides must satisfy the triangle inequality.Alternatively, maybe consider that for a triangle with sides a ≤ b ≤ c, the area is 1. We need to find the possible ranges for a, b, c.Let me start by analyzing side a. Since a is the smallest side, it can't be too large, but how small can it be?First, the triangle inequality: a + b > c. But since c is the largest side, c < a + b. Also, since a ≤ b ≤ c, we have c ≥ b ≥ a.Also, using the area formula. Let's use Heron's formula:Area = √[s(s -a)(s -b)(s -c)] =1, where s=(a + b + c)/2.But this equation is complicated. Alternatively, consider that the area can be written as (1/2)*base*height. If we take a as the base, then the height h_a corresponding to base a must satisfy (1/2)*a*h_a =1 ⇒ h_a =2/a.But the height h_a must be less than or equal to the other sides. Since in a triangle, the height corresponding to a base must be less than or equal to the other sides. For example, in a right-angled triangle, the height is equal to the other side when the angle is right. Otherwise, the height is shorter.But more formally, the height h_a can be expressed in terms of sides b and c and the angle between them. However, this might not be straightforward.Alternatively, using the formula for area in terms of sides and angles: Area = (1/2)*b*c*sinA =1, where A is the angle opposite side a. Since a ≤ b ≤ c, angle A is the smallest angle, opposite the smallest side a.But since sinA ≤1, we have (1/2)*b*c*sinA ≤(1/2)*b*c. Therefore, (1/2)*b*c ≥1 ⇒ b*c ≥2.So, the product of sides b and c must be at least 2. Also, since a ≤ b ≤ c, we have a ≤ b ≤ c.Additionally, the triangle inequalities must hold: a + b > c, a + c > b, b + c > a. But since a ≤ b ≤ c, the most restrictive is a + b > c.So, c < a + b.Also, from the area, we have b*c ≥2, as above.So, for side a, since a ≤ b ≤ c, and c < a + b, and b*c ≥2.We need to find the possible values of a, b, c.First, let's consider side a. What is the minimum possible value of a?Since a is the smallest side, and the area is 1, a can't be zero. Let's see if there's a lower bound.Suppose a approaches zero. Then, since a ≤ b ≤ c, b and c must be at least a, so they would also approach zero, but the area is 1, which is fixed. But as a approaches zero, the product b*c must approach infinity to keep b*c ≥2. Therefore, as a approaches zero, b and c must increase such that their product is at least 2, but also satisfy c < a + b ≈ b (since a approaches zero). So c < b, which contradicts c ≥ b. Therefore, as a approaches zero, c approaches b, and since c < a + b ≈ b, so c < b, which contradicts c ≥ b. Therefore, a cannot be made arbitrarily small. There must be a lower bound on a.Therefore, there must be some minimal value of a such that the conditions are satisfied.To find the minimal a, we can consider the case when a = b = c, which is an equilateral triangle. For an equilateral triangle with side length a, the area is (√3/4)a² =1 ⇒ a²=4/√3 ⇒ a=2/(3^{1/4})≈1.5197. But since in our problem, the sides are a ≤ b ≤ c, not necessarily equal. However, in the case of an equilateral triangle, all sides are equal, so a = b = c. But since the problem allows a ≤ b ≤ c, this is included.But maybe the minimal a occurs when the triangle is degenerate, but since the area is fixed at 1, it can't be degenerate. Alternatively, when the triangle is isoceles with two sides equal.Wait, perhaps the minimal a occurs when the triangle is right-angled, maximizing the area for given sides. Let's think.If we fix sides a and b, the maximum area is (1/2)*a*b, which occurs when the angle between them is 90°. So, for the area to be 1, (1/2)*a*b=1 ⇒ a*b=2. So, in this case, if the triangle is right-angled with legs a and b, then c=√(a² + b²). But since we have a ≤ b ≤ c, this would satisfy a ≤ b ≤ √(a² + b²) only if b ≤ √(a² + b²), which is always true. Wait, but in this case, c=√(a² + b²) is the hypotenuse, which is greater than both a and b, so a ≤ b ≤ c holds.But if we take a right-angled triangle with legs a and b, then c=√(a² + b²), and area=1 implies a*b=2. Also, triangle inequalities are satisfied since a + b > c ⇒ a + b > √(a² + b²). Squaring both sides: (a + b)² > a² + b² ⇒ a² + 2ab + b² > a² + b² ⇒ 2ab >0, which is true. So this is valid.Therefore, in this case, we can set a*b=2, with a ≤ b ≤ √(a² + b²). If we want to find the minimal possible a, we need to minimize a subject to a*b=2 and a ≤ b.Given a ≤ b and a*b=2, then a ≤ √(2), since if a = b, then a²=2 ⇒ a=√2≈1.4142. But since a ≤ b, to minimize a, we can let b approach infinity, but then a=2/b approaches zero. Wait, but earlier we saw that if a approaches zero, c=√(a² + b²)=b*sqrt(1 + (a/b)²)≈b + (a²)/(2b), which would be slightly larger than b. However, the triangle inequality requires that a + b > c ≈b, so a + b > b ⇒ a >0, which is true. But in this scenario, as a approaches zero, b approaches infinity such that a*b=2. However, this seems possible, but then side c≈b also approaches infinity. But the area remains 1. However, this seems contradictory because if sides b and c are approaching infinity, the area should also increase, but in our case, the area is fixed at 1. Wait, no. In a right-angled triangle, the area is (1/2)*a*b=1, so even if a approaches zero and b approaches infinity with a*b=2, the area remains 1. So in this case, the sides a and b can be such that a=2/b, with b≥a, so b≥sqrt(2), because since a ≤ b, then 2/b ≤b ⇒ b²≥2 ⇒ b≥sqrt(2). Therefore, in this case, the minimal possible a is when b is minimal, which is b=sqrt(2), hence a=2/sqrt(2)=sqrt(2). But if we allow b to be larger, then a=2/b can be smaller than sqrt(2). Wait, but if a ≤ b, then a=2/b ≤b ⇒ 2/b ≤b ⇒b²≥2 ⇒b≥sqrt(2). Therefore, the minimal possible a is when b is minimal, i.e., b=sqrt(2), hence a=sqrt(2). But if we take b larger than sqrt(2), then a=2/b would be smaller than sqrt(2). For example, take b=2, then a=1, which is less than sqrt(2)≈1.414. But in this case, a=1 ≤b=2 ≤c=√(1² +2²)=√5≈2.236. This satisfies a ≤b ≤c. Therefore, a can be as small as approaching zero, but wait, in this right-angled triangle scenario, a=2/b, and as b approaches infinity, a approaches zero. However, we need to check if all triangle inequalities are satisfied. For example, take a=1, b=2, c=√5≈2.236. Then, a + b=3 >c≈2.236; a +c≈1 +2.236=3.236> b=2; b +c≈2 +2.236=4.236> a=1. So all inequalities hold. Similarly, if a=0.1, then b=2/0.1=20, c=√(0.1² +20²)≈√(400.01)≈20.00025. Then, a +b=0.1 +20=20.1 >c≈20.00025? No, 20.1 >20.00025, which is true. So the triangle inequality holds. Therefore, even for very small a, as long as b=2/a, and c=√(a² +b²), the triangle inequality a +b >c holds because a +b= a +2/a >√(a² + (2/a)^2 ). Let's verify for a=0.1:a +b=0.1 +20=20.1c=√(0.01 +400)=√400.01≈20.00025So 20.1 >20.00025, which is true. Similarly, for a approaching zero:a +b= a +2/a ≈2/a (since a is very small)c=√(a² + (2/a)^2 )≈√(4/a²)=2/aSo, a +b ≈2/a, and c≈2/a, so a +b ≈c. But since a +b must be >c, we have 2/a >2/a, which is not true. Wait, but actually, a +b =a +2/a, which is slightly larger than 2/a, so a +b =2/a +a >2/a. Therefore, as a approaches zero, a +b approaches 2/a from above, and c approaches 2/a from below, so the inequality a +b >c holds.Therefore, in the right-angled triangle case, even as a approaches zero, the triangle inequality holds. Therefore, theoretically, a can be made arbitrarily small, approaching zero, as long as b=2/a and c=√(a² +b²). However, in reality, the problem states that the triangle has area 1, and sides a ≤b ≤c. Therefore, a can take any value in (0, ∞), but subject to a ≤b ≤c and the triangle inequalities. However, since b=2/a, and c=√(a² +4/a²), but also, we must have b ≤c.Wait, in the right-angled triangle case, c=√(a² +b²), and since b=2/a, then c=√(a² +4/a²). We need to ensure that b ≤c.Is 2/a ≤√(a² +4/a²)?Square both sides:4/a² ≤a² +4/a² ⇒0 ≤a², which is always true. Therefore, b ≤c holds. So in this case, with a right-angled triangle, a can be any positive value, with b=2/a, and c=√(a² +4/a²), and this satisfies a ≤b ≤c and the triangle inequalities. Therefore, theoretically, there's no lower bound on a; it can approach zero. However, this contradicts our earlier thought experiment where we considered that if a approaches zero, then b and c approach infinity, but the area remains 1. But mathematically, this seems possible.Wait, but in the problem statement, part b) asks "What values can the sides a, b, c take?" given that the area is 1 and a ≤b ≤c. So, according to the right-angled triangle case, a can be any positive real number, with b=2/a and c=√(a² +4/a²), as long as a ≤b ≤c. Since b=2/a and a ≤b ⇒a ≤2/a ⇒a² ≤2 ⇒a ≤√2. So, there's an upper bound on a: a ≤√2.Wait, this is a critical point. If we require a ≤b, and b=2/a, then a ≤2/a ⇒a² ≤2 ⇒a ≤√2≈1.4142. Therefore, in the right-angled triangle scenario, a can vary from approaching 0 up to √2, with b=2/a varying from approaching infinity down to 2/√2=√2, and c=√(a² +4/a²).But wait, if a is allowed up to √2, then when a=√2, b=2/√2=√2, so a=b=√2, and c=√( (√2)^2 + (√2)^2 )=√(2+2)=√4=2. So, in this case, we have an isoceles right-angled triangle with legs √2 and hypotenuse 2. The area is (1/2)*√2*√2= (1/2)*2=1, which checks out. And since a=b=√2 ≤c=2, this satisfies a ≤b ≤c.Therefore, in the right-angled triangle case, a can range from approaching 0 up to √2, inclusive. Therefore, the maximum possible value of a is √2, when the triangle is isoceles right-angled with legs √2.But this seems contradictory to the earlier example where a=1, b=2, c=√5, which also satisfies a ≤b ≤c and area=1. In this case, a=1 <√2≈1.414, so that's within the range.But what if we consider non-right-angled triangles? For example, suppose we have a triangle with sides a=2, b=2, c=2 (equilateral). The area would be (√3/4)*(2)^2=√3≈1.732, which is greater than 1. But we need area=1. So, to get area=1, the sides must be smaller. For an equilateral triangle, area=1 ⇒ side length a= (4/√3)^{1/2}≈1.5196. So, in this case, a=1.5196 ≤b=1.5196 ≤c=1.5196. This is valid, but this is a specific case.Therefore, the maximum possible a is when the triangle is equilateral? Let's see. For a given area, the equilateral triangle has the maximum area for a given side length? No, actually, for given side lengths, the equilateral triangle maximizes the area. But here, the area is fixed, so to maximize a (the smallest side), perhaps the equilateral triangle gives the maximum a. Because in other configurations, a can be smaller. For example, in the right-angled triangle, a can be up to √2≈1.414, which is less than the equilateral triangle's a≈1.5196. Therefore, the maximal a is achieved in the equilateral triangle.Wait, let's check this.For an equilateral triangle with area 1, the side length a is given by (√3/4)a²=1 ⇒a²=4/√3 ⇒a=2/(3^{1/4})≈2/1.316≈1.5196.In the right-angled isoceles triangle case, a=√2≈1.4142, which is less than 1.5196. Therefore, the equilateral triangle allows a larger minimal side a. Therefore, the maximum possible value of a is achieved when the triangle is equilateral, so a=2/(3^{1/4})≈1.5196.Therefore, the possible values of a are from approaching 0 up to approximately1.5196. But we need to find the exact value.So, the maximum a occurs when the triangle is equilateral, so a_max=2/(3^{1/4})=2*3^{-1/4}=2/(√[4]{3})=2/(3^{1/4}).Alternatively, using Heron's formula:For an equilateral triangle, s=(a+a+a)/2=3a/2.Area=√[s(s -a)^3]=√[(3a/2)(3a/2 -a)^3]=√[(3a/2)(a/2)^3]=√[(3a/2)(a³/8)]=√[(3a^4)/16]=(a²√3)/4=1 ⇒a²=4/√3 ⇒a=2/(3^{1/4})).Yes, so a_max=2/(3^{1/4}).Therefore, the range for a is (0,2/(3^{1/4})].But we need to confirm if this is indeed the maximum possible a.Suppose there's another triangle with a >2/(3^{1/4}), sides a ≤b ≤c, area=1. Is this possible?Assume a=2, but then even if the triangle is equilateral, area would be (√3/4)*4=√3≈1.732>1. To have area=1, the sides must be smaller. Therefore, if a is larger than the equilateral triangle's side, the area would be larger than 1, which contradicts the given area. Therefore, the maximal a is achieved by the equilateral triangle.Therefore, for part b) 1), side a can take values in (0, 2/(3^{1/4})].Now, moving to side b. What values can side b take?Since a ≤b ≤c, and the area is 1. We need to find the range for b.First, the minimal value of b is at least a, which can be approaching 0, but when a is at its maximum (equilateral triangle), b=a=2/(3^{1/4}).But when a is smaller, b can vary. For example, in the right-angled triangle where a approaches 0, b=2/a approaches infinity. But since a ≤b ≤c, and in the right-angled triangle case, c=√(a² +b²)=√(a² +4/a²). As a approaches 0, c approaches 2/a, so b=2/a and c≈2/a, so b ≤c, which is true.But for a fixed a, the maximal value of b would be when c is as large as possible. Wait, but in reality, for fixed a and b, c must be at least b and satisfy a +b >c. So, c <a +b and c ≥b.Therefore, for fixed a and b, c ∈[b, a +b).But the area is fixed at 1, so we have constraints from both the area and the triangle inequality.Alternatively, consider that for given a and b with a ≤b, we can solve for c such that the area is 1 and the triangle inequalities hold.Using Heron's formula:Area=√[s(s -a)(s -b)(s -c)]=1, where s=(a +b +c)/2.This is complicated, but perhaps we can use the formula for area in terms of sides and angles.Area=(1/2)*a*b*sinγ=1, where γ is the angle between sides a and b. Therefore, sinγ=2/(a*b). Since sinγ ≤1, we have a*b ≥2, as before.Additionally, using the Law of Cosines for angle γ: c²=a² +b² -2ab*cosγ.Since c ≥b, we have:a² +b² -2ab*cosγ ≥b² ⇒a² -2ab*cosγ ≥0 ⇒a ≥2b*cosγ.But since sinγ=2/(a*b), we can relate cosγ and sinγ via cosγ=√(1 -sin²γ)=√(1 -4/(a²b²)).Therefore, the inequality becomes:a ≥2b*√(1 -4/(a²b²)).This is getting complicated. Maybe a better approach is to consider that given a and b with a ≤b and a*b ≥2, we can find c such that c ∈[b, a +b) and the area is 1.But even this might not be straightforward.Alternatively, consider that for given a and b, there's a range of possible c values that satisfy both the triangle inequality and the area condition.But perhaps it's easier to consider the minimal and maximal possible values for b.The minimal value of b occurs when a is at its maximum, which is a=2/(3^{1/4})≈1.5196. In the equilateral triangle case, all sides are equal, so b=a=2/(3^{1/4}).The maximal value of b is when a is at its minimal value (approaching zero), and b=2/a approaches infinity. However, since a must be ≤b, and a approaches zero, b can approach infinity.Therefore, the range for b is [2/(3^{1/4}), ∞).Wait, but this can't be right. Because when a is at its maximum, b is also at its maximum (since in the equilateral triangle, all sides are equal). But in other configurations, when a is smaller, b can be larger.Wait, in the equilateral triangle, a=b=c=2/(3^{1/4})≈1.5196. If we have a triangle with a=1, then b can be from 1 up to ... ?Given a=1, then from the area formula, (1/2)*1*b*sinγ=1 ⇒b*sinγ=2. Since sinγ ≤1, we have b ≥2. But since a ≤b ≤c, and a=1, b≥1. But from the area requirement, b ≥2. So for a=1, b must be ≥2.Wait, let's check this. If a=1, then to have area=1, we have (1/2)*1*b*sinγ=1 ⇒b*sinγ=2. Since sinγ ≤1, then b ≥2. Also, since a=1 ≤b ≤c, and c <a +b=1 +b.So for a=1, b must be ≥2, and c ∈[b,1 +b).Therefore, in this case, b can range from 2 to infinity.Therefore, for a fixed a=1, b ∈[2, ∞). Similarly, for a=0.5, we have (1/2)*0.5*b*sinγ=1 ⇒0.25*b*sinγ=1 ⇒b*sinγ=4. Therefore, b ≥4/sinγ. But sinγ ≤1, so b ≥4. So for a=0.5, b ≥4.But since a ≤b ≤c, and a=0.5, then b ≥0.5. However, due to the area constraint, b must be ≥4. Therefore, for a=0.5, b ∈[4, ∞).Therefore, the minimal value of b for a given a is 2/a (from the area constraint, since (1/2)*a*b*sinγ=1 and sinγ ≤1 ⇒b ≥2/a).Since a ≤b, we have a ≤b ≥2/a ⇒a ≤2/a ⇒a² ≤2 ⇒a ≤√2. Which aligns with our earlier conclusion that a ≤√2.Therefore, for each a ∈(0, √2], the minimal possible b is max(a, 2/a). But since a ≤√2, then 2/a ≥√2 when a ≤√2. For example, when a=1, 2/a=2 ≥1; when a=√2, 2/a=√2.Therefore, for each a ∈(0, √2], the minimal b is 2/a, and since a ≤√2, 2/a ≥√2.Therefore, the minimal value of b across all possible triangles is √2, occurring when a=√2, b=√2, c=2 (the right-angled isoceles triangle). And the maximal value of b is unbounded, approaching infinity as a approaches 0.Thus, the range for b is [√2, ∞).Wait, but in the equilateral triangle case, a=b=c=2/(3^{1/4})≈1.5196, which is greater than √2≈1.4142. Therefore, in that case, b=1.5196, which is greater than √2. Therefore, the minimal value of b is not √2, but rather the minimal b is when a is maximal (equilateral case), b=2/(3^{1/4})≈1.5196. But earlier we saw that when a=√2≈1.4142, b=2/√2=√2≈1.4142, which is less than the equilateral case. But in that right-angled isoceles triangle, a=√2, b=√2, c=2, so b=√2, which is less than the equilateral case's b≈1.5196. Therefore, the minimal possible value of b is √2≈1.4142, and the maximal is infinity.But wait, in the equilateral triangle, b is larger than √2, but there exists triangles where b is smaller than the equilateral's b, like the right-angled isoceles triangle where b=√2≈1.4142, which is less than 1.5196. Therefore, the minimal b is √2, and the maximal a is 2/(3^{1/4})≈1.5196.But this contradicts because in the right-angled isoceles triangle, a=√2≈1.4142 <b=√2≈1.4142. Wait, no, in that case, a=b=√2, since it's isoceles. Therefore, a=√2, b=√2, c=2. So a and b are equal, hence a ≤b ≤c is satisfied.Therefore, in this case, the minimal value of b is √2, achieved when a=√2, b=√2, c=2. But in the equilateral triangle, b≈1.5196 is greater than √2≈1.4142. Therefore, the minimal possible b is √2, and the maximal possible a is achieved in the equilateral triangle.Therefore, for side b, the range is [√2, ∞).But we need to verify.Take a triangle with a=1, b=2, c=√5≈2.236. Here, a=1 <b=2 <c≈2.236. The area is 1. This satisfies b≥2/a=2/1=2. So b=2 is minimal for a=1.Similarly, for a=√2, b=√2, c=2. This gives area=1, and a ≤b ≤c.If we take a=1.5, then minimal b=2/1.5≈1.333, but since a=1.5 ≤b, then b must be at least 1.5. But 2/a≈1.333 <1.5, so the minimal b is max(a, 2/a)=1.5. Therefore, for a=1.5, b≥1.5, and the area=1 requires (1/2)*1.5*b*sinγ=1 ⇒b*sinγ=4/3≈1.333. Since sinγ ≤1, then b≥4/3≈1.333, but since b≥a=1.5, the minimal b is 1.5. Therefore, for a=1.5, the minimal b=1.5, but in this case, the triangle would need to have sinγ= (4/3)/b. If b=1.5, then sinγ= (4/3)/1.5≈0.888, which is valid.Therefore, in general, for a given a, the minimal b is max(a, 2/a). But since a ≤√2, then 2/a ≥√2 when a ≤√2. For example, when a approaches 0, 2/a approaches infinity; when a=√2, 2/a=√2. Therefore, the overall minimal value of b across all possible triangles is when a=√2, giving b=√2. Therefore, the range for b is [√2, ∞).Now, for side c. Since c is the largest side, and c <a +b, and given that a ≤b ≤c, and the area is 1.We need to find the possible values of c.The minimal value of c occurs when c is as small as possible. Since c ≥b, the minimal c is when c=b. Which occurs when the triangle is isoceles with sides a, b, b. The area would be (1/2)*a*b*sinγ=1. In this case, using the Law of Cosines: c²=b²=a² +b² -2ab*cosγ ⇒0=a² -2ab*cosγ ⇒cosγ=a/(2b). Since sinγ=2/(ab), and sin²γ +cos²γ=1, we have [2/(ab)]² +[a/(2b)]²=1 ⇒4/(a²b²) +a²/(4b²)=1 ⇒Multiply both sides by4b²:16/(a²) +a²=4b².But this seems complicated. Alternatively, in the isoceles case with sides a, b, b, the area is 1.The area can be calculated using the formula for an isoceles triangle: (b/4)*√(4b² -a²)=1.Let's derive this. The height h of the isoceles triangle with sides a, b, b is h=√(b² - (a/2)^2). Therefore, area=(a/2)*h=(a/2)*√(b² -a²/4)=1. So:(a/2)*√(b² -a²/4)=1 ⇒a*√(4b² -a²)/4=1 ⇒a√(4b² -a²)=4 ⇒a²(4b² -a²)=16.But since we have the minimal c=b, we can express this as:a²(4b² -a²)=16.But also, from the earlier condition for the area in terms of a and b: (1/2)*a*b*sinγ=1. In this isoceles case, angle γ is the angle between the two sides of length b. So, the angle opposite side a is γ, and the two equal angles are opposite the sides of length b.But perhaps this is getting too involved. Let's consider specific examples.For the right-angled isoceles triangle with a=√2, b=√2, c=2, the area is 1. Here, c=2.For the equilateral triangle with a=b=c≈1.5196, the area is 1. Here, c≈1.5196.But wait, this contradicts the earlier statement that the minimal c is when c=b. In the equilateral triangle, c=b=a≈1.5196, which is less than the c=2 in the right-angled isoceles triangle. Therefore, the minimal c is achieved in the equilateral triangle, not in the right-angled isoceles triangle.Therefore, this suggests that the minimal c occurs when the triangle is equilateral, so c=2/(3^{1/4})≈1.5196.But in the right-angled isoceles triangle, c=2, which is larger.Therefore, the minimal c is achieved in the equilateral triangle, and the maximal c is unbounded, approaching infinity as a approaches 0 and b=2/a approaches infinity.But we need to confirm.In the equilateral triangle, all sides are equal, so c=a≈1.5196. In the right-angled isoceles triangle, c=2. In other triangles, like a=1, b=2, c=√5≈2.236, which is larger than 2.But when a approaches 0, c=√(a² +b²)=√(a² +4/a²)≈√(4/a²)=2/a, which approaches infinity as a approaches 0. Therefore, the maximal c is unbounded.Therefore, the range for c is [2/(3^{1/4}), ∞).But wait, in the equilateral triangle, c=2/(3^{1/4})≈1.5196. In the right-angled isoceles triangle, c=2, which is larger. Therefore, the minimal c is achieved in the equilateral triangle, and c can be as large as desired.But we need to verify if there are triangles with c less than in the equilateral case.Suppose we take a triangle that is acute or obtuse but not equilateral. For example, take a=1.5, b=1.5, c=1.5 (equilateral), area≈1.732>1. To get area=1, we need to reduce the sides. As calculated earlier, the equilateral triangle with area=1 has sides≈1.5196.If we take an isoceles triangle with a=1.5, b=1.5, and c less than 1.5, but this would violate the triangle inequality since a +b >c ⇒1.5 +1.5 >c ⇒c <3. But if we make c very small, the area would decrease. However, in this case, since a and b are fixed, the area depends on the angle between them. For example, if a=1.5, b=1.5, and angle γ between them is very small, the area=(1/2)*1.5*1.5*sinγ≈0. As γ increases, the area increases. To have area=1, we need sinγ= (2)/(1.5*1.5)=2/2.25≈0.888. So, γ≈62.7 degrees. Then, using the Law of Cosines, c²=1.5² +1.5² -2*1.5*1.5*cosγ≈2.25 +2.25 -4.5*cos(62.7°)=4.5 -4.5*0.46≈4.5 -2.07=2.43 ⇒c≈1.56. This is greater than the equilateral case's c≈1.5196. Therefore, this triangle has c≈1.56>1.5196, so the minimal c is indeed achieved in the equilateral triangle.Therefore, the range for c is [2/(3^{1/4}), ∞).But let's check another example. Take a=2/(3^{1/4}), b=2/(3^{1/4}), c=2/(3^{1/4}) (equilateral). Area=1. If we make a slightly smaller, say a=1.5, then b must be at least 1.5, and c would be greater than or equal to b. But the area would require adjustments. However, in this case, the minimal c is achieved only in the equilateral triangle.Therefore, compiling the results:1) Side a can range from approaching 0 up to 2/(3^{1/4})], so a ∈ (0, 2/(3^{1/4})].2) Side b can range from √2 up to infinity, so b ∈ [√2, ∞).3) Side c can range from 2/(3^{1/4}) up to infinity, so c ∈ [2/(3^{1/4}), ∞).But wait, in the equilateral triangle, c=2/(3^{1/4}), and in the right-angled isoceles triangle, c=2. So c must be at least 2/(3^{1/4})≈1.5196. Therefore, the range for c is [2/(3^{1/4}), ∞).But we need to confirm if c can take any value between 2/(3^{1/4}) and infinity.Suppose we want c=3. Is there a triangle with a ≤b ≤c=3 and area=1?Yes. For example, take a=0.5, b=4, c=√(0.5² +4²)=√(16.25)≈4.031, but c≈4.031 >b=4, which violates a ≤b ≤c. So instead, let's take a=0.5, b=4, c must be ≥b=4. Then, using the area formula, (1/2)*0.5*4*sinγ=1 ⇒1*sinγ=1 ⇒sinγ=1 ⇒γ=90°. Therefore, c=√(0.5² +4²)=√(16.25)≈4.031. Therefore, this triangle has a=0.5, b=4, c≈4.031, area=1, and a ≤b ≤c. Therefore, c can be≈4.031, which is greater than 2/(3^{1/4})≈1.5196. Similarly, c can be any value greater than or equal to 2/(3^{1/4}).Therefore, the answers for part b) are:1) Side a: (0, 2/(3^{1/4})]2) Side b: [√2, ∞)3) Side c: [2/(3^{1/4}), ∞)But we should express 2/(3^{1/4}) in a more simplified radical form. Since 3^{1/4}=√[4]{3}, so 2/(3^{1/4})=2*3^{-1/4}=2*3^{3/4}/3= (2*3^{3/4})/3= but maybe it's better to leave it as 2/(3^{1/4}) or 2*3^{3/4}/3.Alternatively, note that 3^{1/4}= (3^{1/2})^{1/2}=√√3, so 2/(√√3)=2/(3^{1/4}).But perhaps it's better to rationalize or express in terms of exponents.But the problem might expect the answer in terms of radicals. Let me compute 3^{1/4}:3^{1/4}= (3^{1/2})^{1/2}=√(√3)=√√3. So 2 divided by √√3 can be written as 2/√√3 =2*√√3/(√3)= but this is getting complicated. Maybe leave it as 2/(3^{1/4}).Therefore, the final answers:a) 1) 0° < A ≤60°, 2) 0° < B <90°, 3) 60° ≤ C <180°.b) 1) 0 < a ≤2/(3^{1/4}), 2) √2 ≤b <∞, 3) 2/(3^{1/4}) ≤c <∞.But let me check units. The problem doesn't specify units, so assuming units are consistent.Confirming with the right-angled isoceles triangle:a=√2, b=√2, c=2, area=1. Check if these values fit into the ranges:a=√2≈1.414 ≤2/(3^{1/4})≈1.5196: yes, since 1.414 <1.5196.b=√2≈1.414 ≥√2: yes.c=2 ≥2/(3^{1/4})≈1.5196: yes.Another example: equilateral triangle with a=2/(3^{1/4})≈1.5196, area=1. Correct.For side a approaching 0, b=2/a approaching infinity, c=√(a² +4/a²)≈2/a approaching infinity. Correct.Therefore, these ranges seem correct.Final Answera) 1) (boxed{0^circ < A leq 60^circ}) 2) (boxed{0^circ < B < 90^circ}) 3) (boxed{60^circ leq C < 180^circ})b) 1) (boxed{0 < a leq dfrac{2}{sqrt[4]{3}}}) 2) (boxed{sqrt{2} leq b < infty}) 3) (boxed{dfrac{2}{sqrt[4]{3}} leq c < infty})</think> SummaryProblem a) Given a triangle (ABC) with angles (A leq B leq C), we need to determine the possible values for each angle.1. Angle (A): - As the smallest angle, (A) must be greater than (0^circ) and at most (60^circ) (since in an equilateral triangle, all angles are (60^circ)). - Range: (0^circ < A leq 60^circ)2. Angle (B): - (B) is between (A) and (C). Since (A leq B leq C) and the sum of angles is (180^circ), (B) can approach (90^circ) but cannot reach it. - Range: (0^circ < B < 90^circ)3. Angle (C): - As the largest angle, (C) must be at least (60^circ) (equilateral triangle) and less than (180^circ). - Range: (60^circ leq C < 180^circ)Problem b) Given a triangle with sides (a leq b leq c) and area 1, we need to determine the possible values for each side.1. Side (a): - The smallest side (a) can approach 0 but is maximized in an equilateral triangle. The maximum value is (frac{2}{sqrt[4]{3}}). - Range: (0 < a leq frac{2}{sqrt[4]{3}})2. Side (b): - The minimal value of (b) occurs in a right-angled isosceles triangle where (b = sqrt{2}). (b) can increase indefinitely. - Range: (sqrt{2} leq b < infty)3. Side (c): - The smallest value of (c) occurs in an equilateral triangle, which is (frac{2}{sqrt[4]{3}}). (c) can increase indefinitely. - Range: (frac{2}{sqrt[4]{3}} leq c < infty) Final Answera) 1) (boxed{0^circ < A leq 60^circ})  2) (boxed{0^circ < B < 90^circ})  3) (boxed{60^circ leq C < 180^circ})b) 1) (boxed{0 < a leq dfrac{2}{sqrt[4]{3}}})  2) (boxed{sqrt{2} leq b < infty})  3) (boxed{dfrac{2}{sqrt[4]{3}} leq c < infty})]

🔑:45. a) It is clear that1) 0<A leqslant 60^{circ},where A is the smallest angle of triangle ABC (angle A leqslant 60^{circ}, because A+B+C=180^{circ}=3 cdot 60^{circ} and A leqslant B leqslant C; angle A=60^{circ}, if triangle ABC is equilateral);2) 0<B<90^{circ},where B is the middle-sized angle of triangle ABC (angle B can be arbitrarily small if triangle ABC is isosceles with a very large angle C at the vertex (Fig. 118, a); angle B can be arbitrarily close to 90^{circ} if the triangle is isosceles with a very small angle A at the vertex (Fig. 118, b); angle B<90^{circ}, because angle B leqslant angle C and left.angle B+angle C<180^{circ}rightrceil)3) 60^{circ} leqslant C<180^{circ}where C is the largest angle of triangle ABC (angle C geqslant 60^{circ}, because A+B+C=180^{circ}=3 cdot 60^{circ} and C geqslant B geqslant A; angle C=60^{circ}, if triangle ABC is equilateral).b) 1) It is clear that the smallest side a of triangle ABC with an area of 1 can be arbitrarily small: if the legs of a right triangle are varepsilon (where varepsilon can be arbitrarily small!) and 2 / varepsilon, then the area of the triangle is 1. The side a can also be arbitrarily large: if two sides of the triangle are M and b geqslant M (where M can be arbitrarily large!), and the angle C between these sides varies, starting from 90^{circ} and approaching 180^{circ}, then the third side c is always the largest, and the area of the triangle, starting from a large value frac{1}{2} b M, decreases without bound, and thus, passing through all values between frac{1}{2} b M and 0, it also passes through the value 1.2) Since the smallest side a of triangle ABC with an area of 1 can be arbitrarily large, its middle side AC=b can also be arbitrarily large. Furthermore, if the triangle has a (fixed) side AC=b and some (arbitrary) side BC=a leqslant b, then the vertex B of the triangle belongs to the (shaded in Fig. 119, a) circle K of radius b centered at point C. Since every point of this circle is at a distance d leqslant b from the line AC, the height BD of triangle ABC does not exceed b, and its area cannot be greater than b^{2} / 2 (it equals b^{2} / 2 if BC=b and BC perp AC). Therefore, the middle side b of triangle ABC with an area of 1 cannot be less than sqrt{2} (otherwise, it would be b^{2} / 2<1).3) It is clear that the largest side c of a triangle with an area of 1 can be arbitrarily large, but it cannot be arbitrarily small. If the side AB=c is fixed and AC=b leqslant c, BC=a leqslant c, then the vertex C of the triangle belongs to the intersection of two circles K_{1} and K_{2} of radius c centered at B and A; therefore, the height CE of the triangle, and thus its area, will be the largest when C coincides with the vertex of the shaded curvilinear triangle in Fig. 119, b, i.e., when ABC is an equilateral triangle with side c and area frac{c^{2} sqrt{3}}{4}. From this, it follows that if the area of triangle ABC is 1, then its largest side c cannot be less than frac{2}{sqrt[4]{3}}=frac{2 sqrt[4]{27}}{3}.Thus, 0<a<infty ; sqrt{2} leqslant b<infty ; frac{2 sqrt[4]{27}}{3}<c<infty.a)b)Fig. 119.

🧠:[<think>Alright, let's tackle this problem. The question is asking which of the given expressions (A, B, C, D) is correct when transformed from left to right. So, essentially, we need to check each pair of expressions and see if they are equal for all values of a and b (assuming b isn't zero in the original fractions, to avoid division by zero). Let me go through each option one by one.Starting with option A: (a/b) = (a²/b²). Hmm. Let's see. If I take the left side, which is a/b, and square both the numerator and the denominator, I get a²/b². But squaring a fraction is not the same as the original fraction unless the fraction is 1 or 0. For example, let's pick a = 2 and b = 3. Then the left side is 2/3, and the right side is (4)/(9). But 2/3 is not equal to 4/9. So A is not correct. Unless there's some restriction on a and b that makes them equal, but since the question doesn't specify, we have to assume they can be any real numbers (except where denominators are zero). So A is out.Next, option B: (a/b) = (2a)/(2b). Let's see. If we multiply both numerator and denominator by 2, we get 2a/2b. But 2a divided by 2b simplifies back to a/b because the 2's cancel out. So, yes, multiplying numerator and denominator by the same non-zero number (in this case, 2) doesn't change the value of the fraction. So B should be correct. Let me test with numbers. Let a = 1 and b = 2. Then left side is 1/2, right side is 2/4, which is also 1/2. So they are equal. Another example: a = 3, b = 5. Left is 3/5, right is 6/10, which simplifies to 3/5. So yes, B seems correct. But let me check the other options just to be thorough.Option C: (a/b) = (a+2)/(b+2). This is interesting. If we add 2 to both numerator and denominator, does that keep the fraction the same? Let's test with numbers. Let a = 2 and b = 2. Then left side is 2/2 = 1, right side is (2+2)/(2+2) = 4/4 = 1. Okay, they are equal here. But what if a = 1 and b = 3. Left is 1/3 ≈ 0.333, right is (1+2)/(3+2) = 3/5 = 0.6. Not equal. So in general, adding the same number to numerator and denominator changes the fraction unless specific conditions are met. Therefore, C is not a correct transformation in general. So C is incorrect.Option D: (-2 + a)/b = - (2 + a)/b. Let me parse this. The left side is (a - 2)/b, and the right side is - (a + 2)/b. Let's compute both sides with some numbers. Let a = 3 and b = 1. Left side: (3 - 2)/1 = 1/1 = 1. Right side: - (3 + 2)/1 = -5. 1 is not equal to -5. Therefore, D is not correct. Alternatively, maybe factor out a negative sign? Let's see: (a - 2)/b versus - (a + 2)/b. To make them equal, (a - 2)/b must equal - (a + 2)/b. Multiply both sides by b (assuming b ≠ 0): a - 2 = - (a + 2). Then, a - 2 = -a - 2. Adding a to both sides: 2a - 2 = -2. Adding 2: 2a = 0 => a = 0. So the equality holds only if a = 0, but not generally. Therefore, D is incorrect in general.Therefore, the only correct transformation is option B. Let me just double-check my reasoning for B. Multiplying numerator and denominator by the same non-zero constant (here, 2) is a valid equivalence transformation for fractions. So yes, (a/b) is equivalent to (2a)/(2b) because multiplying by 2/2 (which is 1) doesn't change the value. So, B is correct. The others are not valid in general.Final Answerboxed{B}</think>To determine which expression is correct when transformed from left to right, we analyze each option:- Option A: (frac{a}{b} = frac{a^2}{b^2}) Squaring both the numerator and the denominator changes the value of the fraction unless (a = 0) or (b = 1). For example, if (a = 2) and (b = 3), (frac{2}{3} neq frac{4}{9}). Therefore, A is incorrect.- Option B: (frac{a}{b} = frac{2a}{2b}) Multiplying both the numerator and the denominator by 2 does not change the value of the fraction. Simplifying (frac{2a}{2b}) gives (frac{a}{b}). Testing with values such as (a = 1) and (b = 2) confirms equality. Therefore, B is correct.- Option C: (frac{a}{b} = frac{a+2}{b+2}) Adding 2 to both the numerator and the denominator changes the fraction. For example, if (a = 1) and (b = 3), (frac{1}{3} neq frac{3}{5}). Therefore, C is incorrect.- Option D: (frac{-2 + a}{b} = -frac{2 + a}{b}) Testing with values such as (a = 3) and (b = 1) shows (frac{1}{1} neq -5). The equation holds only if (a = 0), not generally. Therefore, D is incorrect.Thus, the correct answer is (boxed{B}).,<think>Okay, so I have this problem here, and I need to figure out which of the given expressions is correct when transformed from left to right. The options are A, B, C, D. Let me take each one step by step.Starting with option A: (frac{a}{b} = frac{a^2}{b^2}). Hmm, let's see. If I start with (frac{a}{b}) and square both the numerator and the denominator, does that equal the original fraction? Well, squaring both numerator and denominator would give (left(frac{a}{b}right)^2), right? But that's not the same as (frac{a}{b}) unless (a/b) is 0 or 1. Because if (frac{a}{b}) is 1, then squaring both would still be 1. If it's 0, then squaring would still be 0. But in general, for any other values, this wouldn't hold. For example, if (a = 2) and (b = 1), then (frac{2}{1} = 2), but (frac{4}{1} = 4), which is not equal. So A is not correct. So probably A is incorrect.Moving on to option B: (frac{a}{b} = frac{2a}{2b}). Let's check this. If I take the original fraction (frac{a}{b}) and multiply both numerator and denominator by 2, I get (frac{2a}{2b}). But multiplying numerator and denominator by the same non-zero number should not change the value of the fraction. For example, if (a = 3), (b = 4), then (frac{3}{4} = 0.75) and (frac{6}{8} = 0.75) as well. So yes, (frac{2a}{2b}) simplifies back to (frac{a}{b}) because the 2's cancel out. So B seems correct. But let me check the other options just to be sure.Option C: (frac{a}{b} = frac{a + 2}{b + 2}). Hmm, adding 2 to both numerator and denominator. Let's test with numbers. Let’s say (a = 2) and (b = 3). Then (frac{2}{3}) is approximately 0.666..., and (frac{2 + 2}{3 + 2} = frac{4}{5} = 0.8), which is not equal. So C is not correct. Another example: if (a = 1) and (b = 1), then (frac{1}{1} = 1), and (frac{3}{3} = 1), so in that case, it works. But that's a specific case. Generally, adding the same number to numerator and denominator changes the fraction unless the original fraction was 1. So C is not a correct transformation in general.Option D: (frac{-2 + a}{b} = -frac{2 + a}{b}). Let me parse this. The left side is (frac{a - 2}{b}), right? Because -2 + a is the same as a - 2. The right side is (-frac{2 + a}{b}), which is the same as (frac{-(2 + a)}{b} = frac{-2 - a}{b}). Comparing the two: left side is (frac{a - 2}{b}), right side is (frac{-2 - a}{b}). Are these the same? Let's see. Take specific values: let a = 3, b = 1. Left side: (3 - 2)/1 = 1. Right side: - (2 + 3)/1 = -5. 1 ≠ -5. So they are not equal. Therefore, D is incorrect.Wait, maybe another example. Let a = 0. Then left side: (-2 + 0)/b = -2/b. Right side: - (2 + 0)/b = -2/b. So in this case, they are equal. Hmm. So when a = 0, D is correct. But when a ≠ 0, like in the previous example, they are not. So D is only correct if a = 0, but not in general. Therefore, D is not a valid transformation in general.So summarizing:A: Incorrect, because squaring numerator and denominator doesn't preserve the value unless the fraction is 0 or 1.B: Correct, because multiplying numerator and denominator by the same number (2) preserves the value.C: Incorrect, because adding the same number to numerator and denominator doesn't preserve the fraction except in specific cases.D: Incorrect, because the negative sign applies to both terms in the numerator on the right side, which is not the same as the left side unless a is 0.Therefore, the correct answer is B.Final Answerboxed{B}</think>To determine which expression is correct when transformed from left to right, we analyze each option:Option A: (frac{a}{b} = frac{a^2}{b^2})- Squaring both the numerator and the denominator changes the value of the fraction unless (a/b) is 0 or 1. For example, if (a = 2) and (b = 1), (frac{2}{1} neq frac{4}{1}). Therefore, A is incorrect.Option B: (frac{a}{b} = frac{2a}{2b})- Multiplying both the numerator and the denominator by 2 does not change the value of the fraction. For example, (frac{3}{4} = frac{6}{8}). This is a valid transformation. Therefore, B is correct.Option C: (frac{a}{b} = frac{a + 2}{b + 2})- Adding 2 to both the numerator and the denominator changes the value of the fraction. For example, (frac{2}{3} neq frac{4}{5}). Therefore, C is incorrect.Option D: (frac{-2 + a}{b} = -frac{2 + a}{b})- The left side is (frac{a - 2}{b}) and the right side is (frac{-2 - a}{b}). For (a = 3) and (b = 1), (frac{1}{1} neq -5). Therefore, D is incorrect.Thus, the correct answer is (boxed{B}).]

🔑:To analyze each option step-by-step:Option A: frac{a}{b} = frac{a^2}{b^2}- Squaring both the numerator and the denominator changes the ratio if a and b have different signs. For example, if a = -2 and b = 2, then frac{a}{b} = frac{-2}{2} = -1, but frac{a^2}{b^2} = frac{(-2)^2}{2^2} = 1. Since the values are not equal, this transformation is incorrect.Option B: frac{a}{b} = frac{2a}{2b}- Multiplying both the numerator and the denominator by the same non-zero number does not change the value of the fraction. This can be shown as follows: frac{a}{b} = frac{2 cdot a}{2 cdot b} = frac{2a}{2b}. This transformation is correct because it maintains the equality.Option C: frac{a}{b} = frac{a+2}{b+2}- Adding the same number to both the numerator and the denominator generally changes the value of the fraction. For example, if a = 1 and b = 2, then frac{a}{b} = frac{1}{2}, but frac{a+2}{b+2} = frac{1+2}{2+2} = frac{3}{4}. Since the values are not equal, this transformation is incorrect.Option D: frac{-2+a}{b} = -frac{2+a}{b}- Changing the sign inside the numerator does not equate to multiplying the entire fraction by -1. For example, if a = 3 and b = 2, then frac{-2+a}{b} = frac{1}{2}, but -frac{2+a}{b} = -frac{5}{2}. Since the values are not equal, this transformation is incorrect.Therefore, the correct answer, following the analysis of each option, is:boxed{B}

🧠:[<think>Okay, so there's this problem where 70 students gathered at the station, and they're from three cities: Ljubljana, Celje, and Maribor. The problem gives two conditions: first, the number of students from Maribor and Ljubljana together is equal to the number from Celje. Second, the students from Celje and Maribor together are six times as many as those from Ljubljana. I need to find out how many students are from each city. Hmm, let's break this down step by step.First, let's assign variables to each city to make it easier. Let me denote:- Let L be the number of students from Ljubljana.- Let C be the number from Celje.- Let M be the number from Maribor.The total number of students is 70, so that gives me the first equation:L + C + M = 70.Okay, that's straightforward. Now the problem states that students from Maribor and Ljubljana together were as many as those from Celje. So that means M + L equals C. So that's the second equation:M + L = C.Third condition: students from Celje and Maribor together were six times as many as those from Ljubljana. So C + M equals 6 times L. So:C + M = 6L.Alright, so now I have three equations:1. L + C + M = 702. M + L = C3. C + M = 6LI need to solve this system of equations. Let me see how to approach this. Maybe substitution would work here since some equations can be substituted into others.Starting with equation 2: M + L = C. That means I can express C in terms of M and L. So, C = M + L. Maybe I can substitute this into the other equations to reduce the number of variables.Let's substitute C into equation 1. Equation 1 becomes:L + (M + L) + M = 70.Simplify that: L + M + L + M = 70. Combine like terms: 2L + 2M = 70. Dividing both sides by 2 gives L + M = 35. So, that's equation 4: L + M = 35.Now, equation 3 was C + M = 6L. But since C is M + L (from equation 2), substitute that into equation 3:(M + L) + M = 6L.Simplify that: M + L + M = 6L. Combine like terms: 2M + L = 6L. Subtract L from both sides: 2M = 5L. So, 2M = 5L. Let's write that as equation 5: 2M = 5L.From equation 4, we have L + M = 35. Let's see if we can express one variable in terms of the other. Maybe solve equation 5 for M. From equation 5: 2M = 5L → M = (5/2)L. So, M is 2.5 times L. Hmm, but we need whole numbers since the number of students can't be a fraction. So, L must be a multiple of 2 to make M an integer. Let's keep that in mind.Now, substitute M = (5/2)L into equation 4: L + (5/2)L = 35. Combine like terms: (1 + 5/2)L = 35 → (7/2)L = 35. Multiply both sides by 2/7: L = 35 * (2/7) = 10. So, L is 10. Then M = (5/2)*10 = 25. So, M is 25. Then from equation 2, C = M + L = 25 + 10 = 35. Let me check if these numbers satisfy all equations.Check equation 1: L + C + M = 10 + 35 + 25 = 70. Correct. Check equation 2: M + L = 25 + 10 = 35 = C. Correct. Check equation 3: C + M = 35 + 25 = 60, and 6L = 6*10 = 60. Correct. All equations are satisfied. So, the numbers are Ljubljana: 10, Celje: 35, Maribor: 25.Wait, but just to be thorough, let me double-check my steps. When I substituted C into equation 1, I got L + M = 35, which is straightforward. Then equation 3 substitution led to 2M = 5L, which gave M = (5/2)L, and then substituting into equation 4 gave L = 10, leading to M = 25. Then C = 35. All checks out. So, seems like the solution is correct. The fractions worked out to whole numbers, so no problem there. Since 5/2 times 10 is 25, which is an integer, that's good. So, the final answer is Ljubljana: 10 students, Celje: 35 students, Maribor: 25 students.Final AnswerThe number of students from Ljubljana is boxed{10}, from Celje is boxed{35}, and from Maribor is boxed{25}.</think>Let ( L ) be the number of students from Ljubljana, ( C ) be the number from Celje, and ( M ) be the number from Maribor. We are given the following conditions:1. The total number of students is 70: [ L + C + M = 70 ]2. Students from Maribor and Ljubljana together are as many as those from Celje: [ M + L = C ]3. Students from Celje and Maribor together are six times as many as those from Ljubljana: [ C + M = 6L ]Substituting ( C = M + L ) from the second equation into the first equation:[L + (M + L) + M = 70]Simplifying, we get:[2L + 2M = 70 implies L + M = 35]Next, substitute ( C = M + L ) into the third equation:[(M + L) + M = 6L]Simplifying, we get:[2M + L = 6L implies 2M = 5L implies M = frac{5}{2}L]Substituting ( M = frac{5}{2}L ) into ( L + M = 35 ):[L + frac{5}{2}L = 35 implies frac{7}{2}L = 35 implies L = 10]Then, solving for ( M ):[M = frac{5}{2} times 10 = 25]Finally, solving for ( C ):[C = M + L = 25 + 10 = 35]The number of students from Ljubljana is (boxed{10}), from Celje is (boxed{35}), and from Maribor is (boxed{25}).,<think>Okay, so I need to solve this problem where there are 70 students from three cities: Ljubljana, Celje, and Maribor. The problem gives two conditions: 1. The number of students from Maribor and Ljubljana together is equal to the number from Celje.2. The number of students from Celje and Maribor together is six times the number from Ljubljana.Hmm, let's break this down. Let me assign variables to each city to make it easier. Let's say:- Let L be the number of students from Ljubljana.- Let C be the number of students from Celje.- Let M be the number of students from Maribor.The total number of students is 70, so the first equation I can write is:L + C + M = 70.That's straightforward. Now, the first condition says that students from Maribor and Ljubljana together were as many as those from Celje. So that translates to:M + L = C.Okay, so that's the second equation. The second condition states that students from Celje and Maribor together were six times as many as those from Ljubljana. So that would be:C + M = 6L.Alright, so now I have three equations:1. L + C + M = 702. M + L = C3. C + M = 6LI need to solve for L, C, and M. Let me see how to substitute these equations. Looking at equation 2: M + L = C. So I can substitute C in the other equations with (M + L). Let's try that in equation 3 first. Equation 3 is C + M = 6L. Substituting C from equation 2 gives:(M + L) + M = 6LSimplify that:M + L + M = 6LCombine like terms:2M + L = 6LSubtract L from both sides:2M = 5LSo, 2M = 5L => M = (5/2)LHmm, so M is (5/2) times L. Let me note that down. So M = (5/2)L. Now, let's substitute C from equation 2 into equation 1. Equation 1 is L + C + M = 70. Substituting C with (M + L):L + (M + L) + M = 70Simplify:L + M + L + M = 70Combine like terms:2L + 2M = 70Divide both sides by 2:L + M = 35So, L + M = 35. But from equation 2, C = L + M, so that means C = 35. Wait, that's interesting. So Celje has 35 students. Let me check if that makes sense.If C = 35, then from equation 3: C + M = 6L. So 35 + M = 6L. But we also know that M = (5/2)L. So substituting M into 35 + M = 6L gives:35 + (5/2)L = 6LMultiply both sides by 2 to eliminate the fraction:70 + 5L = 12LSubtract 5L from both sides:70 = 7LDivide both sides by 7:L = 10Okay, so Ljubljana has 10 students. Then, from M = (5/2)L, which is (5/2)*10 = 25. So Maribor has 25 students. Wait, let me check if these numbers add up to 70. Ljubljana is 10, Celje is 35, Maribor is 25. 10 + 35 + 25 = 70. Yes, that's correct. Also, check the first condition: Maribor and Ljubljana together: 25 + 10 = 35, which is equal to Celje. That's good. Second condition: Celje and Maribor together: 35 + 25 = 60, which is six times Ljubljana (10*6=60). That works too. So the solution seems to be Ljubljana: 10, Celje: 35, Maribor:25. But let me just go through the steps again to make sure I didn't make any substitution errors. Starting equations:1. L + C + M = 702. L + M = C3. C + M = 6LFrom equation 2: C = L + M. Substitute into equation 3:(L + M) + M = 6L => L + 2M = 6L => 2M = 5L => M = (5/2)LSubstitute C = L + M into equation 1:L + (L + M) + M = 70 => 2L + 2M = 70 => L + M = 35. But since M = (5/2)L, substitute into L + M = 35:L + (5/2)L = 35 => (7/2)L = 35 => L = 35*(2/7) = 10. Then M = (5/2)*10 = 25, and C = L + M = 10 + 25 = 35. All equations check out, so the answer is correct.Final AnswerThe number of students from Ljubljana is boxed{10}, from Celje is boxed{35}, and from Maribor is boxed{25}.</think>Let ( L ) be the number of students from Ljubljana, ( C ) be the number from Celje, and ( M ) be the number from Maribor. We are given the following equations:1. ( L + C + M = 70 )2. ( L + M = C )3. ( C + M = 6L )From equation 2, we substitute ( C ) with ( L + M ) in equations 1 and 3:Substituting into equation 3:[(L + M) + M = 6L implies L + 2M = 6L implies 2M = 5L implies M = frac{5}{2}L]Substituting into equation 1:[L + (L + M) + M = 70 implies 2L + 2M = 70 implies L + M = 35]Substituting ( M = frac{5}{2}L ) into ( L + M = 35 ):[L + frac{5}{2}L = 35 implies frac{7}{2}L = 35 implies L = 10]Then, we find ( M ):[M = frac{5}{2} times 10 = 25]And substituting ( L ) and ( M ) back into equation 2 to find ( C ):[C = L + M = 10 + 25 = 35]All values satisfy the original equations. Therefore, the number of students from each city is:[boxed{10} text{ (Ljubljana)}, boxed{35} text{ (Celje)}, boxed{25} text{ (Maribor)}]]

🔑:B3. Let x be the number of students from Ljubljana, y the number of students from Celje, and z the number of students from Maribor. Then, x+y+z=70, x+z=y, and y+z=6 x. The solution to the system is x=10, y=35, z=25. There were 10 students from Ljubljana, 35 students from Celje, and 25 students from Maribor.Answer ............................................................................................................

🧠:[<think>Okay, so I need to find the number of four-digit "fancy" numbers where all digits are different. A "fancy" number has an equal number of even and odd digits. Since it's a four-digit number, that means we need exactly 2 even digits and 2 odd digits. Also, the digits must all be different. Let me break this down step by step.First, let's recall that a four-digit number cannot start with zero. So the first digit (thousands place) has to be from 1 to 9. The other three digits (hundreds, tens, units) can be from 0 to 9, but all digits must be distinct.Since the number must have two even and two odd digits, we need to consider cases based on the position of the even and odd digits. However, because the first digit has a restriction (can't be zero), it might be easier to split the problem into two parts: considering whether the first digit is even or odd, and then counting the possibilities accordingly.Let me outline the possible cases:1. Case 1: First digit is even. - Then, we need one more even digit among the remaining three positions. - The other two digits must be odd.2. Case 2: First digit is odd. - Then, we need one more odd digit among the remaining three positions. - The other two digits must be even.I need to calculate the number of valid four-digit numbers for each case and then sum them up.Let's start with Case 1: First digit is even.First, determine how many choices we have for the first digit. Even digits are 0, 2, 4, 6, 8. But since the first digit cannot be zero, the possible even digits for the first position are 2, 4, 6, 8. So that's 4 choices.Now, after choosing the first digit (even), we need to choose three more digits such that exactly one is even and two are odd. But all digits must be distinct.Wait, but actually, since the first digit is already even, we need exactly one more even digit in the remaining three positions, and two odd digits. However, we must ensure that all digits are different and that we don't repeat any digits.So let's proceed step by step.First digit: 4 choices (2,4,6,8).Now, for the remaining three digits, we need 1 even and 2 odd digits.But the even digits available now are 0,2,4,6,8 minus the one already used for the first digit. So if the first digit was, say, 2, then the remaining even digits are 0,4,6,8. Similarly, if the first digit was 4, remaining evens are 0,2,6,8, etc. So in any case, there are 4 remaining even digits (since one even digit is already used in the first position).Similarly, the odd digits are 1,3,5,7,9. Since none have been used yet (the first digit was even), there are 5 odd digits available.So, for the remaining three positions (hundreds, tens, units), we need to place 1 even and 2 odd digits. The positions can be in any order. Let's compute the number of such combinations.First, choose which of the three positions will be the even digit. There are C(3,1) = 3 choices.For each such choice:- The chosen position (even digit) can be filled with any of the remaining 4 even digits (since one even was used in the first digit). But we have to remember that 0 is now allowed in the other positions.Wait, but we need to make sure that digits are unique. Since the first digit is already an even digit (non-zero), the remaining even digits include 0 and the other three even digits not used in the first position.Wait, perhaps a better approach is:Total even digits available initially: 5 (0,2,4,6,8). But since the first digit is even and cannot be 0, we used one of 2,4,6,8. So remaining even digits: 0 and the remaining three from (2,4,6,8). For example, if first digit is 2, remaining evens are 0,4,6,8. So that's 4 even digits left.So, regardless of which even digit we pick first, there are 4 remaining even digits (including 0). So, when we choose the even digit for the remaining three positions, we have 4 choices, but we have to make sure that we don't repeat digits.Similarly, the odd digits: we have 5 odd digits (1,3,5,7,9) all available, since none have been used yet.So, let's break it down:1. Choose the first digit: 4 options (even, non-zero).2. Choose which of the remaining three positions will be even: 3 options.3. Choose the even digit for that position: 4 options (including 0).4. Choose the two odd digits for the remaining two positions. Since we have 5 odd digits available, and they must be distinct, the number of permutations is P(5,2) = 5*4 = 20.But wait, steps 3 and 4 need to account for the fact that digits must be unique across all four digits. However, since the even digits chosen in step 3 are from the remaining evens (which don't overlap with the first digit), and the odd digits are entirely separate, there is no overlap between these choices. Therefore, we can multiply them.Therefore, the total number for Case 1 is:4 (first digit) * [3 (positions for even) * 4 (even digits) * 20 (odd digits)] = 4 * 3 * 4 * 20 = 4 * 3 * 80 = 4 * 240 = 960.Wait, that seems high. Let me check again.Wait, actually, after choosing the first digit (4 options), then:For the remaining three digits:- Choose which position is even: 3 choices.- For that even position: 4 choices (since one even digit is already used, and 0 is allowed here). But wait, actually, the total remaining even digits are 4 (since one was used in the first digit, which was 2,4,6,8, so 0 is still available). So, yes, 4 choices.- Then, for the two odd positions: we need to choose two distinct odd digits. There are 5 odd digits (1,3,5,7,9), so for the first odd digit: 5 choices, second odd digit: 4 choices. So 5*4 = 20.Therefore, for each first digit, the number of combinations is 3 * 4 * 20 = 240.But since there are 4 choices for the first digit, total is 4 * 240 = 960. Okay, that's 960 numbers for Case 1.Now, Case 2: First digit is odd.First, how many choices for the first digit? Odd digits are 1,3,5,7,9. So that's 5 choices.Since the first digit is odd, we need two more odd digits in the remaining three positions, but we need a total of two odd digits (since the number must have two even and two odd). Wait, hold on. Wait, total digits: four. If the first digit is odd, then we need one more odd digit and two even digits in the remaining three positions.Wait, because the total number of odd digits should be two, so we need 1 more odd in the remaining three, and two evens.But let me check again: if first digit is odd, then we need one more odd digit and two even digits in the remaining three positions to make total two odd and two even. Yes.So, similar structure to Case 1.First digit: 5 choices (1,3,5,7,9).Remaining three digits: 1 odd and 2 evens.Again, the remaining three digits must all be distinct and different from the first digit.So, step by step:1. Choose the first digit: 5 options (odd digits).2. Choose which of the remaining three positions will be odd: C(3,1) = 3.3. Choose the odd digit for that position: since one odd digit is already used (the first digit), there are 4 remaining odd digits.4. Choose the two even digits for the remaining two positions. Here, even digits include 0,2,4,6,8. All five even digits are available, but we need to ensure they are distinct and different from each other.Wait, but we need to pick two different even digits. However, since the first digit is odd, none of the even digits have been used yet, so all even digits (including 0) are available. But we have to pick two distinct even digits for the two positions.But here's a catch: if we choose 0 as one of the even digits, that's allowed except for the first digit, but since these are the remaining positions (hundreds, tens, units), 0 is allowed here. However, we need to make sure that the two even digits are distinct and not overlapping with each other or with the first digit (but the first digit is odd, so no overlap).Therefore, the number of ways to choose two distinct even digits is P(5,2) = 5*4 = 20. Since order matters here because the positions are distinct.Wait, but actually, since we are placing the two even digits into two specific positions (after choosing which position is odd), the order does matter. So, if we first pick the two even digits and assign them to the positions, that's 5 choices for the first even digit and 4 for the second. So 5*4=20.Therefore, the total for each first digit in Case 2 is:5 (first digit) * [3 (positions for odd) * 4 (remaining odd digits) * 20 (even digits)] = 5 * 3 * 4 * 20 = 5 * 3 * 80 = 5 * 240 = 1200.Wait, but hold on. Let's check again.First, first digit: 5 choices (odd digits).Then, for the remaining three digits:- Choose which position is odd: 3 choices.- Choose the odd digit: since one odd digit is already used (the first digit), there are 4 left. So 4 choices.- Choose the two even digits: all even digits (0,2,4,6,8) are available, but they have to be distinct. So for the first even position: 5 choices, then for the second even position: 4 choices. So 5*4 = 20.Therefore, for each first digit, the number is 3 * 4 * 20 = 240. Then multiplied by 5 first digits: 5 * 240 = 1200. So 1200 numbers for Case 2.Therefore, total fancy numbers: Case 1 (960) + Case 2 (1200) = 2160.Wait, but let me verify this. Because sometimes when dealing with permutations, it's easy to overcount or undercount.Wait, let's take a small example for Case 1. Suppose first digit is 2 (even). Then, remaining even digits are 0,4,6,8. So 4 choices. The odd digits are 1,3,5,7,9.So, for example, if we choose position 2 (hundreds place) as even: 4 choices (0,4,6,8). Then positions 3 and 4 (tens and units) need to be odd. So 5 choices for first odd, 4 for the second. So 5*4=20. So for each position (3 possibilities), 4*20=80. Then per first digit, 3*4*20=240. Then 4 first digits, so 4*240=960. That seems okay.For Case 2: first digit is 1 (odd). Remaining odd digits: 3,5,7,9 (4 choices). Even digits: 0,2,4,6,8. So if we choose, say, position 2 as odd: 4 choices (3,5,7,9). Then positions 3 and 4 are even: 5*4=20. So for each first digit, 3*4*20=240. Then 5 first digits, so 5*240=1200. That seems okay.Total 960 + 1200 = 2160. Hmm.But wait, let me check if there's an alternative way to compute this to verify.Alternative approach:Total four-digit numbers with all distinct digits: 9 (first digit) * 9 * 8 * 7 = 4536.But we need only those with exactly two even and two odd digits.Alternatively, the number of four-digit numbers with distinct digits, two even and two odd digits.But since the first digit can't be zero, we have to be careful.Another way: choose two even and two odd digits, ensuring that the first digit is non-zero, and all digits are unique.So, split into two cases:1. First digit is even.2. First digit is odd.Which is exactly what we did before.But let's see if we can compute it using combinations.For Case 1: First digit even.We need to choose 2 even digits (one is the first digit) and 2 odd digits.But since the first digit is even and cannot be zero, the first digit is one of 2,4,6,8. Then, we need one more even digit (can be 0,2,4,6,8 excluding the first digit), and two odd digits (1,3,5,7,9).So, the number of ways:Choose first digit: 4 options (2,4,6,8).Then choose one more even digit from the remaining 4 (since we exclude the first digit, and include 0). So 4 choices.Then choose two distinct odd digits from the 5 available: C(5,2) = 10, and arrange them.But also, we have to arrange all four digits. Wait, but actually, the positions matter.Wait, maybe the way to compute is:First, choose the digits, then arrange them.But since the first digit is fixed as even (non-zero), and then we have three more digits where one is even and two are odd.But perhaps:Total number for Case 1:First, choose the first digit: 4.Then, choose 1 even digit from the remaining 4 (0,2,4,6,8 minus the first digit), so 4 choices.Then choose 2 odd digits from 5: C(5,2) = 10.Now, arrange the remaining three digits (1 even, 2 odds) in the three positions. Since the three positions can be arranged in any order, but the even digit can be in any of the three positions. The number of permutations is 3! / (1!2!) = 3 (for the even digit's position) multiplied by the permutations of the two odd digits.Wait, but the two odd digits are distinct, so once we choose their positions, they can be arranged in 2! ways.Wait, let's clarify:We have three positions: hundreds, tens, units.We need to place 1 even digit and 2 odd digits.First, choose which position is even: 3 choices.Then, place the chosen even digit in that position: 1 way (since we already chose which digit it is).Then, place the two odd digits in the remaining two positions: 2! = 2 ways.Therefore, total permutations for the remaining three digits: 3 * 2 = 6.But wait, but the even digit is already chosen (from 4 options) and the two odd digits are chosen (from 10 pairs). Wait, no:Wait, actually, the digits are already selected. So once we have the digits:First digit: fixed.Even digit for the remaining: chosen (4 options).Odd digits: two chosen (10 options).Then, the number of arrangements is 3 (positions for the even digit) * 2! (arrangements of the two odd digits).Therefore, total number for each first digit:4 (even digits) * 10 (odd pairs) * 3 * 2 = 4*10*6=240. Then multiplied by 4 (first digits): 4*240=960. Same as before.Similarly, Case 2: First digit is odd.First digit: 5 choices.Then, choose 1 odd digit from remaining 4: 4 choices.Choose 2 even digits from 5 available (0,2,4,6,8): C(5,2)=10.Then arrange these three digits (1 odd, 2 evens) in the remaining three positions.Number of arrangements: choose position for the odd digit: 3 choices, then arrange the two even digits in the remaining positions: 2!.So total per first digit: 4 * 10 * 3 * 2 = 240. Multiply by 5 first digits: 5*240=1200.Total: 960 + 1200 = 2160. Same as before.So, seems like 2160 is the answer.But let me check if another approach gives the same result.Alternative approach:Total number of four-digit numbers with all distinct digits: 9 * 9 * 8 * 7 = 4536. As mentioned earlier.Number of four-digit numbers with two even and two odd digits, all distinct: ?We can compute this using combinations:First, split based on the number of even and odd digits.But since digits must be unique and the first digit cannot be zero.Alternatively, consider that choosing two even digits and two odd digits, where the first digit is non-zero and all digits are distinct.But it's a bit complex.First, how many ways to choose two even digits and two odd digits for a four-digit number, with all digits distinct and first digit non-zero.But we need to consider whether the first digit is even or odd.If first digit is even: then we have one even digit fixed (non-zero), and another even digit (could be zero), and two odd digits.If first digit is odd: then we have one odd digit fixed, another odd digit, and two even digits (which can include zero).So, in effect, this is the same as our initial case analysis.Alternatively, compute it as:Total number of four-digit numbers with two even and two odd digits, all distinct:= (number with first digit even and 1 more even, 2 odd) + (number with first digit odd and 1 more odd, 2 even)Which is exactly what we computed as 960 + 1200 = 2160.Alternatively, using multinomial coefficients:But I think the case analysis is the correct approach here.Alternatively, let's compute the total number by first selecting the digits and then arranging them, ensuring the first digit is non-zero.Total number of ways to select two even and two odd digits, then arrange them such that the first digit is non-zero and all digits are distinct.But even here, we need to handle the cases where the even digits include zero or not.Wait, maybe:First, select two even digits and two odd digits. Then subtract those arrangements where the first digit is zero (if any).But since first digit cannot be zero, we need to ensure that if one of the even digits is zero, we don't count arrangements where zero is in the first position.But this might complicate.Alternatively, split into two cases:1. The two even digits include zero.2. The two even digits do not include zero.But this might get too involved. Let's try.First, when considering all four-digit numbers with two even and two odd digits, all distinct:Case A: First digit is even.Then, the two even digits include the first digit (non-zero) and another even digit (could be zero). The two odd digits are arbitrary.Case B: First digit is odd.Then, the two odd digits include the first digit and another odd digit. The two even digits are arbitrary, which can include zero.But this again mirrors our original case split.Alternatively, compute the total number as:Total = (number of two even digits sets * number of two odd digits sets) * number of valid permutations (first digit non-zero, all distinct).But we need to account for the fact that when selecting two even digits, if one of them is zero, then we have to subtract the permutations where zero is in the first position.This seems complicated, but let's try:First, compute the number of ways to choose two even digits and two odd digits.Even digits: There are 5 even digits (0,2,4,6,8). So number of ways to choose two even digits: C(5,2) = 10. But if we choose two even digits, one of which could be zero.Odd digits: There are 5 odd digits (1,3,5,7,9). Number of ways to choose two odd digits: C(5,2)=10.Thus, total ways to choose digits: 10 * 10 = 100.But then, for each such combination, we need to arrange the four digits into a four-digit number, with no repetition and first digit non-zero.However, some of the even digits include zero, so we need to adjust for that.So, for each combination of two even digits and two odd digits:If the two even digits do not include zero:Number of such combinations: C(4,2)=6 (choosing two even digits from 2,4,6,8). Then, two odd digits: C(5,2)=10. So total 6*10=60.For each of these 60 combinations, all digits are non-zero. So, when arranging, the first digit can be any of the four digits. So permutations: 4! = 24. But since all digits are unique and non-zero, all permutations are valid. So total numbers: 60 * 24 = 1440.If the two even digits include zero:Number of such combinations: C(4,1)=4 (choosing one non-zero even digit, the other is zero). Then, two odd digits: C(5,2)=10. Total: 4*10=40.For each of these 40 combinations, one of the even digits is zero. So when arranging, we need to exclude permutations where zero is in the first position.Total permutations: 4! = 24. Invalid permutations: those with zero first. How many? Fix zero in first position, then arrange the remaining three digits: 3! = 6. So valid permutations: 24 - 6 = 18.Thus, total numbers for these combinations: 40 * 18 = 720.Therefore, total four-digit numbers with two even and two odd digits, all distinct: 1440 + 720 = 2160.Which matches our previous result. So that confirms the answer is 2160.Therefore, the number of four-digit "fancy" numbers with all different digits is 2160.Final Answerboxed{2160}</think>To determine the number of four-digit "fancy" numbers with different digits, we need to consider numbers with exactly two even and two odd digits. We split the problem into two cases based on whether the first digit is even or odd.Case 1: First digit is even.- The first digit can be 2, 4, 6, or 8 (4 choices).- We need one more even digit (from the remaining 4 even digits, including 0) and two odd digits (from 5 choices).- The number of ways to choose the position for the remaining even digit: ( binom{3}{1} = 3 ).- The number of ways to choose the remaining even digit: 4.- The number of ways to choose and arrange the two odd digits: ( P(5, 2) = 5 times 4 = 20 ).- Total for this case: ( 4 times 3 times 4 times 20 = 960 ).Case 2: First digit is odd.- The first digit can be 1, 3, 5, 7, or 9 (5 choices).- We need one more odd digit (from the remaining 4 odd digits) and two even digits (from 5 choices, including 0).- The number of ways to choose the position for the remaining odd digit: ( binom{3}{1} = 3 ).- The number of ways to choose the remaining odd digit: 4.- The number of ways to choose and arrange the two even digits: ( P(5, 2) = 5 times 4 = 20 ).- Total for this case: ( 5 times 3 times 4 times 20 = 1200 ).Summing both cases, the total number of four-digit "fancy" numbers is:[ 960 + 1200 = 2160 ]Thus, the number of all four-digit "fancy" numbers written with different digits is (boxed{2160}).,<think>Okay, let's try to figure out how to solve this problem. So, we need to find the number of four-digit "fancy" numbers where all digits are different. A "fancy" number here is defined as having an equal number of even and odd digits. Since it's a four-digit number, that means we need exactly 2 even digits and 2 odd digits. Also, all the digits must be different. Hmm, okay. Let's break this down step by step.First, let's recall that a four-digit number can't start with zero. So the thousands place (the first digit) has to be from 1 to 9. The other three digits (hundreds, tens, ones places) can be from 0 to 9, but all digits must be distinct. So, we need to count all four-digit numbers with distinct digits where exactly two digits are even and two are odd.Let me think about how to approach this. Maybe we can separate the problem based on whether the first digit is even or odd because the first digit has restrictions (can't be zero). Then, depending on whether it's even or odd, we can calculate the number of possibilities for the remaining digits accordingly. That seems like a plan.Case 1: The first digit is even.Case 2: The first digit is odd.Let's tackle Case 1 first.Case 1: First digit is even.Since the number must have two even and two odd digits, if the first digit is even, then among the remaining three digits, we need one even and two odd digits. But all digits must be different.First, choose the first digit. Even digits are 0, 2, 4, 6, 8. But since the first digit can't be zero, the possible choices are 2, 4, 6, 8. So there are 4 choices for the first digit.Now, after choosing the first digit, we need to choose three more digits (hundreds, tens, ones places) such that exactly one is even and two are odd, and none of them repeat any digits already used.Wait, but the remaining digits can include zero, right? Because the other positions can have zero. However, we have to remember that digits must be distinct. So, once we choose the first digit, we can't repeat it in the remaining positions.So, let's proceed step by step.First, first digit: 4 choices (even, non-zero).Now, remaining digits: need 1 even and 2 odd digits.First, how many even digits are left? Originally, there are 5 even digits (0,2,4,6,8). We've already used one even digit (the first digit), so there are 4 even digits left. However, zero is still among them unless the first digit was zero, but since the first digit can't be zero, zero is still available. So, 4 even digits remaining.For the odd digits, originally, there are 5 odd digits (1,3,5,7,9). None have been used yet, so 5 odd digits available.So, we need to choose 1 even and 2 odd digits for the remaining three positions. But also, we have to arrange these three digits in the hundreds, tens, and ones places. So, the number of ways to choose and arrange them.First, choose 1 even digit out of the remaining 4 even digits: C(4,1) ways.Then, choose 2 odd digits out of 5: C(5,2) ways.Then, arrange these 3 digits (1 even and 2 odds) in the three positions. Since the digits are distinct, the number of arrangements is 3!.But wait, hold on. Wait, but when we choose the digits, we need to make sure that even digits might include zero, but since the first digit is already non-zero, the even digits here can include zero, but we don't have any restriction on the other positions for zero. So, is there a problem here?Wait, no. Because even if the even digit chosen is zero, as long as we don't place it in the first position, which is already taken care of. So, when we choose 1 even digit from the remaining 4 (which includes zero), and then arrange the three digits (including possibly zero) into the last three positions, that's allowed. So, the calculation should be okay.So, putting it all together for Case 1:Number of ways = 4 (choices for first digit) * [C(4,1) * C(5,2)] * 3!Let me compute this:C(4,1) = 4C(5,2) = 103! = 6So, 4 * (4 * 10) * 6 = 4 * 40 * 6 = 4 * 240 = 960.Wait, is that correct? Wait, no, hold on. Wait, 4 (first digit) * [4 * 10] (choosing 1 even and 2 odds) * 6 (arrangements). So 4 * 40 * 6 = 960. Hmm, okay.But wait, let me double-check. Let's think differently. Maybe we can compute the number of possible numbers in Case 1 as:First digit: 4 choices.Then, for the remaining three digits:We need to choose 1 even and 2 odd digits, all distinct and not equal to the first digit.But since the first digit is even, and we already used one even digit, we have 4 even digits left (including zero) and 5 odd digits.So, selecting 1 even digit: 4 choices.Selecting 2 odd digits: C(5,2) = 10.Then, arranging these three digits (1 even, 2 odd) in the three positions: 3! = 6.So total for Case 1: 4 * 4 * 10 * 6. Wait, that's how I did before: 4 * (4*10*6) = 4*240=960. Okay.But wait, actually, the 4 (first digit) multiplied by (4 * 10 * 6) is 4 * 240 = 960. Yes. That's correct.Case 2: First digit is odd.Similarly, the first digit is odd. Then, since we need two even and two odd digits in total, the remaining three digits must include two even and one odd digit.First, the first digit: odd digits are 1,3,5,7,9. So, 5 choices.Then, remaining three digits: 2 even and 1 odd, all distinct, not repeating the first digit.Original even digits: 5 (0,2,4,6,8). None used yet because the first digit is odd. So, 5 even digits available.Original odd digits: 5 (1,3,5,7,9). One already used (the first digit), so 4 remaining.So, we need to choose 2 even digits from 5 and 1 odd digit from 4. Then arrange them in the three positions.So, number of ways:First digit: 5 choices.Then, choose 2 even digits: C(5,2) = 10.Choose 1 odd digit: C(4,1) = 4.Arrange the 3 digits (2 even, 1 odd): 3! = 6.So total for Case 2: 5 * [10 * 4] * 6 = 5 * 40 * 6 = 5 * 240 = 1200.Wait, that seems straightforward. Let me verify again.First digit: 5 choices (odd digits).Then, choose 2 even digits from 5 available (0,2,4,6,8) and 1 odd digit from remaining 4 odd digits (since first digit is already used). So, C(5,2) * C(4,1) = 10 * 4 = 40. Then, arrange those three digits in the three positions: 3! = 6.Therefore, total for Case 2: 5 * 40 * 6 = 1200.Adding Case 1 and Case 2 together: 960 + 1200 = 2160.Hmm, so is the answer 2160? Wait, but let me think again. Did I account for all possibilities? Let me check for possible mistakes.First, in Case 1: the first digit is even (4 choices: 2,4,6,8). Then, the remaining digits include one even and two odds. The even digit could be zero, but we have to make sure that when we choose the even digits, we account for the fact that the first digit is already used. But in my calculation, since the first digit was even, we subtracted one from the total even digits, leaving 4 (including zero). Then, choosing 1 out of those 4. So that's okay.In Case 2: first digit is odd (5 choices). Then, remaining three digits have two evens (from all 5 even digits, including zero) and one odd (from remaining 4 odds). So, that's correct. Then, arranging all three digits. Since even digits include zero, but as long as we arrange them in the last three positions, which can include zero, that's okay.Another check: total four-digit numbers with distinct digits are 9 * 9 * 8 * 7 = 4536. Our answer is 2160, which is roughly half of that. Which makes sense because we are looking for exactly two even and two odd digits. However, the distribution might not be exactly half because the first digit can't be zero, so maybe slightly different. Wait, 2160 / 4536 ≈ 0.476, which is close to half. So, seems plausible.Alternatively, maybe another way to compute this is using combinations.Total number of four-digit numbers with distinct digits: 9 * 9 * 8 * 7 = 4536. But how many of these have exactly two even and two odd digits?But since the first digit affects the count, maybe we can compute it as:Number of four-digit numbers with distinct digits, two even and two odd digits: sum over the number of possibilities where first digit is even and rest have one even and two odd, plus first digit is odd and rest have two even and one odd. Which is exactly what we did in Case 1 and Case 2.Alternatively, maybe compute using positions.But I think the way we did is correct.Wait, let's think of another approach. Let's think in terms of selecting the digits first, then arranging them.But since the first digit cannot be zero, we have to be careful.To have a four-digit number with distinct digits, two even and two odd.Possible scenarios:Either the number has two even digits (including possibly zero) and two odd digits, with the first digit being even or odd.But the first digit can't be zero. So, we need to split into cases where the first digit is even (non-zero) or odd.Which is the same as our original approach.Alternatively, another way: total number of four-digit numbers with two even and two odd digits, all distinct.First, choose two even digits and two odd digits. But we need to ensure that zero is not in the first position. So, we can compute:Total number of such numbers without considering the first digit: C(5,2) * C(5,2) * 4! (choosing 2 even digits from 5, 2 odd digits from 5, and arranging all 4 digits). But then subtract the cases where the first digit is zero.Wait, but this might be more complicated.Wait, if we proceed this way, we need to adjust for the fact that zero cannot be in the first position.Alternatively, total number of such numbers where two even digits include zero: then, when arranging, zero could be in the first position, which is invalid. So, perhaps this approach is more involved.But let's try.First, total number of ways to choose two even digits and two odd digits, all distinct.Case A: Both even digits are non-zero. There are 4 non-zero even digits (2,4,6,8). So, choosing 2 from them: C(4,2). Then, choosing 2 odd digits from 5: C(5,2). Then, arrange all four digits. Since none of the even digits is zero, the first digit can be any of the four digits. So, total arrangements: 4!.Case B: One even digit is zero, and the other is non-zero. So, choosing 1 non-zero even digit (C(4,1)) and 1 zero. Then, choosing 2 odd digits (C(5,2)). Then, arrange the four digits. However, in this case, zero cannot be in the first position. So, total arrangements: 4! - 3! (total arrangements minus arrangements where zero is first). Because if we fix zero in the first position, the remaining 3 digits can be arranged in 3! ways. So, valid arrangements: 4! - 3! = 24 - 6 = 18.Therefore, total numbers in Case B: C(4,1) * C(5,2) * 18.Case A: C(4,2) * C(5,2) * 24.So, total numbers: [C(4,2)*C(5,2)*24] + [C(4,1)*C(5,2)*18]Let's compute this.C(4,2) = 6, C(5,2) = 10. So first term: 6 * 10 * 24 = 60 * 24 = 1440.C(4,1) = 4, C(5,2) = 10. Second term: 4 * 10 * 18 = 40 * 18 = 720.Total: 1440 + 720 = 2160.Same result as before. So this confirms that the answer is 2160.Therefore, the number of four-digit "fancy" numbers with distinct digits is 2160.Final AnswerThe number of four-digit "fancy" numbers with different digits is boxed{2160}.</think>To determine the number of four-digit "fancy" numbers with different digits, we need to consider numbers with exactly two even digits and two odd digits. We split this into two cases based on whether the first digit is even or odd.Case 1: The first digit is even.- Choices for the first digit: 4 (2, 4, 6, 8)- Remaining digits: 1 even and 2 odd- Remaining even digits: 4 (0, 2, 4, 6, 8 excluding the first digit)- Remaining odd digits: 5 (1, 3, 5, 7, 9)- Number of ways to choose 1 even and 2 odd digits: ( binom{4}{1} times binom{5}{2} )- Number of ways to arrange these digits: ( 3! )- Total for Case 1: ( 4 times left( binom{4}{1} times binom{5}{2} times 3! right) = 4 times (4 times 10 times 6) = 960 )Case 2: The first digit is odd.- Choices for the first digit: 5 (1, 3, 5, 7, 9)- Remaining digits: 2 even and 1 odd- Remaining even digits: 5 (0, 2, 4, 6, 8)- Remaining odd digits: 4 (1, 3, 5, 7, 9 excluding the first digit)- Number of ways to choose 2 even and 1 odd digit: ( binom{5}{2} times binom{4}{1} )- Number of ways to arrange these digits: ( 3! )- Total for Case 2: ( 5 times left( binom{5}{2} times binom{4}{1} times 3! right) = 5 times (10 times 4 times 6) = 1200 )Adding both cases together, the total number of four-digit "fancy" numbers with different digits is:[ 960 + 1200 = 2160 ]Thus, the number of four-digit "fancy" numbers with different digits is (boxed{2160}).]

🔑:Solution. We have 5 even digits: 0,2,4,6,8 and 5 odd digits: 1,3,5,7,9. Two even digits out of 5 can be chosen in C_{5}^{2}=10 ways, i.e., we have the following possibilities{{0,2},{0,4},{0,6},{0,8},{2,4},{2,6},{2,8},{4,6},{4,8},{6,8}} .In the same number of ways, two out of 5 odd digits can be chosen.For each pair (one pair of even and one pair of odd numbers), there are a total of 4!=24 ways to form a sequence of four digits. Therefore, we have a total of 10 cdot 10 cdot 24=2400 four-digit sequences consisting of two even and two odd digits.It remains to count those that start with 0. If a pair of even numbers contains 0, then the number of four-digit numbers starting with 0 is 3!=6. Since we have four even pairs containing 0, it follows that the number of all "special" four-digit sequences starting with 0 is 4 cdot 10 cdot 6=240.Therefore, according to this, the number of all four-digit "special" numbers written with different digits is 2400-240=2160.