🧠:[<think>Okay, let's try to figure out this problem step by step. So, we have an isosceles triangle ABC where AB equals AC, and the base BC is 65 cm. There's a point P on BC such that the perpendicular distances from P to AB and AC are 24 cm and 36 cm respectively. We need to find the area of triangle ABC. The answer choices are given, so maybe that can help check our work later.First, since ABC is isosceles with AB = AC, that tells me the triangle is symmetric with respect to the altitude from A to BC. Let me draw this triangle in my mind. Vertex A is at the top, and BC is the base of length 65 cm. Point P is somewhere along BC. From P, if I drop perpendiculars to AB and AC, their lengths are 24 cm and 36 cm. Hmm, since the triangle is isosceles, maybe we can use coordinate geometry here. Let's place the triangle in a coordinate system to make it easier. Let me set point B at (0,0) and point C at (65,0). Then, since AB = AC, the apex A must lie somewhere along the perpendicular bisector of BC. The midpoint of BC is at (32.5, 0), so the x-coordinate of A is 32.5. Let me denote the coordinates of A as (32.5, h), where h is the height of the triangle. Then, the area of ABC would be (1/2)*base*height = (1/2)*65*h. So, if we can find h, we can find the area.Now, point P is somewhere on BC. Let's denote the coordinates of P as (x, 0), where x is between 0 and 65. The perpendicular distances from P to AB and AC are 24 cm and 36 cm. So, we need to express these distances in terms of x and h, then solve for h.First, let's find the equations of sides AB and AC. For AB: Points A (32.5, h) and B (0,0). The slope of AB is (h - 0)/(32.5 - 0) = h / 32.5. So, the equation of AB is y = (h / 32.5)x.For AC: Points A (32.5, h) and C (65,0). The slope is (0 - h)/(65 - 32.5) = -h / 32.5. So, the equation of AC is y = (-h / 32.5)(x - 65). Let me check that. When x = 65, y = 0, which is correct. Expanding that equation: y = (-h / 32.5)x + (h / 32.5)*65. Since 65 is 2*32.5, that simplifies to y = (-h / 32.5)x + 2h. Now, the distance from point P (x,0) to AB and AC. The formula for the distance from a point (x0,y0) to the line ax + by + c = 0 is |ax0 + by0 + c| / sqrt(a^2 + b^2).First, let's write the equations of AB and AC in standard form.For AB: y - (h / 32.5)x = 0. Multiplying both sides by 32.5 to eliminate the decimal: 32.5y - h x = 0. So, the standard form is h x - 32.5 y = 0. So, a = h, b = -32.5, c = 0.Distance from P (x,0) to AB: |h*x - 32.5*0 + 0| / sqrt(h^2 + (-32.5)^2) = |h x| / sqrt(h^2 + 32.5^2). But since h and x are positive (since coordinates are in the first quadrant), this simplifies to h x / sqrt(h^2 + 32.5^2). According to the problem, this distance is 24 cm.Similarly, for AC: The equation in standard form. Starting from y = (-h / 32.5)x + 2h. Let's rearrange: (h / 32.5)x + y - 2h = 0. Multiply both sides by 32.5 to eliminate the denominator: h x + 32.5 y - 65 h = 0. So, standard form: h x + 32.5 y - 65 h = 0. Therefore, a = h, b = 32.5, c = -65 h.Distance from P (x,0) to AC: |h*x + 32.5*0 -65 h| / sqrt(h^2 + 32.5^2) = |h x -65 h| / sqrt(h^2 + 32.5^2). Since x is between 0 and 65, h x -65 h = h(x -65) is negative, so the absolute value is |h(x -65)| = h(65 - x). Therefore, the distance is h(65 - x) / sqrt(h^2 + 32.5^2). According to the problem, this distance is 36 cm.So, we have two equations:1) h x / sqrt(h^2 + 32.5^2) = 242) h (65 - x) / sqrt(h^2 + 32.5^2) = 36Let me call sqrt(h^2 + 32.5^2) as D to simplify notation. Then:1) h x / D = 242) h (65 - x) / D = 36If I divide equation 2 by equation 1, I get:(h (65 - x) / D) / (h x / D) ) = 36 /24Simplify: (65 - x)/x = 3/2Cross multiplying: 2(65 - x) = 3x => 130 - 2x = 3x => 130 = 5x => x = 26.So, x is 26 cm. So, point P is 26 cm from B and 65 -26 = 39 cm from C.Now, substitute x =26 into equation 1:h *26 / sqrt(h^2 +32.5^2) =24Let me solve for h. Let's write:26 h =24 sqrt(h^2 +32.5^2)Square both sides to eliminate the square root:(26 h)^2 = (24)^2 (h^2 +32.5^2)Compute left side: 676 h^2Right side: 576 (h^2 + 1056.25) => 576 h^2 + 576*1056.25Bring all terms to left side:676 h^2 -576 h^2 -576*1056.25 =0100 h^2 -576*1056.25 =0So, 100 h^2 =576*1056.25Compute 576*1056.25. Let's break this down:576 * 1000 =576,000576 *56.25=?56.25 is 56 +0.25. So 576*56=32,256 and 576*0.25=144. So total 32,256 +144=32,400Therefore, 576*1056.25=576,000 +32,400=608,400So, 100 h^2 =608,400 => h^2 =6,084 => h= sqrt(6,084)Compute sqrt(6084). Let me recall that 78^2=6,084 because 70^2=4,900, 8^2=64, (70+8)^2=70^2 +2*70*8 +8^2=4,900+1,120+64=6,084. So h=78 cm.Therefore, the height h is 78 cm. Then, the area of triangle ABC is (1/2)*65*78.Compute that: 65*78. Let's do 65*70=4,550 and 65*8=520, so total 4,550+520=5,070. Then half of that is 5,070 /2=2,535 cm².Looking at the options, D is 2535. So that should be the answer. Let me check if everything makes sense.Wait, but let me verify once again. Because sometimes when squaring both sides, extraneous solutions can come up, but here since all quantities are positive, it should be okay. Let me confirm h=78. Then, compute sqrt(h² +32.5²)=sqrt(78² +32.5²). Compute 78²=6,084, 32.5²=1,056.25. So total is 6,084 +1,056.25=7,140.25. sqrt(7,140.25). Let's see, 84.5²= (80 +4.5)^2=80² +2*80*4.5 +4.5²=6,400 +720 +20.25=7,140.25. So sqrt(7,140.25)=84.5. Then, check equation 1: h*x/D=78*26 /84.5. Compute 78*26. 70*26=1,820, 8*26=208, total=1,820+208=2,028. Divide by84.5: 2,028 /84.5. Let's compute 84.5*24=2,028. So 2,028 /84.5=24, which matches. Similarly equation 2: h*(65 -26)/D=78*39 /84.5. 78*39. Let's do 70*39=2,730, 8*39=312, total=2,730+312=3,042. Divided by84.5: 3,042 /84.5. 84.5*36=3,042, so that equals 36. Perfect, that checks out. So h=78 is correct.Therefore, area is indeed (1/2)*65*78=2,535. So the answer is D) 2535. That's one of the options. So I think that's correct.Final Answerboxed{D}</think>Given an isosceles triangle (ABC) with (AB = AC) and (BC = 65) cm. A point (P) on (overline{BC}) has perpendicular distances 24 cm and 36 cm to (overline{AB}) and (overline{AC}) respectively. We need to find the area of (triangle ABC).1. Coordinates and Equations: - Place (B) at ((0, 0)), (C) at ((65, 0)), and (A) at ((32.5, h)). - The equations of lines (AB) and (AC) are derived: - (AB: y = frac{h}{32.5}x) - (AC: y = -frac{h}{32.5}(x - 65))2. Distances from (P): - Let (P) be ((x, 0)). The perpendicular distances from (P) to (AB) and (AC) are given by: - Distance to (AB: frac{hx}{sqrt{h^2 + 32.5^2}} = 24) - Distance to (AC: frac{h(65 - x)}{sqrt{h^2 + 32.5^2}} = 36)3. Solving for (x): - Dividing the second equation by the first gives (frac{65 - x}{x} = frac{36}{24} = frac{3}{2}). - Solving (frac{65 - x}{x} = frac{3}{2}) gives (x = 26).4. Solving for (h): - Substitute (x = 26) into the first distance equation: [ frac{26h}{sqrt{h^2 + 32.5^2}} = 24 ] - Squaring both sides and solving gives (h = 78).5. Area Calculation: - The area of (triangle ABC) is (frac{1}{2} times 65 times 78 = 2535) cm(^2).Thus, the area of (triangle ABC) is (boxed{D}).,<think>Okay, so I need to solve this problem about an isosceles triangle ABC where AB equals AC, and the base BC is 65 cm. There's a point P on BC such that the perpendicular distances from P to AB and AC are 24 cm and 36 cm respectively. I need to find the area of triangle ABC, and the answer choices are given. Let me try to visualize this first.Since ABC is isosceles with AB = AC, the triangle is symmetric about the altitude from A to BC. Let me draw this triangle with vertex A at the top, and BC as the base. Point P is somewhere along BC. From P, if I drop perpendiculars to AB and AC, their lengths are 24 cm and 36 cm. Hmm, so these are the heights from P to the two equal sides.I remember that in coordinate geometry, sometimes assigning coordinates can help. Maybe I can place BC on the x-axis for simplicity. Let me set point B at (0,0) and point C at (65,0). Since the triangle is isosceles with AB = AC, the apex A must lie somewhere along the perpendicular bisector of BC. The midpoint of BC is at (32.5, 0), so point A should be at (32.5, h), where h is the height of the triangle. Then the area of ABC would be (1/2)*base*height = (1/2)*65*h. So if I can find h, I can compute the area.Now, point P is somewhere on BC. Let me denote the coordinates of P as (x, 0), where x is between 0 and 65. The problem states that the perpendicular distance from P to AB is 24 cm and to AC is 36 cm. I need to relate these distances to x and h.First, let me find the equations of sides AB and AC. For AB: points A(32.5, h) and B(0,0). The slope of AB is (h - 0)/(32.5 - 0) = h/32.5. So the equation of AB is y = (h/32.5)x.Similarly, for AC: points A(32.5, h) and C(65,0). The slope is (0 - h)/(65 - 32.5) = -h/32.5. The equation of AC is y = (-h/32.5)(x - 65). Simplifying, y = (-h/32.5)x + (65h)/32.5.Now, the distance from a point (x0, y0) to a line ax + by + c = 0 is |ax0 + by0 + c| / sqrt(a^2 + b^2). Let me use this formula to find the perpendicular distances from P(x,0) to AB and AC.First, let's write the equations of AB and AC in standard form.For AB: y = (h/32.5)x => (h/32.5)x - y = 0. Multiply both sides by 32.5 to eliminate the decimal: h x - 32.5 y = 0. So the standard form is h x - 32.5 y = 0.Similarly, for AC: y = (-h/32.5)x + (65h)/32.5. Multiply through by 32.5: 32.5 y = -h x + 65 h. Bring all terms to the left: h x + 32.5 y - 65 h = 0. So the standard form is h x + 32.5 y - 65 h = 0.Now, the distance from P(x,0) to AB is |h*x - 32.5*0| / sqrt(h^2 + (-32.5)^2) = |h x| / sqrt(h^2 + 32.5^2). According to the problem, this distance is 24 cm.Similarly, the distance from P(x,0) to AC is |h*x + 32.5*0 - 65 h| / sqrt(h^2 + 32.5^2) = |h x - 65 h| / sqrt(h^2 + 32.5^2). This distance is given as 36 cm.So we have two equations:1) |h x| / sqrt(h^2 + 32.5^2) = 242) |h (x - 65)| / sqrt(h^2 + 32.5^2) = 36Since x is between 0 and 65, and h is positive (height of the triangle), let's analyze the absolute values.For equation 1: x is between 0 and 65, so h x is non-negative because h > 0 and x >=0. So |h x| = h x.For equation 2: x - 65 is between -65 and 0, so h(x - 65) is negative because h > 0. Therefore |h(x - 65)| = h(65 - x).So we can rewrite the equations without absolute values:1) (h x) / sqrt(h^2 + 32.5^2) = 242) (h (65 - x)) / sqrt(h^2 + 32.5^2) = 36Let me denote D = sqrt(h^2 + 32.5^2). Then equations become:1) h x / D = 242) h (65 - x) / D = 36If I divide equation 1 by equation 2, the D and h terms will cancel out:( h x / D ) / ( h (65 - x) / D ) = 24 / 36Simplify:x / (65 - x) = 2 / 3Cross multiplying:3x = 2(65 - x)3x = 130 - 2x5x = 130x = 26So x is 26 cm. Now, substitute x = 26 back into equation 1:h * 26 / sqrt(h^2 + 32.5^2) = 24Let me solve for h. Let me denote sqrt(h^2 + 32.5^2) as S. Then:26 h / S = 24 => S = (26 h) / 24 = (13 h)/12But S = sqrt(h^2 + 32.5^2). Therefore:(13 h / 12)^2 = h^2 + 32.5^2Compute left side:(169 h^2) / 144 = h^2 + 1056.25Multiply both sides by 144:169 h^2 = 144 h^2 + 144 * 1056.25Compute 144 * 1056.25. Let me compute 1056.25 * 144:1056.25 * 144 = (1000 + 56.25) * 144 = 1000*144 + 56.25*144 = 144,000 + 8,100 = 152,100. Wait, 56.25 * 144. Let me check that again.56.25 * 144: 56 * 144 = (50*144) + (6*144) = 7200 + 864 = 8064; 0.25*144=36. So total is 8064 + 36 = 8100. So 144*1056.25=144*1000 + 144*56.25=144,000 + 8100=152,100.So back to the equation:169 h^2 - 144 h^2 = 152,10025 h^2 = 152,100h^2 = 152,100 / 25 = 6,084Therefore h = sqrt(6,084). Let me compute that.sqrt(6,084). Let's see, 78^2 = 6,084 because 70^2=4,900, 80^2=6,400, so 78^2=(70+8)^2=70^2 + 2*70*8 +8^2=4,900 + 1,120 +64=6,084. So h=78 cm.Therefore, the height h is 78 cm. Then the area of triangle ABC is (1/2)*65*78.Compute 65*78 first. Let's do 60*78=4,680 and 5*78=390, so total is 4,680 + 390 = 5,070. Then half of that is 2,535.Looking at the answer choices, option D is 2535. So the answer should be D.Wait, but let me double-check my steps because the answer is one of the options, but let me confirm.Wait, point P is at x=26. So the distances from P to AB and AC are 24 and 36. Let me verify with h=78.First, compute D = sqrt(h^2 + 32.5^2) = sqrt(78^2 + 32.5^2). Let me compute 78^2=6,084; 32.5^2=1,056.25. So D= sqrt(6,084 + 1,056.25)=sqrt(7,140.25). What is sqrt(7,140.25)?Hmm, 84^2=7,056, 85^2=7,225. So 84.5^2=(84 + 0.5)^2=84^2 + 2*84*0.5 +0.5^2=7,056 +84 +0.25=7,140.25. So D=84.5.Then, check equation 1: h x / D = 78 *26 /84.5. Let's compute 78*26: 70*26=1,820, 8*26=208, so total 2,028. Then divide by 84.5: 2,028 /84.5.Let me compute 84.5 *24=2,028. Because 84 *24=2,016, 0.5*24=12, total 2,016+12=2,028. So 2,028 /84.5=24, which matches the first distance. Similarly, equation 2: h*(65 -x)/D=78*(65 -26)/84.5=78*39/84.5.78*39: 70*39=2,730, 8*39=312, total 3,042. Then 3,042 /84.5. 84.5*36=3,042 (since 84*36=3,024, 0.5*36=18, so 3,024+18=3,042). Therefore, 3,042 /84.5=36. Correct. So h=78 cm is correct. Then area is 1/2*65*78=65*39=2,535. So D is correct.But wait, the answer choices include 2535 as option D, which matches. So I think that's the answer.But just to be thorough, let me think if there's another approach. Maybe using areas.Alternatively, since P is on BC, and the distances to AB and AC are 24 and 36. Maybe the areas of triangles APB and APC can be related?But triangle APB has base AB and height 24, but AB is equal to AC. Wait, but AB is not the base here; actually, APB is a triangle with base BP and height 24 (distance from P to AB). Similarly, APC has base PC and height 36. But the area of APB would be (1/2)*BP*24 and area of APC would be (1/2)*PC*36. Then the sum of these areas would be the area of ABC. Wait, but ABC is isosceles with AB=AC, but the heights from P are different. Wait, but in that case, the area of ABC would be equal to area APB + area APC = (1/2)*BP*24 + (1/2)*PC*36.But BP + PC = BC = 65. Let BP = m, PC = 65 - m. So area ABC = (1/2)*m*24 + (1/2)*(65 - m)*36 = 12 m + 18*(65 - m) = 12m + 1,170 - 18m = -6m + 1,170.But the area of ABC is also (1/2)*65*h. So (1/2)*65*h = -6m + 1,170. But I need another equation relating m and h. Hmm, but how?Alternatively, maybe we can use coordinates again. Wait, but this approach might not lead me anywhere unless I relate m and h. Alternatively, maybe similar triangles?Wait, perhaps the ratio of the distances from P to AB and AC is 24:36 = 2:3. Earlier, we found that x=26, so BP=26, PC=39. So BP/PC=26/39=2/3, which is the same ratio as the distances. Hmm, interesting. Is there a relation in the triangle where the ratio of distances from a point on the base to the two sides is equal to the ratio of the segments of the base?Wait, maybe in an isosceles triangle, the ratio of the distances from a point on the base to the equal sides is equal to the ratio of the segments into which the point divides the base. That seems to be the case here. Because BP/PC=2/3 and the distances are 24/36=2/3. So that's a property I might not have known, but in this case, since we derived it through coordinates, perhaps that's a general property.But in any case, using that we found x=26, and h=78, leading to area 2535, which is one of the options, so I think that's correct.Therefore, the answer should be D) 2535.Final Answerboxed{D}</think>Given an isosceles triangle (ABC) with (AB = AC) and (BC = 65) cm. Point (P) is on (overline{BC}) such that the perpendicular distances from (P) to (overline{AB}) and (overline{AC}) are 24 cm and 36 cm, respectively. We need to find the area of (triangle ABC).1. Coordinate Setup: - Place (B) at ((0, 0)) and (C) at ((65, 0)). The midpoint of (BC) is ((32.5, 0)), so the coordinates of (A) are ((32.5, h)), where (h) is the height of the triangle.2. Equations of Lines: - The equation of line (AB) is (y = frac{h}{32.5}x). - The equation of line (AC) is (y = -frac{h}{32.5}(x - 65)).3. Perpendicular Distances: - The distance from (P(x, 0)) to (AB) is (frac{|hx|}{sqrt{h^2 + 32.5^2}} = 24). - The distance from (P(x, 0)) to (AC) is (frac{|h(65 - x)|}{sqrt{h^2 + 32.5^2}} = 36).4. Solving for (x): - Dividing the equations for the distances gives (frac{x}{65 - x} = frac{24}{36} = frac{2}{3}). - Solving (frac{x}{65 - x} = frac{2}{3}) gives (x = 26).5. Finding (h): - Substituting (x = 26) into the distance equation for (AB): [ frac{26h}{sqrt{h^2 + 32.5^2}} = 24 ] - Let (D = sqrt{h^2 + 32.5^2}), then (26h = 24D). - Squaring both sides and solving gives (h = 78) cm.6. Area Calculation: - The area of (triangle ABC) is (frac{1}{2} times 65 times 78 = 2535) cm(^2).Thus, the area of (triangle ABC) is (boxed{D}).]

🔑:1. Identify the given information and setup the problem: - Triangle ABC is isosceles with AB = AC. - AB + AC = 65 cm and BC = 65 cm. - Point P is on overline{BC} such that the perpendicular distances from P to overline{AB} and overline{AC} are 24 cm and 36 cm, respectively.2. Determine the segments BP and CP: - Let the perpendicular distances from P hit AB at D and AC at E. - Since triangle BDP sim triangle CEP in the ratio 2:3, we have: [ frac{BP}{CP} = frac{2}{3} ] - Let BP = 2k and CP = 3k. Since BP + CP = BC = 65 cm, we get: [ 2k + 3k = 65 implies 5k = 65 implies k = 13 ] - Therefore, BP = 2k = 26 cm and CP = 3k = 39 cm.3. Calculate the lengths BD and CE using the Pythagorean Theorem: - For triangle BDP: [ BD^2 + DP^2 = BP^2 implies BD^2 + 24^2 = 26^2 implies BD^2 + 576 = 676 implies BD^2 = 100 implies BD = 10 text{ cm} ] - For triangle CEP: [ CE^2 + EP^2 = CP^2 implies CE^2 + 36^2 = 39^2 implies CE^2 + 1296 = 1521 implies CE^2 = 225 implies CE = 15 text{ cm} ]4. Set up the equation for AB = AC = x using the Pythagorean Theorem: - For triangle ADP and triangle AEP: [ AD = x - 10 quad text{and} quad AE = x - 15 ] [ (x - 10)^2 + 24^2 = (x - 15)^2 + 36^2 ] [ (x - 10)^2 + 576 = (x - 15)^2 + 1296 ] [ x^2 - 20x + 100 + 576 = x^2 - 30x + 225 + 1296 ] [ x^2 - 20x + 676 = x^2 - 30x + 1521 ] [ 10x + 676 = 1521 ] [ 10x = 845 implies x = 84.5 ]5. Calculate the altitude of triangle ABC from side BC: - Using the Pythagorean Theorem: [ text{Altitude} = sqrt{84.5^2 - 32.5^2} ] [ text{Altitude} = sqrt{7140.25 - 1056.25} = sqrt{6084} = 78 text{ cm} ]6. Calculate the area of triangle ABC: - The area is given by: [ text{Area} = frac{1}{2} times BC times text{Altitude} = frac{1}{2} times 65 times 78 = 65 times 39 = 2535 text{ cm}^2 ]The final answer is boxed{2535}

🧠:[<think>Okay, so I need to show that the number (2^{30}(2^{31} - 1)) is a perfect number. I remember that perfect numbers are those where the sum of their proper divisors equals the number itself. The examples given are 6 and 28, which I know are the first two perfect numbers. The problem also mentions that (2^{31} - 1) is a prime number, which probably is a key part of the proof.First, let me recall some facts about perfect numbers. I think there's a theorem related to even perfect numbers. Maybe all even perfect numbers are of the form (2^{p-1}(2^p - 1)) where (2^p - 1) is a prime, called a Mersenne prime. If that's the case, then if (2^{31} - 1) is prime, then (2^{30}(2^{31} - 1)) would fit that form with (p = 31), right? So maybe the problem is asking me to verify that this formula gives a perfect number, given the corresponding Mersenne prime.But how does that formula work? Why is that the case? Let me try to recall or derive the reasoning.First, let's consider the number (n = 2^{k-1}(2^k - 1)) where (2^k - 1) is prime. Let's try to compute the sum of the divisors of n, and then subtract n itself to see if it equals n.Wait, perfect numbers are equal to the sum of their proper divisors. So, the sum of all divisors except the number itself should equal the number. Alternatively, the sum of all divisors (including the number) should be twice the number, since the number is included. So, if (sigma(n)) denotes the sum of the divisors of n, then for a perfect number, (sigma(n) = 2n).Therefore, to check if (2^{30}(2^{31} - 1)) is perfect, I need to compute the sum of its divisors and verify that the sum is twice the number.Given that (2^{31} - 1) is prime, let's note that the number in question is (n = 2^{30} times (2^{31} - 1)). Since 2 and (2^{31} - 1) are coprime (because 2 is a prime not dividing the other prime (2^{31} - 1)), the sum of divisors function (sigma(n)) can be calculated as (sigma(2^{30}) times sigma(2^{31} - 1)).The sum of divisors for a prime power (p^k) is (1 + p + p^2 + dots + p^k). So, (sigma(2^{30}) = 1 + 2 + 2^2 + dots + 2^{30}). That's a geometric series, so the sum is (2^{31} - 1). Then, since (2^{31} - 1) is prime, (sigma(2^{31} - 1) = 1 + (2^{31} - 1) = 2^{31}).Therefore, (sigma(n) = (2^{31} - 1) times 2^{31} = 2^{31}(2^{31} - 1)). But the number n itself is (2^{30}(2^{31} - 1)), so twice n is (2 times 2^{30}(2^{31} - 1) = 2^{31}(2^{31} - 1)), which matches (sigma(n)). Therefore, since (sigma(n) = 2n), n is perfect.Wait, that seems straightforward, but let me check again step by step to make sure I didn't skip any parts.First, factorize n: (n = 2^{30} times q), where (q = 2^{31} - 1) is prime. Since 2 and q are distinct primes, the divisors of n are of the form (2^a times q^b), where (0 leq a leq 30) and (0 leq b leq 1). Therefore, the sum of divisors (sigma(n)) is the product of the sum of divisors of (2^{30}) and the sum of divisors of q.Sum of divisors of (2^{30}): (sum_{a=0}^{30} 2^a = 2^{31} - 1).Sum of divisors of q (which is prime): (1 + q = 1 + (2^{31} - 1) = 2^{31}).Multiply them together: ((2^{31} - 1)(2^{31}) = 2^{31}(2^{31} - 1)).But n is (2^{30}(2^{31} - 1)), so twice n is (2 times 2^{30}(2^{31} - 1) = 2^{31}(2^{31} - 1)).Therefore, since (sigma(n) = 2n), n is indeed a perfect number.Let me think if there's another way someone might approach this problem, maybe by directly summing the proper divisors.Proper divisors of n would be all divisors except n itself. So, if we can list all divisors of n except n and sum them up, and check if that sum equals n.But n is (2^{30} times q), with q prime. The divisors of n are (2^a times q^b) where (0 leq a leq 30) and (0 leq b leq 1). So, excluding n itself, the proper divisors would have either:1. (0 leq a leq 30) and (b = 0), which are just the powers of 2 up to (2^{30}), or2. (0 leq a leq 30) and (b = 1), but not both a=30 and b=1 (since that would be n itself). Wait, no. If (b = 1), then even with a=0, you get q, which is a proper divisor. So actually, all divisors where either (b = 0) (but excluding 1) or (b = 1) (but not n itself). Wait, no, actually, the proper divisors are all divisors except n. So, the divisors are:- All (2^a) for a from 0 to 30 (so 1, 2, 4, ..., (2^{30}))- All (2^a times q) for a from 0 to 30 (so q, 2q, 4q, ..., (2^{30}q))But n is (2^{30}q), so the second set includes n when a=30. Therefore, the proper divisors are:- All (2^a) for a from 0 to 30 (including 1 but excluding n), but wait, (2^{30}) is a proper divisor as well, since it's not equal to n. Then the first set is 1, 2, 4, ..., (2^{30}), and the second set is q, 2q, 4q, ..., (2^{29}q). Wait, because (2^{30}q) is n, so in the second set, when a=30, that term is n, so we exclude that.Therefore, the proper divisors are:1. The divisors that are powers of 2: (1, 2, 4, dots, 2^{30})2. The divisors that are q multiplied by powers of 2 up to (2^{29}): (q, 2q, 4q, dots, 2^{29}q)So, let's compute the sum of the first set. As before, the sum of (2^a) from a=0 to 30 is (2^{31} - 1). Then, the sum of the second set is q times the sum of (2^a) from a=0 to 29. The sum from a=0 to 29 is (2^{30} - 1). Therefore, the sum of the second set is (q(2^{30} - 1)).Therefore, the total sum of proper divisors is:( (2^{31} - 1) + q(2^{30} - 1) )But q is (2^{31} - 1), so substitute:( (2^{31} - 1) + (2^{31} - 1)(2^{30} - 1) )Factor out (2^{31} - 1):( (2^{31} - 1)[1 + (2^{30} - 1)] = (2^{31} - 1)(2^{30}) )But n is (2^{30}(2^{31} - 1)), so the sum of the proper divisors is exactly n. Therefore, n is perfect.Wait, that's the same conclusion as before. So both methods confirm that the sum of the proper divisors equals the number itself.Therefore, using either the sum of divisors function approach or directly summing the proper divisors, we arrive at the conclusion that (2^{30}(2^{31} - 1)) is a perfect number, given that (2^{31} - 1) is prime.I should check if there are any gaps in this reasoning. For instance, in the first approach, we used the multiplicative property of the sum of divisors function. Since 2 and (2^{31} - 1) are coprime, it's valid to say that (sigma(n) = sigma(2^{30}) times sigma(2^{31} - 1)). That's correct because the sum of divisors function is multiplicative for coprime numbers.Also, in the second approach, breaking down the proper divisors into those that are powers of 2 and those that are q times lower powers of 2 is valid because of the unique prime factorization. Each proper divisor must be a product of the primes in n raised to a power less than or equal to their exponents in n, but not both at maximum. Since n is (2^{30} times q^1), the proper divisors can't have both exponent 30 on 2 and exponent 1 on q, which is exactly n.Therefore, splitting the divisors into those without q and those with q (but not the full power on 2) makes sense. Calculating each sum separately and adding them gives the total sum of proper divisors, which equals n. Therefore, the reasoning holds.Another check: let's take the example given, 6 and 28, and see if they fit this formula.For 6: (6 = 2^{2 - 1}(2^2 - 1) = 2^1 times 3). Here, (2^2 - 1 = 3) is prime, so according to the formula, it should be perfect. Similarly, 28 is (2^{3 - 1}(2^3 - 1) = 2^2 times 7), and (2^3 - 1 = 7) is prime. So both examples fit the formula given. Therefore, the formula works for known cases, which gives more confidence.Therefore, putting it all together, since (2^{31} - 1) is prime, (2^{30}(2^{31} - 1)) is an even perfect number according to the Euclid-Euler theorem, which characterizes all even perfect numbers as those of the form (2^{p-1}(2^p - 1)) where (2^p - 1) is prime. Hence, the given number is perfect.Final Answerboxed{2^{30}left(2^{31} - 1right)} is a perfect number.</think>To show that (2^{30}(2^{31} - 1)) is a perfect number, we start by noting that a natural number (n) is perfect if the sum of its proper divisors equals (n). Given that (2^{31} - 1) is a prime number, we use the known result that even perfect numbers are of the form (2^{p-1}(2^p - 1)) where (2^p - 1) is a prime (a Mersenne prime).1. Factorize the number: The number in question is (n = 2^{30}(2^{31} - 1)). Here, (2^{31} - 1) is a prime number, and 2 and (2^{31} - 1) are coprime.2. Sum of divisors function: The sum of the divisors of (n), denoted (sigma(n)), is multiplicative for coprime numbers. Thus, we calculate: - (sigma(2^{30})): The sum of divisors of (2^{30}) is (1 + 2 + 2^2 + cdots + 2^{30} = 2^{31} - 1). - (sigma(2^{31} - 1)): Since (2^{31} - 1) is prime, the sum of its divisors is (1 + (2^{31} - 1) = 2^{31}).3. Multiply the results: [ sigma(n) = sigma(2^{30}) times sigma(2^{31} - 1) = (2^{31} - 1) times 2^{31} ]4. Compare with twice the number: The number (n) is (2^{30}(2^{31} - 1)). Twice this number is: [ 2 times 2^{30}(2^{31} - 1) = 2^{31}(2^{31} - 1) ] This matches (sigma(n)), confirming that (sigma(n) = 2n), hence (n) is perfect.5. Direct sum of proper divisors: Alternatively, the proper divisors of (n) are: - Powers of 2: (1, 2, 4, ldots, 2^{30}) - Products of (2^a) (for (0 leq a leq 29)) with (2^{31} - 1) Summing these: - Sum of powers of 2: (2^{31} - 1) - Sum of products: ((2^{30} - 1)(2^{31} - 1)) Total sum: [ (2^{31} - 1) + (2^{30} - 1)(2^{31} - 1) = (2^{31} - 1)(1 + 2^{30} - 1) = 2^{30}(2^{31} - 1) ]Both methods confirm that the sum of the proper divisors equals (n), proving (2^{30}(2^{31} - 1)) is perfect.[boxed{2^{30}left(2^{31} - 1right)}],<think>Okay, so I need to show that the number (2^{30}(2^{31} - 1)) is a perfect number. I know that a perfect number is one where the sum of its proper divisors equals the number itself. The problem also mentions that (2^{31} - 1) is a prime number. Hmm, prime numbers like that remind me of Mersenne primes, which are primes of the form (2^p - 1). Maybe that's relevant here.First, let me recall the definition of a perfect number. A natural number (n) is perfect if the sum of all its proper divisors (excluding itself) equals (n). The classic examples are 6 and 28, as given. Now, I remember there's a theorem related to even perfect numbers. I think it says that even perfect numbers can be written in the form (2^{p-1}(2^p - 1)), where (2^p - 1) is a Mersenne prime. Is that right?Let me check that. If (2^p - 1) is prime, then (2^{p-1}(2^p - 1)) is an even perfect number. Conversely, every even perfect number is of that form. So, in this problem, the number given is (2^{30}(2^{31} - 1)). Comparing this to the formula (2^{p-1}(2^p - 1)), it looks like (p = 31) here. Since (2^{31} - 1) is given as prime, then according to the theorem, (2^{30}(2^{31} - 1)) should indeed be a perfect number.But maybe I should prove this from scratch, rather than just citing the theorem, to make sure I understand why it works. Let me try to compute the sum of the proper divisors of (2^{30}(2^{31} - 1)) and show that it equals the number itself.First, note that the number in question is (N = 2^{30} times q), where (q = 2^{31} - 1) is prime. Since (q) is prime, the divisors of (N) are all numbers of the form (2^k) and (2^k times q), where (0 leq k leq 30). But since we're looking for proper divisors, we exclude (N) itself, so (k) goes up to 30 for the (2^k) terms, but when multiplying by (q), (k) can only go up to 29, because (2^{30} times q) is the number itself. Wait, let me clarify that.Wait, no. If (N = 2^{30} times q), then the divisors are all numbers (2^k times q^m), where (0 leq k leq 30) and (0 leq m leq 1). But since (q) is prime, (m) can only be 0 or 1. So the divisors are (2^k) for (0 leq k leq 30) and (2^k times q) for (0 leq k leq 30). However, the divisor (2^{30} times q) is (N) itself, so we need to exclude that. Therefore, the proper divisors are:1. All (2^k) for (0 leq k leq 30)2. All (2^k times q) for (0 leq k leq 29)So, the sum of the proper divisors is the sum of the geometric series (2^0 + 2^1 + dots + 2^{30}) plus the sum (q times (2^0 + 2^1 + dots + 2^{29})).Let me compute the first sum. The sum of a geometric series (2^0 + 2^1 + dots + 2^n = 2^{n+1} - 1). So, the sum from (k=0) to (30) is (2^{31} - 1).Then, the second sum is (q times (2^0 + 2^1 + dots + 2^{29})). Again, that's a geometric series. The sum from (k=0) to (29) is (2^{30} - 1). Therefore, the second sum is (q times (2^{30} - 1)).So, the total sum of proper divisors is ((2^{31} - 1) + q times (2^{30} - 1)). Let's substitute (q = 2^{31} - 1) into the equation.Thus, sum = ((2^{31} - 1) + (2^{31} - 1)(2^{30} - 1)).Factor out ((2^{31} - 1)):sum = ((2^{31} - 1) [1 + (2^{30} - 1)]) = ((2^{31} - 1)(2^{30})).But wait, the original number is (2^{30}(2^{31} - 1)), which is exactly the sum we just calculated. However, hold on—the sum of the proper divisors equals the number itself? That would mean the number is perfect. But wait, isn't the sum of all divisors including the number itself equal to twice the number if it's perfect?Wait, maybe I confused the definitions here. Let me double-check.The sum of all positive divisors (including the number itself) is called the sigma function, denoted (sigma(N)). A number (N) is perfect if (sigma(N) = 2N). Because the sum of all divisors (including N) is equal to twice N, hence the sum of proper divisors is (2N - N = N). So, in that case, if I compute (sigma(N)), it should equal (2N).Let me verify that. So, in my case, (N = 2^{30} times q), where (q) is prime. The divisors of (N) are (2^k times q^m) where (0 leq k leq 30) and (0 leq m leq 1). Therefore, the sum of divisors is (sum_{k=0}^{30} sum_{m=0}^{1} 2^k q^m = sum_{k=0}^{30} 2^k times sum_{m=0}^{1} q^m).The first sum is (2^{31} - 1), as before, and the second sum is (1 + q). Therefore, (sigma(N) = (2^{31} - 1)(1 + q)).But since (q = 2^{31} - 1), substituting gives (sigma(N) = (2^{31} - 1)(1 + 2^{31} - 1) = (2^{31} - 1)(2^{31})).But (N = 2^{30} times q = 2^{30}(2^{31} - 1)). Therefore, (2N = 2^{31}(2^{31} - 1)). Comparing that to (sigma(N) = (2^{31} - 1)(2^{31})), we see that (sigma(N) = 2N), which means that (N) is indeed a perfect number.Therefore, this shows that (2^{30}(2^{31} - 1)) is a perfect number when (2^{31} - 1) is prime.Wait, so in my initial calculation, I had included both the sum of the divisors without the number itself, but actually, the proper way is to compute the sum of all divisors and check if it's twice the number. But through the sigma function, we see that it's straightforward.Alternatively, if I compute the sum of proper divisors directly, which is (sigma(N) - N). So, (sigma(N) - N = 2N - N = N), which is the condition for perfection. Therefore, yes, if (sigma(N) = 2N), then the sum of proper divisors is (N), so (N) is perfect.So, in conclusion, by calculating the sum of divisors using the formula for the sigma function, factoring in the prime nature of (q = 2^{31} - 1), we find that the sum is indeed twice the number, hence the number is perfect.Alternatively, breaking it down step by step:1. Given (N = 2^{30} times (2^{31} - 1)), and (2^{31} - 1) is prime.2. The divisors of (N) are (1, 2, 2^2, ldots, 2^{30}) and (1 times (2^{31} - 1), 2 times (2^{31} - 1), ldots, 2^{30} times (2^{31} - 1)).3. However, the divisor (2^{30} times (2^{31} - 1)) is (N) itself, so proper divisors exclude this.4. Sum of the divisors excluding (N) is sum of (2^0) to (2^{30}) plus sum of ((2^0) to (2^{29}) times (2^{31} - 1)).5. Calculating the first sum: (2^{31} - 1).6. Calculating the second sum: ((2^{30} - 1) times (2^{31} - 1)).7. Total sum: ((2^{31} - 1) + (2^{30} - 1)(2^{31} - 1) = (2^{31} - 1)(1 + 2^{30} - 1) = (2^{31} - 1)(2^{30})).8. Which is equal to (2^{30} times (2^{31} - 1) = N). Therefore, the sum of proper divisors equals (N), so (N) is perfect.So both approaches confirm the result. I think this covers it. The key idea is that because (2^{31} - 1) is prime, the divisors are straightforward to enumerate, and their sum can be computed using geometric series, leading to the conclusion that the sum equals (N), hence the number is perfect.Final Answerboxed{2^{30}left(2^{31} - 1right)} is a perfect number.</think>To show that (2^{30}(2^{31} - 1)) is a perfect number, we start by noting that a perfect number is one where the sum of its proper divisors equals the number itself. Given that (2^{31} - 1) is a prime number, we can use the theorem that states if (2^p - 1) is a prime (a Mersenne prime), then (2^{p-1}(2^p - 1)) is an even perfect number.Here, (N = 2^{30}(2^{31} - 1)). The divisors of (N) are of the form (2^k) and (2^k times (2^{31} - 1)) where (0 leq k leq 30). Proper divisors exclude (N) itself, so we consider:1. The divisors (2^k) for (0 leq k leq 30)2. The divisors (2^k times (2^{31} - 1)) for (0 leq k leq 29)The sum of the geometric series for (2^k) from (k=0) to 30 is (2^{31} - 1). The sum of the geometric series for (2^k) from (k=0) to 29 is (2^{30} - 1). Multiplying this by the prime (2^{31} - 1) gives the second part of the sum.Thus, the total sum of proper divisors is:[(2^{31} - 1) + (2^{31} - 1)(2^{30} - 1)]Factoring out (2^{31} - 1), we get:[(2^{31} - 1) left[1 + (2^{30} - 1)right] = (2^{31} - 1) cdot 2^{30}]This is exactly (N), confirming that the sum of the proper divisors equals (N). Therefore, (2^{30}(2^{31} - 1)) is a perfect number.[boxed{2^{30}left(2^{31} - 1right)}]]

🔑:If 2^{31}-1 is a prime number, its only proper divisor is the number 1. Therefore, the proper divisors of 2^{30}left(2^{31}-1right) are1,2,2^{2}, ldots, 2^{29}, 2^{30},left(2^{31}-1right), 2left(2^{31}-1right), 2^{2}left(2^{31}-1right), ldots 2^{29}left(2^{31}-1right)The sum S of these divisors isS=left[1+2+2^{2}+cdots+2^{29}+2^{30}right]+left(2^{31}-1right)left[1+2+2^{2}+cdots+2^{29}right]In each of the two brackets, the sum S_{n} of a geometric progression with n terms appears, where the first term is 1 and the common ratio is 2: the first bracket is S_{31}, with 31 terms, and the second is S_{30}, with thirty terms. Using the formula for the sum of the terms of a geometric progression, we getS_{31}=frac{2^{31}-1}{2-1}=2^{31}-1 quad text { and } quad S_{30}=frac{2^{30}-1}{2-1}=2^{30}-1Thus, the sum of the proper divisors of 2^{30}left(2^{31}-1right) isS=left(2^{31}-1right)+left(2^{31}-1right)left[2^{30}-1right]=left(2^{31}-1right)left(1+2^{30}-1right)=2^{30}left(2^{31}-1right)Therefore, this sum is equal to 2^{30}left(2^{31}-1right), as we wanted to prove.

🧠:[<think>Okay, so I have this trigonometry problem here. It says that given sin(x + π/12) = -1/4, then cos(5π/6 - 2x) is equal to one of the options: A: 7/8, B: 1/8, C: -7/8, D: -1/8. I need to figure out which one is correct. Let me try to work through this step by step.First, let's recall some trigonometric identities that might be helpful here. The problem involves both sine and cosine functions, and there's an angle addition and subtraction. The key might be to express cos(5π/6 - 2x) in terms of sin(x + π/12) or some other related expression. Let me see.The angle inside the cosine is 5π/6 - 2x. Maybe I can manipulate this expression to relate it to x + π/12. Let me see. Let's note that 5π/6 is equal to π - π/6. So, 5π/6 - 2x = π - π/6 - 2x. Hmm, not sure if that helps immediately. Alternatively, maybe use a cosine of a sum or difference formula.The cosine of a difference can be written as cos(A - B) = cosA cosB + sinA sinB. But in this case, it's cos(5π/6 - 2x). Alternatively, since it's cosine of something minus another thing, maybe we can express this as cosine of a multiple angle or use some other identity.Alternatively, maybe express 5π/6 - 2x in terms of a multiple of (x + π/12). Let's try that. Let's see: Let me denote θ = x + π/12. Then, we know sinθ = -1/4. Let's see if we can express 5π/6 - 2x in terms of θ.Since θ = x + π/12, solving for x gives x = θ - π/12. Then, substitute x into 5π/6 - 2x:5π/6 - 2x = 5π/6 - 2(θ - π/12) = 5π/6 - 2θ + 2*(π/12) = 5π/6 - 2θ + π/6.Combine the π terms: 5π/6 + π/6 = 6π/6 = π. So, this simplifies to π - 2θ. Therefore, cos(5π/6 - 2x) = cos(π - 2θ).Now, using the identity for cos(π - α) which is equal to -cosα. Wait, no, cos(π - α) = -cosα. So, cos(π - 2θ) = -cos(2θ). Therefore, cos(5π/6 - 2x) = -cos(2θ).But θ is x + π/12, so 2θ = 2x + π/6. Therefore, cos(2θ) = cos(2x + π/6). Wait, but maybe this is going in circles. Let me check again.Wait, we found that cos(5π/6 - 2x) = -cos(2θ) where θ = x + π/12. Since we know sinθ = -1/4, we can use the double angle formula for cosine. The double angle formula for cosine is cos(2θ) = 1 - 2sin²θ. Alternatively, cos(2θ) = 2cos²θ - 1. Either one could work here. Since we know sinθ, maybe using the first one is better.So, cos(2θ) = 1 - 2sin²θ. Since sinθ = -1/4, sin²θ = (1/4)² = 1/16. Therefore, cos(2θ) = 1 - 2*(1/16) = 1 - 1/8 = 7/8. Then, cos(5π/6 - 2x) = -cos(2θ) = -7/8. Therefore, the answer would be C: -7/8.Wait, but let me double-check this because sometimes it's easy to make a mistake with signs or identities. Let me go through each step again.Starting with the given equation sin(x + π/12) = -1/4. Let θ = x + π/12. Then, sinθ = -1/4. We need to find cos(5π/6 - 2x).Express 5π/6 - 2x in terms of θ. As before, x = θ - π/12. Then, substituting:5π/6 - 2x = 5π/6 - 2(θ - π/12) = 5π/6 - 2θ + π/6 = (5π/6 + π/6) - 2θ = π - 2θ. So, yes, that step seems correct.Therefore, cos(5π/6 - 2x) = cos(π - 2θ). Then, using the identity cos(π - α) = -cosα, so cos(π - 2θ) = -cos(2θ). Therefore, cos(5π/6 - 2x) = -cos(2θ). That seems correct.Then, since θ is such that sinθ = -1/4, we can compute cos(2θ) using the double angle identity. Let's recall that cos(2θ) = 1 - 2sin²θ. So, plugging sinθ = -1/4:cos(2θ) = 1 - 2*(1/4)^2 = 1 - 2*(1/16) = 1 - 1/8 = 7/8. Therefore, cos(5π/6 - 2x) = -7/8. That would be option C.But wait, let me check if there's another way to compute cos(2θ). Alternatively, if I use cos(2θ) = 2cos²θ - 1. But since we know sinθ, maybe we can find cosθ?But since sinθ = -1/4, cosθ would be sqrt(1 - sin²θ) = sqrt(1 - 1/16) = sqrt(15/16) = sqrt(15)/4. However, we need to consider the sign of cosθ. Since sinθ is negative (-1/4), the angle θ is in either the third or fourth quadrant. But without knowing x, we can't determine which quadrant θ is in. However, the value of cos(2θ) only depends on sin²θ or cos²θ, so actually, regardless of the sign of cosθ, cos(2θ) would still be 1 - 2*(1/16) = 7/8. Because when we square it, the sign disappears. So even if cosθ is negative, squaring it would still give 15/16, so 2*(15/16) - 1 = 30/16 - 16/16 = 14/16 = 7/8. Wait, that gives the same result. So regardless of the quadrant, cos(2θ) is 7/8. Therefore, the answer is -7/8, which is option C. So that seems consistent.But let me verify once again. Let's suppose θ is in the third quadrant where sinθ is negative and cosθ is negative. Then, cos(2θ) would be 2cos²θ - 1 = 2*(15/16) - 1 = 30/16 - 16/16 = 14/16 = 7/8. Alternatively, if θ is in the fourth quadrant, where sinθ is negative and cosθ is positive, then cos²θ is still 15/16, so same result. Therefore, regardless of the quadrant, cos(2θ) is 7/8, so cos(5π/6 - 2x) is -7/8. So the answer is C.Alternatively, maybe there's another approach. Let me try expressing cos(5π/6 - 2x) using angle addition formulas. Let's see.Express cos(5π/6 - 2x) as cos( (5π/6) - 2x ). Let me set A = 5π/6 and B = 2x. Then, using the identity cos(A - B) = cosA cosB + sinA sinB. So,cos(5π/6 - 2x) = cos(5π/6)cos(2x) + sin(5π/6)sin(2x).We can compute cos(5π/6) and sin(5π/6). 5π/6 is in the second quadrant. cos(5π/6) = -√3/2, sin(5π/6) = 1/2. Therefore:= (-√3/2)cos(2x) + (1/2)sin(2x)Hmm, but then we need to find cos(2x) and sin(2x). How?But we know sin(x + π/12) = -1/4. Maybe express x in terms of θ = x + π/12, so x = θ - π/12. Then, 2x = 2θ - π/6. Then, cos(2x) = cos(2θ - π/6). Similarly, sin(2x) = sin(2θ - π/6). Maybe use angle subtraction formulas here.Alternatively, maybe use double angle identities. Let's see.Alternatively, perhaps express cos(2x) and sin(2x) in terms of sin(x + π/12). Let's see. Let's let θ = x + π/12, so x = θ - π/12. Then, 2x = 2θ - π/6. Therefore:cos(2x) = cos(2θ - π/6) = cos(2θ)cos(π/6) + sin(2θ)sin(π/6)Similarly, sin(2x) = sin(2θ - π/6) = sin(2θ)cos(π/6) - cos(2θ)sin(π/6)But this seems complicated, but maybe manageable. Let's try plugging into the expression for cos(5π/6 - 2x):cos(5π/6 - 2x) = (-√3/2)cos(2x) + (1/2)sin(2x)Substituting the expressions for cos(2x) and sin(2x):= (-√3/2)[cos(2θ)cos(π/6) + sin(2θ)sin(π/6)] + (1/2)[sin(2θ)cos(π/6) - cos(2θ)sin(π/6)]Now, let's compute the terms:First, note that cos(π/6) = √3/2, sin(π/6) = 1/2.So,First term:(-√3/2)[cos(2θ)*(√3/2) + sin(2θ)*(1/2)]= (-√3/2)*(√3/2 cos2θ + 1/2 sin2θ)= (-√3/2)*(√3/2) cos2θ + (-√3/2)*(1/2) sin2θ= (-3/4)cos2θ - (√3/4) sin2θSecond term:(1/2)[sin2θ*(√3/2) - cos2θ*(1/2)]= (1/2)*(√3/2 sin2θ - 1/2 cos2θ)= (√3/4 sin2θ - 1/4 cos2θ)Now, combining both terms:First term: -3/4 cos2θ - √3/4 sin2θSecond term: √3/4 sin2θ - 1/4 cos2θAdding them together:(-3/4 cos2θ - √3/4 sin2θ) + (√3/4 sin2θ - 1/4 cos2θ)The -√3/4 sin2θ and +√3/4 sin2θ cancel out.For the cosine terms: -3/4 cos2θ - 1/4 cos2θ = - (3/4 + 1/4) cos2θ = -cos2θ.Therefore, cos(5π/6 - 2x) = -cos2θ. Which is the same as we found earlier! Therefore, this alternative approach also leads us to cos(5π/6 - 2x) = -cos2θ. Therefore, the answer is -7/8, which is option C.Therefore, both methods lead to the same conclusion, so it's likely correct.But let me check once more if I made any miscalculations. Let's confirm the double angle formula step. Since sinθ = -1/4, sin²θ = 1/16, so 1 - 2sin²θ = 1 - 2*(1/16) = 1 - 1/8 = 7/8. So cos2θ = 7/8, hence -cos2θ = -7/8. That's correct.Alternatively, if I use cos2θ = 2cos²θ - 1. We know sinθ = -1/4, so cos²θ = 1 - sin²θ = 15/16, so 2*(15/16) - 1 = 30/16 - 16/16 = 14/16 = 7/8. So same result.Therefore, yes, cos(5π/6 - 2x) = -7/8. So the answer is C.I think that's solid. Both approaches confirm the answer is C. So I can be confident that's the correct choice.Final Answerboxed{C}</think>Given (sinleft(x + frac{pi}{12}right) = -frac{1}{4}), we need to find (cosleft(frac{5pi}{6} - 2xright)).First, let (theta = x + frac{pi}{12}). Then, (sintheta = -frac{1}{4}). We need to express (frac{5pi}{6} - 2x) in terms of (theta):[frac{5pi}{6} - 2x = frac{5pi}{6} - 2left(theta - frac{pi}{12}right) = frac{5pi}{6} - 2theta + frac{pi}{6} = pi - 2theta]Thus, we need to find (cos(pi - 2theta)). Using the identity (cos(pi - alpha) = -cos(alpha)), we have:[cos(pi - 2theta) = -cos(2theta)]Next, we use the double angle formula for cosine. Since (sintheta = -frac{1}{4}), we have (sin^2theta = left(-frac{1}{4}right)^2 = frac{1}{16}). Using the identity (cos(2theta) = 1 - 2sin^2theta):[cos(2theta) = 1 - 2left(frac{1}{16}right) = 1 - frac{1}{8} = frac{7}{8}]Therefore,[cosleft(frac{5pi}{6} - 2xright) = -cos(2theta) = -frac{7}{8}]The final answer is (boxed{C}).,<think>Okay, so I have this trigonometry problem here. Let me try to work through it step by step. The problem says:Given that sin(x + π/12) = -1/4, then cos(5π/6 - 2x) equals one of the options A to D. The options are 7/8, 1/8, -7/8, -1/8. Alright, I need to find the value of the cosine expression given that sine value. Let me think about how to approach this.First, I remember that trigonometric equations often require using identities or angle addition formulas. The given equation is sin(x + π/12) = -1/4. The expression we need to find is cos(5π/6 - 2x). Maybe there's a way to relate these two expressions through some trigonometric identities or angle manipulations.Let me write down what I know. The angle in the sine is (x + π/12), and the angle in the cosine is (5π/6 - 2x). Maybe if I can express the cosine angle in terms of the sine angle, I can use some identity to relate them.Let me see. Let's first analyze the angle inside the cosine: 5π/6 - 2x. Hmm. Maybe I can rewrite this angle in terms of (x + π/12) or something similar. Let me try to manipulate it.Let's compute 5π/6 - 2x. Let's see if we can express this as a multiple of (x + π/12) or maybe another angle. Alternatively, perhaps using a double-angle formula or a co-function identity.Alternatively, maybe express the cosine of that angle using a cosine addition formula or convert it into a sine function since the given equation involves sine. Let me recall that cos(θ) = sin(θ + π/2). Maybe that could be useful here.Alternatively, perhaps express the cosine term in terms of sin(2x) or cos(2x), but since the argument is 5π/6 - 2x, maybe use some kind of angle subtraction formula.Wait, let me think. Let me first note that 5π/6 is equal to π - π/6. So 5π/6 - 2x = π - π/6 - 2x. Then, cos(5π/6 - 2x) = cos(π - π/6 - 2x). Using the identity cos(π - α) = -cos(α), but here it's π - (π/6 + 2x). So cos(π - (π/6 + 2x)) = -cos(π/6 + 2x). Therefore, cos(5π/6 - 2x) = -cos(π/6 + 2x). Hmm, so maybe this simplifies to negative cosine of (2x + π/6). So now, if I can compute cos(2x + π/6), then I can just take the negative of that to get the answer.Alternatively, perhaps using a different identity. Let me check that step again. Because cos(π - θ) = -cosθ, yes, that's correct. So here, θ would be π/6 + 2x. Therefore, cos(5π/6 - 2x) = -cos(2x + π/6). So now, the problem reduces to finding -cos(2x + π/6). So maybe if I can compute cos(2x + π/6), then multiply by -1 to get the desired value.But how to compute cos(2x + π/6)? Let's think. Maybe express this as cos(2(x + π/12)), but wait, 2x + π/6 is equal to 2x + π/6, which is 2x + π/6. Wait, π/6 is 2π/12, so π/6 is equal to 2*(π/12). So, 2x + π/6 = 2x + 2*(π/12) = 2(x + π/12). Oh! That seems useful. Because the given angle in the sine is x + π/12. So, 2(x + π/12) is exactly 2x + π/6. So, cos(2x + π/6) = cos(2*(x + π/12)). Therefore, cos(2x + π/6) = cos(2θ) where θ = x + π/12. So, we can write cos(2θ) where θ is such that sinθ = -1/4.Therefore, since we know sinθ = -1/4, we can use the double-angle identity for cosine to find cos(2θ). The double-angle identity for cosine is cos(2θ) = 1 - 2sin²θ. Alternatively, cos(2θ) = 2cos²θ - 1. Since we know sinθ, maybe the first identity is better because we can compute it directly from sinθ.So, cos(2θ) = 1 - 2sin²θ. Let's compute that. Given that sinθ = -1/4, so sin²θ = (1/4)² = 1/16. Then, 1 - 2*(1/16) = 1 - 2/16 = 1 - 1/8 = 7/8. Therefore, cos(2θ) = 7/8. Therefore, cos(2x + π/6) = 7/8. But remember that the original expression we needed was -cos(2x + π/6), which would be -7/8. Therefore, cos(5π/6 - 2x) = -7/8. Looking at the options, that's option C: -7/8. So is that the answer? Let me check my steps again to be sure.First, the given equation: sin(x + π/12) = -1/4. Let me denote θ = x + π/12. Then sinθ = -1/4. Then, we needed to compute cos(5π/6 - 2x). I rewrote 5π/6 - 2x as π - (π/6 + 2x), then applied the identity cos(π - α) = -cosα, so that becomes -cos(π/6 + 2x). Then, π/6 + 2x can be written as 2x + π/6, which is equal to 2(x + π/12) = 2θ. Therefore, cos(2θ) = 7/8 as computed. Therefore, the original expression is -cos(2θ) = -7/8. That seems correct. So the answer should be C: -7/8.Wait a minute, but hold on. Let me verify that step where I rewrote 5π/6 - 2x as π - (π/6 + 2x). Let's compute π - (π/6 + 2x):π - π/6 - 2x = (6π/6 - π/6) - 2x = (5π/6) - 2x. Yes, that's correct. So 5π/6 - 2x is indeed equal to π - (π/6 + 2x), so cos(5π/6 - 2x) = cos(π - (π/6 + 2x)) = -cos(π/6 + 2x). So that part checks out.Then, π/6 + 2x. Let's express 2x in terms of θ. Since θ = x + π/12, then x = θ - π/12. Therefore, 2x = 2θ - π/6. Therefore, π/6 + 2x = π/6 + 2θ - π/6 = 2θ. So yes, π/6 + 2x = 2θ. Therefore, cos(π/6 + 2x) = cos(2θ). So that part is also correct.Therefore, since θ = x + π/12, and sinθ = -1/4, we can compute cos(2θ) = 1 - 2sin²θ = 1 - 2*(1/16) = 1 - 1/8 = 7/8. Then, cos(5π/6 - 2x) = -cos(2θ) = -7/8, which is option C. So that seems right.But let me just check if there's another approach or if I missed a negative sign somewhere. For example, when I used the identity cos(π - α) = -cosα, that's correct. And since α here is (π/6 + 2x), then cos(π - α) = -cosα. So that gives us the negative sign. Then, the rest follows.Alternatively, maybe I can approach this using angle addition formulas. Let's try that as a verification.We have cos(5π/6 - 2x). Let's use the cosine of a difference identity: cos(A - B) = cosA cosB + sinA sinB. So here, A is 5π/6 and B is 2x. Therefore:cos(5π/6 - 2x) = cos(5π/6)cos(2x) + sin(5π/6)sin(2x).We know that cos(5π/6) = cos(π - π/6) = -cos(π/6) = -√3/2, and sin(5π/6) = sin(π - π/6) = sin(π/6) = 1/2. So substituting these in:cos(5π/6 - 2x) = (-√3/2)cos(2x) + (1/2)sin(2x).Hmm, but this seems more complicated because now we need to find cos(2x) and sin(2x). But perhaps we can relate this to the given equation sin(x + π/12) = -1/4.Alternatively, since we already found that cos(5π/6 - 2x) = -cos(2θ) where θ = x + π/12 and sinθ = -1/4, leading to the result -7/8, maybe this alternative approach is longer. But let's see if we can get the same answer through this method.First, let's compute cos(2x) and sin(2x). To do that, we can express x in terms of θ. Since θ = x + π/12, then x = θ - π/12. Therefore, 2x = 2θ - π/6. Therefore:cos(2x) = cos(2θ - π/6) = cos(2θ)cos(π/6) + sin(2θ)sin(π/6).Similarly, sin(2x) = sin(2θ - π/6) = sin(2θ)cos(π/6) - cos(2θ)sin(π/6).But this seems like it's getting more involved. However, since we already know cos(2θ) = 7/8, we can compute sin(2θ) using the identity sin²(2θ) + cos²(2θ) = 1. So sin(2θ) = ±√(1 - (7/8)²) = ±√(1 - 49/64) = ±√(15/64) = ±√15/8. But since θ is such that sinθ = -1/4, we need to determine the sign of sin(2θ).First, θ = x + π/12. The sine of θ is -1/4, which is negative. So θ is in either the third or fourth quadrant. However, without knowing x, we can't be sure which quadrant θ is in. But perhaps we can determine the quadrant based on the value of cosθ. Since sinθ = -1/4, cosθ = ±√(1 - sin²θ) = ±√(1 - 1/16) = ±√(15/16) = ±√15/4. The sign depends on the quadrant. If θ is in the third quadrant, cosine is negative; if in the fourth, cosine is positive. However, we don't have additional information about θ, so we might need to consider both possibilities. But since we are dealing with cos(2θ) = 7/8, which is positive. Let's recall that cos(2θ) = 7/8. Since 7/8 is positive, 2θ must be in a quadrant where cosine is positive, i.e., first or fourth quadrant. Therefore, θ could be in the first or fourth quadrant? Wait, if 2θ is in the first or fourth quadrant, then θ is in the first or second quadrant (if 2θ is in the first) or θ is in the fourth or third quadrant (if 2θ is in the fourth). But since sinθ is negative, θ is in the third or fourth quadrant. So combining these, θ is in the fourth quadrant (since third quadrant θ would make 2θ in the sixth quadrant? Wait, angles are modulo 2π, so perhaps this is more complicated.Wait, maybe it's simpler. Since cos(2θ) = 7/8 > 0, then 2θ is in the first or fourth quadrant. Therefore, θ is in the first or fourth quadrant when halved. But θ itself has sinθ = -1/4, so θ is in the third or fourth quadrant. Therefore, combining these two, θ must be in the fourth quadrant because if θ is in the third quadrant, 2θ would be in the sixth quadrant (i.e., angle greater than π), where cosine is negative. But since cos(2θ) is positive, 2θ must be in the first or fourth quadrant, but θ is in the third or fourth. If θ is in the third quadrant, 2θ is in the third or fourth? Wait, θ is in third quadrant: π < θ < 3π/2. Then, 2θ is between 2π and 3π, which is equivalent to between 0 and π (subtracting 2π), so 2θ would be in the first or second quadrant, where cosine is positive or negative. Wait, this is getting confusing. Maybe better to proceed with the information we have.We have sinθ = -1/4, so θ is in the third or fourth quadrant. Let's suppose θ is in the fourth quadrant. Then, cosθ is positive, so cosθ = √15/4. Then, sin2θ = 2sinθcosθ = 2*(-1/4)*(√15/4) = -√15/8. Alternatively, if θ is in the third quadrant, cosθ is negative, so cosθ = -√15/4, and sin2θ = 2*(-1/4)*(-√15/4) = √15/8. So sin2θ is positive if θ is in the third quadrant, and negative if θ is in the fourth. But since cos(2θ) is positive 7/8, we can find sin2θ as ±√15/8. However, without more information, we can't determine the exact sign. However, since we need to compute cos(5π/6 - 2x) = (-√3/2)cos(2x) + (1/2)sin(2x). Let's see if we can express this in terms of sin2θ and cos2θ.Wait, earlier we found that 2x = 2θ - π/6. Therefore, cos(2x) = cos(2θ - π/6) and sin(2x) = sin(2θ - π/6). Let me compute those:cos(2θ - π/6) = cos2θ cosπ/6 + sin2θ sinπ/6 = (7/8)(√3/2) + (±√15/8)(1/2) = (7√3)/16 ± (√15)/16.Similarly, sin(2θ - π/6) = sin2θ cosπ/6 - cos2θ sinπ/6 = (±√15/8)(√3/2) - (7/8)(1/2) = (±√45)/16 - 7/16 = (±3√5)/16 - 7/16.Then, substituting back into the expression:cos(5π/6 - 2x) = (-√3/2)[ (7√3)/16 ± (√15)/16 ] + (1/2)[ (±3√5)/16 - 7/16 ].Let me compute each term:First term: (-√3/2) * [ (7√3)/16 ± (√15)/16 ]= (-√3/2) * (7√3 ± √15)/16= (- (7*3 ± √3*√15) ) / 32= (- (21 ± √45) ) / 32= (-21 ± 3√5)/32Second term: (1/2) * [ (±3√5)/16 - 7/16 ]= [ (±3√5 - 7) / 16 ] * (1/2)= (±3√5 - 7)/32Therefore, combining both terms:cos(5π/6 - 2x) = [ (-21 ± 3√5) + (±3√5 - 7) ] / 32Let's consider the possible combinations of signs:Case 1: First term is -21 + 3√5, second term is +3√5 -7Then total numerator: (-21 + 3√5) + (3√5 -7) = -21 -7 + 6√5 = -28 + 6√5Case 2: First term is -21 -3√5, second term is -3√5 -7Numerator: (-21 -3√5) + (-3√5 -7) = -21 -7 -6√5 = -28 -6√5Case 3: First term is -21 + 3√5, second term is -3√5 -7Numerator: (-21 + 3√5) + (-3√5 -7) = -21 -7 = -28Case 4: First term is -21 -3√5, second term is +3√5 -7Numerator: (-21 -3√5) + (3√5 -7) = -21 -7 = -28Wait, so actually, if we consider the ± signs, depending on the sign of sin2θ, the cross terms with √5 might cancel or add. But in cases 3 and 4, the √5 terms cancel out, leading to -28/32 = -7/8. In cases 1 and 2, we have -28 ±6√5 over 32, which are different values. But the answer options are all rational numbers: 7/8, 1/8, -7/8, -1/8. Therefore, the cross terms must cancel out, leading to -28/32 = -7/8. Therefore, regardless of the sign of sin2θ, the √5 terms must cancel. Therefore, this suggests that only cases 3 and 4 are valid, leading to -7/8. Therefore, confirming the previous result.So through this alternative method, we also arrive at -7/8. Therefore, the answer is indeed C: -7/8.Alternatively, another approach: Let's consider that θ = x + π/12, so x = θ - π/12. Then, let's substitute x into the expression 5π/6 - 2x:5π/6 - 2x = 5π/6 - 2(θ - π/12) = 5π/6 - 2θ + π/6 = (5π/6 + π/6) - 2θ = π - 2θ.Therefore, cos(5π/6 - 2x) = cos(π - 2θ) = -cos(2θ). Since cos(π - α) = -cosα. Therefore, this directly gives us cos(5π/6 - 2x) = -cos(2θ). Since we already found cos(2θ) = 7/8, then the answer is -7/8. This is a much simpler derivation.Wow, that's a much more straightforward approach. So by substituting x in terms of θ, we can express the angle inside the cosine as π - 2θ, leading directly to -cos(2θ). Therefore, that's why we get -7/8. Therefore, this confirms that the answer is indeed C: -7/8.Therefore, after going through different methods and verifying each step, I can be confident that the correct answer is C: -7/8.Final Answerboxed{C}</think>Given (sinleft(x + frac{pi}{12}right) = -frac{1}{4}), we need to find (cosleft(frac{5pi}{6} - 2xright)).First, let (theta = x + frac{pi}{12}). Then, (sintheta = -frac{1}{4}). We need to express (cosleft(frac{5pi}{6} - 2xright)) in terms of (theta).Rewriting the angle inside the cosine:[frac{5pi}{6} - 2x = pi - left(frac{pi}{6} + 2xright)]Using the identity (cos(pi - alpha) = -cos(alpha)), we get:[cosleft(frac{5pi}{6} - 2xright) = -cosleft(frac{pi}{6} + 2xright)]Next, we express (frac{pi}{6} + 2x) in terms of (theta):[frac{pi}{6} + 2x = 2left(x + frac{pi}{12}right) = 2theta]Thus,[cosleft(frac{pi}{6} + 2xright) = cos(2theta)]Using the double-angle identity for cosine:[cos(2theta) = 1 - 2sin^2theta]Given (sintheta = -frac{1}{4}), we find:[sin^2theta = left(-frac{1}{4}right)^2 = frac{1}{16}]Substituting this into the identity:[cos(2theta) = 1 - 2left(frac{1}{16}right) = 1 - frac{1}{8} = frac{7}{8}]Therefore,[cosleft(frac{5pi}{6} - 2xright) = -cos(2theta) = -frac{7}{8}]The final answer is (boxed{C}).]

🔑:Given sin(x+frac{π}{12})=-frac{1}{4}, we are asked to find the value of cos(frac{5π}{6}-2x).First, we can use the identity cos(a-b)=cos(π-b-a) for any angles a and b. Therefore, we have:[cos(frac{5π}{6}-2x) = cos(π-frac{π}{6}-2x)]Next, we recognize that cos(π-theta) = -cos(theta) for any angle theta. Applying this to our equation gives us:[cos(π-frac{π}{6}-2x) = -cos(frac{π}{6}+2x)]We also know the double angle identity for sine, which is sin^2(theta) = frac{1-cos(2theta)}{2}. However, in this context, we need to use the identity cos(theta) = 1 - 2sin^2(theta) to relate the cosine function to the given sine value. Therefore, we have:[-cos(frac{π}{6}+2x) = -[1 - 2sin^2(x+frac{π}{12})]]Substituting the given sin(x+frac{π}{12}) = -frac{1}{4} into the equation, we get:[-[1 - 2(-frac{1}{4})^2] = -[1 - 2(frac{1}{16})] = -[1 - frac{1}{8}] = -frac{7}{8}]Therefore, the value of cos(frac{5π}{6}-2x) is -frac{7}{8}, which corresponds to choice C.So, the final answer is boxed{C}.

🧠:[<think>Okay, so the problem is about finding the maximum volume of a cone that's inscribed in a sphere, and then comparing that volume to the volume of the sphere. Hmm, let me think.First, I need to visualize this. There's a sphere, and inside it, we're inscribing a cone. The cone has to fit perfectly inside the sphere. So, the cone's tip might be at the bottom of the sphere, and the base of the cone would touch the sphere's inner surface. Or maybe the cone is positioned such that both the tip and the edge of the base are on the sphere's surface. I guess the key here is to figure out the dimensions of the cone that would give the maximum volume.Let me recall some geometry formulas. The volume of a cone is (1/3)πr²h, where r is the radius of the base and h is the height. The volume of the sphere is (4/3)πR³, where R is the sphere's radius. So, the problem is asking for the ratio of the maximum cone volume to the sphere's volume.Since the cone is inscribed in the sphere, the sphere's radius must relate to the cone's dimensions. Let me set up a coordinate system. Let's imagine the sphere is centered at the origin (0,0,0) with radius R. If the cone is inscribed, its tip could be at the bottom of the sphere, which would be at (0,0,-R), and the base of the cone would be a circle lying somewhere on the sphere's surface. Wait, but the height of the cone would then be the distance from the tip at (0,0,-R) to the base, which is a plane cutting through the sphere. The base of the cone must lie on the sphere's surface, so the edge of the base would be a point on the sphere.Alternatively, maybe the cone is oriented so that its apex is at the north pole of the sphere (0,0,R) and the base is a circle on the sphere's surface. Hmm, that might make the math easier. Let me consider that scenario. Let’s assume the sphere has radius R, centered at the origin. The cone has its apex at (0,0,R) and its base is a circle lying on the sphere's surface. The height of the cone would then be the distance from (0,0,R) to the base's plane. Let's call the height of the cone h. Wait, if the apex is at (0,0,R), then the height of the cone would be R plus the z-coordinate of the base's plane. Wait, no. Let's think again.If the base of the cone is a circle lying on the sphere, then the center of the base would be along the central axis of the cone, which is the line from the apex to the center of the base. Let's say the apex is at (0,0,R), and the base's center is at (0,0,z), where z is less than R because it's going down towards the center. Then the height of the cone would be R - z. The radius of the cone's base, let's call it r, would satisfy the equation of the sphere. Because any point on the base's circumference is on the sphere's surface. So, the coordinates of such a point would be (x, y, z) where x² + y² + z² = R². Since the base is a circle in the plane z = z0, the radius of the base of the cone would be sqrt(R² - z0²). Wait, because x² + y² = R² - z0², so the radius r of the base is sqrt(R² - z0²).But the height h of the cone is from the apex at (0,0,R) to the base at (0,0,z0), so h = R - z0. Therefore, z0 = R - h. Hmm, substituting that into the radius expression, r = sqrt(R² - (R - h)²). Let me compute that: (R - h)² = R² - 2Rh + h², so R² - (R² - 2Rh + h²) = 2Rh - h². Therefore, r = sqrt(2Rh - h²). Therefore, the volume of the cone is (1/3)πr²h = (1/3)π(2Rh - h²)h = (1/3)π(2Rh² - h³). Now, to find the maximum volume, we need to take the derivative of this volume with respect to h and set it to zero. Let me set up the volume function:V(h) = (1/3)π(2Rh² - h³)Taking derivative with respect to h:V'(h) = (1/3)π(4Rh - 3h²)Set derivative equal to zero:4Rh - 3h² = 0h(4R - 3h) = 0Solutions are h=0 (which gives zero volume) and 4R - 3h = 0 => h = (4R)/3Wait, but h is the height from the apex at (0,0,R) to the base. If h = 4R/3, then the base center is at z0 = R - h = R - 4R/3 = -R/3. So, the base is located at z = -R/3, which is below the center of the sphere. That seems okay, because the cone would extend from the top of the sphere down to below the center.But let's check if that gives a maximum. The second derivative of V(h) would be V''(h) = (1/3)π(4R - 6h). At h = 4R/3, V''(h) = (1/3)π(4R - 6*(4R/3)) = (1/3)π(4R - 8R) = (1/3)π(-4R) which is negative, so it is indeed a maximum.Therefore, the maximum volume is when h = 4R/3, and r = sqrt(2R*(4R/3) - (4R/3)^2) = sqrt( (8R²/3) - (16R²/9) ) = sqrt( (24R²/9 - 16R²/9) ) = sqrt(8R²/9) = (2R√2)/3.Therefore, plugging back into the volume formula:V_max = (1/3)π*(2Rh² - h³) where h = 4R/3.Let me compute 2Rh² = 2R*(16R²/9) = 32R³/9h³ = (64R³)/27So, 32R³/9 - 64R³/27 = (96R³ - 64R³)/27 = 32R³/27Therefore, V_max = (1/3)π*(32R³/27) = 32πR³/(81)Now, the volume of the sphere is (4/3)πR³. So, the ratio of the maximum cone volume to the sphere's volume is:(32πR³/81) / (4πR³/3) = (32/81)/(4/3) = (32/81)*(3/4) = (8/27) ≈ 0.296.Therefore, the maximum volume of the inscribed cone is 8/27 times the volume of the sphere.Wait, but let me double-check these calculations. Let's recalculate the volume:V_max = (1/3)πr²h. We found h = 4R/3, r = 2R√2/3.So, r² = (8R²)/9. Then:V_max = (1/3)π*(8R²/9)*(4R/3) = (1/3)*(8*4)/(9*3) * πR³ = (32/81)πR³. Yep, that's correct. Sphere volume is (4/3)πR³. So the ratio is (32/81)/(4/3) = (32/81)*(3/4) = 8/27. Correct.So the answer is that the maximum volume of the cone inscribed in the sphere is 8/27 of the sphere's volume. So the ratio is 8/27.But wait, I remember that in some problems, the maximum volume of a cone inscribed in a sphere is 8/27 of the sphere's volume. So this seems familiar. Therefore, the answer should be 8/27, which is approximately 0.296, so the cone's maximum volume is 8/27 times the sphere's volume.</think>You will be given a problem. Please reason step by step, and put your final answer within boxed{}:8. The maximum volume of a cone inscribed in a sphere compared to the volume of the sphere is qquadThe maximum volume of a cone inscribed in a sphere compared to the volume of the sphere is boxed{dfrac{8}{27}}.,<think>Okay, so I need to figure out the maximum volume of a cone that can be inscribed in a sphere and then compare that volume to the volume of the sphere. Let me start by visualizing this. Imagine a sphere with radius R, and inside it, there's a cone touching the sphere's surface. The cone has a certain height and base radius. The goal is to find the dimensions of the cone that would give the largest possible volume and then see what fraction of the sphere's volume that is.First, let me recall the formulas involved. The volume of a cone is (1/3)πr²h, where r is the base radius and h is the height. The volume of the sphere is (4/3)πR³. So, I need to maximize the cone's volume in terms of R and then divide it by the sphere's volume.Since the cone is inscribed in the sphere, the tip of the cone is at the bottom of the sphere, and the base of the cone is somewhere inside the sphere. Let me set up a coordinate system where the center of the sphere is at the origin (0,0,0). If the cone's tip is at the bottom, which would be at (0,0,-R), and the base of the cone is a circle on some plane z = h - R, where h is the height from the tip to the base. Wait, maybe I need to adjust that.Alternatively, let's consider the height of the cone as the distance from its base to its tip. If the tip is at the bottom of the sphere, then the height h of the cone would extend upward from there. Let me think. The sphere has radius R, so the center is at 0. If the cone's tip is at the bottom point (-R,0,0) assuming 3D coordinates, but maybe it's easier to use a vertical axis. Let me use a vertical axis where the sphere is centered at the origin (0,0,0), so the top point is (0,0,R) and the bottom point is (0,0,-R). If the cone is inscribed, its tip is at the bottom point (0,0,-R), and its base is a circle lying on the sphere's surface. Wait, no. If the cone is inscribed, then all points of the cone should be inside or on the sphere. So the base of the cone must be a circle on the sphere's surface, and the tip is at the bottom. Wait, maybe not necessarily. Maybe the cone's apex is at the bottom of the sphere, and the base is a circle inside the sphere. But then how is it inscribed? Maybe the cone touches the sphere at its base circumference. So the base of the cone is a circle lying on the sphere's surface, and the apex is at the bottom. That makes sense. So the cone is inscribed such that its base is a great circle of the sphere? No, because a great circle would be the equator, but if the apex is at the bottom, the height would be 2R. Wait, but that might not be the case. Let me think again.Alternatively, maybe the cone is inscribed such that its apex is at the north pole of the sphere, and the base is a circle somewhere on the sphere's surface. Wait, but then the height would be from the north pole to the base. Let me try to model this.Let me consider a sphere of radius R. Let's place the sphere with its center at the origin (0,0,0). Suppose the cone has its apex at the south pole (0,0,-R) and its base is a circle lying on the sphere's surface at some height z = h above the apex. So the height of the cone would be H = h + R, but wait, if the apex is at (0,0,-R) and the base is at z = k, then the height of the cone is k - (-R) = k + R. But since the base is on the sphere, the points on the base must satisfy the sphere's equation x² + y² + z² = R². Let me denote the base of the cone as lying at z = a, where a is between -R and R. Then, the height H of the cone is a - (-R) = a + R.The radius r of the cone's base can be found from the sphere's equation. At height z = a, the radius of the circular section is sqrt(R² - a²). Therefore, r = sqrt(R² - a²).So the volume V of the cone is (1/3)πr²H = (1/3)π(R² - a²)(a + R).Now, we need to maximize this volume as a function of a. So let's write V(a) = (1/3)π(R² - a²)(a + R).To find the maximum, take the derivative of V with respect to a, set it to zero, and solve for a.First, expand the expression inside V(a):(R² - a²)(a + R) = R²(a + R) - a²(a + R) = R²a + R³ - a³ - R a².So V(a) = (1/3)π(R²a + R³ - a³ - R a²).Now, take the derivative V'(a):V'(a) = (1/3)π(R² + 0 - 3a² - 2R a).Simplify:V'(a) = (1/3)π(R² - 3a² - 2R a).Set V'(a) = 0:R² - 3a² - 2R a = 0.Multiply both sides by -1:3a² + 2R a - R² = 0.This is a quadratic equation in terms of a:3a² + 2R a - R² = 0.Solve using quadratic formula:a = [-2R ± sqrt((2R)² - 4*3*(-R²))]/(2*3)Compute discriminant:(4R²) + 12R² = 16R²So sqrt(16R²) = 4RThus,a = [-2R ± 4R]/6Two solutions:a = (-2R + 4R)/6 = (2R)/6 = R/3a = (-2R - 4R)/6 = (-6R)/6 = -RBut a = -R corresponds to the apex of the cone, which would make the height H = -R + R = 0, so the volume is zero. So the valid solution is a = R/3.Therefore, the maximum volume occurs when the base of the cone is at z = R/3.Now, compute the volume at a = R/3.First, H = a + R = R/3 + R = 4R/3The radius r = sqrt(R² - a²) = sqrt(R² - (R²/9)) = sqrt((9R² - R²)/9) = sqrt(8R²/9) = (2√2 R)/3Volume V = (1/3)πr²H = (1/3)π*(8R²/9)*(4R/3)Calculate step by step:(8R²/9)*(4R/3) = (32 R³)/27Multiply by (1/3)π:V = (32 R³/27)*(1/3)π = (32/81)πR³Wait, that seems incorrect. Wait, no:Wait, (8R²/9)*(4R/3) = (8*4)/(9*3) R³ = 32/27 R³Then multiplying by (1/3)π:(32/27 R³) * (1/3)π = 32/(81) π R³Yes, so V = (32/81)πR³Now, the volume of the sphere is (4/3)πR³So the ratio of the cone's maximum volume to the sphere's volume is (32/81 π R³) / (4/3 π R³) = (32/81)/(4/3) = (32/81)*(3/4) = (96/324) = (8/27)Simplify 96/324: divide numerator and denominator by 12: 8/27.So the maximum volume of the cone inscribed in the sphere is 8/27 times the volume of the sphere.Therefore, the answer should be 8/27.</think>You will be given a problem. Please reason step by step, and put your final answer within boxed{}:8. The maximum volume of a cone inscribed in a sphere compared to the volume of the sphere is qquadThe maximum volume of a cone inscribed in a sphere is boxed{dfrac{8}{27}} times the volume of the sphere.]

🔑:Consider the cone S H inscribed in the sphere O, with S H=h,begin{array}{l}O S=R, H B=r Rightarrow(h-R)^{2}+r^{2}=R^{2} Rightarrow h^{2}-2 h R+r^{2}=0 Rightarrow r^{2}=h(2 R-h) .end{array}Then V_{S H}=frac{1}{3} pi r^{2} h=frac{1}{3} pi h^{2}(2 R-h)leqslant frac{pi}{6}left(frac{h+h+4 R-2 h}{3}right)^{3}=frac{32}{81} pi R^{3}, equality holds when h=4 R-2 h Rightarrow h=frac{4 R}{3}. Therefore, the required volume ratio is frac{32}{81} times frac{3}{4}=frac{8}{27}.

🧠:[<think>Okay, so I need to find the angle DAE in triangle ABC where the sides are given as a, b, c = 29, 21, 20 respectively. Wait, let me make sure I get the notation right. In triangle ABC, usually, side a is opposite vertex A, side b is opposite vertex B, and side c is opposite vertex C. So that would mean BC is side a = 29, AC is side b = 21, and AB is side c = 20. Hmm, let me confirm that. Yes, in standard notation, side a is BC, side b is AC, and side c is AB. So the triangle has BC = 29, AC = 21, AB = 20.Now, points D and E are on segment BC such that BD = 8, DE = 12, EC = 9. So BC is divided into three parts: BD = 8, DE = 12, EC = 9. Let me check that these add up to BC. 8 + 12 + 9 = 29, which matches BC = 29. So D is 8 units from B, then E is another 12 units from D, and then E is 9 units away from C. So starting from B, the points are D at 8, then E at 8 + 12 = 20, so EC is 9, which totals 29. That makes sense.We need to find angle DAE. So points A, D, E. So angle at point A between points D and E. Wait, no. Wait, angle DAE is the angle at point A formed by the lines AD and AE. Wait, no, hold on. Angle DAE is the angle at point A between points D and E? Wait, no. Wait, angle DAE is the angle at point A between points D and E. Wait, but D and E are on BC, so angle DAE is the angle at A formed by lines AD and AE. So we have triangle ADE, with vertex at A, and we need to find that angle. So maybe we can use coordinates or vectors or trigonometry laws to compute this angle.First, maybe I should draw the triangle ABC. Let me sketch it mentally. BC is the base with length 29. BD = 8, DE = 12, EC = 9. So D is closer to B, E is closer to C. Points D and E divide BC into three segments. Then AD and AE are cevians from A to D and E. The angle between them at A is DAE, which we need to find.Possible approaches:1. Use coordinates. Assign coordinates to points B, C, then find coordinates of A using the given side lengths. Then find coordinates of D and E, compute vectors AD and AE, then find the angle between them using the dot product formula.2. Use the Law of Cosines in triangle ADE. But for that, we need to know the lengths of AD, AE, and DE. DE is given as 12. But we need to find AD and AE. To find AD and AE, perhaps use Stewart's theorem on triangles ABD and AEC.Let me consider both approaches.First, coordinates. Let me set up coordinate system. Let me place point B at (0,0), point C at (29,0). Then BC is along the x-axis. Then point D is at (8,0), since BD = 8. Then E is at (8 + 12,0) = (20,0). So D is (8,0), E is (20,0). Now, we need to find coordinates of point A. Since AB = 20, AC = 21. So point A is somewhere above the x-axis. Let’s denote coordinates of A as (x, y). Then distance from A to B is sqrt((x - 0)^2 + (y - 0)^2) = 20, and distance from A to C is sqrt((x - 29)^2 + y^2) = 21. So we have two equations:x² + y² = 20² = 400,(x - 29)² + y² = 21² = 441.Subtract the first equation from the second:(x - 29)² + y² - x² - y² = 441 - 400Expanding (x - 29)²: x² - 58x + 841 - x² = 41Simplify: -58x + 841 = 41Then -58x = 41 - 841 = -800Thus x = (-800)/(-58) = 800/58 = 400/29 ≈ 13.7931.Then substitute x back into first equation to find y:(400/29)² + y² = 400So y² = 400 - (160000)/841Compute 400 in terms of 841 denominator: 400 = 400 * 841 / 841 = 336,400 / 841Thus y² = (336,400 - 160,000)/841 = 176,400 / 841Therefore y = sqrt(176,400 / 841) = sqrt(176,400)/sqrt(841) = 420 / 29 ≈ 14.4828.So coordinates of A are (400/29, 420/29).Then coordinates of D are (8, 0) and E are (20, 0). Now, vectors AD and AE can be computed.Vector AD is from A to D: (8 - 400/29, 0 - 420/29)Compute 8 as 232/29, so 232/29 - 400/29 = (-168)/29Similarly, 0 - 420/29 = -420/29Thus vector AD = (-168/29, -420/29)Similarly, vector AE is from A to E: (20 - 400/29, 0 - 420/29)20 is 580/29, so 580/29 - 400/29 = 180/290 - 420/29 = -420/29Thus vector AE = (180/29, -420/29)Now, the angle between vectors AD and AE can be found using the dot product formula:cosθ = (AD · AE) / (|AD| |AE|)First compute the dot product AD · AE:(-168/29)(180/29) + (-420/29)(-420/29)= [(-168)(180) + (420)(420)] / (29²)Compute numerator:-168*180: 168*180 = 30,240. So -30,240.420*420 = 176,400Sum: -30,240 + 176,400 = 146,160Thus the dot product is 146,160 / (29²) = 146,160 / 841Now compute |AD| and |AE||AD| = sqrt( (-168/29)^2 + (-420/29)^2 ) = (1/29)sqrt(168² + 420²)Similarly, |AE| = sqrt( (180/29)^2 + (-420/29)^2 ) = (1/29)sqrt(180² + 420²)Compute 168²: 168*168 = 28,224420² = 176,400So |AD| = (1/29)sqrt(28,224 + 176,400) = (1/29)sqrt(204,624) Similarly, sqrt(204,624). Let me calculate sqrt(204,624). Let's see:204,624 divided by 144 is 1421. So sqrt(204,624) = sqrt(144*1421) = 12*sqrt(1421). Hmm, not helpful. Alternatively, perhaps compute sqrt(204,624). Let me check 452² = 204,304. 453² = 452² + 2*452 +1 = 204,304 + 904 + 1 = 205,209. So sqrt(204,624) is between 452 and 453. But maybe exact value isn't needed since we are dealing with ratios.Similarly, |AE|:180² = 32,400420² = 176,400Sum: 32,400 + 176,400 = 208,800sqrt(208,800). Similarly, sqrt(208,800). Let's see, 456² = 207,936. 457² = 456² + 2*456 +1 = 207,936 + 912 +1 = 208,849. So sqrt(208,800) is between 456 and 457. Again, exact value not needed.But perhaps when we compute |AD| |AE|, the 1/29 * 1/29 = 1/841, so the denominator will be 841 multiplied by sqrt(204,624) * sqrt(208,800). But maybe these can be simplified.Wait, but perhaps instead of computing decimal approximations, we can factor the numbers to see if there's a common factor.Wait, let's look at 204,624 and 208,800. Let's see:204,624 divided by 48: 204,624 / 48 = 4263. So maybe 204,624 = 48 * 4263. Hmm, not sure.Wait, 204,624 = 168² + 420². Let me factor 168 and 420:168 = 12*14 = 12*14420 = 12*35So 168 = 12*14, 420 = 12*35. Then 168² + 420² = (12²)(14² + 35²) = 144*(196 + 1225) = 144*(1421). So sqrt(168² + 420²) = 12*sqrt(1421)Similarly, 180 and 420:180 = 60*3, 420 = 60*7. So 180² + 420² = (60²)(3² + 7²) = 3600*(9 + 49) = 3600*58. So sqrt(180² + 420²) = 60*sqrt(58)Therefore |AD| = (1/29)*12*sqrt(1421) = (12/29)*sqrt(1421)|AE| = (1/29)*60*sqrt(58) = (60/29)*sqrt(58)Then the product |AD||AE| = (12/29)(60/29)sqrt(1421*58) = (720/841)sqrt(1421*58)Now, let's compute the numerator of cosθ: 146,160 /841So cosθ = (146,160 /841) / (720/841 * sqrt(1421*58)/1 )Wait, that's messy. Let me see:cosθ = [146,160 /841] / [ (720/841) * sqrt(1421*58) ]The 841 in the numerator and denominator cancels:cosθ = 146,160 / (720 * sqrt(1421*58))Simplify 146,160 / 720: 146,160 ÷ 720. Let's divide numerator and denominator by 10: 14,616 /72. 14,616 ÷72: 72*200=14,400, so 14,616 -14,400=216. 216 ÷72=3. So total is 203. So 146,160 /720 = 203.Therefore cosθ = 203 / sqrt(1421*58)Compute 1421*58:1421 * 50 = 71,0501421 *8 = 11,368Total: 71,050 +11,368=82,418So cosθ = 203 / sqrt(82,418)Compute sqrt(82,418). Let's check 287² = 82,369. 288² = 82,944. So sqrt(82,418) is between 287 and 288. Let's see how close:287² = 82,36982,418 -82,369=49So sqrt(82,418)=287 + 49/(2*287) + ... ≈287 + 0.085=287.085. But maybe exact value is needed? Hmm.Wait, 82,418 = 2*41,209. Is 41,209 a square? Let me check: 203² = 41,209. Yes! 200²=40,000, 203²=40,000 + 600 +9=40,609? Wait, no. Wait 200²=40,000, 3²=9, 2*200*3=1,200. So (200 +3)^2=200² + 2*200*3 +3²=40,000 +1,200 +9=41,209. Yes! So sqrt(41,209)=203. Therefore sqrt(82,418)=sqrt(2*41,209)=sqrt(2)*sqrt(41,209)=sqrt(2)*203=203√2.Therefore, cosθ=203/(203√2)=1/√2.Therefore, θ=45 degrees or π/4 radians.Wow! So the angle DAE is 45 degrees.Wait, that's a nice result. Let me check my steps again to make sure I didn't make a mistake.First, coordinates assigned correctly: B(0,0), C(29,0). Then D(8,0), E(20,0). Coordinates of A found correctly: solving x² + y² =400 and (x-29)^2 + y²=441. Subtract to get linear equation, solved x=400/29. Then y=420/29. Vectors AD and AE computed as (-168/29, -420/29) and (180/29, -420/29). Dot product computed as (-168)(180) + (-420)(-420) = -30,240 + 176,400 =146,160. Then |AD| and |AE| computed via sqrt(168² +420²) and sqrt(180² +420²). Factored out to 12*sqrt(1421) and 60*sqrt(58). Then product |AD||AE| becomes 720/841*sqrt(1421*58). Then numerator 146,160/841. Then ratio gives 203 / sqrt(82,418). Then sqrt(82,418)=sqrt(2*41,209)=sqrt(2)*203. So cosθ=1/√2, so θ=45°. All steps seem correct. So angle DAE is 45 degrees. So the answer is 45°, which in boxed form is boxed{45^circ}.Alternatively, maybe using Stewart’s theorem.Let me try Stewart’s theorem to compute lengths AD and AE, then use the Law of Cosines in triangle ADE.Stewart’s theorem: For a triangle with side lengths a, b, c, and a cevian of length d to a side of length m divided into segments of length m and n, the formula is b²m + c²n = a(d² + mn)So for triangle ABC, with BC = 29. For point D on BC with BD=8, DC=21 (since BC=29, DE=12, EC=9, but wait, BD=8, DE=12, EC=9. So from B: BD=8, DE=12, EC=9. So D is 8 from B, E is 20 from B (8+12). So for point D: BD=8, DC=21 (since BC=29, 29-8=21). Wait, but EC=9, so DC is DE + EC =12 +9=21. Yes, so DC=21. So for Stewart’s theorem applied to point D in triangle ABC: sides AB=20, AC=21, BC=29. The segments BD=8, DC=21. So Stewart’s theorem gives:AB² * DC + AC² * BD = BC*(AD² + BD*DC)So 20² *21 +21² *8 =29*(AD² +8*21)Compute left side: 400*21 +441*8 = 8400 + 3528 = 11,928Right side:29*(AD² +168)So 11,928 =29AD² +29*168Compute 29*168: 168*30=5040, minus 168=5040-168=4872So 11,928 -4872=29AD²11,928-4,872=7,056Thus AD²=7,056/29=243.3103... So AD= sqrt(243.3103)= approx 15.6. But exact value is sqrt(7056/29). Wait, 7056=29*243.3103? Wait, but 7056 divided by 29: 29*243=29*(240+3)=29*240=6,960 +29*3=87 → 7,047. Then 7,056-7,047=9. So 7056=29*243 +9. So 7056=29*243 +9. Therefore, AD²=7056/29=243 +9/29. So AD= sqrt(243 +9/29). Hmm, not a nice number, but perhaps we can find AE similarly.For point E on BC: BE=20 (BD + DE=8+12=20), EC=9. So applying Stewart’s theorem to point E in triangle ABC:AB² * EC + AC² * BE = BC*(AE² + BE*EC)So 20²*9 +21²*20=29*(AE² +20*9)Compute left side:400*9 +441*20=3,600 +8,820=12,420Right side:29*(AE² +180)So 12,420=29AE² +29*180=29AE² +5,220Thus 12,420 -5,220=7,200=29AE²Therefore AE²=7,200/29≈248.2759, so AE= sqrt(7200/29). Again, sqrt(7200/29)=sqrt(7200)/sqrt(29)= (60*sqrt(2))/sqrt(29). Wait, 7200= 3600*2=60²*2, so sqrt(7200)=60√2. So AE=60√2/sqrt(29). Similarly, AD= sqrt(7056/29). Wait, 7056=84², right? 84*84=7,056. Yes. So sqrt(7056)=84. So AD=84/sqrt(29). Therefore, AD=84/sqrt(29), AE=60√2/sqrt(29). Then DE=12. So in triangle ADE, sides AD=84/sqrt(29), AE=60√2/sqrt(29), DE=12.Then angle DAE can be found using the Law of Cosines:DE²=AD² + AE² -2AD*AE*cosθSo 12² = (84²/29) + (60²*2/29) -2*(84/sqrt(29))*(60√2/sqrt(29))*cosθCompute left side: 144.Right side: (7056 + 7200)/29 -2*(84*60√2)/29 *cosθCompute numerators:7056 +7200=14,256Thus 14,256/29 - (2*84*60√2)/29 *cosθ =144Multiply both sides by 29:14,256 - 2*84*60√2 *cosθ =144*29=4,176Therefore:14,256 -4,176=2*84*60√2 *cosθ10,080=2*84*60√2 *cosθSimplify RHS:2*84=168, 168*60=10,080. So 10,080√2 *cosθ=10,080Divide both sides by10,080:√2 *cosθ=1Thus cosθ=1/√2Therefore θ=45°, as before. So angle DAE is 45 degrees. So even with Stewart's theorem and Law of Cosines, we get the same result. Hence, the answer is 45 degrees. Therefore, the answer is boxed{45^circ}Final Answerboxed{45^circ}</think>To find the angle ( angle DAE ) in triangle ( ABC ) with sides ( a = 29 ), ( b = 21 ), and ( c = 20 ), where points ( D ) and ( E ) divide ( BC ) into segments ( BD = 8 ), ( DE = 12 ), and ( EC = 9 ), we can use coordinate geometry and vector analysis.1. Assign Coordinates: - Place ( B ) at ( (0, 0) ) and ( C ) at ( (29, 0) ). - Points ( D ) and ( E ) are at ( (8, 0) ) and ( (20, 0) ) respectively.2. Find Coordinates of ( A ): - Using the distances ( AB = 20 ) and ( AC = 21 ), solve for ( A ): [ begin{cases} x^2 + y^2 = 400 (x - 29)^2 + y^2 = 441 end{cases} ] - Solving these equations, we find ( A ) at ( left( frac{400}{29}, frac{420}{29} right) ).3. Vectors ( overrightarrow{AD} ) and ( overrightarrow{AE} ): - ( overrightarrow{AD} = left( 8 - frac{400}{29}, 0 - frac{420}{29} right) = left( -frac{168}{29}, -frac{420}{29} right) ) - ( overrightarrow{AE} = left( 20 - frac{400}{29}, 0 - frac{420}{29} right) = left( frac{180}{29}, -frac{420}{29} right) )4. Dot Product and Magnitudes: - Dot product ( overrightarrow{AD} cdot overrightarrow{AE} ): [ left( -frac{168}{29} cdot frac{180}{29} right) + left( -frac{420}{29} cdot -frac{420}{29} right) = frac{146160}{841} ] - Magnitudes ( |overrightarrow{AD}| ) and ( |overrightarrow{AE}| ): [ |overrightarrow{AD}| = frac{1}{29} sqrt{168^2 + 420^2} = frac{12sqrt{1421}}{29} ] [ |overrightarrow{AE}| = frac{1}{29} sqrt{180^2 + 420^2} = frac{60sqrt{58}}{29} ]5. Calculate ( cos theta ): - Using the dot product formula: [ cos theta = frac{overrightarrow{AD} cdot overrightarrow{AE}}{|overrightarrow{AD}| |overrightarrow{AE}|} = frac{146160 / 841}{(12sqrt{1421}/29)(60sqrt{58}/29)} = frac{203}{sqrt{82418}} = frac{1}{sqrt{2}} ] - Therefore, ( theta = 45^circ ).Thus, the angle ( angle DAE ) is ( boxed{45^circ} ).,<think>Okay, so I need to find the angle DAE in triangle ABC where the sides are given as a = 29, b = 21, c = 20. Wait, actually, in triangle notation, usually a is BC, b is AC, and c is AB. Let me confirm that. Yeah, typically, in triangle ABC, side a is opposite vertex A, so BC = a, AC = b, AB = c. But the problem says "a, b, c = 29, 21, 20 respectively". So BC = 29, AC = 21, AB = 20. Let me note that down.So, ABC has BC = 29, AC = 21, AB = 20. Then points D and E are on BC such that BD = 8, DE = 12, EC = 9. So BC is divided into three segments: BD = 8, DE = 12, EC = 9. Let me visualize this. Starting from B, moving 8 units to D, then 12 units to E, then 9 units to C. So total BC = 8 + 12 + 9 = 29, which matches the given length. Good.Now, we need to find angle DAE. So points D and E are on BC, and A is the opposite vertex. So angle at A between points D and E. Hmm. So angle DAE is the angle at A formed by connecting A to D and A to E. To find this angle, perhaps we can use coordinates or vectors or trigonometry laws like the Law of Cosines or Law of Sines. Let me think.First, maybe coordinate geometry. Let me place the triangle in a coordinate system. Let's set point B at (0,0) and point C at (29,0), since BC is the base. Then point A is somewhere above the x-axis. Given that AB = 20 and AC = 21, so coordinates of A can be found using the distance formula.Let me denote coordinates:- Let B = (0, 0)- Let C = (29, 0)- Let A = (x, y)Then, the distance from A to B is 20, so:√[(x - 0)^2 + (y - 0)^2] = 20 ⇒ x² + y² = 400 ...(1)Similarly, distance from A to C is 21:√[(x - 29)^2 + (y - 0)^2] = 21 ⇒ (x - 29)^2 + y² = 441 ...(2)Subtract equation (1) from equation (2):(x - 29)^2 + y² - (x² + y²) = 441 - 400 ⇒ x² - 58x + 841 + y² - x² - y² = 41 ⇒ -58x + 841 = 41 ⇒ -58x = 41 - 841 = -800 ⇒ x = (-800)/(-58) = 800/58 = 400/29 ≈ 13.7931Then substitute x back into equation (1):(400/29)² + y² = 400 ⇒ y² = 400 - (160000/841) = (400*841 - 160000)/841 = (336400 - 160000)/841 = 176400/841 ⇒ y = √(176400/841) = 420/29 ≈ 14.4828So coordinates of A are (400/29, 420/29). Let me check:Distance from A to B: sqrt((400/29)^2 + (420/29)^2) = sqrt((160000 + 176400)/841) = sqrt(336400/841) = sqrt(400) = 20. Good.Distance from A to C: sqrt((400/29 - 29)^2 + (420/29)^2) = sqrt((400/29 - 841/29)^2 + (420/29)^2) = sqrt((-441/29)^2 + (420/29)^2) = sqrt((194481 + 176400)/841) = sqrt(370881/841) = sqrt(441) = 21. Correct.So coordinates are correct. Now, points D and E on BC:BC is from (0,0) to (29,0). BD = 8, so D is at (8, 0). Then DE = 12, so E is at (8 + 12, 0) = (20, 0). Wait, BD = 8, DE = 12, EC = 9. So starting from B at (0,0):- BD = 8: D is at (8, 0)- DE = 12: E is at (8 + 12 = 20, 0)- EC = 9: From E to C is 9 units, which is 20 + 9 = 29, matching point C at (29,0). Perfect.So coordinates:- D = (8, 0)- E = (20, 0)- A = (400/29, 420/29)Now, angle DAE is the angle at point A between points D and E. To find angle DAE, we can compute vectors AD and AE, then use the dot product formula to find the angle between them.First, let's find coordinates of vectors AD and AE.Vector AD is from A to D: D - A = (8 - 400/29, 0 - 420/29) = ( (232 - 400)/29, -420/29 ) = ( (-168)/29, -420/29 )Vector AE is from A to E: E - A = (20 - 400/29, 0 - 420/29) = ( (580 - 400)/29, -420/29 ) = ( 180/29, -420/29 )Now, the angle between vectors AD and AE can be found using:cosθ = (AD · AE) / (|AD| |AE| )First, compute the dot product AD · AE:= [ (-168/29)(180/29) + (-420/29)(-420/29) ]= [ (-168 * 180) / 841 + (420 * 420) / 841 ]Calculate numerator:-168 * 180 = -30240420 * 420 = 176400So total numerator: (-30240 + 176400)/841 = 146160 / 841Now compute |AD| and |AE|:|AD| = sqrt( (-168/29)^2 + (-420/29)^2 ) = sqrt( (28224 + 176400)/841 ) = sqrt(204624 / 841) = sqrt(243.36) ?Wait, let's compute 204624 divided by 841:204624 ÷ 841. Since 29² = 841. Let's do 204624 ÷ 841.First, 841 × 243 = 841 × 200 = 168200; 841 × 40 = 33640; 841 × 3 = 2523. So total 168200 + 33640 = 201,840 + 2523 = 204,363. Which is more than 204,624. Hmm. Maybe I miscalculated.Wait, 841 × 243 = (800 × 243) + (41 × 243) = 194,400 + 9,963 = 204,363. So 204,363. Then 204,624 - 204,363 = 261. So 243 + 261/841. So sqrt(243 + 261/841). Wait, maybe it's better to keep as fractions.Wait, |AD| is sqrt( ( (-168)^2 + (-420)^2 ) / 29² ) = sqrt( (28224 + 176400)/841 ) = sqrt(204624 / 841 ) = sqrt(204624)/sqrt(841) = sqrt(204624)/29.What is sqrt(204624)? Let's compute 204624. Let's divide by 16: 204624 ÷16 = 12789. So sqrt(204624) = 4*sqrt(12789). Hmm, maybe factor 12789.Alternatively, maybe notice that 168 and 420 are multiples of 42: 168 = 42 * 4, 420 = 42 * 10. Then, 168² + 420² = (42²)(4² + 10²) = 1764*(16 + 100) = 1764*116. Then sqrt(1764*116) = 42*sqrt(116). 116 factors into 4*29, so sqrt(116) = 2*sqrt(29). Thus, sqrt(1764*116) = 42*2*sqrt(29) = 84*sqrt(29). Therefore, |AD| = 84*sqrt(29)/29 = 84/29 * sqrt(29) = 84/√29. Wait, but 84^2 = 7056, 84^2 * 29 = 7056*29 = 204,624. So sqrt(204624) = 84*sqrt(29). Therefore, |AD| = 84*sqrt(29)/29 = (84/29)*sqrt(29) = 84/√29? Wait, no. Wait, sqrt(204624) = 84*sqrt(29). So sqrt(204624)/29 = 84*sqrt(29)/29. So |AD| = (84/29) * sqrt(29) = (84/sqrt(29))? Wait, that's not simplified. Wait, 84*sqrt(29)/29 can be written as 84/√29 * (sqrt(29)/sqrt(29)) = 84/√29. Wait, maybe better to leave as 84*sqrt(29)/29. Similarly for |AE|.Compute |AE|:Vector AE is (180/29, -420/29). So its magnitude is sqrt( (180/29)^2 + (-420/29)^2 ) = sqrt( (32400 + 176400)/841 ) = sqrt(208800/841). Let's see.Again, 180 and 420. 180 = 60*3, 420 = 60*7. So 180² + 420² = 60²*(9 + 49) = 3600*58 = 208,800. So sqrt(208800) = sqrt(3600*58) = 60*sqrt(58). Therefore, |AE| = 60*sqrt(58)/29.So |AD| = 84*sqrt(29)/29, |AE| = 60*sqrt(58)/29.Now, going back to the dot product:AD · AE = 146160 / 841Therefore, cosθ = (146160 / 841) / ( (84*sqrt(29)/29) * (60*sqrt(58)/29) )First, compute the denominator:(84 * 60 * sqrt(29) * sqrt(58)) / (29 * 29) ) = (5040 * sqrt(29*58)) / 841Note that 29*58 = 29*29*2 = 841*2. So sqrt(841*2) = 29*sqrt(2). Therefore:sqrt(29*58) = sqrt(29*29*2) = 29*sqrt(2)Therefore, denominator = (5040 * 29*sqrt(2)) / 841But 841 is 29², so:= (5040 * 29 * sqrt(2)) / (29²) ) = (5040 / 29) * sqrt(2)Therefore, cosθ = (146160 / 841) / ( (5040 / 29) * sqrt(2) )Simplify numerator and denominator:Numerator: 146160 / 841Denominator: (5040 / 29) * sqrt(2)So cosθ = (146160 / 841) * (29 / 5040) * (1 / sqrt(2))Simplify:146160 / 5040 = Let's compute 5040 * 29 = 146160. Wait, 5040 * 29 = 5040*30 - 5040 = 151200 - 5040 = 146160. So 146160 / 5040 = 29.Therefore, cosθ = (29) / 841 * (29) / sqrt(2). Wait, wait:Wait, 146160 / 5040 = 29, so:cosθ = (29) / 841 * 29 / sqrt(2) ?Wait, let's re-examine:Wait, the expression is (146160 / 841) * (29 / 5040) * (1 / sqrt(2))But 146160 / 5040 = 29, so:= (29) * (29 / 841) * (1 / sqrt(2)) )Because 146160 / 5040 = 29, and then 29 / 841 is from 29 / 841 (since 146160 / 841 is 146160/(29²)), but wait:Wait, 146160 / 841 = (146160 ÷ 29) / (841 ÷ 29) = 5040 / 29. Wait, let me compute 146160 ÷ 29:29 × 5000 = 145,000. 146,160 - 145,000 = 1,160. 29 × 40 = 1,160. So 146,160 ÷ 29 = 5000 + 40 = 5040. Therefore, 146160 / 841 = 5040 / 29.Therefore, cosθ = (5040 / 29) * (29 / 5040) * (1 / sqrt(2)) ) = (1) * (1 / sqrt(2)) ) = 1 / sqrt(2)Therefore, cosθ = 1 / √2 ⇒ θ = 45 degrees.So angle DAE is 45 degrees, or π/4 radians.Wait, that seems straightforward. Let me check again.So, the dot product of AD and AE came out to be 146160/841, and the product of their magnitudes was (84*sqrt(29)/29)*(60*sqrt(58)/29) = (84*60*sqrt(29*58))/29² = 5040*sqrt(1682)/841. Wait, 29*58 = 1682, which is 29*29*2 = 841*2, so sqrt(1682) = 29*sqrt(2). Therefore, product of magnitudes is 5040*29*sqrt(2)/841 = 5040*sqrt(2)/29. Then the dot product is 146160/841 = 5040/29. So cosθ = (5040/29) / (5040*sqrt(2)/29) = 1/sqrt(2). Therefore, θ = 45°. So angle DAE is 45 degrees.That's interesting. So regardless of the coordinates, the angle comes out as 45°. I wonder if there's a more elegant way to see this without coordinates, maybe using some properties of triangles or angle bisectors or something else. But since the coordinate method worked and gave 45°, which is a nice angle, that's probably the answer.Let me see if there's another approach. Maybe using Stewart's theorem to find lengths AD and AE, then using the Law of Cosines on triangle DAE.First, let's try Stewart's theorem. Stewart's theorem relates the length of a cevian in a triangle to the lengths of the sides. For a triangle with sides a, b, c, and a cevian of length d dividing the side a into m and n, Stewart's formula is: b²m + c²n = a(d² + mn)So, applying Stewart's theorem to find AD and AE.First, find AD. In triangle ABC, with BC = 29, AB = 20, AC = 21. Point D divides BC into BD = 8 and DC = 21 (since BC is 29, BD = 8, so DC = 29 - 8 = 21). Wait, no, BD = 8, DE = 12, EC = 9. So from B to D is 8, D to E is 12, E to C is 9. So DC would be DE + EC = 12 + 9 = 21. So BD = 8, DC = 21.Wait, but for Stewart's theorem, if we want to find AD, then in triangle ABC, with BC = 29, BD = 8, DC = 21. So Stewart's theorem gives:AB² * DC + AC² * BD = BC * (AD² + BD * DC)So plugging in:20² * 21 + 21² * 8 = 29 * (AD² + 8 * 21)Compute left side:400 * 21 = 8400441 * 8 = 3528Total: 8400 + 3528 = 11928Right side: 29*(AD² + 168)So 11928 = 29AD² + 29*168Compute 29*168: 168*30 = 5040; 5040 - 168 = 4872Thus, 11928 = 29AD² + 4872Subtract 4872: 11928 - 4872 = 7056 = 29AD²Therefore, AD² = 7056 / 29 ⇒ AD = sqrt(7056/29) = 84 / sqrt(29) ≈ 15.62Similarly, compute AE using Stewart's theorem. Point E divides BC into BE = 20 (BD + DE = 8 + 12 = 20) and EC = 9. So in triangle ABC, with BC = 29, BE = 20, EC = 9.Applying Stewart's theorem:AB² * EC + AC² * BE = BC * (AE² + BE * EC)Left side:20² * 9 + 21² * 20 = 400*9 + 441*20 = 3600 + 8820 = 12420Right side: 29*(AE² + 20*9) = 29*(AE² + 180)Therefore:12420 = 29AE² + 5220Subtract 5220: 12420 - 5220 = 7200 = 29AE² ⇒ AE² = 7200 / 29 ⇒ AE = sqrt(7200/29) = (60*sqrt(2))/sqrt(29) ≈ 60*1.4142/5.385 ≈ 15.70Wait, but in the coordinate method, |AD| was 84*sqrt(29)/29 ≈ 84/5.385 ≈ 15.6, and |AE| was 60*sqrt(58)/29 ≈ 60*7.615/29 ≈ 456.9/29 ≈ 15.75. Close enough. So Stewart's theorem gives the same lengths. Good.Now, to find angle DAE, we can use the Law of Cosines in triangle DAE. But we need the lengths of AD, AE, and DE.Wait, DE is given as 12, right? Because BD = 8, DE = 12, EC = 9. So DE = 12. Wait, DE is the segment between D and E on BC, which is 12. But in triangle DAE, DE is the side opposite angle DAE. Wait, no. In triangle DAE, the sides are AD, AE, and DE. So if we know all three sides, we can use Law of Cosines to find the angle at A.So DE is 12. So in triangle DAE:AD = 84/sqrt(29) ≈ 15.62AE = 60*sqrt(2)/sqrt(29) ≈ 15.70DE = 12So angle DAE is between AD and AE, with opposite side DE = 12.Law of Cosines:DE² = AD² + AE² - 2*AD*AE*cos(angle DAE)So plugging in:12² = (84²/29) + (60²*2/29) - 2*(84/sqrt(29))*(60*sqrt(2)/sqrt(29))*cosθCompute each term:Left side: 144Right side:First term: (7056/29) + (7200/29) = (7056 + 7200)/29 = 14256/29Second term: -2*(84*60*sqrt(2)/29)*cosθ = -2*(5040*sqrt(2)/29)*cosθ = - (10080*sqrt(2)/29)*cosθTherefore:144 = 14256/29 - (10080*sqrt(2)/29)*cosθMultiply both sides by 29:144*29 = 14256 - 10080*sqrt(2)*cosθCalculate 144*29:144*30 = 4320; 4320 - 144 = 4176So:4176 = 14256 - 10080*sqrt(2)*cosθSubtract 14256:4176 - 14256 = -10080*sqrt(2)*cosθ ⇒ -10080 = -10080*sqrt(2)*cosθDivide both sides by -10080:1 = sqrt(2)*cosθ ⇒ cosθ = 1/√2 ⇒ θ = 45°So same result. Therefore, angle DAE is 45 degrees. This confirms the result from the coordinate method.Alternatively, is there a property or theorem that directly gives this angle as 45°? Maybe trigonometric identities or recognizing some ratio.Alternatively, maybe using coordinates was more straightforward here. Both methods confirm the angle is 45°, so the answer is 45 degrees.Final Answerboxed{45^circ}</think>To find the angle ( angle DAE ) in triangle ( ABC ) with sides ( a = 29 ), ( b = 21 ), ( c = 20 ), where points ( D ) and ( E ) lie on segment ( BC ) with ( BD = 8 ), ( DE = 12 ), and ( EC = 9 ):1. Coordinate Setup: - Place ( B ) at ( (0, 0) ) and ( C ) at ( (29, 0) ). - Determine coordinates of ( A ) using the distance formula: - ( AB = 20 ) and ( AC = 21 ). - Solving the system of equations for ( A ): [ x^2 + y^2 = 400 quad text{and} quad (x - 29)^2 + y^2 = 441 ] - Coordinates of ( A ) are found to be ( left( frac{400}{29}, frac{420}{29} right) ).2. Coordinates of ( D ) and ( E ): - ( D ) is at ( (8, 0) ). - ( E ) is at ( (20, 0) ).3. Vectors Calculation: - Vectors ( overrightarrow{AD} ) and ( overrightarrow{AE} ): [ overrightarrow{AD} = left( 8 - frac{400}{29}, 0 - frac{420}{29} right) = left( frac{-168}{29}, frac{-420}{29} right) ] [ overrightarrow{AE} = left( 20 - frac{400}{29}, 0 - frac{420}{29} right) = left( frac{180}{29}, frac{-420}{29} right) ]4. Dot Product and Magnitudes: - Dot product ( overrightarrow{AD} cdot overrightarrow{AE} ): [ left( frac{-168}{29} cdot frac{180}{29} right) + left( frac{-420}{29} cdot frac{-420}{29} right) = frac{-30240 + 176400}{841} = frac{146160}{841} ] - Magnitudes ( |overrightarrow{AD}| ) and ( |overrightarrow{AE}| ): [ |overrightarrow{AD}| = sqrt{left( frac{-168}{29} right)^2 + left( frac{-420}{29} right)^2} = frac{84sqrt{29}}{29} ] [ |overrightarrow{AE}| = sqrt{left( frac{180}{29} right)^2 + left( frac{-420}{29} right)^2} = frac{60sqrt{58}}{29} ]5. Cosine of the Angle: - Using the dot product formula: [ cos theta = frac{overrightarrow{AD} cdot overrightarrow{AE}}{|overrightarrow{AD}| |overrightarrow{AE}|} = frac{frac{146160}{841}}{left( frac{84sqrt{29}}{29} right) left( frac{60sqrt{58}}{29} right)} = frac{1}{sqrt{2}} ] - Therefore, ( theta = 45^circ ).The final answer is ( boxed{45^circ} ).]

🔑:45 o Solution Put AD = x, AE = y. By repeated use of the cosine rule we have, cos B = (29 2 + 20 2 - 21 2 )/(2 · 20 · 29) = 20/29, x 2 = 20 2 + 8 2 - 2 · 8 · 20 · cos B = 7056/29, cos C = (21 2 + 29 2 - 20 2 )/(2 · 21 · 29) = 21/29, y 2 = 21 2 + 9 2 - 2 · 21 · 9 · cos C = 7200/29. Finally, cos DAE = (x 2 + y 2 - 12 2 )/(2 · x · y) = 1/√2. So angle DAE = 45 o . 15th Irish 2002 © John Scholes [email protected] 15 June 2003 Last updated/corrected 15 Jun 03

🧠:[<think>Okay, let's tackle this problem step by step. So, we have two curves: C₁ is the hyperbola x² - y² = 1, and C₂ is the parabola y² = 2px. They intersect at points M and N. The circumcircle of triangle OMN passes through point P(7/2, 0). We need to find the equation of C₂, which means we need to find the value of p.First, let's recall that the circumcircle of a triangle passes through all three vertices of the triangle. Here, the triangle is OMN, where O is the origin. So, the circumcircle passes through O, M, N, and also the point P(7/2, 0). Wait, but if the circumcircle already passes through O, M, N, then P must be another point on the same circle. So, this gives us an additional condition that point P lies on the circumcircle of triangle OMN. Therefore, we can use the power of a point or some circle equation conditions to relate these points.First, we need to find the coordinates of points M and N where C₁ and C₂ intersect. To find the intersection points, let's solve the two equations simultaneously.Given:1. x² - y² = 1 (C₁)2. y² = 2px (C₂)We can substitute y² from the second equation into the first equation. Let's do that.Substitute y² = 2px into x² - y² = 1:x² - (2px) = 1So, x² - 2px - 1 = 0This is a quadratic equation in x. Let's denote this as:x² - 2px - 1 = 0We can solve this quadratic for x using the quadratic formula:x = [2p ± √(4p² + 4)] / 2Simplify the discriminant:√(4p² + 4) = √[4(p² + 1)] = 2√(p² + 1)Therefore, the solutions for x are:x = [2p ± 2√(p² + 1)] / 2 = p ± √(p² + 1)So, the x-coordinates of points M and N are x = p + √(p² + 1) and x = p - √(p² + 1). Now, to find the corresponding y-coordinates, we can use y² = 2px.For x = p + √(p² + 1):y² = 2p(p + √(p² + 1)) = 2p² + 2p√(p² + 1)Similarly, for x = p - √(p² + 1):y² = 2p(p - √(p² + 1)) = 2p² - 2p√(p² + 1)But since y² must be non-negative, we need to check if 2p² - 2p√(p² + 1) is non-negative. Let's see:2p² - 2p√(p² + 1) = 2p(p - √(p² + 1))But note that √(p² + 1) > |p|, so if p is positive, then p - √(p² + 1) is negative, so y² would be negative, which is impossible. Similarly, if p is negative, then p - √(p² + 1) is also negative, so again y² would be negative. Therefore, the solution x = p - √(p² + 1) gives a negative y², which is not possible. Therefore, this solution is extraneous and must be discarded. Hence, there seems to be only one intersection point? But the problem states that they intersect at points M and N. Hmm, this is confusing.Wait, maybe I made a mistake here. Let's double-check.Wait, if x = p - √(p² + 1), then even if p is positive, let's compute x:Suppose p is positive. Then √(p² + 1) > p, so x = p - √(p² + 1) is negative. Then y² = 2p x would be 2p times a negative number, which would be negative. But y² can't be negative, so that's impossible. If p is negative, then let's see. Let p = -k where k is positive. Then x = -k - √(k² + 1), which is negative, so y² = 2*(-k)*x = -2k*x. Since x is negative, y² is positive. Wait, hold on, maybe if p is negative, x = p - √(p² + 1) could give a positive y².Wait, let's suppose p is negative. Let p = -k, k > 0.Then, x = p - √(p² + 1) = -k - √(k² + 1). That's negative, so y² = 2p x = 2*(-k)*(-k - √(k² + 1)) = 2k(k + √(k² + 1)), which is positive. Therefore, in this case, both x = p + √(p² + 1) and x = p - √(p² + 1) would give positive y² if p is negative. Wait, let's check that.Wait, when p is negative, let's take p = -k, k > 0.First solution: x = p + √(p² + 1) = -k + √(k² + 1). Let's see if this is positive or negative. Since √(k² + 1) > k, so this is positive. Therefore, x = -k + √(k² + 1). Then y² = 2p x = 2*(-k)*(-k + √(k² + 1)) = 2k(k - √(k² + 1)). Now, since √(k² + 1) > k, this would be 2k times a negative number, so y² would be negative. Which is impossible. Therefore, even with p negative, x = p + √(p² + 1) gives a negative y². Wait, that's confusing. Let me recast this.Wait, if p is negative, then 2p x would be positive only if x is negative, right? Because 2p x = 2*(-k)*x. So if x is negative, then 2*(-k)*x = positive.So, for x = p + √(p² + 1) when p is negative:x = p + √(p² + 1) = -k + √(k² + 1). Since √(k² + 1) > k, then x is positive. So 2p x = 2*(-k)*(positive) = negative, which is impossible for y².Therefore, only x = p - √(p² + 1) would result in y² positive when p is negative, because x is negative (as p - √(p² + 1) when p is negative would be negative). Let's see:If p is negative, then x = p - √(p² + 1) is negative (since both terms are negative). Then y² = 2p x = 2*(negative)*(negative) = positive. So in this case, x = p - √(p² + 1) gives a valid y². So perhaps when p is negative, we have two intersection points with real coordinates?Wait, this is getting complicated. Let's re-examine the quadratic equation.We have x² - 2px - 1 = 0. The solutions are x = [2p ± √(4p² + 4)] / 2 = p ± √(p² + 1). So regardless of the value of p, these are two real roots. However, when we substitute back into y² = 2px, we must have y² ≥ 0.Therefore, for each solution x, we have y² = 2p x. Therefore, 2p x must be non-negative.So for each x solution, we need 2p x ≥ 0. Therefore, for each x, the product p x must be non-negative.So let's analyze the two roots:x₁ = p + √(p² + 1)x₂ = p - √(p² + 1)First, note that √(p² + 1) is always positive, regardless of p.Case 1: p > 0Then x₁ = p + √(p² + 1) is positive, so 2p x₁ is positive (since p >0 and x₁ >0). Thus, y² is positive, so valid.x₂ = p - √(p² + 1). Since √(p² +1) > p (because p² +1 > p²), so x₂ is negative. Then 2p x₂ = 2p*(negative) = negative. Thus, y² would be negative, which is invalid. Therefore, when p >0, only x₁ gives a valid intersection point.Case 2: p < 0Let p = -k, where k >0.Then x₁ = -k + √(k² + 1). Since √(k² + 1) > k, x₁ is positive. Then 2p x₁ = 2*(-k)*(positive) = negative, so y² is invalid.x₂ = -k - √(k² + 1) which is negative. Then 2p x₂ = 2*(-k)*(negative) = positive, so y² is valid. Therefore, when p <0, only x₂ gives a valid intersection point.Wait, so in both cases, whether p is positive or negative, only one intersection point is valid? But the problem states that C₁ and C₂ intersect at points M and N, so two points. This suggests that there must be two intersection points. Therefore, perhaps p must be such that both solutions x₁ and x₂ give y² ≥0.But from the above analysis, that seems possible only if both x₁ and x₂ have 2p x ≥0. Let's check:Suppose p >0. Then x₁ = p + √(p² +1) is positive, so 2p x₁ positive. x₂ = p - √(p² +1) is negative, so 2p x₂ negative. Therefore, only x₁ is valid.If p <0. Then x₁ = p + √(p² +1) is p + something larger than |p|. If p is negative, say p = -k, then x₁ = -k + √(k² +1). Since √(k² +1) > k, x₁ is positive. Then 2p x₁ = 2*(-k)*(positive) = negative, invalid. x₂ = p - √(p² +1) = -k - √(k² +1), which is negative. Then 2p x₂ = 2*(-k)*(negative) = positive. So x₂ is valid.So in both cases, only one valid intersection point. But the problem says they intersect at two points M and N. So this suggests that maybe p =0? But if p=0, then C₂ becomes y²=0, which is the x-axis. Then intersection with C₁: x² - 0 =1 => x=±1. So points (1,0) and (-1,0). But y²=0 implies y=0, so those are two points. So when p=0, there are two intersection points. However, p=0 would make C₂ degenerate (the x-axis), but the problem states "curve C₂: y²=2px". If p=0, it's the x-axis, which is a valid curve (a line). But in that case, the circumcircle of triangle OMN would be the circle passing through O(0,0), M(1,0), N(-1,0). But those three points are colinear on the x-axis, so the circle would be undefined unless they are non-colinear. But three colinear points can't form a circle unless it's a line, which isn't a circle. Hence, p=0 is invalid because the three points would be colinear, so no circumcircle. Therefore, the problem must have p ≠0, but then according to our previous analysis, only one intersection point. But the problem says two points M and N. This is a contradiction.Wait, so perhaps there's a mistake in my reasoning. Let me check again.Wait, when p=0, we get two points, but they are on the x-axis, making the triangle OMN degenerate. For p≠0, according to previous reasoning, only one valid intersection point. But the problem states that curves C₁ and C₂ intersect at points M and N. Therefore, maybe the curves actually intersect at two points even when p≠0. Maybe my analysis is wrong.Wait, let's take a specific example. Let's pick p=1.C₁: x² - y² =1C₂: y²=2xSolving:x² - (2x) =1 => x² -2x -1=0. Solutions x=(2 ± √(4 +4))/2=(2±√8)/2=1±√2So x=1+√2 and x=1-√2. Now, check y²=2x.For x=1+√2, y²=2(1+√2) which is positive.For x=1-√2, since √2 ≈1.414, so 1-√2≈-0.414, which is negative. Then y²=2*(1-√2)≈-0.828, which is negative. Therefore, only x=1+√2 gives a valid point. Therefore, only one intersection point when p=1.Similarly, if p is negative, say p=-1.C₂: y²= -2xIntersection with C₁: x² - y²=1 => x² - (-2x)=1 =>x² +2x -1=0. Solutions x=(-2±√(4 +4))/2=(-2±√8)/2=-1±√2x=-1+√2≈0.414, y²=-2x≈-0.828 invalid.x=-1-√2≈-2.414, y²=-2x≈4.828, positive. Therefore, only x=-1-√2 gives valid y. Therefore, only one intersection point.So this seems like for p≠0, there's only one intersection point. But when p=0, two intersection points but degenerate. So the problem statement must have a typo? Or perhaps there's something wrong here.Wait, but maybe in complex plane, there are two intersections, but in real plane, only one. But the problem talks about points M and N, so they must be real points. Therefore, perhaps the problem is assuming that p is such that both intersection points are real? But according to the above, unless p=0, there's only one real intersection. But when p=0, it's degenerate.Wait, this is confusing. Let's check the equations again.C₁: x² - y² =1 is a hyperbola.C₂: y²=2px is a parabola. The parabola opens to the right if p>0 and to the left if p<0.A hyperbola and a parabola can intersect at up to 4 points. But in this case, substituting, we end up with a quadratic equation, which can have two solutions. However, due to the substitution, perhaps some solutions are extraneous or complex.Wait, perhaps the problem is correct, and my previous analysis is missing something. Let me try again.Given x² - y²=1 and y²=2px. Substitute y²=2px into the hyperbola equation:x² - 2px =1, so x² -2px -1=0. The discriminant is (2p)^2 +4=4p² +4=4(p² +1). Which is always positive, so two real roots. Therefore, two real x-values. However, when we plug back into y²=2px, we have to ensure that y² is non-negative.Therefore, for each root x, y²=2px must be non-negative. Therefore, for each root x, 2px ≥0.Therefore, for each x, either p and x are both positive or both negative.Given that x² -2px -1=0, let's analyze the roots.The roots are x = [2p ± √(4p² +4)]/2 = p ± √(p² +1).So for each p:If p is positive:x₁ = p + √(p² +1) >0, so 2p x₁ >0, valid.x₂ = p - √(p² +1). Since √(p² +1) > p (as p is positive), x₂ is negative. So 2p x₂ <0, invalid.Therefore, only x₁ is valid.If p is negative:x₁ = p + √(p² +1). Since √(p² +1) > |p|, if p is negative, then p + √(p² +1) = √(p² +1) + p. Since p is negative, this is √(p² +1) - |p|. Because p² = |p|², so √(p² +1) = √(|p|² +1). Let k = |p|. Then x₁ = √(k² +1) - k. Since √(k² +1) > k, x₁ is positive. Therefore, 2p x₁ = 2*(-k)*(positive) = negative, which makes y² negative, invalid.x₂ = p - √(p² +1). Since p is negative, this is -k - √(k² +1), which is negative. Therefore, 2p x₂ = 2*(-k)*(-k - √(k² +1)) = 2k(k + √(k² +1)), which is positive. Therefore, y² is positive, valid.Therefore, regardless of whether p is positive or negative, only one valid intersection point. Therefore, the problem's statement that they intersect at points M and N must be incorrect unless there's a different interpretation.Wait, but the problem says "intersects at points M and N". So maybe even though algebraically there are two roots, only one gives real points. But the problem says two points, so perhaps complex points? But the circumcircle is in the real plane. So this is confusing.Alternatively, maybe the problem is considering multiplicities? If the curves are tangent to each other, they intersect at one point with multiplicity two. But the problem says "points M and N", implying two distinct points.Alternatively, perhaps there's a mistake in the problem statement, but since it's an exam problem, we need to find a way.Wait, but maybe I made a mistake in assuming real intersection points. Wait, the quadratic equation x² -2px -1=0 has two real roots for x, but when substituting back into y²=2px, we have to have y² non-negative. Therefore, even though there are two real x-values, only those x-values for which 2px is non-negative will give real y-values. Therefore, depending on the sign of p, only one x gives real y. Hence, only one intersection point. Therefore, the problem's statement might be incorrect. However, since it's given, perhaps there's another way.Wait, maybe the curves C₁ and C₂ actually intersect at two real points when p is a particular value, and the circumcircle passes through P(7/2, 0). So even though for general p they intersect once, maybe for a specific p they intersect twice. Wait, but according to the previous analysis, unless p=0, which gives two points but degenerate, otherwise only one point. So this is a contradiction.Wait, let's check when p=0, as mentioned before. Then C₂ is y²=0, the x-axis. Intersection with C₁: x²=1, so points (1,0) and (-1,0). Then triangle OMN is the line segment from (0,0) to (1,0) to (-1,0), which is colinear, so no circumcircle. Therefore, p=0 is invalid.Alternatively, maybe the problem assumes that the curves intersect at two points with the same x-coordinate but different y-coordinates? Wait, but for a given x, y²=2px gives two y-values: y=±√(2px). So if 2px is positive, then two real y-values. Therefore, for each valid x (where 2px ≥0), there are two points: (x, √(2px)) and (x, -√(2px)). Therefore, even if there is only one x-value, there are two points M and N, symmetric about the x-axis.Ah! That must be it! I think I missed that. So when we have x = p + √(p² +1) for p>0, then y²=2p x is positive, so y=±√(2p x). Therefore, the two points M and N are (x, √(2p x)) and (x, -√(2p x)). Similarly, if p is negative, x is negative and y²=2p x is positive (since both p and x are negative), so again y=±√(2p x). Therefore, regardless of the sign of p, there are two intersection points, symmetric about the x-axis.Therefore, M and N are (x, y) and (x, -y). So they lie symmetrically above and below the x-axis. Therefore, the triangle OMN is a triangle with vertices at O(0,0), M(x, y), N(x, -y). Therefore, this triangle is isoceles with base MN and vertex at O.Therefore, the circumcircle of triangle OMN must pass through these three points. Additionally, it must pass through point P(7/2, 0). Therefore, we need to find p such that the circumcircle of OMN passes through (7/2, 0).Okay, so now that we have clarified that M and N are (x, y) and (x, -y), where x is p + √(p² +1) for p>0 or x is p - √(p² +1) for p<0, but with y=±√(2p x). Wait, but for p>0, x is positive, so y²=2p x is positive, giving real y. For p<0, x is negative, so y²=2p x=2*(negative)*(negative)=positive, so also real y. So regardless of p, there are two points M and N: (x, y) and (x, -y).But in either case, the points M and N are symmetric about the x-axis. Therefore, the triangle OMN is isoceles with base MN and vertex at O. The circumcircle of this triangle can be determined by finding the circle passing through O, M(x, y), N(x, -y), and P(7/2, 0).Let's proceed step by step.First, let's parameterize the points. Let’s denote the coordinates of M as (a, b) and N as (a, -b). Since they lie on both curves, we have:For curve C₁: a² - b² =1For curve C₂: b² =2 p aSo from C₂: b²=2pa, so we can substitute into C₁:a² - 2pa =1 => a² -2pa -1=0Which is the same quadratic as before. Therefore, solving for a:a = [2p ± √(4p² +4)] /2 = p ± √(p² +1)But as we saw earlier, depending on the sign of p, only one solution is valid (the other gives a negative y²). Wait, but if we take p positive, then a = p + √(p² +1), which is positive, so y²=2pa is positive, so valid. If p is negative, then a = p - √(p² +1), which is negative, and y²=2pa=2p*(negative)=positive (since p is negative). So in both cases, a is either p + √(p² +1) (for p>0) or p - √(p² +1) (for p<0). So let's denote a as:a = p + √(p² +1) if p>0a = p - √(p² +1) if p<0But maybe we can just keep a as p + √(p² +1) regardless, and see if that works. Wait, but for p negative, p + √(p² +1) is positive, but y²=2pa=2p*(positive)=negative, which is invalid. Therefore, a must be p - √(p² +1) for p<0. So perhaps we need to split into cases.Alternatively, perhaps since the problem is asking for the equation of C₂, which is y²=2px, regardless of the value of p, we can proceed by considering the general case.Let’s denote a = p + sgn(p)√(p² +1). Wait, maybe not. Alternatively, let's use the fact that a and p are related by a² - 2pa -1=0. So from this quadratic equation, we can express p in terms of a:From a² - 2pa -1=0,2pa = a² -1Therefore, p = (a² -1)/(2a)So p can be expressed in terms of a. Therefore, we can parameterize everything in terms of a.Given that M is (a, b) and N is (a, -b), with b²=2pa, and a² - b²=1. From these, we can write b²=2pa, and substituting into a² - b²=1 gives a² -2pa=1, which is the same as above.So, given that, we can write p=(a² -1)/(2a)Now, the circumcircle of triangle OMN passes through point P(7/2, 0). So, the circle passing through O(0,0), M(a,b), N(a,-b), and P(7/2,0).We need to find the equation of the circle passing through these four points.But since three non-colinear points determine a circle, and P is a fourth point on the same circle. So, we can use the three points O, M, and P to determine the circle, then verify that N also lies on it. Or since O, M, N, P are all on the circle, we can set up equations.Let’s recall the general equation of a circle: x² + y² + Dx + Ey + F =0.Since the circle passes through O(0,0), substituting x=0, y=0 gives 0 +0 +0 +0 +F=0 => F=0.Therefore, the equation simplifies to x² + y² + Dx + Ey =0.Now, the circle passes through M(a, b) and N(a, -b), and P(7/2, 0).First, plug in M(a, b):a² + b² + D a + E b =0Plug in N(a, -b):a² + b² + D a - E b =0Plug in P(7/2, 0):(7/2)² + 0² + D*(7/2) + E*0 =0 => (49/4) + (7D)/2 =0So, from P's equation:49/4 + (7D)/2 =0 => multiply both sides by 4:49 +14D=0 => 14D= -49 => D= -49/14= -7/2Therefore, D= -7/2Now, substitute D back into the equations from M and N.For M(a, b):a² + b² + (-7/2)a + E b =0For N(a, -b):a² + b² + (-7/2)a - E b =0Subtract the two equations:(a² + b² -7a/2 + E b) - (a² + b² -7a/2 - E b) =0 -0Which simplifies to:2 E b =0 => E b=0So, either E=0 or b=0.But b cannot be zero because points M and N are distinct and not on the x-axis (since if b=0, then M and N would coincide at (a,0), but C₁ and C₂ would intersect at a single point on the x-axis, which would not form a triangle). Therefore, b≠0, so E=0.Therefore, E=0.So the equation of the circle is x² + y² - (7/2)x =0, or multiplying by 2 to eliminate fractions:2x² +2y² -7x=0Now, check if this circle passes through both M(a, b) and N(a, -b). Let's check with M(a, b):2a² +2b² -7a=0But from earlier, we have from the circle equation: a² + b² - (7/2)a=0, which multiplied by 2 gives 2a² +2b² -7a=0, which is the same as above. Similarly for N(a, -b). Therefore, the circle equation is correct.But we also have that points M(a, b) lie on both C₁ and C₂.From C₁: a² - b²=1From C₂: b²=2paFrom the circle equation: 2a² +2b² -7a=0So let's use these equations to solve for a and p.First, from C₁ and C₂:We have a² - b²=1 and b²=2pa. Substitute b² into the first equation:a² -2pa=1 => a² -2pa -1=0We can express p in terms of a as p=(a² -1)/(2a)From the circle equation: 2a² +2b² -7a=0. Substitute b²=2pa:2a² +2*(2pa) -7a=0 => 2a² +4pa -7a=0Factor out a:a(2a +4p -7)=0Since a≠0 (because a=0 would imply from C₁: -b²=1, impossible), so 2a +4p -7=0Therefore:2a +4p -7=0 => 2a +4p =7But we have p=(a² -1)/(2a). Substitute this into the above equation:2a +4*( (a² -1)/(2a) ) =7Simplify:2a + (4(a² -1))/(2a) =7Simplify the second term:4/(2a)*(a² -1)=2/a*(a² -1)Thus:2a + (2(a² -1))/a =7Multiply both sides by a to eliminate denominator:2a² +2(a² -1)=7aExpand:2a² +2a² -2=7aCombine like terms:4a² -7a -2=0So quadratic equation:4a² -7a -2=0Solve for a:a = [7 ±√(49 +32)]/8 = [7 ±√81]/8 = [7±9]/8Thus:a=(7+9)/8=16/8=2a=(7-9)/8=(-2)/8= -1/4So two possible solutions for a: a=2 and a=-1/4Now, we need to check which of these are valid based on the earlier relations.Recall that a is the x-coordinate of points M and N. Since C₁ is x² - y²=1, x²=1 + y²≥1. Therefore, |a|≥1. Therefore, a=2 is valid (|2|≥1), but a=-1/4 is invalid (|-1/4|<1). Therefore, discard a=-1/4. Therefore, a=2.Now, compute p using p=(a² -1)/(2a):p=(4 -1)/(4)=3/4Therefore, p=3/4Therefore, the equation of curve C₂ is y²=2*(3/4)x= (3/2)x, so y²= (3/2)xBut let's verify this.Given p=3/4, then C₂:y²=2*(3/4)x= (3/2)xThen, intersection points with C₁: x² - y²=1Substitute y²= (3/2)x:x² - (3/2)x =1 => multiply by 2:2x² -3x -2=0Solve:x=(3±√(9 +16))/4=(3±√25)/4=(3±5)/4Thus, x=(8)/4=2 or x=(-2)/4=-1/2But x=-1/2 would give y²=(3/2)*(-1/2)= -3/4, invalid. Therefore, valid x=2, y²= (3/2)*2=3 => y=±√3. Therefore, points M(2, √3) and N(2, -√3)Now, check if the circumcircle of triangle OMN passes through P(7/2,0).We can compute the equation of the circle passing through O(0,0), M(2,√3), N(2,-√3), and check if P(7/2,0) lies on it.First, the general equation of the circle: x² + y² + Dx + Ey + F=0. Since it passes through O(0,0), F=0.So equation: x² + y² + Dx + Ey=0Plug in M(2,√3):4 +3 +2D + E√3=0 =>7 +2D +E√3=0 ...(1)Plug in N(2,-√3):4 +3 +2D -E√3=0 =>7 +2D -E√3=0 ...(2)Subtract equation (2) from equation (1):(7 +2D +E√3) - (7 +2D -E√3) =0 -0=>2E√3=0 => E=0Then from equation (1):7 +2D=0 => D= -7/2Therefore, the equation is x² + y² - (7/2)x=0Check if P(7/2,0) lies on this circle:(7/2)² +0² - (7/2)(7/2)= (49/4) - (49/4)=0. Yes, it satisfies.Therefore, the circle equation is correct, and p=3/4 is the correct value.Therefore, the equation of curve C₂ is y²=2*(3/4)x= (3/2)x, which can be written as y²= (3/2)x. However, typically, equations are written without fractions in the coefficients, so multiplying both sides by 2:2y²=3x. But the standard form of a parabola is y²=4ax, so here it's y²= (3/2)x, which is equivalent to y²=4*(3/8)x, but maybe they prefer the form with denominator cleared. However, the problem asks for the equation of curve C₂, which was given as y²=2px. Therefore, substituting p=3/4, the equation is y²=2*(3/4)x= (3/2)x. So the answer is y²=3x/2, but written in boxed form.Alternatively, multiply both sides by 2 to get 2y²=3x, but since the original equation is y²=2px, p=3/4. So the equation is y²=2*(3/4)x, which is y²=(3/2)x. So both forms are acceptable, but likely the problem expects y²=3x/2 or y²=(3/2)x. In Chinese textbooks, often fractions are written with horizontal lines, but in LaTeX, it's frac{3}{2}x.But let's confirm once more.We found p=3/4, so C₂ is y²=2*(3/4)x=3/2 x. Therefore, the equation is y²= (3/2)x. To write this neatly, we can write it as y² = frac{3}{2}x, which is the same as 2y² =3x if we clear the fraction. But since the original form is y²=2px, where p=3/4, the answer should be y²=2*(3/4)x=3/2x, so yes, y²= frac{3}{2}x.Therefore, the answer is boxed{y^2 = dfrac{3}{2}x}Final Answerboxed{y^2 = dfrac{3}{2}x}</think>Given the curves ( C_1: x^2 - y^2 = 1 ) and ( C_2: y^2 = 2px ), we need to find the equation of ( C_2 ) such that the circumcircle of (triangle OMN) passes through the point ( Pleft(frac{7}{2}, 0right) ).1. Finding Intersection Points: - Substitute ( y^2 = 2px ) into ( x^2 - y^2 = 1 ): [ x^2 - 2px = 1 implies x^2 - 2px - 1 = 0 ] - Solving the quadratic equation: [ x = p pm sqrt{p^2 + 1} ] - For real ( y ), ( 2px geq 0 ). This gives valid ( x ) values depending on the sign of ( p ).2. Coordinates of Points ( M ) and ( N ): - Points ( M ) and ( N ) are ((a, b)) and ((a, -b)) where ( a = p + sqrt{p^2 + 1} ) for ( p > 0 ) or ( a = p - sqrt{p^2 + 1} ) for ( p < 0 ).3. Circumcircle of (triangle OMN): - The circumcircle passes through ( O(0,0) ), ( M(a, b) ), ( N(a, -b) ), and ( Pleft(frac{7}{2}, 0right) ). - General circle equation: ( x^2 + y^2 + Dx + Ey + F = 0 ). Since it passes through the origin, ( F = 0 ). - Substituting ( Pleft(frac{7}{2}, 0right) ): [ left(frac{7}{2}right)^2 + 0^2 + D left(frac{7}{2}right) = 0 implies frac{49}{4} + frac{7D}{2} = 0 implies D = -frac{7}{2} ] - Substituting ( M(a, b) ) and ( N(a, -b) ): [ a^2 + b^2 - frac{7}{2}a = 0 quad text{and} quad E = 0 ] - Using ( a^2 - b^2 = 1 ) and ( b^2 = 2pa ): [ 2a^2 + 4pa - 7a = 0 implies 4a^2 - 7a - 2 = 0 ] - Solving the quadratic equation for ( a ): [ a = 2 quad (text{valid solution}) ] - Calculating ( p ): [ p = frac{a^2 - 1}{2a} = frac{4 - 1}{4} = frac{3}{4} ]4. Equation of ( C_2 ): - Substituting ( p = frac{3}{4} ): [ y^2 = 2 left(frac{3}{4}right) x implies y^2 = frac{3}{2} x ]Thus, the equation of curve ( C_2 ) is (boxed{y^2 = dfrac{3}{2}x}).,<think>Okay, let me try to figure this problem out step by step. So, we have two curves: C₁ is the hyperbola x² - y² = 1, and C₂ is the parabola y² = 2px. They intersect at points M and N. The circumcircle of triangle OMN passes through another point P(7/2, 0). We need to find the equation of C₂, which basically means finding the value of p.First, let's recall that the circumcircle of a triangle passes through all three vertices of the triangle. Here, the triangle is OMN, where O is the origin. The circumcircle passes through O, M, N, and also the point P(7/2, 0). So, the circumcircle must pass through four points: O, M, N, and P. However, three non-collinear points determine a unique circle, so if P is also on that circle, it must satisfy the equation derived from O, M, N.So, the strategy is probably to find the coordinates of points M and N by solving the intersection of C₁ and C₂, then find the equation of the circumcircle of triangle OMN, and then plug in the point P into that equation to solve for p.Let me start by finding the points of intersection between C₁ and C₂.Given:C₁: x² - y² = 1C₂: y² = 2pxSince C₂ gives y² in terms of x, I can substitute that into C₁'s equation.Substitute y² = 2px into x² - y² = 1:x² - (2px) = 1Which simplifies to:x² - 2px - 1 = 0This is a quadratic equation in x. Let's solve for x:x = [2p ± √( (2p)² + 4 )]/2Wait, the discriminant would be (2p)^2 + 4*1*1 = 4p² + 4 = 4(p² + 1)Therefore, x = [2p ± 2√(p² + 1)]/2 = p ± √(p² + 1)So the x-coordinates of points M and N are p + √(p² + 1) and p - √(p² + 1). Let's denote these as x₁ and x₂.Then, the corresponding y-coordinates can be found from C₂: y² = 2px, so y = ±√(2px). So for each x, we have two y-values, but since M and N are intersection points, I need to check if both signs are valid.Wait, but since both curves are symmetric with respect to the x-axis, if (x, y) is an intersection point, then (x, -y) is also an intersection point. Therefore, the points M and N are probably symmetric across the x-axis. So maybe they have the same x-coordinate and opposite y-coordinates? Wait, but solving x² - y²=1 and y²=2px, so substituting gives x² - 2px =1, which is quadratic in x, so two solutions for x. So actually, there are two different x-values, each with two y-values (positive and negative). But since the problem mentions points M and N, maybe there are two distinct points? Wait, the hyperbola x² - y²=1 and the parabola y²=2px. Depending on the value of p, they can intersect at two or four points? Hmm.But the quadratic equation in x we solved gives two x-values. For each x-value, since y²=2px, each x gives two y-values (positive and negative) provided that 2px is non-negative. Since y² can't be negative, so 2px must be ≥ 0. So, if p is positive, then x must be ≥ 0. If p is negative, then x must be ≤ 0. But our hyperbola x² - y² =1 has x² ≥1, so x ≥1 or x ≤ -1. So, depending on the sign of p, the intersections will be in different regions.But let's not get bogged down here. Let's proceed with the algebra.So, the x-coordinates of the intersection points are x₁ = p + √(p² + 1) and x₂ = p - √(p² + 1). Let's compute these.First, compute √(p² + 1). So, regardless of p, √(p² +1) is a positive real number. Therefore, x₁ = p + √(p² +1) and x₂ = p - √(p² +1).Now, let's check the possible values of x₁ and x₂.If p is positive, then x₁ = p + √(p² +1) is definitely positive. x₂ = p - √(p² +1). Let's see, √(p² +1) > √(p²) = |p|. If p is positive, then √(p² +1) > p, so x₂ = p - √(p² +1) is negative. Therefore, in this case, x₂ would be negative. But since for the parabola y²=2px, if p is positive, then x must be non-negative. Therefore, x₂ would give a negative x, which would make y²=2px negative, which is impossible. So, if p is positive, x₂ is invalid. Therefore, only x₁ is valid, and then y would be ±√(2p x₁). So, in that case, the intersection points would be two points: (x₁, √(2p x₁)) and (x₁, -√(2p x₁)), which are symmetric over the x-axis.Similarly, if p is negative, then 2px would require x to be non-positive. The hyperbola x² - y²=1 has x² ≥1, so x ≥1 or x ≤ -1. If p is negative, then x must be ≤0. So x ≤ -1. Then, x₁ = p + √(p² +1). If p is negative, say p = -k where k >0, then x₁ = -k + √(k² +1). Since √(k² +1) > k, so x₁ = -k + √(k² +1) is positive. But then x would be positive, which is invalid for p negative. So x₁ would be invalid, and x₂ = p - √(p² +1) = -k - √(k² +1), which is negative. Therefore, x₂ would be valid for p negative, and then y²=2p x₂ would be positive because p and x₂ are both negative, so 2p x₂ is positive. So, again, two points: (x₂, √(2p x₂)) and (x₂, -√(2p x₂)).Therefore, regardless of the sign of p, there are two intersection points, symmetric over the x-axis. So, points M and N are (x, y) and (x, -y) if p is positive, with x positive; or (x, y) and (x, -y) with x negative if p is negative.But in the problem statement, they just mention points M and N. So, perhaps M and N are these two points with the same x-coordinate and opposite y-coordinates. Then, triangle OMN would have vertices at O(0,0), M(x, y), and N(x, -y). That triangle would be isosceles with base MN and vertex at O.Alternatively, if the intersections have different x-coordinates, but that seems not possible as we saw earlier. Wait, but when we solved the equations, we had two x-values: x₁ and x₂. But depending on the sign of p, one of them is valid and the other isn't. Wait, maybe I made a mistake there. Let me check again.Suppose p is positive. Then, the parabola y²=2px opens to the right. The hyperbola x² - y²=1 has branches on the right and left. The right branch of the hyperbola could intersect the parabola at two points (x, y) and (x, -y) since the parabola is symmetric about the x-axis. Similarly, the left branch of the hyperbola can't intersect the parabola because the parabola is only on the right side when p is positive. If p is negative, the parabola opens to the left, and the left branch of the hyperbola could intersect the parabola at two points (x, y) and (x, -y). So, in either case, there are two points of intersection, symmetric about the x-axis. Therefore, points M and N must be these two points, (x, y) and (x, -y). Therefore, their x-coordinates are the same? Wait, but when we solved x² - 2px -1 =0, we got two different x solutions. Wait, no. Wait, let me double-check.Wait, no. The quadratic equation in x is x² - 2px -1=0. The solutions are x = [2p ± √(4p² +4)]/2 = [2p ± 2√(p² +1)]/2 = p ± √(p² +1). So, two different x-values. Wait, but earlier reasoning suggested that if p is positive, one x is positive and the other is negative, but if p is negative, one x is negative and the other is positive. Wait, let's take p positive first.Let p=1. Then, x₁ =1 + √(1 +1)=1 + √2≈2.414, which is positive. x₂=1 - √2≈-0.414, which is negative. But for p=1, parabola opens to the right, so x must be ≥0, so x₂ is invalid. Therefore, only x₁ is valid. So, in that case, only two points: (x₁, sqrt(2p x₁)) and (x₁, -sqrt(2p x₁)). Wait, but the quadratic equation gives two x-values. But only one of them is valid. Therefore, perhaps in this case, the curves intersect at two points with the same x-coordinate but different y's? Wait, no. Wait, no, substitution gives that for each x, there is a y²=2px. If the x is same, then y would be ±sqrt(2px). So, if x is same, then two points. If x is different, then different x's. Wait, perhaps the curves intersect at two points with different x's? Wait, but in the case of p=1, x₂ is negative, which would lead to y²=2*1*(-0.414)≈-0.828, which is invalid. Therefore, only x₁ is valid, leading to two points. Therefore, perhaps in general, the curves intersect at two points (x₁, y₁) and (x₁, -y₁). Therefore, points M and N are these two points. Therefore, triangle OMN would have vertices at O(0,0), M(x₁, y₁), N(x₁, -y₁). Then, the circumcircle of triangle OMN would pass through these three points, and also P(7/2,0). Therefore, we need to find the equation of the circle passing through O, M, N, and P, and then solve for p.Alternatively, since O, M, N, and P are on the same circle, the circle must pass through these four points. But three points (O, M, N) determine a circle, so P must lie on that circle.Alternatively, since triangle OMN is isosceles with OM=ON (since M and N are symmetric over x-axis), the circumcircle can be found, and then we can enforce that P lies on it.So, let's proceed step by step.First, let's denote the coordinates of M and N.From earlier, the x-coordinate of the valid intersection point(s):If p is positive, the valid x is x₁ = p + √(p² +1). Then, y²=2p x₁, so y=±√(2p x₁). Therefore, points M and N are (x₁, √(2p x₁)) and (x₁, -√(2p x₁)).Similarly, if p is negative, the valid x is x₂ = p - √(p² +1). Since p is negative, let me write p = -k where k >0. Then, x₂ = -k - √(k² +1), which is negative. Then, y²=2p x₂ = 2*(-k)(x₂) = -2k x₂. But x₂ is negative, so -2k x₂ is positive. So y=±√(-2k x₂). Thus, points M and N would be (x₂, ±√(-2k x₂)) where k = -p.But perhaps, regardless of the sign of p, we can express points M and N as (x, y) and (x, -y), where x is the valid x-coordinate (positive if p is positive, negative if p is negative), and y = sqrt(2p x) if p is positive, or y = sqrt(-2p |x|) if p is negative. Wait, maybe we can just work with variables.Let me suppose p is positive. Then, x = p + sqrt(p² +1), and y = sqrt(2p x). Let's denote x = p + sqrt(p² +1). Then, y² = 2p x = 2p(p + sqrt(p² +1)).But maybe we can keep it symbolic for now.So, points O(0,0), M(x, y), N(x, -y). Then, the circumcircle of triangle OMN.To find the circumcircle passing through O(0,0), M(x, y), N(x, -y), and P(7/2, 0).Since the circle passes through O, M, N, and P, we can set up the equation of the circle and plug in these points.The general equation of a circle is x² + y² + D x + E y + F = 0. Since O(0,0) is on the circle, plugging in (0,0) gives 0 + 0 + 0 + 0 + F =0 => F=0. So the equation simplifies to x² + y² + D x + E y =0.Now, points M(x, y) and N(x, -y) lie on this circle. Let's plug in M(x, y):x² + y² + D x + E y =0Similarly, plug in N(x, -y):x² + y² + D x + E (-y) =0Subtracting these two equations:[ x² + y² + D x + E y ] - [ x² + y² + D x - E y ] =0 -0 => 2 E y =0Since this must hold for the specific y value of point M, which is non-zero (because otherwise M and N would coincide with O, but the hyperbola x² - y²=1 at y=0 gives x²=1, so x=±1. But substituting into C₂: y²=2p x, which would be 0=2p x, so x=0. But x=±1 and x=0 is a contradiction, so y cannot be zero. Therefore, 2 E y =0 implies E=0.Therefore, the equation of the circle becomes x² + y² + D x =0.Now, we also know that point P(7/2, 0) is on this circle. Plugging in (7/2, 0):(7/2)² + 0² + D*(7/2) =0Calculates to:49/4 + (7D)/2 =0Multiply both sides by 4 to eliminate denominators:49 + 14D =0 => 14D = -49 => D = -49/14 = -7/2.Therefore, the equation of the circle is x² + y² - (7/2)x =0.Now, this circle must also pass through points M(x, y) and N(x, -y). Let's verify this with point M(x, y):x² + y² - (7/2)x =0But from the equation of the hyperbola C₁: x² - y² =1 => y² =x² -1.So substitute y² =x² -1 into the circle equation:x² + (x² -1) - (7/2)x =0Simplify:x² + x² -1 - (7/2)x =0 => 2x² - (7/2)x -1 =0Multiply both sides by 2 to eliminate fraction:4x² -7x -2=0So, 4x² -7x -2=0. Let's solve for x:x = [7 ±√(49 +32)] /8 = [7 ±√81]/8 = [7 ±9]/8So, x=(7+9)/8=16/8=2 or x=(7-9)/8=-2/8=-1/4.But we need to check if these x-values correspond to points on both curves C₁ and C₂.First, x=2:From hyperbola C₁: x² - y²=1 => 4 - y²=1 => y²=3 => y=±√3. Then, from parabola C₂: y²=2p x => 3=2p*2 => 3=4p => p=3/4.Second, x=-1/4:From hyperbola C₁: x² - y²=1 => (1/16) - y²=1 => y²=1 -1/16=15/16. Then, from parabola C₂: y²=2p x => 15/16=2p*(-1/4) => 15/16= -p/2 => p= -15/8.But earlier, when we found points M and N, we considered that if p is positive, x is positive, and if p is negative, x is negative. So, here, when x=2, p=3/4 (positive), which is valid. When x=-1/4, p=-15/8 (negative), which would require x to be negative. However, let's verify whether in this case, the intersection points are valid.First, for p=3/4: parabola is y²=2*(3/4)x = (3/2)x. Then, intersection with hyperbola x² - y²=1 gives x² - (3/2)x=1 => x² - (3/2)x -1=0. Multiply by 2: 2x² -3x -2=0. Solving: x=(3±√(9+16))/4=(3±5)/4. So x=(8)/4=2 or x=(-2)/4=-1/2. But when p=3/4, parabola is opening to the right, so x must be non-negative. So x=2 is valid, and x=-1/2 is invalid. Therefore, only x=2 is valid. Therefore, points M and N are (2, ±√(3)).Similarly, for p=-15/8: parabola is y²=2*(-15/8)x= (-15/4)x. Therefore, x must be ≤0. The intersection with hyperbola x² - y²=1: x² - (-15/4 x)=1 => x² + (15/4)x -1=0. Multiply by 4: 4x² +15x -4=0. Solutions: x=(-15 ±√(225 +64))/8=(-15±√289)/8=(-15±17)/8. So x=(2)/8=1/4 or x=(-32)/8=-4. Since p is negative, x must be ≤0, so x=-4 is valid, x=1/4 is invalid. So intersection points are (-4, y), with y²=2*(-15/8)*(-4)=2*(15/8)*4= (30/8)*4=120/8=15. So y=±√15. Therefore, points M and N would be (-4, ±√15). Then, the circumcircle of triangle OMN (with points O(0,0), M(-4, √15), N(-4, -√15)) would pass through P(7/2, 0). Let's check.But wait, in our previous steps, when we derived x=-1/4, we found p=-15/8. But in this case, the valid x is x=-4. So there seems to be a discrepancy. Wait, this suggests that my previous substitution might have an error.Wait, let me retrace. Earlier, when we substituted the hyperbola equation into the circle equation, we arrived at 4x² -7x -2=0, leading to x=2 and x=-1/4. However, when we check x=2, p=3/4, which is valid, but when x=-1/4, we get p=-15/8, but in reality, when p=-15/8, the intersection x is -4, not -1/4. Therefore, there's a mistake here.So, what's the problem?Let me check the substitution step again.We had the circle equation: x² + y² - (7/2)x =0.For point M(x, y) on the hyperbola C₁: x² - y²=1 => y²=x² -1.Substituting into the circle equation:x² + (x² -1) - (7/2)x =0Which simplifies to:2x² - (7/2)x -1=0Multiply both sides by 2: 4x² -7x -2=0Solutions x=(7±√(49 +32))/8=(7±9)/8=2 or -1/4. So, x=2 and x=-1/4. But when x=-1/4, let's see if this is on the hyperbola and parabola.For x=-1/4, hyperbola gives y²=x² -1=1/16 -1= -15/16, which is impossible. Therefore, x=-1/4 is not a valid solution. Therefore, only x=2 is valid.Therefore, the only valid solution is x=2, leading to p=3/4. So the equation of C₂ is y²=2*(3/4)x= (3/2)x. Therefore, the answer would be y²= (3/2)x.But wait, but in that case, when p=3/4, the parabola is y²=(3/2)x, and the hyperbola x² - y²=1 intersect at x=2, y=±√3. Then, the circumcircle of triangle OMN, which is O(0,0), M(2,√3), N(2,-√3). Let's check if the circumcircle passes through P(7/2,0).The equation of the circle was found to be x² + y² - (7/2)x=0. Plugging in P(7/2,0):(7/2)^2 +0^2 - (7/2)*(7/2)= (49/4) - (49/4)=0. So yes, it satisfies. Therefore, P is on the circle.Therefore, the equation of C₂ is y²=2px with p=3/4, so y²= (3/2)x. Therefore, the answer is y²= (3/2)x, or written as y²= 3x/2. But perhaps rationalized as 2y²=3x. However, the problem says "the equation of curve C₂", and C₂ is given as y²=2px. So substituting p=3/4, the equation is y²=2*(3/4)x= (3/2)x. So the answer is y²= (3/2)x, which can be written as y²= frac{3}{2}x. In boxed form, that's boxed{y^2 = dfrac{3}{2}x}.But let's check the other possibility when p=-15/8. Even though in our previous steps, x=-1/4 led to inconsistency, maybe there's another way. Wait, if p=-15/8, then the parabola is y²=2*(-15/8)x= (-15/4)x. Therefore, x must be negative. Then, solving C₁ and C₂:x² - y²=1 and y²= (-15/4)x. Substitute:x² - (-15/4)x=1 => x² + (15/4)x -1=0. Multiply by 4: 4x² +15x -4=0.Solutions: x=(-15 ±√(225 +64))/8=(-15±17)/8. So x=(2)/8=1/4 or x=(-32)/8=-4. Since x must be negative, only x=-4 is valid. Then y²= (-15/4)*(-4)=15. So y=±√15. Therefore, points M(-4,√15) and N(-4,-√15). Then, the circumcircle of triangle OMN.Let's find the equation of the circle passing through O(0,0), M(-4,√15), N(-4,-√15), and check if P(7/2,0) is on it.Using the general circle equation x² + y² + Dx + Ey + F=0. Since O is on the circle, F=0. So equation becomes x² + y² + Dx + Ey=0.Plugging in M(-4,√15):(-4)^2 + (√15)^2 + D*(-4) + E*(√15)=0 =>16 +15 -4D + E√15=0 =>31 -4D + E√15=0 --- (1)Plugging in N(-4,-√15):(-4)^2 + (-√15)^2 + D*(-4) + E*(-√15)=0 =>16 +15 -4D - E√15=0 =>31 -4D - E√15=0 --- (2)Subtracting equation (2) from equation (1):[31 -4D + E√15] - [31 -4D - E√15] =0 => 2E√15=0 => E=0.Substituting E=0 into equation (1):31 -4D=0 => D=31/4.Therefore, the equation of the circle is x² + y² + (31/4)x=0.Now, check if P(7/2,0) is on this circle:(7/2)^2 +0^2 + (31/4)*(7/2)= 49/4 + (217)/8= (98/8 +217/8)=315/8≈39.375 ≠0. Therefore, P is not on this circle.Therefore, p=-15/8 is invalid. Therefore, the only valid solution is p=3/4, leading to equation y²=(3/2)x.Therefore, the answer is boxed{y^2 = dfrac{3}{2}x}.Final Answerboxed{y^2 = dfrac{3}{2}x}</think>Given the curves ( C_1: x^2 - y^2 = 1 ) and ( C_2: y^2 = 2px ) intersecting at points ( M ) and ( N ), we need to find the equation of ( C_2 ) such that the circumcircle of ( triangle OMN ) passes through the point ( Pleft(frac{7}{2}, 0right) ).1. Find intersection points ( M ) and ( N ): - Substitute ( y^2 = 2px ) into ( x^2 - y^2 = 1 ): [ x^2 - 2px = 1 implies x^2 - 2px - 1 = 0 ] - Solving the quadratic equation: [ x = frac{2p pm sqrt{4p^2 + 4}}{2} = p pm sqrt{p^2 + 1} ] - For valid ( x ) values, ( 2px geq 0 ). If ( p > 0 ), ( x = p + sqrt{p^2 + 1} ) is valid. If ( p < 0 ), ( x = p - sqrt{p^2 + 1} ) is valid.2. Determine the circumcircle of ( triangle OMN ): - Points ( M ) and ( N ) are symmetric about the x-axis, so the triangle ( OMN ) is isosceles. - The general equation of the circle passing through ( O(0,0) ), ( M(x, y) ), and ( N(x, -y) ) is: [ x^2 + y^2 + Dx + Ey = 0 ] - Since ( E = 0 ) (due to symmetry), the equation simplifies to: [ x^2 + y^2 + Dx = 0 ]3. Incorporate point ( Pleft(frac{7}{2}, 0right) ): - Substituting ( Pleft(frac{7}{2}, 0right) ) into the circle equation: [ left(frac{7}{2}right)^2 + 0^2 + D left(frac{7}{2}right) = 0 implies frac{49}{4} + frac{7D}{2} = 0 implies 49 + 14D = 0 implies D = -frac{7}{2} ] - The circle equation becomes: [ x^2 + y^2 - frac{7}{2}x = 0 ]4. Verify intersection points: - For ( M(x, y) ) on both ( C_1 ) and ( C_2 ): [ x^2 - y^2 = 1 implies y^2 = x^2 - 1 ] - Substituting ( y^2 ) into the circle equation: [ x^2 + (x^2 - 1) - frac{7}{2}x = 0 implies 2x^2 - frac{7}{2}x - 1 = 0 implies 4x^2 - 7x - 2 = 0 ] - Solving for ( x ): [ x = frac{7 pm 9}{8} implies x = 2 text{ (valid)}, quad x = -frac{1}{4} text{ (invalid)} ] - For ( x = 2 ), ( y^2 = 3 ) and ( p = frac{3}{4} ).Thus, the equation of curve ( C_2 ) is (boxed{y^2 = dfrac{3}{2}x}).]

🔑:8. y^{2}=frac{3}{2} x.As shown in Figure 2, let points Mleft(x_{0}, y_{0}right), Nleft(x_{0},-y_{0}right), and T be the intersection of M N with the x-axis. Then Tleft(x_{0}, 0right).Since points O, M, P, N are concyclic, by the intersecting chords theorem, we havebegin{array}{l}O T cdot T P=M T cdot T N Rightarrow x_{0}left(frac{7}{2}-x_{0}right)=y_{0}^{2}=2 p x_{0}left(x_{0}>0right) .end{array}Solving this, we get x_{0}=frac{7}{2}-2 p,y_{0}^{2}=2 p x_{0}=7 p-4 p^{2} text {. }Substituting into the equation of curve C_{1}, we getbegin{array}{l}left(frac{7}{2}-2 pright)^{2}-left(7 p-4 p^{2}right)=1 Rightarrow 32 p^{2}-84 p+45=0 .end{array}Solving this, we get p=frac{15}{8} (discard) or p=frac{3}{4}.Thus, the equation of curve C_{2} is y^{2}=frac{3}{2} x.