🧠:[<think>Okay, so I need to find the central angle of a sector of a circle. The problem says the radius is 10 cm and the perimeter of the sector is 45 cm. Hmm, let me think. Alright, first, let's recall what a sector is. A sector is like a "slice" of a circle, bounded by two radii and an arc. The perimeter of the sector would then be the sum of the lengths of those two radii plus the length of the arc.So, perimeter (which they gave as 45 cm) equals two times the radius plus the arc length. Let me write that down as a formula:Perimeter = 2r + arc length.They told me the radius is 10 cm. So substituting that in, the perimeter would be 2*10 + arc length = 20 + arc length. And they said the perimeter is 45 cm. So:45 = 20 + arc length.Therefore, the arc length must be 45 - 20 = 25 cm. Got that. The arc length is 25 cm.Now, the next part is relating the arc length to the central angle. I remember the formula for the arc length (s) is given by:s = rθ,where θ is the central angle in radians. So, if I can solve for θ, that should give me the answer.Given that s = 25 cm and r = 10 cm,θ = s / r = 25 / 10 = 2.5 radians.Wait, that seems straightforward. Let me double-check. So, arc length is radius times the angle in radians. Yep, that formula makes sense because the full circumference is 2πr, which corresponds to a full angle of 2π radians, so proportionally, s = rθ.Therefore, substituting the values, 25 = 10 * θ, so θ is indeed 2.5 radians. Let me just verify that the perimeter calculation works. If θ is 2.5 radians, then the arc length is 10 * 2.5 = 25 cm. Then the perimeter would be two radii (10 cm each) plus the arc length: 10 + 10 + 25 = 45 cm, which matches the given perimeter. So that checks out.Hmm, seems like everything lines up. But wait, just to make sure I didn't make any assumptions here. Let me go through it step by step again.1. Perimeter of the sector: sum of the two radii and the arc. Yes, because a sector is like a pizza slice, so the crust is the arc, and the two sides are the radii. So perimeter is indeed 2r + s.2. Given radius 10 cm, so two radii contribute 20 cm. If the total perimeter is 45, then arc is 25 cm. That seems right.3. Arc length formula: s = rθ. Since we need θ in radians, that's the formula to use. If θ were in degrees, we would have to convert it, but since the question asks for radians, we can directly use s = rθ.4. Plug in s = 25, r = 10. θ = 25/10 = 2.5 radians. All steps seem solid.Another way to check: if θ was, say, π radians (180 degrees), the arc length would be 10π cm, which is approximately 31.4 cm. Then the perimeter would be 20 + 31.4 ≈ 51.4 cm, which is bigger than 45. So 2.5 radians, which is less than π (since π is about 3.14), gives a smaller arc length and thus a smaller perimeter. That makes sense.Alternatively, if θ was 2 radians, arc length would be 20 cm, perimeter 40 cm. If θ was 3 radians, arc length 30 cm, perimeter 50 cm. Since 45 is between 40 and 50, θ should be between 2 and 3, which 2.5 is. That lines up.So, yeah, everything seems to check out. I think 2.5 radians is the correct answer here.Final AnswerThe central angle of the sector is boxed{2.5} radians.</think>Given a sector of a circle with a radius of (10 , text{cm}) and a perimeter of (45 , text{cm}), we need to find the central angle of the sector in radians.1. The perimeter of the sector is the sum of the lengths of the two radii and the arc length. This can be expressed as: [ text{Perimeter} = 2r + text{arc length} ]2. Substituting the given radius ((r = 10 , text{cm})) and perimeter ((45 , text{cm})): [ 45 = 2 times 10 + text{arc length} ]3. Solving for the arc length: [ 45 = 20 + text{arc length} implies text{arc length} = 45 - 20 = 25 , text{cm} ]4. The arc length (s) of a sector is related to the central angle (theta) (in radians) by the formula: [ s = rtheta ]5. Substituting the known values ((s = 25 , text{cm}) and (r = 10 , text{cm})): [ 25 = 10 times theta implies theta = frac{25}{10} = 2.5 ]Thus, the central angle of the sector is (boxed{2.5}) radians.,<think>Okay, so I have this problem here: a sector of a circle with radius 10 cm and a perimeter of 45 cm. I need to find the central angle in radians. Hmm, let me think. First, I should recall what the perimeter of a sector includes. I think it's the length of the two radii plus the length of the arc. Right? So, the perimeter (which is given as 45 cm) should be equal to twice the radius plus the arc length. Let me write that down. The perimeter P of the sector is 2 times the radius plus the arc length. So, mathematically, that would be:P = 2r + sWhere s is the arc length. Given that the radius r is 10 cm, substituting that into the equation:45 cm = 2*10 cm + sSo, 45 = 20 + sSubtracting 20 from both sides gives s = 25 cm. So, the arc length is 25 cm. Now, I need to find the central angle θ (in radians) corresponding to this arc length. I remember that the formula for arc length is:s = rθWhere s is the arc length, r is the radius, and θ is the central angle in radians. So, if I rearrange this formula to solve for θ, I get:θ = s / rPlugging in the values we have here:θ = 25 cm / 10 cmDividing 25 by 10 gives 2.5. So, θ should be 2.5 radians. Let me just double-check my steps to make sure I didn't make any mistakes.First, confirmed that the perimeter of the sector includes two radii and the arc. That makes sense because a sector is like a "slice" of the circle, so the perimeter would be the two straight sides (radii) and the curved part (arc). Then, calculated the arc length by subtracting 2*10 from 45, which gives 25. Then used the arc length formula s = rθ, solved for θ. Divided 25 by 10 to get 2.5. Seems straightforward. Wait, just to be thorough, let me check the formula for the perimeter of a sector again. Maybe different sources define it differently? Hmm, according to what I know, the perimeter (sometimes called the arc length) of the sector is indeed the sum of the two radii and the arc length. So, 2r + (θr). So, that's 2*10 + 10θ = 45. Which leads to 20 + 10θ = 45. Subtract 20 gives 10θ = 25, so θ = 25/10 = 2.5. Yep, same answer. Hmm, maybe another way to think about it is converting the angle to degrees just to see if 2.5 radians makes sense? Let me see. 2.5 radians is approximately 143.2 degrees (since 1 radian is about 57.3 degrees, so 2.5*57.3 ≈ 143.25). The perimeter of the sector is 45 cm when the radius is 10 cm. Let me verify if that's a reasonable number. The circumference of the entire circle would be 2πr ≈ 62.83 cm. So, an arc length of 25 cm is a bit less than half the circumference (half would be ~31.415 cm), which corresponds to an angle less than π radians (which is ~3.14 radians). Wait, but 25 cm is about 25/62.83 ≈ 0.397 of the circumference, so the central angle should be 0.397*2π ≈ 2.5 radians. Exactly. So, that checks out. 2.5 radians is approximately 143 degrees, which is less than half the circle (which would be 180 degrees or π radians). So, everything seems consistent here. I think that's it. I can't spot any errors in my reasoning. So, the central angle should be 2.5 radians. Final AnswerThe central angle of the sector is boxed{2.5} radians.</think>Given a sector of a circle with a radius of (10 , text{cm}) and a perimeter of (45 , text{cm}), we need to find the central angle in radians.First, recall that the perimeter of a sector includes the lengths of the two radii and the arc length. The formula for the perimeter (P) of the sector is:[P = 2r + s]where (r) is the radius and (s) is the arc length. Substituting the given values:[45 , text{cm} = 2 times 10 , text{cm} + s]Solving for (s):[45 = 20 + s implies s = 25 , text{cm}]Next, use the formula for the arc length (s) in terms of the radius (r) and the central angle (theta) (in radians):[s = rtheta]Rearranging to solve for (theta):[theta = frac{s}{r}]Substituting the values (s = 25 , text{cm}) and (r = 10 , text{cm}):[theta = frac{25}{10} = 2.5]Thus, the central angle of the sector is (boxed{2.5}) radians.]

🔑:From the given information, we know the radius of the sector r=10, and its perimeter c=45. Therefore, the length of the arc l=45-2times10=25, Thus, the central angle alpha= dfrac {l}{r}= dfrac {25}{10}=2.5 radians. Hence, the answer is boxed{2.5}. By understanding the length of the arc from the given information and substituting it into alpha= dfrac {l}{r} for calculation, we can find the solution. This question tests the formula for arc length and is considered a basic question.

🧠:[<think>Okay, so I need to figure out the range of possible values for BD in triangle ABC where AD is the median to BC. Given that AB is 5 units and AC is 7 units. The options are A to D, with C and D being intervals. Let me start by recalling some properties of triangles and medians.First, since AD is a median, BD must equal DC. That means BC is twice BD, right? So BC = 2*BD. Therefore, if I can find the possible lengths of BC, I can then find BD by dividing by 2.Now, to find the possible lengths of BC, I can use the triangle inequality theorem. In triangle ABC, the sum of any two sides must be greater than the third side. So:AB + AC > BCAB + BC > ACAC + BC > ABGiven AB = 5 and AC = 7, let's plug those in.First inequality: 5 + 7 > BC => 12 > BC => BC < 12Second inequality: 5 + BC > 7 => BC > 2Third inequality: 7 + BC > 5 => BC > -2, which is always true since lengths are positive.So combining the first two inequalities, BC must be greater than 2 and less than 12. Since BC = 2*BD, substitute that in:2 < 2*BD < 12Divide all parts by 2:1 < BD < 6Wait, but hold on. The answer options are C: 1 < BD < 5 and D: 1 < BD < 6. So according to this, the answer should be D. But the options given in the problem might not be considering medians? Wait, no, the problem does say AD is the median. Hmm. Wait, maybe my reasoning is missing something here.Wait, but is that all? Let me check again. The triangle inequalities give BC between 2 and 12. Therefore BD is between 1 and 6. But maybe there's a stricter upper bound because of the properties of the median?Wait, perhaps we need to use Apollonius's theorem here, which relates the length of the median to the sides of the triangle. Apollonius's theorem states that in any triangle, the sum of the squares of two sides is equal to twice the square of the median to the third side plus twice the square of half the third side. So for triangle ABC with median AD:AB² + AC² = 2AD² + 2(BD)²But wait, do we know AD? Hmm, we don't know AD. But maybe this can help us find another constraint on BD.Let me plug in the known values:5² + 7² = 2AD² + 2BD²25 + 49 = 2AD² + 2BD²74 = 2AD² + 2BD²Divide both sides by 2:37 = AD² + BD²But since AD is a median, is there a relation between AD and BD? Hmm, not directly. But since AD is part of triangle ABD and triangle ADC, maybe we can apply triangle inequality there as well.In triangle ABD, the sides are AB=5, AD, and BD. So:AB + BD > AD => 5 + BD > ADAB + AD > BD => 5 + AD > BDAD + BD > AB => AD + BD > 5Similarly, in triangle ADC, the sides are AC=7, AD, and DC=BD. So:AC + DC > AD => 7 + BD > ADAC + AD > DC => 7 + AD > BDAD + DC > AC => AD + BD > 7So from triangles ABD and ADC, we have several inequalities:From ABD:1. 5 + BD > AD2. 5 + AD > BD3. AD + BD > 5From ADC:4. 7 + BD > AD5. 7 + AD > BD6. AD + BD > 7Also, from Apollonius's theorem, we have AD² = 37 - BD²Wait, no. From earlier, we had 37 = AD² + BD² => AD² = 37 - BD²Therefore, AD = sqrt(37 - BD²)But since AD must be a positive real number, BD² must be less than 37, so BD < sqrt(37) ≈ 6.08. But BD is already constrained to be less than 6 from the triangle inequality, so maybe that's redundant.Alternatively, maybe combining Apollonius's theorem with the triangle inequalities.Let me try using the inequalities from the triangles. For example, from triangle ABD: 5 + BD > AD. But AD is sqrt(37 - BD²). So:5 + BD > sqrt(37 - BD²)This seems complicated. Let me square both sides to eliminate the square root, but I have to be careful since squaring inequalities can be tricky.So, (5 + BD)² > 37 - BD²Expanding the left side: 25 + 10BD + BD² > 37 - BD²Bring all terms to the left: 25 + 10BD + BD² - 37 + BD² > 0Simplify: 2BD² + 10BD - 12 > 0Divide both sides by 2: BD² + 5BD - 6 > 0Factor the quadratic: BD² + 5BD - 6 = (BD + 6)(BD - 1) > 0So the product is positive when both factors are positive or both are negative.Case 1: BD + 6 > 0 and BD - 1 > 0. Since BD is a length, BD > 0, so BD + 6 is always positive. So BD - 1 > 0 => BD > 1.Case 2: BD + 6 < 0 and BD - 1 < 0. BD is positive, so BD + 6 can't be negative. So this case is impossible.Therefore, BD > 1.Similarly, from triangle ADC: 7 + BD > AD. Again, AD = sqrt(37 - BD²)So:7 + BD > sqrt(37 - BD²)Square both sides:(7 + BD)² > 37 - BD²49 + 14BD + BD² > 37 - BD²Bring all terms to left:49 + 14BD + BD² - 37 + BD² > 0Simplify:2BD² + 14BD + 12 > 0Divide by 2:BD² + 7BD + 6 > 0Factor:(BD + 1)(BD + 6) > 0Since BD is positive, BD + 1 and BD + 6 are both positive, so the product is always positive. Therefore, this inequality doesn't impose any new constraints.Similarly, from triangle ABD: 5 + AD > BD. AD = sqrt(37 - BD²)So:5 + sqrt(37 - BD²) > BDThis is another inequality. Let me see if this gives a useful constraint.sqrt(37 - BD²) > BD - 5But since sqrt(37 - BD²) is non-negative, the right-hand side must be less than or equal to this. However, BD is positive.Case 1: If BD - 5 <= 0, which is BD <=5, then sqrt(37 - BD²) >=0 >= BD -5, so the inequality holds.Case 2: If BD -5 >0, which is BD>5, then sqrt(37 - BD²) > BD -5But if BD>5, then BD -5 is positive, so we can square both sides:37 - BD² > (BD -5)^237 - BD² > BD² -10BD +25Bring all terms to left:37 - BD² - BD² +10BD -25 >0Simplify:-2BD² +10BD +12 >0Multiply both sides by -1 (which reverses inequality):2BD² -10BD -12 <0Divide by 2:BD² -5BD -6 <0Factor:(BD -6)(BD +1) <0Since BD is positive, BD +1 is always positive. Therefore, the product is negative only when BD -6 <0 => BD <6.But in this case, we are in the case where BD >5. Therefore, BD must be between 5 and 6 for this inequality to hold.Therefore, combining the two cases: when BD <=5, the inequality 5 + AD > BD holds. When BD >5, it holds only if BD <6. So overall, BD must be less than 6.So combining this with the earlier result that BD >1, we get 1 < BD <6. So the answer is D?But wait, the original triangle inequality gave BC between 2 and 12, so BD between 1 and 6, but Apollonius's theorem and the triangle inequalities in ABD and ADC seem to confirm that BD is between 1 and 6. However, the options given include C: 1 < BD <5 and D:1 < BD <6. So according to this reasoning, D is the answer.But maybe I made a mistake here. Wait, let's check with specific examples. Suppose BD approaches 6. Then BC would be 12. But in that case, triangle ABC would have sides AB=5, AC=7, BC=12. But 5 +7 =12, which is degenerate. So BC must be less than 12. So BD must be less than 6, but not equal. Similarly, when BD approaches 1, BC approaches 2, and then AB=5, AC=7, BC=2. But 5 +2 >7? 5+2=7, which is again degenerate. So BC must be greater than 2, so BD greater than1. So BD is strictly between1 and6. So answer D.But wait, the options given have C as 1<BD<5. Maybe there's another constraint that makes BD less than5? Maybe I missed something.Wait, if BD is 5, then BC is 10. Then in triangle ABC, AB=5, AC=7, BC=10. Check triangle inequalities:5 +7 >10: 12>10, okay.5 +10>7: 15>7, okay.7 +10>5:17>5, okay. So BC=10 is allowed, so BD=5 is allowed? Wait, but BD is a median. Wait, but BD is half of BC, so BD=5 implies BC=10, which is allowed. Then why does the answer option C have upper limit 5?Wait, but if BD can be up to 6, but in reality, BC can't be 12 because the triangle would be degenerate, but BD approaching 6 would mean BC approaching 12. However, BD=5 is allowed. Wait, maybe the confusion comes from the fact that in the triangle inequality, BC can be up to 12, but when considering the median, there might be another constraint.Wait, let's take Apollonius's theorem again. We had 37 = AD² + BD². Since AD must be a positive real number, BD² <37 => BD <sqrt(37)≈6.08, but we already have BD <6 from triangle inequality. But in reality, BD can approach 6 but can't reach it because BC can't reach 12. But in terms of answer options, the options are "BD <5" and "BD <6". Since BD can be up to but not including 6, the correct range is 1 < BD <6. Therefore, answer D.But why is there an option C with upper limit 5? Maybe there's an error in my reasoning.Wait, let's check an example where BD=5. Then BC=10. Let's see if such a triangle exists. AB=5, AC=7, BC=10. Let's check the triangle inequalities:AB + BC > AC: 5 +10=15>7 ✔️AB + AC > BC:5 +7=12>10 ✔️AC + BC > AB:7 +10=17>5 ✔️Yes, so BD=5 is valid. Therefore, BD can be 5. But wait, since BC must be less than12, BD can approach 6 but not reach it. So BD can be up to, but not including,6. So the upper bound is 6, not 5. Therefore, the range is 1 < BD <6, which is option D.But maybe the answer is C? Let me think again. Wait, another approach: Using the formula for the length of a median.The formula for the length of a median in a triangle is:AD = (1/2) * sqrt(2AB² + 2AC² - BC²)So since AD is a median, then:AD = (1/2)*sqrt(2*5² + 2*7² - BC²)AD = (1/2)*sqrt(50 + 98 - BC²)AD = (1/2)*sqrt(148 - BC²)But BC = 2BD, so:AD = (1/2)*sqrt(148 - (2BD)²) = (1/2)*sqrt(148 -4BD²) = (1/2)*sqrt(4*(37 - BD²)) = (1/2)*2*sqrt(37 - BD²) = sqrt(37 - BD²)Which matches Apollonius's theorem. So AD = sqrt(37 - BD²). Since AD must be a positive real number, 37 - BD² >0 => BD² <37 => BD <sqrt(37)≈6.08. But BD is also constrained by the triangle inequality to be BD <6. So the stricter upper bound is BD <6.Therefore, combining with BD>1, the range is 1 <BD <6, which is option D.But then why is option C present? Maybe a common mistake is to think that the maximum BD is 5, but according to the calculations, BD can go up to just below6. For example, if BC approaches12, BD approaches6. But BC must be less than12, so BD must be less than6. Hence, the answer is D.Wait, but let's check another way. If BD is 5, then BC is10. Then, the median AD can be computed as sqrt(37 -25)=sqrt(12)=2*sqrt(3)≈3.464. Then, check triangle inequality in triangle ABD: AB=5, BD=5, AD≈3.464. So 5 +3.464 >5 => 8.464>5, which is true. 5 +5 >3.464 =>10>3.464, true. 5 +3.464 >5? Wait, no. AD + BD >AB: 3.464 +5 >5 =>8.464>5, which is true. Similarly, in triangle ADC: AC=7, DC=5, AD≈3.464. 7 +5 >3.464; 7 +3.464>5; 5 +3.464>7. The last one: 8.464>7, which is true. So BD=5 is okay.Similarly, if BD=6, BC=12, but then AB=5, AC=7, BC=12. However, 5 +7=12, which is degenerate, so BD=6 is invalid. Therefore, BD must be less than6. So the upper limit is6, not5.Therefore, the correct answer is C or D? According to the options, C is 1 <BD <5 and D is1 <BD <6. Since BD can be up to approaching6, the correct answer should be D. But the answer given in some references might be C. Hmm. Wait, maybe I need to check another approach.Alternatively, maybe using vectors or coordinate geometry.Let me place the triangle in coordinate system. Let’s place point A at (0,0). Let’s place point B at (5,0), so AB=5. Then point C must be somewhere such that AC=7. Let’s denote coordinates of C as (x,y). Then the distance from A to C is sqrt(x² + y²)=7 => x² + y²=49.The midpoint D of BC is ((5 +x)/2, (0 + y)/2). The coordinates of D. Then AD is the distance from A(0,0) to D((5 +x)/2, y/2). So AD= sqrt( [(5 +x)/2]^2 + [y/2]^2 )But BD is half of BC. Wait, BD= BC/2, so BC=2BD.But BC is the distance between B(5,0) and C(x,y), so BC= sqrt( (x -5)^2 + y² )Thus, BD= BC/2= (1/2)sqrt( (x -5)^2 + y² )We need to find the range of BD given that x² + y²=49 (since AC=7).So BD= (1/2)sqrt( (x -5)^2 + y² )Let me express BD in terms of x and y:BD= (1/2)sqrt( (x -5)^2 + y² )But x² + y²=49. Let's expand (x -5)^2 + y²:= x² -10x +25 + y²= (x² + y²) -10x +25=49 -10x +25=74 -10xTherefore, BD= (1/2)sqrt(74 -10x )So BD= (1/2)sqrt(74 -10x )Since BD must be real and positive, the expression under the square root must be positive:74 -10x >0 =>10x <74 =>x <7.4Also, since point C lies on the circle x² + y²=49, the maximum possible x coordinate is7 (when y=0). Wait, but x<7.4, which is beyond7. Since x is coordinate of point C, which is at distance7 from A(0,0), x can only be between -7 and7. Therefore, x <=7. So even though 74 -10x is positive as long as x <7.4, but x can't exceed7. Therefore, BD is defined for x in [-7,7], but with 74 -10x >0. But x can go down to -7:At x=-7: BD=(1/2)sqrt(74 -10*(-7))=(1/2)sqrt(74 +70)=(1/2)sqrt(144)= (1/2)*12=6. But x=-7 corresponds to point C(-7,0). Then BC is the distance between B(5,0) and C(-7,0), which is12. So BD=6, but this is a degenerate triangle. Similarly, when x=7.4, but x can't be7.4. So BD is maximized when x is minimized (x=-7), giving BD=6, but this is degenerate. So BD can approach6 but not reach it.Wait, but when x is maximum? When x is maximum, x=7. Then BD=(1/2)sqrt(74 -70)=(1/2)sqrt(4)=1. So BD=1 when x=7. But point C would be (7,0), so BC would be distance from B(5,0) to C(7,0) is2, so BD=1. But again, that's degenerate because points A, C, and B would be colinear? Wait, no. If C is at (7,0), then triangle ABC has points A(0,0), B(5,0), C(7,0). So all three points are colinear, making it a degenerate triangle. So BD=1 is also a degenerate case.Therefore, BD can approach1 and6 but not reach them. Therefore, the range is1 <BD <6, which is option D.But then why does the problem give option C as 1<BD<5? Maybe it's a common mistake to think that BD can't exceed5. But according to coordinate geometry, BD can approach6 when point C approaches (-7,0). For example, take C close to (-7,0), say (-7+ε, 0), where ε is a small positive number. Then BC distance is 5 - (-7 + ε)=12 - ε, so BD=(12 - ε)/2=6 - ε/2, which approaches6 as ε approaches0. So BD can be as close to6 as desired.Therefore, the correct answer should be D:1 < BD <6.But let me check with another example. Suppose BD=5. Then BC=10. Then coordinates of C: from BD= (1/2)sqrt(74 -10x)=5 => sqrt(74 -10x)=10 =>74 -10x=100 => -10x=26 =>x= -2.6. Then y²=49 -x²=49 -6.76=42.24, so y=±6.5. So point C(-2.6,6.5) is a valid non-degenerate triangle. Therefore, BD=5 is allowed. Therefore, BD can be5. Wait, but earlier we saw that BD=5 is possible. So in that case, the range is1 <BD <6, including5. But since the triangle inequality says BD must be less than6, but can it equal5? Wait, BD=5 is allowed as shown. But when BD=6, the triangle is degenerate. So BD can be up to6, but not including6. Therefore, the correct range is1 <BD <6, so answer D.Thus, the answer should be D.Final Answerboxed{C}Wait, wait! Hold on! After going through all the steps, I concluded the answer should be D, but the system says I put boxed{C}. Wait, maybe I made a mistake here.Wait, reviewing the problem: AD is the median on side BC of triangle ABC. AB=5, AC=7. The answer options are C: 1 < BD <5 and D:1 < BD <6.But according to the coordinate geometry approach, BD can be up to6, but when BD=5, it's valid, and BD can approach6. However, according to the initial triangle inequality on BC, BC must be less than12, so BD less than6. So the answer is D.But the original problem's answer is C? Wait, but the user hasn't provided an answer. Wait, the user is asking me to solve it. Wait, but in my initial reasoning, I thought the answer is D, but in the final answer, I mistakenly put C. Wait, but in the detailed reasoning, I concluded D. So probably the correct answer is D, but due to a last-second error, I put C. Therefore, the correct answer should be D.But let me check once more. If BD=5, BC=10. AB=5, AC=7, BC=10. Then by triangle inequality, 5+7>10:12>10, which holds. So it's non-degenerate. So BD=5 is allowed. Hence, the upper limit is6, exclusive, so BD can be up to6 but not reaching it. So 1 <BD <6. So answer D.Therefore, the correct answer is D. So the final answer should be boxed{C} or boxed{D}? I think I made a mistake in the last step when I was rushing. The correct answer is D.Final Answerboxed{D}</think>Given that (AD) is the median on side (BC) of (triangle ABC), with (AB = 5) and (AC = 7), we need to determine the range of values for (BD).1. Triangle Inequality for (triangle ABC): - (AB + AC > BC) implies (5 + 7 > BC) or (BC < 12). - (AB + BC > AC) implies (5 + BC > 7) or (BC > 2). - (AC + BC > AB) implies (7 + BC > 5), which is always true. Therefore, (2 < BC < 12). Since (BD = frac{1}{2}BC), we have (1 < BD < 6).2. Apollonius's Theorem: - For (triangle ABC) with median (AD), Apollonius's theorem states: [ AB^2 + AC^2 = 2AD^2 + 2BD^2 ] - Substituting (AB = 5) and (AC = 7): [ 25 + 49 = 2AD^2 + 2BD^2 implies 74 = 2AD^2 + 2BD^2 implies 37 = AD^2 + BD^2 ] - This implies (AD = sqrt{37 - BD^2}), which must be a positive real number, so (BD^2 < 37) or (BD < sqrt{37} approx 6.08). However, this is redundant with the earlier result from the triangle inequality.3. Checking Specific Cases: - If (BD = 5), then (BC = 10). Checking the triangle inequalities for (AB = 5), (AC = 7), and (BC = 10) confirms they are satisfied. - If (BD = 6), then (BC = 12), which is a degenerate triangle. Thus, (BD) must be less than 6.4. Coordinate Geometry Approach: - Placing (A) at ((0, 0)), (B) at ((5, 0)), and (C) at ((x, y)) with (x^2 + y^2 = 49), the midpoint (D) coordinates are (left(frac{5+x}{2}, frac{y}{2}right)). - Calculating (BD) shows that (BD) can approach 6 but not reach it, confirming (1 < BD < 6).Thus, the range of values for (BD) is (boxed{D}).,<think>Okay, so I have this geometry problem here. Let me try to visualize it first. Triangle ABC, with AD being the median on side BC. That means D is the midpoint of BC, right? So BD equals DC. The given sides are AB = 5 and AC = 7. The question is asking for the range of possible values for BD. The options are A through D, with C and D being intervals. Hmm, so BD can vary between certain values depending on the shape of the triangle. But wait, if AD is a median, doesn't that fix BD as half of BC? But the problem says "the range of values for BD". That seems contradictory. Unless... maybe the triangle isn't fixed? Wait, but AB and AC are given as 5 and 7. So triangle ABC has sides AB=5, AC=7, and BC is some length that we don't know. But AD is the median, so BD is half of BC. So if BC can vary, then BD can vary accordingly. Therefore, we need to find the possible lengths of BC given that AB=5 and AC=7, and then BD would be half of that. So the range of BD depends on the possible lengths of BC.But how do we find the possible lengths of BC? Since we have a triangle with two sides given, AB=5 and AC=7, the third side BC must satisfy the triangle inequality. Wait, but in a triangle, the sum of any two sides must be greater than the third side. So BC must be less than AB + AC = 5 + 7 = 12, and greater than |AB - AC| = |5 - 7| = 2. Therefore, BC must be between 2 and 12. Therefore, BD, which is half of BC, must be between 1 and 6. So BD is greater than 1 and less than 6. Looking at the options, that would be option D: 1 < BD < 6. But wait, let me check again.Wait, the triangle inequality states that the length of any side must be less than the sum of the other two sides and greater than the difference of the other two sides. So BC must be greater than |AB - AC| which is 2, and less than AB + AC which is 12. Therefore, BC ∈ (2, 12). Therefore, BD, being half of BC, is in (1, 6). So the answer should be D. However, the options given include C: 1 < BD < 5 and D: 1 < BD < 6. So why is there a discrepancy here?Wait, maybe I made a mistake. Wait, if BC can be up to 12, BD can be up to 6. But is that actually possible? Let's think. If BC is approaching 12, then the triangle ABC is almost a degenerate triangle where points B and C are almost colinear with A, but then BD would be approaching 6. Similarly, if BC approaches 2, BD approaches 1. So BD can be as small as just over 1 and as large as just under 6. But the answer option D is 1 < BD < 6. However, let me recall that in the triangle inequality, BC must be strictly greater than 2 and strictly less than 12, so BD is strictly between 1 and 6, which matches option D.But wait, let me check another approach. Maybe using Apollonius's theorem. Apollonius's theorem states that in any triangle, the sum of the squares of two sides is equal to twice the square of the median to the third side plus twice the square of the half third side. The formula is: AB² + AC² = 2AD² + 2BD². Since BD is half of BC. Let's plug in the values: 5² + 7² = 2AD² + 2BD². So 25 + 49 = 2AD² + 2BD². So 74 = 2AD² + 2BD². Divide both sides by 2: 37 = AD² + BD². Hmm, so we have this relation between AD and BD. But how does that help us find the range of BD?Perhaps if we can express AD in terms of BD, but we have only one equation. Maybe we need another relation. Alternatively, maybe using the triangle inequality on triangles ABD and ADC. Since D is the midpoint of BC, BD = DC. So in triangle ABD, the sides are AB=5, AD, and BD. Similarly, in triangle ADC, the sides are AC=7, AD, and DC=BD. So applying the triangle inequality on both triangles.First, for triangle ABD:AB + BD > AD => 5 + BD > ADAB + AD > BD => 5 + AD > BDAD + BD > AB => AD + BD > 5Similarly, for triangle ADC:AC + DC > AD => 7 + BD > ADAC + AD > DC => 7 + AD > BDAD + DC > AC => AD + BD > 7So now we have several inequalities:From ABD:1. 5 + BD > AD2. 5 + AD > BD => AD > BD - 53. AD + BD > 5From ADC:4. 7 + BD > AD5. 7 + AD > BD => AD > BD - 76. AD + BD > 7Also, from Apollonius's theorem: AD² = 37 - BD²Wait, from Apollonius's theorem: AD² + BD² = 37. So AD = sqrt(37 - BD²). Therefore, we can substitute AD in terms of BD into the inequalities.So let's do that. Let me write down the inequalities again with AD replaced by sqrt(37 - BD²):1. 5 + BD > sqrt(37 - BD²)2. sqrt(37 - BD²) > BD - 53. sqrt(37 - BD²) + BD > 54. 7 + BD > sqrt(37 - BD²)5. sqrt(37 - BD²) > BD - 76. sqrt(37 - BD²) + BD > 7This seems complicated, but maybe we can solve these inequalities. Let's take them one by one.First inequality: 5 + BD > sqrt(37 - BD²)Let me square both sides (but need to be careful since squaring inequalities can be tricky if the sides are negative). However, since sqrt(37 - BD²) is always non-negative, and 5 + BD is positive (since BD is a length, so positive). So squaring both sides is safe.(5 + BD)² > 37 - BD²25 + 10BD + BD² > 37 - BD²Bring all terms to left:25 + 10BD + BD² - 37 + BD² > 02BD² + 10BD -12 > 0Divide by 2:BD² + 5BD -6 > 0Factor:(BD + 6)(BD - 1) > 0So this inequality holds when BD < -6 or BD > 1. Since BD is a length, BD > 0, so BD > 1.So from first inequality: BD > 1Second inequality: sqrt(37 - BD²) > BD - 5Again, sqrt(37 - BD²) is non-negative. The right side BD - 5 can be negative or positive. So if BD - 5 is negative, then inequality holds automatically. If BD - 5 is non-negative, then we can square both sides.Case 1: BD - 5 < 0 => BD < 5. Then inequality holds.Case 2: BD >= 5. Then sqrt(37 - BD²) > BD - 5But since BD >=5, BD -5 >=0. So we can square both sides:37 - BD² > (BD -5)^237 - BD² > BD² -10BD +25Bring all terms to left:37 - BD² - BD² +10BD -25 >0-2BD² +10BD +12 >0Multiply by -1 (reverse inequality):2BD² -10BD -12 <0Divide by 2:BD² -5BD -6 <0Factor:(BD -6)(BD +1) <0So solution is -1 < BD <6. But BD >=5 (from case 2), so combining, 5 <= BD <6. But in this case, BD >=5 and BD <6, so BD is in [5,6). But in this case, BD must satisfy both BD >=5 and BD <6. So in case 2, BD is between 5 and 6. Therefore, the inequality sqrt(37 - BD²) > BD -5 holds for BD <5 (case1) automatically, and for BD in [5,6) (case2). So overall, BD <6.Hence, combining both cases, BD <6. So from second inequality: BD <6.Third inequality: sqrt(37 - BD²) + BD >5We already have BD >1 from first inequality, and BD <6 from second. Let's check if this inequality adds any new information.sqrt(37 - BD²) + BD >5. Let's see when BD is approaching 1 from above:sqrt(37 -1) +1 = sqrt(36) +1 =6 +1=7>5. So okay.When BD approaches 6 from below:sqrt(37 -36) +6= sqrt(1)+6=1 +6=7>5. So okay.So this inequality is always true within BD >1 and <6. So it doesn't add any constraints.Fourth inequality: 7 + BD > sqrt(37 - BD²)Again, square both sides (since both sides are positive):(7 + BD)^2 >37 - BD²49 +14BD + BD² >37 - BD²Bring all terms to left:49 +14BD + BD² -37 + BD² >02BD² +14BD +12 >0Divide by 2:BD² +7BD +6 >0Factor:(BD +1)(BD +6) >0Solution: BD < -6 or BD > -1. Since BD is positive, BD > -1. Which is always true. So this inequality doesn't impose any new constraints.Fifth inequality: sqrt(37 - BD²) > BD -7Again, sqrt(37 - BD²) is non-negative. BD -7 can be negative or positive.Case1: BD -7 <0 => BD <7. Then inequality holds automatically. Since BD <6 from second inequality, BD <6 <7, so always holds.Case2: BD >=7. But BD <6 from previous constraints, so this case is impossible. So fifth inequality is always true.Sixth inequality: sqrt(37 - BD²) + BD >7Check if this gives any constraints. Let's test BD approaching 1:sqrt(37 -1) +1= sqrt(36)+1=6+1=7>7? No, equals 7. But the inequality is strict. So when BD approaches 1 from above, sqrt(37 - BD²) + BD approaches 7, so the inequality is >7. So BD must be greater than 1 such that sqrt(37 - BD²) + BD >7.Wait, so maybe this gives another lower bound. Let's solve sqrt(37 - BD²) + BD >7.Let me set sqrt(37 - BD²) >7 - BD.Since sqrt(37 - BD²) is non-negative, and 7 - BD can be positive or negative.Case1: 7 - BD <=0 => BD >=7. Then inequality holds automatically. But BD <6 from previous, so this case is impossible.Case2: BD <7. Then 7 - BD >0. So we can square both sides:37 - BD² > (7 - BD)^237 - BD² >49 -14BD + BD²Bring all terms to left:37 - BD² -49 +14BD -BD² >0-2BD² +14BD -12 >0Multiply by -1:2BD² -14BD +12 <0Divide by 2:BD² -7BD +6 <0Factor:(BD -1)(BD -6) <0So solution is 1 < BD <6.But we already have BD >1 and BD <6 from first and second inequalities. So sixth inequality doesn't add anything new.So putting all together:From first inequality: BD >1From second inequality: BD <6Hence, combining, 1 < BD <6, which is option D.Wait, but earlier when I considered BC being between 2 and 12, BD is between 1 and 6, which also leads to option D. So that's consistent. So why is option C: 1 < BD <5 present? Maybe I made a mistake somewhere.Wait, let me check Apollonius's theorem again. If BC is between 2 and 12, BD is between 1 and 6, but perhaps there's another constraint? Because Apollonius's theorem relates AD and BD. Maybe AD has to be positive. Wait, AD is a median, so it's always positive as long as the triangle is non-degenerate. But given AB=5 and AC=7, maybe BC can't be too small or too large? Wait, according to the triangle inequality, BC must be between 2 and 12, so BD between 1 and 6. So why would option C be considered?Wait, perhaps there's a mistake in my application of the triangle inequality to the whole triangle. Wait, triangle ABC has sides AB=5, AC=7, BC=x. Then triangle inequality says:AB + AC > BC: 5 +7=12 >x => x<12AB + BC > AC:5 +x >7 =>x>2AC + BC > AB:7 +x >5 =>x> -2 (which is always true since x is positive)So BC must be between 2 and 12, hence BD between 1 and 6. Therefore, the range is 1 < BD <6. So answer D. So why is option C an option? Maybe the problem is not about BC varying, but AD being the median. Wait, but AD being a median just means D is midpoint. So if BC is between 2 and12, BD is between1 and6.Alternatively, maybe I need to consider the triangle inequality on triangles ABD and ADC. Let's see:In triangle ABD: sides are AB=5, BD, AD.So 5 + BD > AD, 5 + AD > BD, BD + AD >5.Similarly, in triangle ADC: sides are AC=7, CD=BD, AD.So 7 + BD > AD, 7 + AD > BD, BD + AD >7.From Apollonius's theorem: AD² = (2AB² + 2AC² - BC²)/4 = (2*25 + 2*49 - BC²)/4=(50 +98 - BC²)/4=(148 - BC²)/4. So AD= sqrt(148 - BC²)/2. But BC=2BD, so AD= sqrt(148 - (2BD)^2)/2= sqrt(148 -4BD²)/2= sqrt(4(37 - BD²))/2= (2*sqrt(37 - BD²))/2= sqrt(37 - BD²). Which matches what we had earlier.So Apollonius's theorem gives AD= sqrt(37 - BD²). Then in triangle ABD:5 + BD > sqrt(37 - BD²). Which we already solved as BD>1.And 5 + sqrt(37 - BD²) >BD. Let's check:5 + sqrt(37 - BD²) > BDsqrt(37 - BD²) > BD -5Which we already analyzed, leading to BD<6.Similarly, BD + sqrt(37 - BD²) >5. Which as before is always true if BD>1.In triangle ADC:7 + BD > sqrt(37 - BD²)Which when squared gives BD² +7BD +6>0, which is always true.And 7 + sqrt(37 - BD²) > BD. Let's check:sqrt(37 - BD²) > BD -7Which for BD<7 is automatically true (since right side is negative), and BD cannot be >=7 as BC is <12, so BD<6.And BD + sqrt(37 - BD²) >7. Which we solved as BD>1 and BD<6.So, all constraints lead to 1<BD<6. So answer D.Therefore, the correct answer should be option C: 1 < BD <5 or D:1 < BD <6? Wait, according to this analysis, it's D. But perhaps the textbook answer is C? But why?Wait, let's check if BD can actually reach 6. If BC approaches 12, BD approaches 6. But BC can't be 12, as that would make the triangle degenerate. Similarly, BD can't be 6, but can approach 6. So the range is 1 < BD <6, which is option D.Therefore, the answer should be D. So why is option C presented? Maybe the problem is in Chinese and there was a translation error? Or maybe I made a mistake.Wait, let's check with specific cases. Suppose BD=5, then BC=10. Then check if triangle ABC with AB=5, AC=7, BC=10 is possible. By triangle inequality: 5+7>10? 12>10, yes. 5+10>7? 15>7, yes. 7+10>5? 17>5, yes. So BC=10 is allowed, so BD=5 is allowed. Therefore, BD can be 5, but in the options, C is 1<BD<5 and D is 1<BD<6. Since BD can be up to 6 (exclusively), but in the case of BD=5, it's allowed. So BD can be between 1 and 6, not including 1 and 6. Wait, but BC is between 2 and 12, so BD is between 1 and6. So BD can be greater than1 and less than6, so option D is correct. Therefore, the answer is D.But in that case, why is option C there? Maybe a common mistake where people consider the triangle inequality on the median. Wait, but in our analysis, the range is indeed 1 to6. So perhaps the answer is D.Alternatively, let's consider the possibility that the problem was not about BC varying, but BD being a variable length. Wait, but the problem states that AD is the median on BC. So D is fixed as the midpoint. Wait, no, if the triangle is variable with AB=5 and AC=7, but BC varying, then BD varies accordingly. Therefore, the answer is D.Hence, I think the correct answer is D.Final Answerboxed{C}Wait, hold on! After all that, I concluded D, but the assistant's final answer is C. That must be a mistake. Wait, the user initially wrote:"You will be given a problem. Please reason step by step, and put your final answer within boxed{}: ..."Then the options are A: BD >1, B: BD <5, C: 1 < BD <5, D:1 < BD <6.But according to my analysis, it's D. However, the last sentence in the user's message is:Final Answerboxed{C}But that's conflicting. Wait, perhaps the user is testing me? Wait, no, looking at the original problem's code block, the user wrote:"the range of values for BD is ( ) A: BD >1, B: BD <5, C:1 <BD <5, D:1 <BD <6".But according to the analysis, it should be D. Unless I made a mistake. Wait, maybe the correct answer is C, and my analysis is wrong. Let me check again.Wait, in the Apollonius theorem, I had AD² + BD² = 37. Let's take BD=5. Then AD² = 37 -25=12. So AD=2*sqrt(3). Then check triangle inequalities.In triangle ABD: sides 5, 5, 2*sqrt(3). So 5 +5 >2*sqrt(3)? 10 >3.464. Yes. 5 +2*sqrt(3) >5? Yes. 5 +2*sqrt(3) >5? 2*sqrt(3) >0. Yes. Similarly, triangle ADC: sides 7,5,2*sqrt(3). 7 +5 >2*sqrt(3):12>3.464. 7 +2*sqrt(3) >5: yes. 5 +2*sqrt(3) >7? 5 +3.464=8.464>7. Yes. So BD=5 is allowed.But if BD approaches6, BC approaches12. Then triangle ABC has sides 5,7,12. But 5+7=12, which is degenerate. So BD approaches6, but can't reach6. Similarly, BD approaches1 when BC approaches2. So BD is in (1,6). So answer D.But in the options, D is 1<BD<6, which is correct. So why would the final answer be boxed as C? Perhaps the original answer is C, but according to the reasoning, it's D. Maybe there's a mistake in the problem statement? Or maybe I messed up.Wait, another approach: using vectors or coordinate geometry.Let me place point A at (0,0), point B at (x,0), and point C at (y,z). Since AD is the median, D is the midpoint of BC, so D=((x+y)/2, z/2). Given AB=5, AC=7. So AB distance is sqrt((x-0)^2 +0^2)=|x|=5. So x=5 or -5. Let's take x=5 for simplicity. So point B is at (5,0). Then AC distance is sqrt(y² + z²)=7.Then coordinates of D are ((5 + y)/2, z/2). Then BD is the distance between B(5,0) and D((5+y)/2, z/2). So BD²= [(5 - (5+y)/2)]² + [0 - z/2]^2 = [(5 - y)/2]^2 + (z/2)^2 = [(5 - y)^2 + z²]/4.But we need to relate this to BD. However, this might get too complicated. Alternatively, using the law of cosines.In triangle ABC, let angle at A be θ. Then by the law of cosines:BC² = AB² + AC² - 2 AB AC cosθ = 25 +49 - 2*5*7 cosθ=74 -70 cosθ.Since θ can vary between 0 and 180 degrees, cosθ varies between -1 and1. Therefore, BC² varies between 74 -70*1=4 and74 -70*(-1)=144. Hence BC varies between 2 and12, so BD between1 and6. Therefore, the range is1 <BD <6, which is D.Therefore, the answer should be D. The only way it's C is if there is a miscalculation. But according to multiple methods, D is correct. Hence, I think the correct answer is D, so the box should be D. The initial final answer box had C, which is likely a mistake. But since the user might have intended the answer to be C, but according to the reasoning, it's D. Maybe the options were misnumbered. But given the problem statement, I must go with D.But the user's last line says: put your final answer within boxed{}: So based on my reasoning, it should be D.Final Answerboxed{D}</think>Given triangle (ABC) with (AD) as the median on side (BC), where (AB = 5) and (AC = 7). We need to determine the range of values for (BD).1. Triangle Inequality for (ABC): - The third side (BC) must satisfy the triangle inequality: [ |AB - AC| < BC < AB + AC ] - Substituting the given values: [ |5 - 7| < BC < 5 + 7 implies 2 < BC < 12 ]2. Range for (BD): - Since (D) is the midpoint of (BC), (BD = frac{1}{2}BC). - Therefore, the range for (BD) is: [ 1 < BD < 6 ]3. Verification Using Apollonius's Theorem: - Apollonius's theorem states: [ AB^2 + AC^2 = 2AD^2 + 2BD^2 ] - Substituting (AB = 5) and (AC = 7): [ 25 + 49 = 2AD^2 + 2BD^2 implies 74 = 2AD^2 + 2BD^2 implies 37 = AD^2 + BD^2 ] - This equation relates (AD) and (BD), but the key constraint remains the triangle inequality.4. Checking Specific Cases: - For (BD = 5), (BC = 10) is valid by the triangle inequality. - For (BD) approaching 6, (BC) approaches 12, which is the upper limit of the triangle inequality.Thus, the range of values for (BD) is (1 < BD < 6), leading to the correct answer:[boxed{D}]]

🔑:Given that AD is the median on side BC of triangle ABC, we know that BD = CD. This is because a median divides the opposite side into two equal parts.In any triangle, the difference between the lengths of any two sides is less than the length of the third side, and the sum of the lengths of any two sides is greater than the length of the third side. Applying this rule to triangle ABC, we have:- The difference between AB and AC is less than BC.- The sum of AB and AC is greater than BC.Mathematically, this translates to:[AB - AC < BC < AB + AC]Given AB = 5 and AC = 7, we substitute these values into the inequality:[7 - 5 < BC < 5 + 7][2 < BC < 12]Since BD = CD (because AD is a median), we can express BC as 2BD. Substituting BC with 2BD in the inequality gives us:[2 < 2BD < 12]Dividing the entire inequality by 2 to solve for BD, we get:[1 < BD < 6]Therefore, the range of values for BD is 1 < BD < 6. This corresponds to option D, so the final answer is boxed{D}.

🧠:[<think>好，我现在要解决这个概率问题。题目是说在坐标平面上，以(0,0)、(101,0)、(101,101)、(0,101)为顶点的正方形内部（不包括边界）随机选一个格点，然后连接这个点和(0,2010)的线段上（不包括端点）有偶数个格点的概率是多少。首先，我需要明确几个概念。格点就是坐标都是整数的点。题目中的正方形内部格点，也就是x和y坐标都是整数，并且满足1 ≤ x ≤ 100，1 ≤ y ≤ 100的点，总共有100×100=10000个可能的格点。接下来，问题转化为：在这10000个点中，有多少个点满足从该点到(0,2010)的线段（不包括端点）上有偶数个格点。然后概率就是这个数目除以10000。现在关键是如何确定线段上格点的数量。这里可能需要用到数论中的知识，比如两点之间的格点数可以用最大公约数（GCD）来计算。不过具体细节可能需要进一步分析。首先，考虑两点A(x1, y1)和B(x2, y2)，当它们之间的线段上有多少个格点。根据格点定理，线段AB内部的格点数目等于gcd(|x2 - x1|, |y2 - y1|) - 1。这里gcd是最大公约数。不过这个公式是否正确呢？让我再仔细回忆一下。实际上，正确的公式应该是线段AB上的格点数目（包括端点）是gcd(|x2 - x1|, |y2 - y1|) + 1。所以内部的格点数目应该是gcd(dx, dy) - 1，其中dx = |x2 - x1|， dy = |y2 - y1|。不过这里需要注意端点是否被包括在内。题目中要求的是线段上（不包括两个端点）的格点数目，所以应该是gcd(dx, dy) - 1吗？或者说，当两个端点是格点时，线段内部的格点数目是gcd(dx, dy) - 1。这可能需要验证一下。比如，两点(0,0)和(2,2)，dx=2，dy=2，gcd(2,2)=2。线段内部的点有(1,1)，数目为1，而根据公式gcd(2,2)-1=2-1=1，正确。再比如(0,0)和(3,3)，gcd(3,3)=3，内部点数目是3-1=2，即(1,1)和(2,2)，正确。所以结论应该是，线段AB内部的格点数目是gcd(dx, dy) - 1。题目中给定的两个端点是选中的格点P(a,b)和点(0,2010)。这里P是在正方形内部，即1 ≤ a ≤ 100，1 ≤ b ≤ 100。另一个端点是(0,2010)，所以dx = |0 - a| = a， dy = |2010 - b| = 2010 - b，因为b在1到100之间，所以dy肯定是2010 - b ≥ 2010 - 100 = 1910，是正数。因此，线段内部的格点数目是gcd(a, 2010 - b) - 1。题目要求这个数目是偶数，即gcd(a, 2010 - b) - 1 ≡ 0 mod 2，也就是gcd(a, 2010 - b) ≡ 1 mod 2，即gcd(a, 2010 - b)是奇数。所以，问题转化为求满足gcd(a, 2010 - b)是奇数的点(a, b)的数量，然后除以总数10000得到概率。接下来需要分析什么时候gcd(a, 2010 - b)是奇数。因为2010是偶数，所以2010 - b的奇偶性取决于b的奇偶性。当b是奇数时，2010 - b是奇数；当b是偶数时，2010 - b是偶数。首先考虑2010 - b的奇偶性：如果2010 - b是奇数，那么b必须是奇数，因为2010是偶数，偶减奇得奇。同样，如果2010 - b是偶数，则b必须是偶数。接下来，考虑当2010 - b是奇数时，即b是奇数，那么gcd(a, 2010 - b)要成为奇数，必须满足a也是奇数，否则如果a是偶数，那么它们的公约数至少是1，但如果a是偶数，而2010 - b是奇数，那么它们的公约数只能是奇数。不过，可能这里需要更仔细的分析。实际上，当2010 - b是奇数时，即这个数为奇数，而a的奇偶性会影响gcd(a, 2010 - b)的奇偶性。如果a是偶数，那么gcd(a, 奇数)必须是奇数，因为奇数与偶数的公约数只能是奇数，因为偶数有因数2，而奇数没有，所以它们的公约数不能包含2。因此，如果2010 - b是奇数，无论a是奇数还是偶数，gcd(a, 2010 - b)都是奇数吗？比如，a是偶数，例如a=2，2010 - b=3，gcd(2,3)=1，是奇数；a=4，2010 - b=5，gcd(4,5)=1，奇数；所以确实，当2010 - b为奇数时，无论a是奇数还是偶数，gcd(a, 2010 - b)都是奇数，因为2010 - b是奇数，而任何数和一个奇数的公约数都是奇数，因为如果公约数包含2的话，那这个奇数就不可能被2整除，所以公约数必定是奇数。因此，当2010 - b是奇数时，gcd(a, 2010 - b)一定是奇数，所以无论a是奇数还是偶数，都满足条件。而当2010 - b是偶数时，即b是偶数，那么2010 - b是偶数，此时我们需要gcd(a, 2010 - b)是奇数。但2010 - b是偶数，所以如果a是偶数的话，gcd(a, 2010 - b)至少是2，所以是偶数；而如果a是奇数的话，那么gcd(a, 2010 - b)就是奇数，因为奇数和偶数的最大公约数是奇数。因为2010 - b是偶数，可以表示为2k，而a是奇数的话，那么gcd(a, 2k)=gcd(a, k)。因为a是奇数，所以和2互质，因此gcd(a, 2k)=gcd(a, k)，而k= (2010 - b)/2，这里k是一个整数。所以，当2010 - b是偶数时，即b是偶数，此时如果a是奇数的话，gcd(a, 2010 - b)是奇数；如果a是偶数的话，gcd(a, 2010 - b)是偶数。因此，总结两种情况：当b是奇数时，2010 - b是奇数，此时不管a是奇数还是偶数，gcd(a, 2010 - b)都是奇数，所以这种情况下满足条件，对应的线段内部有偶数个格点；当b是偶数时，2010 - b是偶数，此时只有当a是奇数时，gcd(a, 2010 - b)才是奇数，所以这种情况下满足条件的a是奇数；现在需要计算满足条件的点的数量：首先，总共有100个可能的b值（1到100），其中奇数和偶数各占一半，即50个奇数，50个偶数。对于每个b值，a的范围是1到100，共100个可能的a值，其中奇数和偶数各50个。当b是奇数时，对应的50个b值，每个b对应100个a，其中所有a都满足条件，所以这种情况下的点数为50×100=5000个；当b是偶数时，对应的50个b值，每个b对应50个奇数的a，所以这种情况下满足条件的点数为50×50=2500个；所以总共有满足条件的点数是5000 + 2500 = 7500个；因此，所求的概率是7500 / 10000 = 3/4；不过，我需要再仔细检查一下，是否我的分析正确。首先，当b是奇数时，2010 - b是奇数，所以无论a是奇数还是偶数，gcd(a, 2010 - b)都是奇数。这个是对的，因为奇数与任何数的公约数都是奇数，对吗？例如，a是偶数，比如a=2，2010 - b=3，那么gcd(2,3)=1，是奇数；a是奇数，比如a=3，2010 - b=3，gcd(3,3)=3，奇数；当b是偶数时，2010 - b是偶数，比如b=2，那么2010 - b=2008，此时如果a是偶数，比如a=4，那么gcd(4,2008)至少是4，是偶数；如果a是奇数，比如a=3，那么gcd(3,2008)= gcd(3,2008). 2008 ÷3=669余1，所以gcd(3,2008)=1，奇数。所以，当b是偶数时，只有当a是奇数时，gcd(a,2010 - b)是奇数，而当a是偶数时，gcd(a,2010 - b)是偶数，因此这种情况下满足条件的a是奇数，共有50个。因此，总满足的点数是：当b为奇数时，每个b对应的100个a都满足，共有50×100=5000；当b为偶数时，每个b对应的50个a满足，共有50×50=2500；总和为5000+2500=7500；概率7500/10000=3/4=0.75；所以答案应该是3/4，即概率为3/4。不过，可能我的分析哪里有问题吗？再考虑另一个例子，比如当b=100（偶数），则2010 - b=1910，这是一个偶数。当a是奇数，比如a=1，gcd(1,1910)=1，奇数，满足条件；当a=3，gcd(3,1910)=gcd(3,1910%3)=gcd(3,1910-3×636=1910-1908=2)=gcd(3,2)=1，奇数；当a=5，gcd(5,1910)=5，是奇数吗？不，5是奇数，所以gcd=5，奇数，满足条件；当a=2，gcd(2,1910)=2，偶数，不满足条件；当a=4，gcd(4,1910)=2，偶数，不满足条件；是的，当a是奇数时，gcd(a, 偶数)是奇数，当a是偶数时，gcd是偶数；因此，我的分析是对的。因此满足条件的点数是7500，概率3/4。不过，题目中的点是在正方形内部，即1 ≤ x ≤ 100，1 ≤ y ≤ 100，所以每个坐标都是整数且在这个范围内，共有100×100=10000个点。所以，结论是概率3/4，即3/4的概率。不过，是否有其他因素需要考虑？比如，当线段上的内部格点数目为偶数时，对应的是gcd(a,2010 -b) -1为偶数，即gcd(a,2010 -b)为奇数；而当gcd(a,2010 -b)为偶数时，内部的格点数目是偶数吗？比如，如果gcd是2，则内部数目是2 -1=1，奇数；如果gcd是4，内部数目是4-1=3，奇数；所以当gcd是偶数时，数目是奇数；当gcd是奇数时，数目是gcd -1，即偶数；因此，当且仅当gcd(a,2010 -b)是奇数时，内部格点数目为偶数；所以，我的分析正确，需要计算这样的点数，即当gcd(a,2010 -b)是奇数时的点数；所以，结果是3/4的概率。因此，我认为答案是3/4，也就是概率为3/4，用boxed{dfrac{3}{4}}表示。不过，再仔细检查一遍：当b是奇数时，2010 -b是奇数，所以不管a是奇数还是偶数，gcd(a,奇数)是否为奇数？是的，例如，a是偶数，gcd(2,3)=1；a是奇数，gcd(3,5)=1，都是奇数；当b是偶数时，2010 -b是偶数，所以只有当a是奇数时，gcd(a,偶数)是奇数；而a是奇数时，由于偶数的因数是2，而奇数不含因数2，所以gcd(a,偶数)=gcd(a, 偶数/2^k)，其中k是2的幂次，所以最后的结果是gcd(a,奇数)，而a是奇数，所以gcd(a,奇数)还是奇数；例如，a=3，2010 -b=2008=2^3×251，那么gcd(3,2008)=gcd(3,251)，251除以3余2，所以gcd(3,251)=1，奇数；如果a=5，2010 -b=2008，则gcd(5,2008)=gcd(5,2008%5)=gcd(5,3)=1，奇数；所以确实，当b是偶数且a是奇数时，gcd(a,2010 -b)是奇数，满足条件；当b是偶数且a是偶数时，gcd(a,2010 -b)至少是2，所以是偶数，不满足条件；而当b是奇数时，无论a奇偶，gcd都是奇数；因此，满足条件的点的数量：对于b是奇数的情况，共50个b，每个b对应100个a，因此50×100=5000；对于b是偶数的情况，共50个b，每个b对应50个奇数a，因此50×50=2500；总和7500，概率7500/10000=3/4；没错，所以答案应该是3/4。答案boxed{dfrac{3}{4}}</think>首先，我们需要确定在正方形内部（不包括边界）的格点数目，即1 ≤ x ≤ 100，1 ≤ y ≤ 100，共有100×100=10000个可能的格点。接下来，我们需要确定连接点(0,2010)和随机选出的格点P(a, b)的线段上（不包括端点）有偶数个格点的条件。根据数论中的结论，线段内部的格点数目为gcd(a, 2010 - b) - 1。题目要求这个数目是偶数，即gcd(a, 2010 - b) - 1 ≡ 0 (mod 2)，即gcd(a, 2010 - b) 必须是奇数。分析2010 - b的奇偶性：- 当b是奇数时，2010 - b为奇数，此时无论a是奇数还是偶数，gcd(a, 2010 - b)都是奇数。- 当b是偶数时，2010 - b为偶数，此时只有当a是奇数时，gcd(a, 2010 - b)才是奇数。计算满足条件的点的数量：- 当b是奇数时，共有50个b值，每个b对应100个a值，满足条件的点数为50×100=5000个。- 当b是偶数时，共有50个b值，每个b对应50个奇数a，满足条件的点数为50×50=2500个。总共有满足条件的点数5000 + 2500 = 7500个，概率为7500/10000 = 3/4。最终答案：boxed{dfrac{3}{4}},<think>好的，我现在要解决的问题是：在坐标平面上，给定一个顶点为(0,0)、(101,0)、(101,101)和(0,101)的正方形内部（不包括边界）随机选取一个格点，求连接该点与点(0,2010)的线段（不包括端点）上有偶数个格点的概率是多少。首先，我需要理解题目中的关键概念和条件。题目中的格点是指x和y坐标都是整数的点。所选取的格点位于正方形内部，所以其x和y坐标都是整数，且满足1 ≤ x ≤ 100，1 ≤ y ≤ 100。接下来，需要确定连接这个点与点(0,2010)的线段上（不包括端点）有多少个格点，并且计算这种情况的概率。题目要求的是线段上有偶数个格点的概率，那么首先我需要找到一个方法来判断给定两个点时，它们之间的线段上有多少个格点。然后，根据这个数目是否为偶数，统计符合条件的点数目，再计算概率。首先，回忆一下两个格点之间的线段上的格点数目计算方法。根据数论中的贝祖定理，两点之间的线段上的格点数目（不包括端点）等于这两个点的坐标差的绝对值的最大公约数减一。具体来说，如果两点分别为(a,b)和(c,d)，则线段上的格点数目为gcd(|c - a|, |d - b|) - 1。其中gcd表示最大公约数。不过，这里需要注意，这里的端点是被排除在外的，所以公式中的结果就是中间点的数量。例如，如果gcd是1，那么中间没有格点；如果gcd是2，中间有1个格点，以此类推。不过，题目中的两个点中的一个是(0,2010)，另一个点位于正方形内部，即坐标为(x,y)，其中1 ≤ x ≤ 100，1 ≤ y ≤ 100。因此，我们需要计算这两个点之间的线段上的格点数目，即gcd(|x - 0|, |y - 2010|) - 1 = gcd(x, |y - 2010|) - 1。因为x和y都是正整数，且y ≤ 100，所以|y - 2010| = 2010 - y，也是正整数。于是，线段上的格点数目为gcd(x, 2010 - y) - 1。题目要求的是这个数目为偶数的情况的概率。即，我们需要计算当gcd(x, 2010 - y) - 1为偶数时，即gcd(x, 2010 - y)为奇数时，此时数目为奇数减1是否为偶数？不，这里的数目是gcd(x, 2010 - y) - 1，所以当gcd为奇数时，数目是奇数减1，即偶数；当gcd为偶数时，数目是偶数减1，即奇数。因此，当gcd(x, 2010 - y)为奇数时，格点数目为偶数；当gcd为偶数时，格点数目为奇数。所以问题转化为求在满足条件的x和y下，gcd(x, 2010 - y)为奇数的概率。因此，问题转换为求在x ∈ [1,100]，y ∈ [1,100]的情况下，gcd(x, 2010 - y)为奇数的点的数量占总点的数量的比例，即概率。接下来，我需要分析如何计算这样的点的数量。首先，注意到2010是一个具体的数值，我们可以先对其进行分解。2010 = 2 × 3 × 5 × 67。这说明2010是一个偶数，且它包含2作为质因数。因此，2010 - y的奇偶性取决于y的奇偶性。如果y是偶数，那么2010 - y是偶数；如果y是奇数，那么2010 - y是奇数。例如，当y为偶数时，2010是偶数，减去偶数y，结果还是偶数；当y为奇数时，2010减奇数是奇数。因此，2010 - y的奇偶性由y的奇偶性决定。接下来，考虑gcd(x, 2010 - y)的奇偶性。如果这个gcd是奇数，那么它必须满足x和2010 - y都是奇数，或者其中至少有一个是奇数，但它们的最大公约数没有2这个因子。不过更准确的说法是，当且仅当x和2010 - y的最大公约数中没有因子2时，gcd(x, 2010 - y)是奇数。换句话说，要么x和2010 - y都是奇数，或者它们的公共因子中没有2。但更系统的方法是考虑将x和2010 - y中的因子2提取出来。假设x = 2^k * a，其中a是奇数，k ≥ 0；同理，2010 - y = 2^m * b，其中b是奇数，m ≥ 0。那么，gcd(x, 2010 - y) = 2^{min(k,m)} * gcd(a, b)。因此，gcd(x, 2010 - y)为奇数的条件是min(k,m) = 0，即k=0或m=0，或者更准确地说，当且仅当x和2010 - y中至少有一个是奇数，或者两者都是偶数，但在这种情况下，它们的gcd中的2的指数是min(k,m)，如果min(k,m) ≥ 1，那么gcd就包含因子2，即为偶数。因此，gcd为奇数当且仅当x和2010 - y没有公共的因子2，即要么x和2010 - y都是奇数，或者其中一个为奇数，另一个为偶数，但它们的gcd中没有2。不过，这种情况下，如果其中一个是奇数，另一个是偶数，那么它们的gcd必然是奇数，因为偶数那个含有因子2，而奇数那个不含，所以gcd中不会有2。例如，x是奇数，2010 - y是偶数，则gcd(x, 2010 - y) = gcd(奇数，偶数)，此时gcd(x, 2010 - y) = gcd(奇数， 偶数/2^k)，但这里可能需要更仔细的分析。实际上，如果x是奇数，而2010 - y是偶数，那么2010 - y可以被2整除，但x不能被2整除，所以它们的gcd中不可能有2的因子，因此gcd(x, 2010 - y)一定是奇数。同样地，如果x是偶数，2010 - y是奇数，即y是奇数，此时2010 - y是奇数，x是偶数，它们的gcd也是奇数，因为奇数与偶数的gcd只能是奇数。而当x和2010 - y都是偶数时，它们的gcd至少含有因子2，所以gcd(x, 2010 - y)一定是偶数。而当x和2010 - y都是奇数时，它们的gcd也可能是奇数或1，但无论如何都是奇数。因此，总结一下，gcd(x, 2010 - y)为奇数当且仅当x和2010 - y不同时为偶数。换句话说，当且仅当x是奇数，或者2010 - y是奇数（即y是奇数），或者两者都是奇数。不过更准确地说，当x和2010 - y中至少有一个是奇数时，gcd(x, 2010 - y)为奇数；当两者都是偶数时，gcd(x, 2010 - y)为偶数。因此，我们需要计算在x ∈ [1,100]，y ∈ [1,100]的情况下，x和2010 - y中至少有一个是奇数的点的数量，即当x为奇数或者y为奇数的情况，或者两者都是奇数的情况。不过，这里需要仔细分析，因为可能存在x和2010 - y同时为偶数的情况，这时候gcd是偶数。因此，当且仅当x和2010 - y中至少有一个是奇数时，gcd是奇数，此时对应的格点数目为偶数。所以，我们需要计算满足x是奇数或y是奇数的点的数量，再除以总点数（100×100=10000）。不过，这里有一个问题：2010 - y的奇偶性由y的奇偶性决定，因为2010是偶数，所以当y是奇数时，2010 - y是奇数；当y是偶数时，2010 - y是偶数。因此，2010 - y的奇偶性与y的奇偶性相反。或者说，2010 - y的奇偶性等同于y的奇偶性取反，因为偶数减奇数是奇数，偶数减偶数是偶数。因此，当y是奇数时，2010 - y是奇数；当y是偶数时，2010 - y是偶数。因此，2010 - y的奇偶性与y的奇偶性相反。因此，当y是奇数时，2010 - y是奇数；当y是偶数时，2010 - y是偶数。因此，x和2010 - y中至少有一个是奇数的条件等价于x是奇数或者y是奇数。因为当y是奇数时，2010 - y是奇数，无论x是什么，此时至少有一个是奇数（即2010 - y是奇数）。当y是偶数时，2010 - y是偶数，此时如果x是奇数，那么x是奇数，所以至少有一个是奇数；如果x是偶数，则两者都是偶数，此时gcd(x, 2010 - y)是偶数。因此，满足条件的点数等于x是奇数或者y是奇数的点的数量，再减去x和y都是奇数的点的数量（因为这些点被重复计算了两次）。根据容斥原理，满足条件的点数等于（x奇数的数量×y任意的数量） + （y奇数的数量×x任意的数量） - （x和y都是奇数的数量）。不过更准确的，应该是：满足条件的点数 = (x为奇数的数目 × y为任意的数目) + (y为奇数的数目 × x为任意的数目) - (x为奇数且y为奇数的数目)。不过这里可能更简单的方式是直接计算。在x ∈ [1,100]中，x为奇数的数目是50个（1到100中有50个奇数，50个偶数）。同样，y ∈ [1,100]中，y为奇数的数目也是50个，偶数50个。根据上述分析，当x为奇数或者y为奇数时，满足条件。因此，满足条件的点的数量为：数目 = (奇数x的数量 × 所有y) + (奇数y的数量 × 所有x) - (奇数x且奇数y的数量)但这样计算是不对的，正确的计算应该是使用容斥原理：数目 = 奇数x的数量 × y的数量 + y为奇数的数量 × x的数量 - 奇数x且奇数y的数量 ×1不过其实正确的公式应该是：数目 = (奇数x的数量 × 总y数) + (奇数y的数量 × 总x数) - (奇数x且奇数y的数量)不过这可能也不太对，正确的应该是满足x为奇数或者y为奇数的点的数量，即：数目 = 奇数x的数量 × 总y数 + 奇数y的数量 × 总x数 - 奇数x且奇数y的数量 × 1但实际上，正确的计算应该是：数目 = (奇数x的数量 × 总y数) + (奇数y的数量 × 总x数) - (奇数x且奇数y的数量 × 2) ?不，应该更简单一点，因为满足x为奇数或y为奇数的点的数量是：数目 = 奇数x的数量 × y的总数 + 偶数x的数量 × 奇数y的数量或者，更直接地，数目 = 总点数 - 同时x为偶数且y为偶数的点的数量。这是因为当且仅当x为偶数和y为偶数时，才会出现x和2010 - y都为偶数的情况，此时gcd为偶数。其他情况下，x或y至少有一个是奇数，此时gcd为奇数。因此，满足条件的数目等于总点数减去同时x和y都为偶数的点的数量。因此，总点数是100×100=10000。同时x和y都为偶数的点的数量是50×50=2500，因为在x的范围内有50个偶数（2,4,...,100），y的范围内也有50个偶数。因此，同时x和y都是偶数的点的数量是50×50=2500。因此，满足条件的数目为10000 - 2500 = 7500。因此，概率就是7500 / 10000 = 3/4 = 0.75。不过这里需要验证一下这个结论是否正确，因为在上述分析中，我们假设当且仅当x和y都为偶数时，gcd(x,2010 - y)为偶数，此时线段上的格点数目为奇数，否则为偶数。因此，题目所求的是线段上有偶数个格点的概率，即当gcd(x,2010 - y)为奇数时的情况，对应数目为7500，因此概率是3/4。不过，我需要仔细检查这个结论是否正确，是否存在例外情况。例如，当x和y都为偶数时，x是偶数，y是偶数，那么2010 - y是偶数，因为y是偶数，所以2010 - y = 2010 - 偶数 = 偶数。因此，x和2010 - y都是偶数，所以gcd(x, 2010 - y)至少是2，所以gcd是偶数，因此这种情况下gcd(x, 2010 - y)是偶数，对应的格点数目为偶数 -1，即奇数，因此不符合条件，即线段上有奇数个格点，因此我们需要排除这些情况，即当x和y都是偶数时，格点数目为奇数，其他情况下格点数目为偶数。因此，满足线段上有偶数个格点的条件是当且仅当x和y不都为偶数，即数目为10000 - 2500 = 7500，概率为3/4。这看起来是对的，但需要进一步验证是否在x和y不都为偶数的情况下，gcd(x, 2010 - y)确实是奇数。例如，假设x是奇数，y是偶数，那么2010 - y是偶数（因为y是偶数，2010是偶数，减去偶数还是偶数）。此时x是奇数，2010 - y是偶数。那么gcd(x, 2010 - y) = gcd(奇数, 偶数). 因为奇数不能被2整除，所以这个gcd实际上是gcd(奇数, 偶数/2^k), 其中k是使得偶数能被2^k整除的最大指数。例如，如果x=3，2010 - y=4，则gcd(3,4)=1，是奇数。如果x=5，2010 - y=6，gcd(5,6)=1，奇数。如果x=3，2010 - y=8，gcd(3,8)=1，奇数。如果x=3，2010 - y=2，gcd(3,2)=1，奇数。因此，当x是奇数，2010 - y是偶数时，gcd(x, 2010 - y)一定是奇数，对吗？是的，因为x是奇数，而2010 - y是偶数，所以可以写成2*m，其中m是整数。那么gcd(奇数, 2*m) = gcd(奇数, m)，因为奇数与2互质。例如，gcd(3, 4) = gcd(3, 2) = 1。同样，gcd(5, 6)=1，因为6=2*3，而gcd(5,3)=1。因此，此时gcd(x,2010 - y)等于gcd(x, m)，而m = (2010 - y)/2，这里x是奇数，m是整数。因此，gcd(x, m)的结果可能为奇数或1，但无论如何，因为x是奇数，所以这个gcd是奇数。同样，当x是偶数，y是奇数时，2010 - y是奇数，因为y是奇数，2010是偶数，所以2010 - y是奇数。此时x是偶数，2010 - y是奇数。那么gcd(x, 2010 - y) = gcd(偶数, 奇数). 因为奇数和偶数中的最大公约数只能是奇数，因为偶数可以分解为2^k * 奇数部分，而奇数部分可能与另一个奇数有公约数，但2^k部分不会被另一个数整除，所以gcd(偶数, 奇数) = gcd(奇数部分, 奇数), 即奇数。例如，x=4，2010 - y=3，则gcd(4,3)=1；x=6，2010 - y=5，gcd(6,5)=1；x=8，2010 - y=7，gcd(8,7)=1；x=10，2010 - y=9，gcd(10,9)=1；如果x=6，2010 - y=9，gcd(6,9)=3，奇数。所以在这种情况下，gcd是奇数。当x和y都是奇数时，x是奇数，y是奇数，所以2010 - y是奇数（因为y是奇数，2010是偶数，减去奇数得到奇数）。因此，x和2010 - y都是奇数，此时它们的gcd也可能是奇数，比如gcd(3,5)=1，gcd(3,9)=3，都是奇数。因此，无论x和y是否都是奇数，只要不同时为偶数，gcd(x,2010 - y)都是奇数，对应的线段上的格点数目是奇数减1，即偶数，符合条件。因此，只有当x和y都是偶数时，gcd(x,2010 - y)是偶数，此时线段上的格点数目为偶数减1，即奇数，不符合条件。所以满足条件的点数是总点数减去x和y都是偶数的点数，即10000 - 2500 = 7500，对应的概率是7500/10000=3/4。不过，这里可能存在一个误区，就是是否所有x和y都是偶数的情况下的gcd都是偶数，而其他情况下都是奇数。比如是否存在某些x和y不都是偶数，但gcd(x,2010 - y)仍然是偶数的可能性？例如，假设x是偶数，y是奇数，那么2010 - y是奇数，所以gcd(偶数, 奇数)一定是奇数，对吗？例如，x=4，y=3，则2010 - y=2007，gcd(4,2007)=1，是奇数。再比如x=6，y=5，2010 - y=2005，gcd(6,2005)=1，奇数。或者x=10，y=1，2010 - y=2009，gcd(10,2009)=1，奇数。但假设x=6，y=3，2010 - y=2007，gcd(6,2007)=3，是奇数吗？不，3是奇数，是的。所以当x是偶数，y是奇数时，gcd(x,2010 - y)确实是奇数。再比如，x是奇数，y是偶数，那么2010 - y是偶数，所以gcd(奇数, 偶数)=奇数，如gcd(3,4)=1，gcd(5,6)=1，gcd(9,8)=1，gcd(15,20)=5，5是奇数。所以无论哪种情况，只要x和y不同时为偶数，gcd都是奇数，因此线段上的格点数目是偶数，符合题目要求的条件。因此，结论应该是概率为3/4，即75%。不过，这里需要注意，题目中的点是在正方形内部，排除边界，即x从1到100，y从1到100，都是整数。因此，总共有100×100=10000个格点。而x和y都是偶数的点数目是50×50=2500，所以符合条件的点数目是10000 - 2500 = 7500，概率是7500/10000=3/4。不过，我突然想到一个问题，这里的结论是否与2010的具体数值有关？或者说，是否因为2010是偶数，导致当y为偶数时，2010 - y是偶数，而当y为奇数时，2010 - y是奇数？如果题目中的点是另一个数，比如2011，那么结论会不会不同？不过在这个问题中，2010是给定的，所以这个结论是适用的。因此，最终的答案应该是3/4，即概率为3/4，对应的分数形式是boxed{dfrac{3}{4}}。不过，为了确保正确，我可以再通过一些例子来验证。例如，考虑x=2，y=2（都是偶数），此时2010 - y=2008，gcd(2,2008)=2，所以格点数目为2-1=1，奇数，符合条件为不满足，所以不计入。如果x=1，y=1（都是奇数），2010 - y=2009，gcd(1,2009)=1，数目为0，0是偶数，计入符合条件。如果x=1，y=2（x奇，y偶），2010 - y=2008，gcd(1,2008)=1，数目为0，偶数，计入。如果x=2，y=1（x偶，y奇），2010 - y=2009，gcd(2,2009)=1，数目为0，偶数，计入。如果x=2，y=4（都是偶），2010 - y=2006，gcd(2,2006)=2，数目为1，奇数，不计入。如果x=6，y=4（都是偶），2010 - y=2006，gcd(6,2006)=2，数目为1，奇数，不计入。如果x=3，y=4（x奇，y偶），2010 - y=2006，gcd(3,2006)=1，数目为0，计入。如果x=4，y=5（x偶，y奇），2010 - y=2005，gcd(4,2005)=1，数目为0，计入。如果x=6，y=9（x偶，y奇），2010 - y=2001，gcd(6,2001)=3，数目为2，偶数，计入。看来所有非双偶的情况都是符合条件的，而双偶情况则不符合。因此，结论正确，概率为3/4。答案boxed{dfrac{3}{4}}</think>首先，我们需要确定连接点(0, 2010)和正方形内部格点(x, y)（其中1 leq x leq 100，1 leq y leq 100）的线段上（不包括端点）的格点数目。根据数论中的贝祖定理，两点之间的线段上的格点数目（不包括端点）等于这两个点的坐标差的绝对值的最大公约数减一。即，线段上的格点数目为gcd(x, 2010 - y) - 1。我们需要计算这个数目为偶数的概率。当且仅当gcd(x, 2010 - y)为奇数时，数目gcd(x, 2010 - y) - 1为偶数。接下来分析gcd(x, 2010 - y)的奇偶性：1. 当y为偶数时，2010 - y为偶数；当y为奇数时，2010 - y为奇数。2. gcd(x, 2010 - y)为奇数的条件是x和2010 - y中至少有一个是奇数。即，当x为奇数或y为奇数时，gcd(x, 2010 - y)为奇数。计算满足条件的点数：- 在1 leq x leq 100中，有50个奇数。- 在1 leq y leq 100中，有50个奇数。- 同时满足x和y都为偶数的点数为50 times 50 = 2500。总点数为100 times 100 = 10000。满足条件的点数为总点数减去同时为偶数的点数，即10000 - 2500 = 7500。因此，概率为frac{7500}{10000} = frac{3}{4}。最终答案：boxed{dfrac{3}{4}}]

🔑:9. frac{3}{4}9. Note that on the line segment joining (a, b) and (c, d), where a, b, c, d are integers, the number of lattice points (excluding the endpoints) is exactly 1 less than the H.C.F. of c-a and d-b.Hence there is an even number of lattice points on the line segment (excluding the endpoints) joining (0,2010) and (a, b) if and only if the H.C.F. of a and 2010-b is odd, which is true if and only if at least one of a, b is odd.Now there are 100 choices for each of a and b (namely, 1,2, ldots, 100 ). Hence the probability that both are even is frac{1}{2} times frac{1}{2}=frac{1}{4}, and so the required probability is 1-frac{1}{4}=frac{3}{4}.

🧠:[<think>Okay, so I need to solve this problem where we pick a real number b from the interval [-17, 17], and then find the probability that the equation x⁴ + 25b² = (4b² - 10b)x² has at least two distinct real solutions. Then express that probability as a fraction m/n where m and n are coprime, and find m + n. Hmm, okay, let's break this down step by step.First, let me rewrite the equation to make it clearer. The original equation is:x⁴ + 25b² = (4b² - 10b)x².Maybe I can move all terms to one side to make it a standard quartic equation. Let's subtract (4b² - 10b)x² from both sides:x⁴ - (4b² - 10b)x² + 25b² = 0.Hmm, this looks like a quadratic in terms of x². So maybe I can substitute y = x², which is a common technique. Let's try that.Let y = x². Then the equation becomes:y² - (4b² - 10b)y + 25b² = 0.So now it's a quadratic equation in y. To find real solutions for x, we need real solutions for y, and then for each positive y, we get two real solutions for x (positive and negative square roots), and for y=0, we get one real solution. But since the problem asks for at least two distinct real solutions, we need to have either two distinct positive y solutions, which would give four real x solutions (two from each y), or one positive y and one zero y, which would give three real x solutions (two from the positive y and one from y=0), or maybe two different positive y's? Wait, but if y=0 is a solution, then substituting back into the equation:If y = 0, then 0 - (4b² - 10b)*0 + 25b² = 25b² = 0. So 25b² = 0 implies b = 0. So unless b=0, y=0 is not a solution. So y=0 is only a solution if b=0. Let me check that.If b=0, then original equation becomes x⁴ + 0 = (0 - 0)x², which simplifies to x⁴ = 0. So x=0 is the only solution, with multiplicity four. But the problem says "at least two distinct real solutions," so even though there are four roots, they are all the same. So in that case, it's just one distinct real solution. So for b=0, the equation does not satisfy the condition. So in most cases, y=0 is not a solution unless b=0, and even then, it doesn't help. So maybe I can ignore y=0 for now.Therefore, to have real solutions for x, we need real, non-negative solutions for y. Because y = x² must be non-negative.So, the quadratic in y is:y² - (4b² - 10b)y + 25b² = 0.For this quadratic to have real solutions, the discriminant must be non-negative. Let's compute the discriminant D:D = [-(4b² - 10b)]² - 4*1*25b²Simplify:D = (4b² - 10b)² - 100b².Let me compute (4b² - 10b)² first:(4b² - 10b)² = (4b²)² - 2*(4b²)*(10b) + (10b)² = 16b⁴ - 80b³ + 100b².So then D = 16b⁴ - 80b³ + 100b² - 100b² = 16b⁴ - 80b³.Factor out 16b³:Wait, 16b⁴ - 80b³ = 16b³(b) - 80b³ = 16b³(b - 5). So:D = 16b³(b - 5).Therefore, the discriminant D = 16b³(b - 5).For the quadratic in y to have real solutions, we need D ≥ 0. So:16b³(b - 5) ≥ 0.Since 16 is positive, this simplifies to:b³(b - 5) ≥ 0.So we need to find the values of b where b³(b - 5) ≥ 0.Let's analyze the inequality b³(b - 5) ≥ 0.First, note that b³ is positive when b > 0, negative when b < 0, and zero when b = 0.So, factor the expression:b³(b - 5) = b² * b * (b - 5).Wait, but maybe it's easier to consider the sign of each factor.So, the critical points are at b = 0 and b = 5.Let me create a sign chart.For b < 0:- b³ is negative (since b is negative and cubed remains negative)- (b - 5) is negative (since b < 0 < 5)- So, product of two negatives is positive. So for b < 0, the expression is positive.At b = 0:The expression is 0.For 0 < b < 5:- b³ is positive (since b is positive)- (b - 5) is negative (since b < 5)- Product is positive * negative = negative.For b > 5:- b³ is positive- (b - 5) is positive- Product is positive * positive = positive.Therefore, the expression b³(b - 5) is ≥ 0 when:- b ≤ 0 or b ≥ 5.But note that at b = 0, the discriminant is zero, so the quadratic in y has a repeated root.Similarly, at b = 5, discriminant is zero as well.But even if the discriminant is non-negative, the quadratic in y may have solutions, but we need to check if those solutions for y are non-negative, since y = x² must be non-negative.So, even if the quadratic in y has real roots, we need those roots to be non-negative for x to have real solutions. So even if D ≥ 0, we need to ensure that y ≥ 0.So, let's break this into cases based on the discriminant.Case 1: D ≥ 0, which happens when b ≤ 0 or b ≥ 5.Case 2: D < 0, which is when 0 < b < 5. In this case, the quadratic in y has no real solutions, so the original equation has no real solutions.But we need to check when there are at least two distinct real solutions. So, even if there are real roots for y, we need to check whether the quadratic in y has at least one positive solution (since each positive y gives two real x solutions, so one positive y gives two real x solutions, two positive y's give four real x solutions). However, if both y solutions are positive and distinct, then x has four distinct real solutions. If both y solutions are positive and equal (double root), then x has two distinct real solutions (each with multiplicity two, but distinct). If one y is positive and one is negative, then we discard the negative one and only have two real x solutions. If both y solutions are negative, then there are no real x solutions.So, let's analyze the quadratic equation for y:y² - (4b² - 10b)y + 25b² = 0.We can denote coefficients as:A = 1, B = -(4b² - 10b), C = 25b².So the quadratic is Ay² + By + C = 0.We can use Vieta's formula: sum of roots y1 + y2 = (4b² - 10b), product y1 * y2 = 25b².Also, since discriminant D = 16b³(b - 5).We need to check for the roots y1 and y2 when D ≥ 0 (i.e., b ≤ 0 or b ≥ 5):First, note that the product of the roots y1 * y2 = 25b², which is non-negative since b² is non-negative. Therefore, the product is always non-negative. So either both roots are positive, both are negative, or both are zero.But 25b² = 0 only when b = 0, so in that case, both roots would be zero? Let's check when b=0.If b=0, the quadratic becomes y² - 0*y + 0 = y² = 0, so y=0 is a double root. Then x²=0 gives x=0, so only one real solution, which is repeated. But since the problem states "at least two distinct real solutions," this case (b=0) doesn't satisfy the condition.Similarly, when b ≠ 0, the product of roots is 25b² > 0, so the roots are both positive or both negative. Since the product is positive and the sum is (4b² - 10b). Let's see.Sum of roots y1 + y2 = 4b² - 10b.So if both roots are positive, then their sum must be positive. If both roots are negative, their sum must be negative. Since the product is positive, so they have the same sign.Therefore:If sum > 0, then both roots are positive.If sum < 0, then both roots are negative.If sum = 0, then roots are zero and ... but product would be 25b², which would need to be zero, but only when b=0.Wait, sum is 4b² - 10b.So when is 4b² - 10b positive?4b² - 10b > 0Factor:2b(2b - 5) > 0So critical points at b=0 and b=5/2=2.5.Sign chart:For b < 0: 2b is negative, (2b -5) is negative, product is positive.For 0 < b < 2.5: 2b positive, (2b -5) negative, product negative.For b > 2.5: both positive, product positive.Therefore, 4b² - 10b > 0 when b < 0 or b > 2.5.Similarly, 4b² - 10b < 0 when 0 < b < 2.5.So, sum of roots:- For b < 0 or b > 2.5: sum positive, so both roots positive.- For 0 < b < 2.5: sum negative, so both roots negative.But remember, we are in the case where D ≥ 0, i.e., b ≤ 0 or b ≥ 5. Wait, but we have to reconcile this with the discriminant condition.Wait, discriminant D ≥ 0 when b ≤ 0 or b ≥ 5. So in those ranges, the quadratic in y has real roots.So for b ≤ 0:If b < 0: sum of roots is 4b² -10b. Since b is negative, 4b² is positive, -10b is positive (since b is negative), so sum is positive + positive = positive.Thus, when b < 0, the quadratic has two positive roots (since sum positive and product positive). Therefore, each positive root for y gives two real roots for x, so total four real roots. However, if the roots are distinct, then there are four distinct real roots. If the roots are equal (double root), then x² = y would give two real roots each with multiplicity two. Wait, but the problem says "at least two distinct real solutions." So if there's a double root in y, then x would have two distinct real solutions each with multiplicity two. So that's two distinct real solutions. But the problem says "at least two distinct real solutions," so even if there's a double root in y, leading to two distinct real solutions (each repeated), that still counts, right? Wait, but actually, if y has a double root, then x² = y would have two real roots, each with multiplicity two, but they are still just two distinct real numbers. So for example, if y = k (double root), then x = sqrt(k) and x = -sqrt(k), each with multiplicity two. So the equation would have two distinct real solutions, each with multiplicity two. Therefore, the equation would have two distinct real solutions. So that satisfies "at least two distinct real solutions."Wait, but in the case when y has two distinct positive roots, then x would have four distinct real solutions (two from each y). If y has a double positive root, then x has two distinct real solutions. If y has two negative roots, then no real solutions. But in the case when discriminant D ≥ 0, which is when b ≤ 0 or b ≥5.Wait, let's parse this again.When discriminant D ≥ 0 (b ≤ 0 or b ≥5):- For b < 0: sum of roots is positive (4b² -10b, which is positive since b is negative, 4b² is positive, -10b is positive), product is positive (25b²). Therefore, both roots positive. So two distinct positive roots (since discriminant is positive? Wait, when D > 0, roots are distinct. When D=0, roots are same.But D = 16b³(b -5). For b < 0: b is negative, so b³ is negative, (b -5) is negative (since b <0 <5). So D = 16 * negative * negative = positive. Therefore, discriminant D is positive for b < 0. So two distinct real roots. Both positive (sum positive, product positive). Therefore, x has four distinct real solutions.For b =0: D =0, quadratic becomes y² -0 +0 = y²=0, so y=0 double root. Then x=0 double root, so only one distinct real solution. So doesn't satisfy the condition.For b ≥5: D =16b³(b -5). If b >5, then b³ positive, (b -5) positive, so D positive. If b=5, D=0.For b >5: discriminant D positive, so two distinct real roots. Sum is 4b² -10b. Let's check if sum is positive. For b >5, 4b² -10b = 2b(2b -5). Since b >5, 2b >10, 2b -5 >10 -5=5, so positive. Therefore, sum positive, product positive (25b²). Therefore, two distinct positive roots. Therefore, x has four distinct real solutions.For b=5: discriminant D=0. Quadratic equation becomes y² - (4*25 -10*5)y +25*25= y² - (100 -50)y +625= y² -50y +625=0. Which factors as (y -25)^2=0. So y=25 double root. Then x²=25, so x=5 and x=-5, each with multiplicity two. So two distinct real solutions. So even though it's a double root in y, we still get two distinct real solutions for x. So that counts.Therefore, for the discriminant D ≥0:- When b <0: two distinct positive roots for y, leading to four distinct real x solutions. So satisfies the condition.- When b=0: one real solution (x=0) with multiplicity four. Doesn't satisfy.- When b >5: two distinct positive roots for y, leading to four distinct real x solutions. Satisfies.- When b=5: one positive double root for y (25), leading to two distinct real x solutions (5 and -5) with multiplicity two. So satisfies.So in total, for b <0 or b ≥5, the equation has at least two distinct real solutions (four if b <0 or b >5, two if b=5). But when b=5, it still counts. However, when b=0, it doesn't. So the valid interval is b ∈ [-17, 0) union [5,17]. But we need to check if in those intervals, all values of b satisfy the condition.Wait, but wait, the original interval is [-17,17]. So we need to consider b ∈ [-17,17]. The acceptable values for b are b ∈ [-17,0) ∪ [5,17], since in those intervals, the equation has at least two distinct real solutions.But wait, when b is in [-17,0), discriminant D is positive, leading to two distinct positive y's, hence four distinct real x's.When b is in [5,17], D is positive for b >5 (so four distinct real x's) and D=0 at b=5, leading to two distinct real x's. So both intervals contribute.But we need to check if for all b ∈ [-17,0) ∪ [5,17], the equation has at least two distinct real solutions, and for other values of b (i.e., (0,5)), it doesn't. Therefore, the measure (length) of the interval where the equation has at least two distinct real solutions is the length of [-17,0) plus the length of [5,17]. Since intervals are continuous, the length from -17 to 0 is 17 units, and from 5 to17 is 12 units. So total favorable length is 17 +12=29. The total interval length is 17 - (-17)=34. So the probability would be 29/34. But wait, wait, that seems too straightforward. Wait, but let me check again.Wait, interval is from -17 to 17, which is length 34. The favorable intervals are from -17 to 0, which is length 17, and from 5 to17, which is 12. So total favorable is 17 +12=29. Therefore, probability is 29/34. But the problem says m and n are relatively prime positive integers such that m/n is the probability. Then m=29, n=34. Are 29 and 34 coprime? 29 is prime, 34=2*17. 29 and 34 share no common factors, so yes. Therefore m +n=29+34=63.Wait, but hold on. Let me verify again. Because the analysis might have a mistake. When we say that for b in [-17,0) and [5,17], the equation has at least two distinct real solutions. But we need to check if there's any exception in those intervals. For example, when b is in [-17,0), is there any b where the quadratic in y has negative roots? Wait, no. Because for b <0, sum of roots is positive (4b² -10b is positive, since b is negative, 4b² is positive and -10b is positive). Product is 25b² which is positive. Therefore, both roots positive. Therefore, for all b <0, the quadratic in y has two distinct positive roots (since D>0), leading to four distinct real roots for x. So yes, that's correct.For b >=5, same logic. Sum is positive, product positive, D >=0 (D>0 for b>5, D=0 at b=5). So for b >5, two distinct positive roots leading to four real x's. At b=5, one positive double root leading to two real x's. So that's okay.Therefore, the favorable interval is indeed [-17,0) and [5,17], which is total length 17 +12=29. The total interval is 34, so probability is 29/34. Therefore m=29, n=34. GCD of 29 and 34 is 1, so m +n=63.But wait, before I conclude, let me check if when b=5, the equation actually has two distinct real solutions. Let's substitute b=5 into the original equation.Original equation: x⁴ +25*(5)^2 = (4*(5)^2 -10*5)x².Compute:Left-hand side: x⁴ +25*25 = x⁴ +625.Right-hand side: (4*25 -50)x² = (100 -50)x² =50x².So equation: x⁴ +625=50x² => x⁴ -50x² +625=0.Let y=x², so y² -50y +625=0.Compute discriminant: 2500 -2500=0. So y=(50)/2=25. So y=25 double root. Therefore, x²=25 =>x=5, x=-5. Each with multiplicity two. So equation has two distinct real solutions (5 and -5), each with multiplicity two. So the problem states "at least two distinct real solutions," which is satisfied here. So yes, b=5 is included.Similarly, let's check a value in [-17,0). Take b=-1.Equation: x⁴ +25*(-1)^2=(4*(-1)^2 -10*(-1))x² =>x⁴ +25= (4 +10)x² =>x⁴ +25=14x² =>x⁴ -14x² +25=0.Let y=x²: y² -14y +25=0.Discriminant: 196 -100=96>0. So roots y=(14 ±sqrt(96))/2=7 ± (4*sqrt(6))/2=7 ±2*sqrt(6). Both positive. Therefore, x²=7 +2√6 and x²=7 -2√6. Both positive, so x=±sqrt(7 +2√6) and x=±sqrt(7 -2√6). So four distinct real solutions. So yes, that works.What about b=10 (which is in [5,17]):Equation: x⁴ +25*(10)^2=(4*(10)^2 -10*10)x² =>x⁴ +2500=(400 -100)x² =>x⁴ +2500=300x² =>x⁴ -300x² +2500=0.Let y=x²: y² -300y +2500=0.Discriminant: 90000 -10000=80000. sqrt(80000)=282.84271247... So roots y=(300 ±282.8427)/2. So approximately (300 +282.8427)/2≈582.8427/2≈291.42135 and (300 -282.8427)/2≈17.1573/2≈8.5786. Both positive, so x=±sqrt(291.42135) and ±sqrt(8.5786). So four distinct real solutions. So that works.What about b=2.5, which is in (0,5). Since b=2.5 is in the interval where D <0 (since 0 <2.5 <5), so quadratic in y has no real solutions, so equation x⁴ +25*(2.5)^2=(4*(2.5)^2 -10*2.5)x².Compute:Left-hand side: x⁴ +25*(6.25)=x⁴ +156.25.Right-hand side: (4*6.25 -25)x²=(25 -25)x²=0. So equation becomes x⁴ +156.25=0. Which implies x⁴= -156.25. No real solutions. So that's correct.So in summary, the favorable interval is indeed [-17,0) union [5,17], length 17 +12=29. Total interval 34. Probability 29/34. So m=29, n=34, m +n=63.But wait, hold on. Wait, in the interval [-17, 0), the length is 17, but when you include 0, but since 0 is excluded, the length is still 17. From -17 to 0 is 17 units, regardless of inclusion. Similarly, from 5 to17 is 12 units. So total 29. So the probability is 29/34. Therefore, the answer is 29 +34=63.But just to double-check, maybe I made an error in the discriminant or somewhere else. Let me verify discriminant calculation again.Original equation: x⁴ +25b²=(4b² -10b)x².Moving all terms to left: x⁴ - (4b² -10b)x² +25b²=0.Let y=x²: y² - (4b² -10b)y +25b²=0.Discriminant D= [4b² -10b]^2 -4*1*25b².Which is 16b⁴ -80b³ +100b² -100b²=16b⁴ -80b³=16b³(b -5). Yes, that's correct.Then D >=0 when b³(b -5) >=0.Which is when b <=0 or b >=5. Correct.So intervals where D >=0 are b <=0 or b >=5, which in the interval [-17,17] correspond to [-17,0] and [5,17]. But since at b=0, the equation has only one real solution (x=0), which doesn't satisfy "at least two distinct real solutions". So need to exclude b=0. So the favorable interval is [-17,0) union [5,17].But in terms of measure (length), whether we include the endpoint or not doesn't affect the length. So [-17,0) has length 17, [5,17] has length12. So total 29. So yes, probability 29/34. Hence, m +n=63.Final Answerboxed{63}</think>Given the equation (x^4 + 25b^2 = (4b^2 - 10b)x^2), we need to determine the probability that it has at least two distinct real solutions when (b) is selected from the interval ([-17, 17]).First, rewrite the equation as a quadratic in (y = x^2):[y^2 - (4b^2 - 10b)y + 25b^2 = 0]The discriminant of this quadratic equation is:[D = (4b^2 - 10b)^2 - 4 cdot 25b^2 = 16b^4 - 80b^3 = 16b^3(b - 5)]For the quadratic to have real solutions, the discriminant must be non-negative:[16b^3(b - 5) geq 0]This inequality holds when (b leq 0) or (b geq 5). We need to check if the roots (y) are non-negative:- For (b < 0), the quadratic has two distinct positive roots, leading to four distinct real solutions for (x).- For (b = 0), the equation has a repeated root (y = 0), leading to one real solution (x = 0), which does not satisfy the condition.- For (b geq 5), the quadratic has two distinct positive roots (for (b > 5)) or a double positive root (for (b = 5)), leading to four or two distinct real solutions for (x), respectively.Thus, the valid intervals for (b) are ([-17, 0)) and ([5, 17]). The length of these intervals is:- ([-17, 0)) has length (17)- ([5, 17]) has length (12)Total favorable length: (17 + 12 = 29). The total interval length is (34). The probability is (frac{29}{34}), where (m = 29) and (n = 34). Therefore, (m + n = 29 + 34 = 63).[boxed{63}],<think>Okay, so I need to solve this problem where a real number b is randomly selected from the interval [-17, 17]. Then, I have to find the probability that the equation x⁴ + 25b² = (4b² - 10b)x² has at least two distinct real solutions. The answer should be in the form of a fraction m/n where m and n are coprime positive integers, and then find m + n. First, let me parse the problem step by step. The equation given is a quartic (degree 4) equation in terms of x, but maybe I can rearrange it to make it look more manageable. Let me try to bring all terms to one side:x⁴ + 25b² - (4b² - 10b)x² = 0That simplifies to:x⁴ - (4b² - 10b)x² + 25b² = 0Hmm, quartic equations can be tricky, but this one seems like a quadratic in terms of x². So maybe I can substitute y = x² and turn it into a quadratic equation in y:y² - (4b² - 10b)y + 25b² = 0Yes, that seems right. So let me write that substitution:Let y = x². Then the equation becomes:y² - (4b² - 10b)y + 25b² = 0Now, for the original equation to have at least two distinct real solutions, the equation in y must have at least one positive real solution (since y = x² must be non-negative, and each positive y gives two real x solutions: ±√y). If the quadratic in y has two distinct positive solutions, then the original equation would have four real solutions, but since the question only asks for at least two distinct real solutions, we need at least one positive real solution for y. However, if the quadratic has one positive and one negative solution, then the original equation would have two real solutions (from the positive y). If both solutions are positive, then four real solutions. If the quadratic has a repeated positive root, then two real solutions (since x² = y would give two identical roots each). But the problem states "at least two distinct real solutions". So I need to ensure that the quadratic in y has at least one positive real root, and that when considering multiplicity, they are distinct. Wait, but if the quadratic has a repeated root that is positive, then x² would have one solution with multiplicity two, leading to x having two real solutions each with multiplicity one? Wait, no. Let's think carefully.Suppose the quadratic in y has a repeated root y = k. Then x² = k would have two real solutions if k > 0, each of multiplicity 1 (since x² - k = 0 factors into (x - √k)(x + √k)). So even if y has a repeated root, as long as it's positive, x would have two distinct real solutions. But if y has a repeated root at zero, then x² = 0 would lead to x = 0 with multiplicity 2. But in that case, the original equation would have only one real solution (x=0) with multiplicity 2, but the problem says "at least two distinct real solutions", so multiplicity doesn't count for distinctness. Therefore, even if the quadratic in y has a repeated positive root, the original equation would have two distinct real solutions (since x = ±√k). So in that case, even a repeated root in y (if positive) is acceptable. However, if the repeated root in y is zero, then x would only have one real solution (x=0), which is a problem.Therefore, for the original equation to have at least two distinct real solutions, the quadratic in y must have at least one positive real root (counted with multiplicity). So if there's one positive root (either distinct or repeated), then the original equation will have at least two distinct real roots. But we need to check if the quadratic in y has at least one positive real root.Therefore, to restate the problem: For the quadratic equation y² - (4b² - 10b)y + 25b² = 0, we need to find the values of b in [-17, 17] such that there is at least one positive real root. Then, the probability would be the length of such b values divided by the total interval length (34).So, first, let me analyze the quadratic equation in y: y² - (4b² - 10b)y + 25b² = 0.To have real roots, the discriminant must be non-negative. Let's compute the discriminant D:D = [-(4b² - 10b)]² - 4 * 1 * 25b²= (4b² - 10b)² - 100b²Let me compute (4b² - 10b)²:(4b² - 10b)² = (4b²)^2 - 2*4b²*10b + (10b)^2 = 16b⁴ - 80b³ + 100b²Therefore, D = 16b⁴ - 80b³ + 100b² - 100b² = 16b⁴ - 80b³So D = 16b⁴ - 80b³ = 16b³(b - 5)Therefore, the discriminant D = 16b³(b - 5). For the quadratic in y to have real roots, we need D ≥ 0.So 16b³(b - 5) ≥ 0. Since 16 is positive, we can ignore it. So the inequality becomes:b³(b - 5) ≥ 0Let me analyze the sign of this expression. Let's factor it:b³(b - 5) = b^3(b - 5)Note that b³ has the same sign as b. So:- If b > 5: then b³ > 0, (b - 5) > 0, so the product is positive.- If b = 5: the expression is 0.- If 0 < b < 5: b³ > 0, (b - 5) < 0, so the product is negative.- If b = 0: the expression is 0.- If b < 0: b³ < 0, (b - 5) < 0, so the product is positive.Therefore, D ≥ 0 when b ∈ (-∞, 0] ∪ [5, ∞). But since our b is restricted to [-17, 17], so the valid intervals for D ≥ 0 are [-17, 0] ∪ [5, 17].But even if the discriminant is non-negative, we need to ensure that at least one root of the quadratic in y is positive. So even if there are real roots, they need to be positive for x² to have real solutions.So first, we need to split the problem into two cases based on the intervals where D ≥ 0: [-17, 0] and [5, 17].But perhaps we need to check in each interval whether the roots are positive.Wait, but maybe there's a better way. Let's recall that for a quadratic equation ay² + by + c = 0, the conditions for having at least one positive root can be determined by checking the signs of the coefficients or using the quadratic formula.In our case, the quadratic equation is y² - (4b² - 10b)y + 25b² = 0. Let's denote A = 1, B = -(4b² - 10b), C = 25b².Then, the quadratic equation is Ay² + By + C = 0. So A = 1, B = -(4b² - 10b), C = 25b².To have at least one positive root, we can use the following criteria:1. If C/A < 0, then the quadratic has one positive and one negative root.2. If C/A = 0, then one root is 0, and the other is -B/A. If -B/A is positive, then there is a positive root.3. If C/A > 0, then both roots have the same sign. In that case, we need to check if the sum of the roots, -B/A, is positive. If both roots are positive, then their sum is positive and their product is positive. If both roots are negative, their sum is negative and product is positive.In our case, C/A = 25b² / 1 = 25b², which is always non-negative (since b is real). So C/A ≥ 0 for all real b. So cases 1 and 2: C/A is 0 when b = 0. Let's handle that first.Case 1: b = 0.If b = 0, the quadratic equation becomes y² - (0 - 0)y + 0 = y² = 0. So the only root is y = 0 with multiplicity 2. Then, x² = 0 implies x = 0 with multiplicity 2. So the original equation has only one distinct real solution (x=0), which doesn't satisfy "at least two distinct real solutions". Therefore, b = 0 is excluded.Case 2: b ≠ 0.Then, C/A = 25b² > 0. So both roots have the same sign. Therefore, to have at least one positive root, both roots need to be positive. Therefore, the sum of the roots must be positive. The sum of the roots is -B/A = (4b² - 10b). So for the sum to be positive:4b² - 10b > 0Let me solve this inequality:4b² - 10b > 0Factor:2b(2b - 5) > 0So, critical points at b = 0 and b = 5/2.Test intervals:- For b < 0: 2b is negative, (2b - 5) is negative. Negative * negative = positive. So inequality holds.- For 0 < b < 5/2: 2b is positive, (2b - 5) is negative. Positive * negative = negative. Inequality does not hold.- For b > 5/2: 2b is positive, (2b - 5) is positive. Positive * positive = positive. Inequality holds.Therefore, 4b² - 10b > 0 when b ∈ (-∞, 0) ∪ (5/2, ∞). But since our b is in [-17, 17], adjusting intervals: [-17, 0) ∪ (5/2, 17].But remember, we also required that the quadratic has real roots, which we found earlier happens when b ∈ [-17, 0] ∪ [5, 17]. So combining both conditions:We need real roots (D ≥ 0): b ∈ [-17, 0] ∪ [5, 17]ANDSum of roots positive: 4b² -10b > 0 => b ∈ [-17, 0) ∪ (5/2, 17]Therefore, intersecting these two intervals:For the discriminant D ≥ 0 and sum positive:- In [-17, 0], D is non-negative, and sum is positive (since for b < 0, sum is positive). But wait, when b is in [-17, 0), the sum 4b² -10b is positive (since 4b² is positive and -10b is positive as b is negative). So yes, sum is positive. However, at b = 0, sum is 0, but we already considered b=0 as a separate case where the quadratic has a double root at y=0, which doesn't give distinct real solutions. So in [-17, 0), we have both D ≥ 0 and sum positive. Therefore, the roots are positive. So the quadratic has two positive roots (since product is positive and sum is positive). Therefore, in this interval, original equation has four real solutions (two from each positive y root). So at least two distinct real solutions.In [5, 17], D is non-negative (since b >=5), and sum 4b² -10b. Let's check if the sum is positive here. For b >=5, 4b² -10b is definitely positive (since 4b² grows faster than 10b). So sum is positive. Therefore, in [5,17], quadratic has two positive roots, leading to four real solutions for x. So at least two distinct real solutions.However, in the interval (5/2, 5), what's the situation? Wait, discriminant D is negative there, right? Because D >=0 only when b <=0 or b >=5. So in (5/2,5), D is negative, so quadratic has no real roots. Therefore, even though sum is positive in (5/2,5), since there are no real roots, the original equation has no real solutions. Therefore, we have to be careful. So combining the two conditions:We need D >=0 (so real roots) AND sum positive (so roots positive). Therefore, the valid intervals are [-17,0] and [5,17].But wait, when b is in [5,17], the discriminant is non-negative, and sum is positive. Similarly, when b is in [-17,0], discriminant is non-negative and sum is positive. However, in between, from 0 to 5, discriminant is negative, so no real roots, so original equation has no real solutions.But wait, let me confirm with specific test values.Take b = -1 (in [-17,0)):Compute discriminant D =16*(-1)^3*(-1 -5)=16*(-1)*(-6)=96>0. So real roots.Sum of roots is 4*(-1)^2 -10*(-1)=4 +10=14>0. Product is 25*(-1)^2=25>0. So two positive roots. Therefore, original equation has four real solutions.Take b=1 (in (0,5)):Discriminant D=16*(1)^3*(1-5)=16*1*(-4)= -64 <0. So no real roots. Original equation has no real solutions.Take b=6 (in [5,17]):D=16*(6)^3*(6-5)=16*216*1=3456>0. Sum=4*(6)^2 -10*6=144-60=84>0. Product=25*(6)^2=900>0. So two positive roots. Original equation has four real solutions.Therefore, the valid interval for b is [-17,0] and [5,17]. However, we need to check if in the entire interval [-17,0], the quadratic in y has real roots and positive roots, which we have confirmed.But wait, when b=0, the quadratic equation becomes y² +0 +0 = y²=0, so only root y=0, which is non-negative but gives x=0 as a repeated root. However, the problem states "at least two distinct real solutions". So when b=0, the equation has only one distinct real solution (x=0) with multiplicity 2. Therefore, b=0 should be excluded from the valid interval.Similarly, when b=5, let's check:At b=5, discriminant D=16*(5)^3*(5-5)=0. So quadratic has a repeated root. Compute the root:The quadratic equation is y² - (4*(5)^2 -10*5)y +25*(5)^2 = y² - (100 -50)y + 625 = y² -50y +625=0Solving: y = [50 ± sqrt(2500 -2500)]/2 = 50/2=25. So repeated root y=25. Then x²=25 gives x=±5, which are two distinct real solutions. So even though the quadratic in y has a repeated root, the original equation has two distinct real solutions. So b=5 is acceptable.Similarly, at b=0, the equation gives x=0 (only one solution), which is not acceptable. Therefore, the valid intervals are [-17,0) and [5,17]. Because at b=0, the equation doesn't satisfy the condition, but at b=5, it does.Therefore, total valid interval length is length of [-17,0) which is 17 - (-17) = wait, no. Wait, the interval [-17,0) is from -17 to 0, which is 17 units long. The interval [5,17] is from 5 to 17, which is 12 units long. So total valid length is 17 +12=29. But the total interval is [-17,17], which is 34 units long. Therefore, the probability is 29/34. But wait, that would be if the valid intervals are [-17,0) and [5,17]. But let me confirm:Wait, the interval [-17,0] is 17 units (from -17 to 0, inclusive). But we have to exclude b=0. So [-17,0) is 17 units (since in measure theory, the single point at 0 doesn't affect the length). Similarly, [5,17] is 12 units. So total measure is 17 +12=29. Therefore, probability is 29/34. But wait, in the problem statement, it says "randomly selected from the interval [-17,17]". So the length is 17 - (-17)=34. So 29/34 simplifies to... 29 and 34 are coprime? 29 is a prime number, 34=2×17, so yes. Therefore, m=29, n=34, m +n=63. But wait, hold on. Wait, no, wait a second. Let me double-check my reasoning.Wait, when b is in [-17,0), we have valid solutions, and when b is in [5,17], also valid. The total measure is 17 (from -17 to 0) plus 12 (from 5 to 17) equals 29. The total interval is 34. So probability is 29/34, which reduces to 29/34, since 29 is prime. Therefore, m=29, n=34, m +n=63.But hold on. Let me check another edge case. For example, take b approaching 0 from the left. Let b = -ε, where ε is a small positive number. Then the quadratic equation becomes y² - (4ε² +10ε)y +25ε² =0. Discriminant D = [4ε² +10ε]^2 -100ε². Let me compute that:(4ε² +10ε)^2 = 16ε⁴ +80ε³ +100ε²Then D =16ε⁴ +80ε³ +100ε² -100ε²=16ε⁴ +80ε³. As ε approaches 0, D approaches 0. Therefore, near b=0 from the left, the quadratic in y has two real roots approaching 0. However, since sum of roots is 4b² -10b, which is 4ε² +10ε, which approaches 0 as ε approaches 0. But wait, when b approaches 0 from the left, sum approaches 0 and product approaches 0. Wait, let's compute the roots:Using quadratic formula:y = [ (4b² -10b) ± sqrt(D) ] / 2But since D=16b³(b -5), when b approaches 0 from the left, D=16b³(b -5). Since b is approaching 0 from the left (negative), b³ is negative, and (b -5) is negative, so D is positive (negative * negative=positive). So sqrt(D) is real.But let's plug in b approaching 0 from the left. Let b = -ε, ε >0 small.Then:Sum of roots = 4b² -10b =4ε² +10εProduct =25b²=25ε²Then the quadratic equation is y² - (4ε² +10ε)y +25ε²=0.Using quadratic formula:y = [4ε² +10ε ± sqrt((4ε² +10ε)^2 -100ε²)] / 2We already computed sqrt(D) as sqrt(16ε⁴ +80ε³). Let me factor sqrt(16ε⁴ +80ε³)=sqrt(16ε³(ε +5)). So sqrt(16ε³(ε +5))=4ε^(3/2)sqrt(ε +5). But as ε approaches 0, this is approximately 4ε^(3/2)*sqrt(5). Therefore, the sqrt(D) term is of order ε^(3/2), which is negligible compared to the linear term in ε in the sum.Therefore, the two roots are approximately:y ≈ [10ε ± 0]/2 =5ε and [4ε² +10ε ± 4ε^(3/2)sqrt(5)] /2 ≈ [10ε]/2=5ε and [4ε² +10ε]/2 ≈5ε +2ε². So both roots approach 0 as ε approaches 0. Therefore, when b approaches 0 from the left, both roots approach 0. Therefore, the original equation x² = y would have solutions approaching 0. But since the roots are distinct (as D is positive for b <0), we have two distinct positive roots for y near 0, leading to four real solutions for x, approaching ±√(5ε) and ±√(5ε +2ε²). As ε approaches 0, these solutions approach 0, but remain distinct. Therefore, even as b approaches 0 from the left, the original equation has four distinct real solutions. However, at b=0, all these solutions collapse to x=0, which is a single solution. Therefore, the point b=0 must be excluded. Hence, the interval is indeed [-17,0) and [5,17].Therefore, the total measure is 17 (from -17 to 0) plus 12 (from 5 to17) equals 29. So probability is 29/34. Thus, m=29, n=34, and m +n=63.Wait, but wait another moment. Let me check if there are any other points where the equation might not have two distinct real solutions even though the quadratic in y has positive roots. For example, suppose the quadratic in y has two distinct positive roots, but when solving x² = y, the roots could result in repeated x solutions. But since x² = y1 and x² = y2 with y1 ≠ y2, then the solutions would be ±√y1 and ±√y2, which are distinct as long as y1 ≠ y2. Therefore, unless y1 = y2, but in that case, the quadratic would have a repeated root. So, if quadratic in y has a repeated positive root, then x² = y would give ±√y, which are two distinct solutions. Therefore, even with a repeated root in y, as long as it's positive, we get two distinct real solutions for x. Therefore, in the interval [5,17], even at b=5 where quadratic has a repeated root y=25, leading to x=±5, which are two distinct real solutions. Therefore, even at the endpoint b=5, we have two distinct real solutions, so b=5 is included. Similarly, in the interval [-17,0), approaching b=0, we have two distinct real solutions, but at b=0, it's only one. Therefore, the valid interval is [-17,0) union [5,17]. Since measure of [-17,0) is 17, and [5,17] is 12, total 29. Therefore, the probability is 29/34. So m +n=29+34=63.But hold on, I just need to confirm once more if there are any other constraints or mistakes in my reasoning.Wait, when b is in [-17,0), the quadratic in y has two distinct positive roots (since D >0, sum >0, product >0). Therefore, x² = y1 and x² = y2 gives four distinct real solutions (since y1 ≠ y2 and positive). Therefore, the original equation has four distinct real solutions. Thus, satisfies "at least two distinct real solutions".When b is in [5,17], same reasoning: quadratic in y has two distinct positive roots (D >0 except at b=5, where D=0 but repeated root y=25 which is positive). So in [5,17], even with b=5, quadratic has a repeated positive root, leading to two distinct real solutions x=±5. So original equation has two distinct real solutions. So the interval [5,17] is entirely valid.Therefore, combining both intervals [-17,0) and [5,17], the measure is 17 +12=29, over total 34. So probability 29/34. Therefore, m=29, n=34, which are coprime since 29 is prime and doesn't divide 34. So m +n=63.Therefore, the answer is boxed{63}.Final Answerboxed{63}</think>Given the equation (x^4 + 25b^2 = (4b^2 - 10b)x^2), we need to determine the probability that it has at least two distinct real solutions when (b) is randomly selected from the interval ([-17, 17]).First, we rewrite the equation as a quadratic in (y = x^2):[y^2 - (4b^2 - 10b)y + 25b^2 = 0]For this quadratic equation to have real roots, the discriminant must be non-negative:[Delta = (4b^2 - 10b)^2 - 4 cdot 25b^2 = 16b^3(b - 5)]The discriminant (Delta geq 0) when (b in [-17, 0] cup [5, 17]).Next, we check if the quadratic equation has at least one positive root. The product of the roots is (25b^2), which is non-negative. For the roots to be positive, the sum of the roots (4b^2 - 10b) must be positive:[4b^2 - 10b > 0 implies 2b(2b - 5) > 0]This inequality holds when (b in (-infty, 0) cup (frac{5}{2}, infty)). Combining this with the intervals where the discriminant is non-negative, we get:- For (b in [-17, 0)), the discriminant is non-negative and the sum is positive.- For (b in [5, 17]), the discriminant is non-negative and the sum is positive.At (b = 0), the equation has only one real solution (x = 0), which is excluded. At (b = 5), the quadratic has a repeated positive root, leading to two distinct real solutions.The valid intervals for (b) are ([-17, 0)) and ([5, 17]). The lengths of these intervals are 17 and 12, respectively. The total length is (17 + 12 = 29), and the length of the entire interval ([-17, 17]) is 34.Thus, the probability is (frac{29}{34}). The integers (m = 29) and (n = 34) are coprime, so (m + n = 29 + 34 = 63).[boxed{63}]]

🔑:1. Start by rearranging the given equation: [ x^4 + 25b^2 = (4b^2 - 10b)x^2 ] This can be rewritten as: [ x^4 - (4b^2 - 10b)x^2 + 25b^2 = 0 ] Let ( y = x^2 ). Then the equation becomes a quadratic in ( y ): [ y^2 - (4b^2 - 10b)y + 25b^2 = 0 ]2. For the quadratic equation ( y^2 - (4b^2 - 10b)y + 25b^2 = 0 ) to have at least two distinct real solutions, its discriminant must be non-negative: [ Delta = (4b^2 - 10b)^2 - 4 cdot 1 cdot 25b^2 ] Simplify the discriminant: [ Delta = (4b^2 - 10b)^2 - 100b^2 ] [ Delta = 16b^4 - 80b^3 + 100b^2 - 100b^2 ] [ Delta = 16b^4 - 80b^3 ] Factor out the common term: [ Delta = 16b^3(b - 5) ]3. For the discriminant to be positive: [ 16b^3(b - 5) > 0 ] This inequality holds when ( b ) is in the intervals: [ b < 0 quad text{or} quad b > 5 ]4. Determine the probability that ( b ) falls within these intervals. The interval for ( b ) is ([-17, 17]). The intervals where the discriminant is positive are ( b < 0 ) and ( b > 5 ).5. Calculate the lengths of these intervals: - For ( b < 0 ), the interval is ([-17, 0)), which has a length of 17. - For ( b > 5 ), the interval is ((5, 17]), which has a length of (17 - 5 = 12).6. The total length of the interval where the discriminant is positive is: [ 17 + 12 = 29 ]7. The total length of the interval ([-17, 17]) is 34. Therefore, the probability that ( b ) falls within the desired intervals is: [ frac{29}{34} ]8. Since ( m ) and ( n ) are relatively prime positive integers such that ( frac{m}{n} = frac{29}{34} ), we have ( m = 29 ) and ( n = 34 ).9. The value of ( m + n ) is: [ 29 + 34 = 63 ]The final answer is ( boxed{63} ).

🧠:[<think>Okay, so I have this problem here with vectors OA and OB, both unit vectors in the xy-plane, right? Their coordinates are given as OA = {p, q, 0} and OB = {r, s, 0}. Both of these vectors form a 45-degree angle with another vector OC = {1, 1, 1}. The question has two parts: first, find p + q and pq, and second, find the acute angle between OA and OB. Let me try to figure this out step by step.Starting with part (1). Since OA is a unit vector, its magnitude should be 1. So, sqrt(p² + q² + 0²) = 1. That simplifies to p² + q² = 1. Similarly, OB is a unit vector, so r² + s² = 1, but maybe we'll need that later for part (2). Now, OA forms a 45-degree angle with OC. The formula for the angle θ between two vectors u and v is cosθ = (u · v)/(|u||v|). Here, θ is 45 degrees, so cos45° = √2/2. Let's apply this formula to vectors OA and OC.The dot product OA · OC = p*1 + q*1 + 0*1 = p + q. The magnitude of OA is 1 (since it's a unit vector), and the magnitude of OC is sqrt(1² + 1² + 1²) = sqrt(3). Therefore, cos45° = (p + q)/(1 * sqrt(3)) => √2/2 = (p + q)/sqrt(3). Multiplying both sides by sqrt(3) gives p + q = sqrt(3) * √2 / 2. Wait, sqrt(3) times sqrt(2) is sqrt(6), so p + q = sqrt(6)/2. Hmm, that's interesting. So that's the value of p + q. The problem asks for both p + q and pq. So we have p + q = sqrt(6)/2. Now, how to find pq?We already know that p² + q² = 1. Let's recall that (p + q)² = p² + 2pq + q². So if we compute (sqrt(6)/2)², that should be equal to p² + q² + 2pq. Plugging the numbers in:(sqrt(6)/2)² = 6/4 = 3/2. And p² + q² is 1. So 3/2 = 1 + 2pq. Subtracting 1 from both sides gives 1/2 = 2pq, so pq = 1/4. Wait, so pq is 1/4? That seems straightforward. Let me check that again.Yes, (p + q)² = 3/2, and p² + q² = 1. Then 3/2 - 1 = 2pq = 1/2, so pq = 1/4. That makes sense. So part (1) is done: p + q = sqrt(6)/2 and pq = 1/4.Moving on to part (2), finding the acute angle between OA and OB. Let's denote this angle as φ. Since OA and OB are both unit vectors, the dot product OA · OB = |OA||OB|cosφ = 1*1*cosφ = cosφ. So cosφ = OA · OB = p*r + q*s + 0*0 = pr + qs. So we need to find pr + qs.But we don't know r and s. However, we do know that OB is also a unit vector forming a 45-degree angle with OC. So similar to OA, the dot product OB · OC = r + s, and the same calculation as before would give r + s = sqrt(6)/2 as well. Similarly, r² + s² = 1, so using the same logic, rs = 1/4. Wait, so similar to OA, OB has r + s = sqrt(6)/2 and rs = 1/4. But how does that help us find pr + qs? Hmm. Maybe there's a way to relate pr + qs with the other variables. Let me think. Let's consider expanding (p + q)(r + s). That would be pr + ps + qr + qs. We know p + q = sqrt(6)/2 and r + s = sqrt(6)/2, so their product is (sqrt(6)/2)^2 = 6/4 = 3/2. Therefore, pr + ps + qr + qs = 3/2. But we need pr + qs. If we can find ps + qr, then we can subtract that from 3/2 to get pr + qs.Alternatively, maybe we can find another expression. Let me recall that (p + q)(r + s) = pr + ps + qr + qs = 3/2. Also, if we consider (p - q)(r - s), that would be pr - ps - qr + qs. Not sure if that helps. Alternatively, maybe we can find pr + qs through some other means.Alternatively, maybe we can use the values of pq and rs. Since pq = 1/4 and rs = 1/4. Hmm, not sure. Wait, perhaps if we consider the sum of OA and OB or something like that. Alternatively, maybe think in terms of variables. Let me consider that p and q are variables satisfying p + q = sqrt(6)/2 and pq = 1/4. Similarly, r and s satisfy r + s = sqrt(6)/2 and rs = 1/4. So both OA and OB have components that are roots of the same quadratic equation. So perhaps p and q are roots of x² - (sqrt(6)/2)x + 1/4 = 0, and similarly for r and s. Therefore, the solutions would be [sqrt(6)/2 ± sqrt( (6/4) - 1 )]/2. Let me compute that discriminant: (6/4 - 1) = 6/4 - 4/4 = 2/4 = 1/2. So sqrt(1/2) = sqrt(2)/2. Therefore, p and q are [sqrt(6)/2 ± sqrt(2)/2]/2. Wait, dividing by 2 again. So that's [sqrt(6) ± sqrt(2)]/4. So p = [sqrt(6) + sqrt(2)]/4 and q = [sqrt(6) - sqrt(2)]/4, or vice versa. Similarly for r and s.But since OA and OB are different vectors, maybe r and s are similar, but perhaps different permutations? Wait, but since they are both in the xy-plane and both form 45 degrees with OC, maybe OA and OB are symmetric in some way. But I'm not sure.Alternatively, maybe we can parameterize OA and OB. Let's think of OA as a vector in the xy-plane making a 45-degree angle with OC. Let me recall that OC is {1,1,1}, so the projection of OC onto the xy-plane is {1,1,0}, which makes a 45-degree angle with the x and y axes. Wait, but OA is a vector in the xy-plane. The angle between OA and OC is 45 degrees. Hmm. The angle between OA and OC can be found using the dot product, which we already did. So OA is in the xy-plane, so its z-component is zero. OC has a z-component of 1, so the angle between OA and OC is determined by their dot product over the product of their magnitudes.But maybe we can visualize this. OC is going up at 45 degrees from the xy-plane, right? Because it's {1,1,1}, so in 3D space. The angle between OA (which is in the xy-plane) and OC is 45 degrees. So perhaps OA is aligned in a certain direction in the xy-plane such that when you project OC onto OA, the cosine of the angle is √2/√3 (since cos45° = √2/2 and |OC| is sqrt(3)), so the dot product p + q equals sqrt(6)/2 as we found before.But how does this help us find pr + qs? Hmm. Let me think. Since both OA and OB are in the xy-plane and have the same p + q and r + s, which are both sqrt(6)/2, and their products pq and rs are 1/4. So both pairs (p, q) and (r, s) are solutions to the equation x² - (sqrt(6)/2)x + 1/4 = 0. Therefore, p and q are the roots of this equation, and similarly for r and s. So, since this quadratic has two roots, which are [sqrt(6) ± sqrt(2)]/4, as calculated before. Therefore, OA and OB could be any two vectors with components ( [sqrt(6) + sqrt(2)]/4 , [sqrt(6) - sqrt(2)]/4 , 0 ) and ( [sqrt(6) - sqrt(2)]/4 , [sqrt(6) + sqrt(2)]/4 , 0 ), but since they are different vectors, maybe OA and OB are these two different possibilities. Wait, but then their dot product would be pr + qs. Let's compute that.If OA is ( [sqrt(6) + sqrt(2)]/4 , [sqrt(6) - sqrt(2)]/4 , 0 ) and OB is ( [sqrt(6) - sqrt(2)]/4 , [sqrt(6) + sqrt(2)]/4 , 0 ), then pr + qs would be:[ (sqrt(6) + sqrt(2))/4 * (sqrt(6) - sqrt(2))/4 ] + [ (sqrt(6) - sqrt(2))/4 * (sqrt(6) + sqrt(2))/4 ]Let me compute each term:First term: (sqrt(6) + sqrt(2))(sqrt(6) - sqrt(2)) = (sqrt(6))^2 - (sqrt(2))^2 = 6 - 2 = 4. Then divided by 4*4 = 16, so 4/16 = 1/4.Second term is the same: (sqrt(6) - sqrt(2))(sqrt(6) + sqrt(2)) = 4, so also 4/16 = 1/4.Therefore, pr + qs = 1/4 + 1/4 = 1/2. Therefore, cosφ = 1/2, so φ = 60 degrees. Wait, because cosφ = 1/2, the angle is 60°, which is acute. So the acute angle between OA and OB is 60 degrees. That seems to be the answer.But wait, is this the only possibility? Let me check. Suppose OA and OB are both the same vector, but the problem states both OA and OB are unit vectors forming 45° with OC, but they might not be the same vector. However, in the problem, they are asking for the angle between OA and OB. If OA and OB are different vectors constructed as above, then their angle is 60°, but maybe there's another way to get a different angle. Wait, but given the constraints, maybe all such vectors OA and OB must satisfy this angle. Let me see.Alternatively, perhaps there's a more straightforward way to compute pr + qs without parameterizing p, q, r, s. Let's think algebraically. We have:From OA: p + q = sqrt(6)/2 and pq = 1/4.From OB: r + s = sqrt(6)/2 and rs = 1/4.We need to compute pr + qs.Let me consider that pr + qs can be written as p*r + q*s. Hmm. Let's try to express this in terms of (p + q)(r + s). We already know that (p + q)(r + s) = pr + ps + qr + qs = 3/2. If we can find ps + qr, then subtract that from 3/2 to get pr + qs. So we need to find ps + qr.Alternatively, consider (p - q)(r - s) = pr - ps - qr + qs. Not sure if that helps. Alternatively, think of ps + qr as p*s + q*r. Let's see. If we can express this in terms of other known quantities.Wait, maybe we can use the fact that (p + q)(r + s) = pr + ps + qr + qs = 3/2.But we also have p² + q² = 1 and r² + s² = 1. Hmm. Let me also note that (p + q)^2 = p² + 2pq + q² = 1 + 2*(1/4) = 1 + 1/2 = 3/2. Which matches our earlier result.Similarly for (r + s)^2 = 3/2. But how to relate pr + qs with these?Wait, let's consider the product (p*r + q*s). Let me think of vectors OA = (p, q, 0) and OB = (r, s, 0). The dot product is pr + qs. We need to find this value.Alternatively, since OA and OB are both vectors in the xy-plane, making 45° with OC. Let me think about their relationship to OC. The projection of OC onto the xy-plane is (1,1,0), which is a vector of length sqrt(2). The angle between OA and OC is 45°, so maybe OA is somehow related to the projection of OC onto the xy-plane. But since OA is a unit vector, maybe OA is the normalized projection? Wait, the projection of OC onto the xy-plane is (1,1,0), which has length sqrt(2). If we normalize that, we get (1/sqrt(2), 1/sqrt(2), 0). But then, the angle between this vector and OC would be computed as:Dot product: (1/sqrt(2), 1/sqrt(2), 0) · (1,1,1) = (1/sqrt(2) + 1/sqrt(2) + 0) = 2/sqrt(2) = sqrt(2). The magnitude of the normalized projection vector is 1, and |OC| is sqrt(3), so cosθ = sqrt(2)/sqrt(3) ≈ 0.816, which is about 35.26 degrees, not 45. So that's not the case.Therefore, OA isn't simply the normalized projection. So back to the problem. Since OA and OB are arbitrary unit vectors in the xy-plane forming 45° with OC, but not necessarily aligned with the projection. Therefore, there are infinitely many such vectors OA and OB in the xy-plane, but given the constraints of p + q and pq, we can find their specific components.But since we were able to parameterize OA and OB as having components [sqrt(6) ± sqrt(2)]/4, and by taking different combinations for OA and OB, their dot product ends up as 1/2, leading to 60°. Therefore, regardless of how OA and OB are chosen (as different solutions to the quadratic), the angle between them is 60°, which is acute. Therefore, the answer is 60°.Alternatively, maybe there's another approach using vector algebra without solving quadratics. Let's consider that OA and OB are two different unit vectors in the xy-plane each forming 45° with OC. Let's consider the angle between OA and OB. Let me recall that the angle between OA and OC is 45°, same for OB and OC. Maybe we can use the formula involving the angles between OA, OB, and OC.But I think the earlier approach is more straightforward. Since OA and OB are both in the xy-plane, their z-components are zero. The vector OC has components (1,1,1). The angle between OA and OC is determined by the dot product, which is p + q, and we found p + q = sqrt(6)/2. Similarly for OB. Then, by parameterizing OA and OB as solutions to the quadratic equation, we found that their dot product is 1/2, leading to an angle of 60°. Therefore, the answer to part (2) is 60 degrees. Let me check if this makes sense. If two vectors each form a 45° angle with a third vector, the angle between them can vary depending on their orientation. But in this specific case, since OA and OB are constrained to the xy-plane and OC has a z-component, there might be only one possible angle between OA and OB. Our calculation shows it's 60°, which seems plausible.Alternatively, maybe there's a geometric interpretation. Imagine OC is a space diagonal of a cube with edges of length 1. The vectors OA and OB are unit vectors in the xy-plane making 45° with OC. But I think the algebraic approach is more precise here.In summary, for part (1), p + q = sqrt(6)/2 and pq = 1/4. For part (2), the angle between OA and OB is 60 degrees. I think that's the answer.Final Answer(1) ( p + q = boxed{dfrac{sqrt{6}}{2}} ) and ( pq = boxed{dfrac{1}{4}} );(2) The acute angle ( angle AOB ) measures ( boxed{60^circ} ).</think>Given the problem where overrightarrow{OA} and overrightarrow{OB} are unit vectors in the xy-plane, both forming a 45^circ angle with overrightarrow{OC} = {1, 1, 1}:1. Finding p + q and pq: - Since overrightarrow{OA} is a unit vector, we have p^2 + q^2 = 1. - The angle between overrightarrow{OA} and overrightarrow{OC} is 45^circ. Using the dot product formula: [ cos 45^circ = frac{overrightarrow{OA} cdot overrightarrow{OC}}{|overrightarrow{OA}| |overrightarrow{OC}|} ] [ frac{sqrt{2}}{2} = frac{p + q}{sqrt{3}} ] Solving for p + q gives: [ p + q = frac{sqrt{6}}{2} ] - Using the identity (p + q)^2 = p^2 + 2pq + q^2: [ left(frac{sqrt{6}}{2}right)^2 = 1 + 2pq ] [ frac{3}{2} = 1 + 2pq implies 2pq = frac{1}{2} implies pq = frac{1}{4} ]2. Finding the acute angle angle AOB: - Both overrightarrow{OA} and overrightarrow{OB} are unit vectors with p + q = frac{sqrt{6}}{2} and r + s = frac{sqrt{6}}{2}, and pq = frac{1}{4} and rs = frac{1}{4}. - The dot product overrightarrow{OA} cdot overrightarrow{OB} = pr + qs needs to be found. Considering the quadratic roots for p, q and r, s: [ p, q = frac{sqrt{6} pm sqrt{2}}{4}, quad r, s = frac{sqrt{6} pm sqrt{2}}{4} ] - Calculating the dot product for different combinations gives: [ pr + qs = frac{1}{4} + frac{1}{4} = frac{1}{2} ] - The angle phi between overrightarrow{OA} and overrightarrow{OB} is: [ cos phi = frac{1}{2} implies phi = 60^circ ] Final Answer(1) ( p + q = boxed{dfrac{sqrt{6}}{2}} ) and ( pq = boxed{dfrac{1}{4}} );(2) The acute angle ( angle AOB ) measures ( boxed{60^circ} ).,<think>Okay, so I have this problem here with vectors OA and OB, both unit vectors in the xy-plane, right? They form a 45-degree angle with vector OC, which is {1,1,1}. I need to find p + q and pq for part (1), and then the angle between OA and OB for part (2). Let me try to break this down step by step.First, OA is a unit vector, so its components {p, q, 0} must satisfy p² + q² = 1. Similarly, OB is also a unit vector, so r² + s² = 1. Both OA and OB form a 45-degree angle with OC. The angle between two vectors can be found using the dot product formula: the cosine of the angle is equal to the dot product of the vectors divided by the product of their magnitudes.Since OC is {1,1,1}, its magnitude is sqrt(1² + 1² + 1²) = sqrt(3). OA is a unit vector, so its magnitude is 1. The angle between OA and OC is 45 degrees, so the dot product of OA and OC should equal |OA||OC|cos(45°). Let's write that out:OA · OC = p*1 + q*1 + 0*1 = p + q. The right side is |OA||OC|cos(45°) = 1 * sqrt(3) * (√2/2) = sqrt(3)/sqrt(2) = sqrt(6)/2. Therefore, p + q = sqrt(6)/2. Wait, but the problem says both OA and OB form 45-degree angles with OC. So similarly, for OB, we have r + s = sqrt(6)/2 as well? Hmm, but part (1) only asks for p + q and pq. So maybe I can focus just on OA first?Wait, but maybe there's more to it. Since both OA and OB are in the xy-plane and form 45 degrees with OC, maybe they have similar properties. Let me check the first part again.The problem states both OA and OB form a 45-degree angle with OC. So yes, both OA · OC = sqrt(6)/2 and OB · OC = sqrt(6)/2. But the first part only asks about OA's p + q and pq, so maybe I don't need to consider OB for part (1). Let me confirm.Problem 14: (1) Find the values of p + q and pq; (2) Find the size of angle AOB (acute angle). So part (1) is about OA, since OA is {p, q, 0} and OB is {r, s, 0}, but (1) asks for p + q and pq, which are components of OA. So maybe part (1) is just related to OA. Wait, but the problem says "both form a 45° angle with OC". So OA and OB each individually form 45° with OC. So each of their dot products with OC is sqrt(6)/2. So OA · OC = p + q = sqrt(6)/2, and OB · OC = r + s = sqrt(6)/2. But since part (1) is only about p + q and pq, perhaps we can just use OA's components.So OA is {p, q, 0}, unit vector: p² + q² = 1. OA · OC = p + q = sqrt(6)/2. Therefore, p + q is given directly as sqrt(6)/2. Wait, but the problem says "Find the values of p + q and pq". So maybe p + q is sqrt(6)/2, and pq can be found using the equation p² + q² = 1 and (p + q)² = p² + 2pq + q². So let's compute that.(p + q)² = (sqrt(6)/2)^2 = (6/4) = 3/2. But p² + q² = 1, so 1 + 2pq = 3/2. Therefore, 2pq = 3/2 - 1 = 1/2, so pq = 1/4. So part (1) is p + q = sqrt(6)/2 and pq = 1/4. That seems straightforward. Let me check if that makes sense.If p + q = sqrt(6)/2 ≈ 1.2247, and p² + q² = 1. Let's see, for example, suppose p and q are both equal. Then p = q = sqrt(6)/4 ≈ 0.6124, then p² + q² = 2*(6/16) = 12/16 = 3/4 ≠ 1. So they can't be equal. But if pq = 1/4, then the quadratic equation would be x² - (sqrt(6)/2)x + 1/4 = 0. Let's compute the discriminant: (sqrt(6)/2)^2 - 4*(1/4) = (6/4) - 1 = (6/4 - 4/4) = 2/4 = 1/2. So the roots are [sqrt(6)/2 ± sqrt(1/2)] / 2. Hmm, sqrt(1/2) is sqrt(2)/2, so the roots would be [sqrt(6) ± sqrt(2)] / 4. So p and q are [sqrt(6) + sqrt(2)]/4 and [sqrt(6) - sqrt(2)]/4. Let me check their squares:For [sqrt(6) + sqrt(2)]/4: square is (6 + 2 + 2*sqrt(12))/16 = (8 + 4*sqrt(3))/16 = (2 + sqrt(3))/4 ≈ (2 + 1.732)/4 ≈ 0.933. Similarly, the other term is [sqrt(6) - sqrt(2)]^2 /16 = (6 + 2 - 2*sqrt(12))/16 = (8 - 4*sqrt(3))/16 = (2 - sqrt(3))/4 ≈ (2 - 1.732)/4 ≈ 0.067. Adding both, 0.933 + 0.067 = 1. That works. So p and q are those values, and their sum is sqrt(6)/2 and product is 1/4. So part (1) seems okay.Now part (2) is the angle between OA and OB. Since OA and OB are both unit vectors, the angle between them can be found by the dot product: OA · OB = |OA||OB|cosθ = cosθ. So if I can find OA · OB, then θ = arccos(OA · OB). OA · OB is pr + qs. Hmm, but I need to find pr + qs. But how?Wait, OA is {p, q, 0}, OB is {r, s, 0}, so their dot product is pr + qs. Both OA and OB are unit vectors in the xy-plane forming 45 degrees with OC. Also, OA and OB are different vectors, probably not the same. So maybe we can find pr + qs in terms of p + q, r + s, and other known quantities.We know from part (1) that p + q = sqrt(6)/2 and similarly, since OB is also a unit vector forming 45 degrees with OC, then r + s = sqrt(6)/2. Also, OA and OB are both unit vectors, so p² + q² = 1 and r² + s² = 1.If I consider OA · OB = pr + qs. Let's see if we can express this in terms of (p + q)(r + s) - (ps + qr). Wait, (p + q)(r + s) = pr + ps + qr + qs. So pr + qs = (p + q)(r + s) - (ps + qr). But I don't know what ps + qr is. Hmm.Alternatively, maybe since OA and OB both lie in the xy-plane, and both form 45 degrees with OC, which is {1,1,1}, we can relate their direction to OC's projection onto the xy-plane. The projection of OC onto the xy-plane is {1,1,0}, which is a vector in the xy-plane. The angle between OA and OC is 45 degrees. But OC is in 3D, so maybe we need another approach.Alternatively, perhaps since OA and OB are both in the xy-plane and form 45 degrees with OC, which is out of the plane, their projections onto OC have a certain relationship. Wait, maybe we can consider the angle between OA and OC. The angle is 45 degrees, so OA · OC = |OA||OC|cos45. Which we already used to find p + q = sqrt(6)/2. Similarly for OB.But how does that help with OA · OB?Alternatively, maybe think of OA and OB as vectors in the xy-plane making some angle with the projection of OC onto the xy-plane. The projection of OC onto the xy-plane is {1,1,0}, which is a vector in the xy-plane. The angle between OA and {1,1,0} can be found. Wait, but the angle given is with OC, not its projection. Hmm. So that might not directly help.Wait, the angle between OA and OC is 45 degrees. Let me recall that the angle between OA and OC is determined by their dot product. So OA is in the xy-plane, OC has a z-component. So OA · OC = p + q + 0*1 = p + q. The magnitude of OC is sqrt(3). So OA · OC = |OA||OC|cos45 => p + q = sqrt(3) * 1 * (sqrt(2)/2) = sqrt(6)/2, which is what we had.Similarly for OB.So OA and OB are two different unit vectors in the xy-plane, each making a 45-degree angle with OC. Since they're in the xy-plane, their z-components are zero. Therefore, their relationship to OC is through their x and y components. So OA and OB both have p + q = sqrt(6)/2 and r + s = sqrt(6)/2. Also, they are unit vectors, so p² + q² = 1 and r² + s² = 1.But how to find the angle between OA and OB, which is the angle between {p, q, 0} and {r, s, 0}. To find that angle, we need the dot product pr + qs.Wait, if we can find pr + qs, then we can compute the cosine of the angle as pr + qs, since both vectors are unit vectors.So, how can we find pr + qs? Let's think.We have OA and OB both in the xy-plane, each satisfying p + q = sqrt(6)/2 and r + s = sqrt(6)/2. Also, p² + q² = 1 and r² + s² = 1. Let's consider OA and OB as two vectors in the plane with these constraints.Is there a relation between pr + qs and p + q, r + s, and maybe other terms?Alternatively, let me consider that OA and OB are two vectors in the xy-plane each making a 45-degree angle with OC. Since OC is in 3D, but OA and OB are in 2D. Maybe their projections onto OC have something to do with it? Wait, maybe not.Alternatively, perhaps think of OA and OB as vectors in the plane such that the angle between OA and OC is 45 degrees, similarly for OB. But OC is a 3D vector. The angle between OA and OC is determined by the dot product formula. Since OA is in the xy-plane, the z-component of OA is zero, so OA · OC = p + q. Similarly for OB.So, for both OA and OB, p + q = r + s = sqrt(6)/2.But how to relate pr + qs?Alternatively, let me note that (p + q)(r + s) = pr + ps + qr + qs = pr + qs + (ps + qr). So pr + qs = (p + q)(r + s) - (ps + qr). We know (p + q)(r + s) = (sqrt(6)/2)^2 = 6/4 = 3/2. So pr + qs = 3/2 - (ps + qr). But we need to find ps + qr.Alternatively, perhaps we can find (p - q)(r - s) = pr - ps - qr + qs. Hmm, but that might not help directly. Wait, let's see:We have p² + q² = 1 and r² + s² = 1. Let's compute (p² + q²)(r² + s²) = 1*1 = 1. By the Cauchy-Schwarz inequality, (pr + qs)^2 ≤ (p² + q²)(r² + s²) = 1, so pr + qs ≤ 1. But we need the actual value.Alternatively, maybe think of pr + qs as the dot product of OA and OB. Since OA and OB are both in the xy-plane, maybe we can parameterize them.Let me consider OA as a vector in the xy-plane with p + q = sqrt(6)/2 and p² + q² = 1. Let me solve for p and q.Let p + q = sqrt(6)/2. Let me call this S = sqrt(6)/2. Then pq = 1/4 from part (1). So the quadratic equation is x² - Sx + pq = 0, which is x² - (sqrt(6)/2)x + 1/4 = 0. The solutions are [sqrt(6)/2 ± sqrt( (6/4) - 1 )]/2 = [sqrt(6)/2 ± sqrt( (6 - 4)/4 )]/2 = [sqrt(6)/2 ± sqrt(2)/2]/2 = [sqrt(6) ± sqrt(2)]/4. So p and q are [sqrt(6) + sqrt(2)]/4 and [sqrt(6) - sqrt(2)]/4. Similarly, r and s would be the same, but since OA and OB can be different vectors, perhaps OB is the other permutation? Wait, but OA and OB are both unit vectors with the same p + q and r + s. So maybe OA is [sqrt(6) + sqrt(2)]/4 in the x-component and [sqrt(6) - sqrt(2)]/4 in the y-component, and OB is the other way around? Or maybe they are different vectors.But actually, there are infinitely many vectors in the xy-plane that satisfy p + q = sqrt(6)/2 and p² + q² = 1. Because once you fix p + q and p² + q², pq is fixed, but the actual values of p and q can be swapped or rotated somehow? Wait, no. If p + q and pq are fixed, then p and q are uniquely determined up to permutation. So the only possible vectors OA are { [sqrt(6)+sqrt(2)]/4 , [sqrt(6)-sqrt(2)]/4 , 0 } and { [sqrt(6)-sqrt(2)]/4 , [sqrt(6)+sqrt(2)]/4 , 0 }.Wait, but if OA and OB are different vectors, then OB could be the other permutation. So if OA is the first one, then OB is the second one, and vice versa. Therefore, their dot product would be pr + qs. Let me compute that.Let me denote OA as {a, b, 0}, where a = [sqrt(6) + sqrt(2)]/4 and b = [sqrt(6) - sqrt(2)]/4. Then OB would be {b, a, 0}. Then OA · OB = a*b + b*a = 2ab. Since ab is equal to pq, which is 1/4. So OA · OB = 2*(1/4) = 1/2. Therefore, the cosine of the angle between OA and OB is 1/2, so the angle is 60 degrees. Wait, but is that the case?Wait, let's compute a and b:a = [sqrt(6) + sqrt(2)]/4 ≈ (2.449 + 1.414)/4 ≈ 3.863/4 ≈ 0.9658b = [sqrt(6) - sqrt(2)]/4 ≈ (2.449 - 1.414)/4 ≈ 1.035/4 ≈ 0.2588So OA ≈ {0.9658, 0.2588, 0}, OB ≈ {0.2588, 0.9658, 0}. Their dot product is (0.9658)(0.2588) + (0.2588)(0.9658) ≈ 0.25 + 0.25 = 0.5. So yes, OA · OB = 0.5, so the angle is arccos(0.5) = 60 degrees.But wait, the problem says "acute angle", which 60 degrees is acute. So is the answer 60 degrees? That seems possible, but let me verify if there's another possibility.But if OA and OB are arbitrary unit vectors in the xy-plane satisfying p + q = sqrt(6)/2 and r + s = sqrt(6)/2, but not necessarily permutations of each other, could there be other angles? Hmm, let's see.Wait, if OA and OB are both vectors in the xy-plane with p + q = sqrt(6)/2 and r + s = sqrt(6)/2, but not necessarily permutations, then their dot product pr + qs can vary. Wait, but given that p + q = S and r + s = S, and p² + q² = 1, r² + s² = 1, then pr + qs can be determined.Wait, let me use the variables:Let S = sqrt(6)/2.For OA: p + q = S, p² + q² = 1. So pq = (S² - 1)/2 = (6/4 - 1)/2 = (2/4)/2 = (1/2)/2 = 1/4.Similarly for OB: r + s = S, r² + s² = 1, so rs = 1/4.Now, OA · OB = pr + qs.We need to find pr + qs. Let me see if we can express this in terms of S and other terms.Let me consider that OA · OB = pr + qs. Let's express p and q in terms of S. From p + q = S and pq = 1/4, we can consider p and q as roots of the quadratic equation x² - Sx + 1/4 = 0. Similarly for r and s.But OA and OB are different vectors, so their roots could be different? Wait, no. If OA and OB are both solutions to x² - Sx + 1/4 = 0, then their components are the same. But since OA and OB are different vectors, maybe they have different components. Wait, but the roots are fixed once S and pq are fixed. So the components p, q and r, s are the same pair, possibly swapped. Therefore, OA and OB are either the same vector or swapped components.But the problem doesn't specify that OA and OB are different vectors, but part (2) asks for the angle between them, so they must be different. So swapping gives a different vector. Therefore, OA and OB have components {a, b} and {b, a}, leading to OA · OB = ab + ba = 2ab = 2*(1/4) = 1/2. Therefore, the angle is 60 degrees.Alternatively, suppose OA and OB are arbitrary vectors satisfying p + q = S and r + s = S, with pq = rs = 1/4. Then pr + qs can be written as p*r + q*s. Let me see if we can find this expression in terms of S and other variables.We have:(p + q)(r + s) = pr + ps + qr + qs = S^2But pr + qs = (p + q)(r + s) - (ps + qr) = S^2 - (ps + qr)So pr + qs = S² - (ps + qr)But ps + qr can be related to other terms. Let's compute (p - q)(r - s) = pr - ps - qr + qs. Therefore, (p - q)(r - s) = (pr + qs) - (ps + qr). So (pr + qs) - (ps + qr) = (p - q)(r - s). But not sure if that helps.Alternatively, consider (p + q)^2 = p² + 2pq + q² = 1 + 2*(1/4) = 1 + 1/2 = 3/2 = S², which checks out.Similarly, (r + s)^2 = 3/2.But how to relate pr + qs?Wait, perhaps we can use the identity:(p² + q²)(r² + s²) = (pr + qs)^2 + (ps - qr)^2This is similar to the Brahmagupta-Fibonacci identity. So since p² + q² = 1 and r² + s² = 1, then 1*1 = (pr + qs)^2 + (ps - qr)^2. Therefore, (pr + qs)^2 = 1 - (ps - qr)^2.But we don't know ps - qr. However, if OA and OB are the same vector, then ps - qr would be zero, but they are different vectors. Wait, but in our case, OA and OB are either {a, b} and {b, a}, so ps - qr would be b*a - a*b = 0. Wait, if OA is {a, b} and OB is {b, a}, then ps - qr is b*b - a*a. Wait, no:Wait, OA is {p, q} = {a, b}, OB is {r, s} = {b, a}. Then ps - qr = a*a - b*b. Wait, actually:ps = p*s = a*aqr = q*r = b*bSo ps - qr = a² - b².But a and b are roots of x² - Sx + 1/4 = 0, so a = [S + sqrt(S² - 1)]/2, and b = [S - sqrt(S² - 1)]/2.So a² - b² = (a - b)(a + b) = (sqrt(S² - 1))*(S). Since S = sqrt(6)/2, S² = 6/4 = 3/2. So sqrt(S² - 1) = sqrt(3/2 - 1) = sqrt(1/2) = 1/sqrt(2). Therefore, a² - b² = (1/sqrt(2))*(sqrt(6)/2) = sqrt(6)/(2*sqrt(2)) = sqrt(3)/2. Therefore, (ps - qr) = sqrt(3)/2.Wait, then (pr + qs)^2 = 1 - (ps - qr)^2 = 1 - (3/4) = 1/4. Therefore, pr + qs = ±1/2. But since OA and OB are both in the xy-plane and forming 45-degree angles with OC, which is in the positive octant, probably the angle between OA and OB is acute, so pr + qs is positive. Therefore, pr + qs = 1/2. Therefore, the cosine of the angle is 1/2, so the angle is 60 degrees.Therefore, the answer for part (2) is 60 degrees, or π/3 radians.But let me check again with the previous approach where OA and OB are permutations, leading to pr + qs = 1/2. This seems consistent. Also, using the Brahmagupta-Fibonacci identity gives the same result. Therefore, the angle AOB is 60 degrees.Therefore, part (1): p + q = sqrt(6)/2 and pq = 1/4; part (2): the angle is 60 degrees.Final Answer(1) ( p + q = boxed{dfrac{sqrt{6}}{2}} ) and ( pq = boxed{dfrac{1}{4}} );(2) The acute angle ( angle AOB ) is ( boxed{60^circ} ).</think>Given the problem, we need to find the values of ( p + q ) and ( pq ) for part (1), and the size of the acute angle ( angle AOB ) for part (2). Part (1)1. Unit Vector Condition: Since ( overrightarrow{OA} ) is a unit vector, we have ( p^2 + q^2 = 1 ).2. Dot Product with ( overrightarrow{OC} ): The angle between ( overrightarrow{OA} ) and ( overrightarrow{OC} = {1, 1, 1} ) is ( 45^circ ). The dot product formula gives: [ overrightarrow{OA} cdot overrightarrow{OC} = p + q = |overrightarrow{OA}||overrightarrow{OC}| cos 45^circ ] Since ( |overrightarrow{OA}| = 1 ) and ( |overrightarrow{OC}| = sqrt{3} ), we get: [ p + q = sqrt{3} cdot frac{sqrt{2}}{2} = frac{sqrt{6}}{2} ]3. Finding ( pq ): Using the identity ( (p + q)^2 = p^2 + 2pq + q^2 ): [ left( frac{sqrt{6}}{2} right)^2 = 1 + 2pq implies frac{3}{2} = 1 + 2pq implies 2pq = frac{1}{2} implies pq = frac{1}{4} ] Part (2)1. Dot Product of ( overrightarrow{OA} ) and ( overrightarrow{OB} ): We need to find ( overrightarrow{OA} cdot overrightarrow{OB} = pr + qs ).2. Using Symmetry and Previous Results: Both ( overrightarrow{OA} ) and ( overrightarrow{OB} ) are unit vectors with ( p + q = r + s = frac{sqrt{6}}{2} ) and ( pq = rs = frac{1}{4} ). The quadratic equation for both vectors has roots ( frac{sqrt{6} + sqrt{2}}{4} ) and ( frac{sqrt{6} - sqrt{2}}{4} ).3. Calculating the Dot Product: If ( overrightarrow{OA} ) and ( overrightarrow{OB} ) are permutations of these roots, their dot product is: [ overrightarrow{OA} cdot overrightarrow{OB} = left( frac{sqrt{6} + sqrt{2}}{4} right) left( frac{sqrt{6} - sqrt{2}}{4} right) + left( frac{sqrt{6} - sqrt{2}}{4} right) left( frac{sqrt{6} + sqrt{2}}{4} right) = frac{1}{2} ]4. Finding the Angle: The cosine of the angle is ( frac{1}{2} ), so the angle is ( 60^circ ). Final Answers(1) ( p + q = boxed{dfrac{sqrt{6}}{2}} ) and ( pq = boxed{dfrac{1}{4}} );(2) The acute angle ( angle AOB ) is ( boxed{60^circ} ).]

🔑:14. (1) |overrightarrow{O A}|=1, i.e., p^{2}+q^{2}=1,|overrightarrow{O C}|=sqrt{3}. Also, overrightarrow{O A} and overrightarrow{O C} form a 45^{circ} angle, so p+q=1 cdot sqrt{3} cdot cos 45^{circ}, thus p+q=frac{sqrt{6}}{2}, p q=frac{1}{4};(2) cos theta=frac{overrightarrow{O A} cdot overrightarrow{O B}}{|overrightarrow{O A}||overrightarrow{O B}|}=p r+q s=frac{1}{4}+frac{1}{4}=frac{1}{2} cdot angle A O B=60^{circ}.

🧠:[<think>Okay, let's see. I need to convert -5/3 π radians to degrees and -135 degrees to radians. Hmm, okay, unit conversion. Right, so radians to degrees and vice versa. Let me start with the first one: converting -5/3 π radians to degrees. First, I remember that the conversion factor between radians and degrees is 180 degrees equals π radians. So, to convert radians to degrees, you multiply by 180/π. But wait, let me think again. If I have X radians, then in degrees that's X multiplied by (180/π). So yes, that's right. So for -5/3 π radians, multiply by 180 over π to get degrees. Let me write that down:-5/3 π radians * (180 degrees / π radians) Oh, the π cancels out. So that simplifies to -5/3 * 180 degrees. Then, compute that. Let's calculate:First, 180 divided by 3 is 60. So 5 times 60 is 300. Then, since it's negative, it's -300 degrees? Wait, wait. Let me check. If it's -5/3 π, then multiplying by 180/π gives (-5/3)*180. So (-5)*180 /3. Yes, 180 divided by 3 is 60, so -5 * 60 = -300. So that's -300 degrees. Hmm, does that make sense? Let me think. π radians is 180 degrees, so 5/3 π is like 1 and 2/3 π. So that's 180 + 2/3 * 180. 2/3 of 180 is 120, so 180 + 120 = 300. But since it's negative, it's -300 degrees. So yeah, that seems right. So the first answer is -300 degrees.Now, the second part: converting -135 degrees to radians. Alright, for this, the conversion factor is π radians over 180 degrees. So you multiply the degrees by π/180 to get radians. Let me write that:-135 degrees * (π radians / 180 degrees)The degrees unit cancels, so we have -135 * π / 180. Let me simplify that fraction. Both numerator and denominator are divisible by 45. 135 divided by 45 is 3, 180 divided by 45 is 4. So that's -3/4 π. So -3π/4 radians. Wait, let me check again. -135 degrees. So 135 degrees is 3/4 of 180 degrees, which is π radians. So yes, 135 degrees is 3π/4 radians, so -135 degrees would be -3π/4 radians. That seems correct.But just to verify, let me do the calculation step by step. -135 * π / 180. Let's factor out 45 from 135 and 180. 135 = 45*3, 180=45*4. So - (45*3) * π / (45*4) = -3π/4. Yep, that checks out. So the second answer is -3π/4 radians.Wait a moment, just to make sure I didn't mix up the signs. Since the original angle is negative, so the converted one should also be negative. So -135 degrees becomes -3π/4 radians. Correct. Because direction is preserved; negative degrees correspond to clockwise rotation, same as negative radians. So that's right.So summarizing:-5/3 π radians is equal to -300 degrees.-135 degrees is equal to -3π/4 radians.I think that's it. Let me just verify once again. For the first conversion:Multiply by 180/π: (-5π/3)*(180/π) = (-5/3)*180 = -5*60 = -300. Correct.Second conversion:Multiply by π/180: (-135)*(π/180) = (-135/180)*π = (-3/4)*π. Correct.Yup, that seems right. No mistakes noticed.Final AnswerThe equivalent of -frac{5}{3}pi in degrees is boxed{-300^{circ}} and -135^{circ} in radians is boxed{-frac{3}{4}pi}.</think>To convert (-frac{5}{3}pi) radians to degrees, we use the conversion factor (frac{180^circ}{pi}):[-frac{5}{3}pi times frac{180^circ}{pi} = -frac{5}{3} times 180^circ = -5 times 60^circ = -300^circ]To convert (-135^circ) to radians, we use the conversion factor (frac{pi}{180}):[-135^circ times frac{pi}{180} = -frac{135pi}{180} = -frac{3pi}{4}] Final AnswerThe equivalent of (-frac{5}{3}pi) in degrees is (boxed{-300^{circ}}) and (-135^{circ}) in radians is (boxed{-frac{3}{4}pi}).,<think>Okay, so I need to convert -5/3 π radians to degrees and -135 degrees to radians. Hmm, let's start with the first part: converting -5π/3 to degrees. I remember that to convert from radians to degrees, you multiply by 180/π. So, let me set that up. The original angle is -5π/3. Multiply that by 180/π to get degrees. Let's see:(-5π/3) * (180/π) = (-5/3) * 180. The π cancels out, which is good. Now, let's compute (-5/3)*180. First, 180 divided by 3 is 60, so 5 times 60 is 300. Since it's negative, it'll be -300. So, -5π/3 radians is equal to -300 degrees. Wait, is that right? Let me check again. Yes, π radians is 180 degrees, so 1 radian is 180/π degrees. So, multiplying by 180/π converts radians to degrees. So, yes, (-5π/3)*(180/π) is indeed (-5/3)*180, which is -300. That seems correct. But angles in degrees can also be represented as positive angles by adding 360 degrees. So, -300 degrees is the same as -300 + 360 = 60 degrees. But the question just says "convert", so maybe we can leave it as -300 degrees. Unless specified otherwise, I think both are correct, but since the original angle was negative, maybe the answer should also be negative. Hmm. Let me check a reference. Wait, when converting angles, the sign is preserved unless specified otherwise. So, if it's negative in radians, it's negative in degrees. So, -5π/3 radians is indeed -300 degrees. So, that's the answer.Okay, moving on to the second part: converting -135 degrees to radians. To convert degrees to radians, we multiply by π/180. So:-135 degrees * (π/180) = ?Let me compute that. First, let's note that 135 and 180 are both divisible by 45. 135 ÷ 45 = 3, 180 ÷ 45 = 4. So, 135/180 = 3/4. Therefore, 135 degrees is 3π/4 radians. But since it's -135 degrees, it'll be -3π/4 radians. Let me check the calculation again:-135 * (π/180) = (-135/180)π = (-3/4)π. Yes, that simplifies to -3π/4. So, -135 degrees is equal to -3π/4 radians. Alternatively, angles can also be expressed in positive terms by adding 2π, but again, since the original angle was negative, unless instructed otherwise, the converted angle should also be negative. So, -3π/4 radians is the correct answer. Wait, but sometimes angles are represented between 0 and 2π or 0 and 360. But since the problem just says "convert", without specifying a particular range, I think the answer is just the direct conversion, preserving the sign. So, double-checking both conversions:For the first one:-5π/3 radians * (180/π) = -5/3 * 180 = -5 * 60 = -300 degrees. Correct.For the second one:-135 degrees * (π/180) = -135/180 * π = -3/4 * π = -3π/4 radians. Correct.Therefore, the final answers are -300 degrees and -3π/4 radians.Final AnswerThe angle -frac{5}{3}pi radians is equivalent to boxed{-300^{circ}} and -135^{circ} is equivalent to boxed{-frac{3}{4}pi} radians.</think>To convert (-frac{5}{3}pi) radians to degrees, we use the conversion factor (frac{180^circ}{pi}):[-frac{5}{3}pi times frac{180^circ}{pi} = -frac{5}{3} times 180^circ = -5 times 60^circ = -300^circ]To convert (-135^circ) to radians, we use the conversion factor (frac{pi}{180}):[-135^circ times frac{pi}{180} = -frac{135}{180}pi = -frac{3}{4}pi] Final AnswerThe angle (-frac{5}{3}pi) radians is equivalent to (boxed{-300^{circ}}) and (-135^{circ}) is equivalent to (boxed{-frac{3}{4}pi}) radians.]

🔑:To convert -frac{5}{3}pi to degrees, we use the conversion formula between radians and degrees. Specifically, we multiply by frac{180^{circ}}{pi}:-frac{5}{3}pi = -frac{5pi}{3} times frac{180^{circ}}{pi} = -300^{circ}To convert -135^{circ} to radians, we again use the conversion formula, this time multiplying by frac{pi}{180^{circ}}:-135^{circ} = -135^{circ} times frac{pi}{180^{circ}} = -frac{3}{4}piTherefore, the answers are -300^{circ} and -frac{3}{4}pi, respectively. This problem involves directly applying the conversion formulas between radians and degrees, testing basic knowledge of angle conversions.Hence, the answers are boxed{-300^{circ}} and boxed{-frac{3}{4}pi}.