🧠:[<think>Okay, let's see. I need to figure out how many vertical pairs in this 2x1007 table have the second row number being a multiple of the first row number. The first row has numbers 1 to 1007, and the second row has 1008 to 2014. Each pair is (k, 1007 + k) for k from 1 to 1007. So the question is: for how many k is 1007 + k a multiple of k?Let me formalize that. We need to find the number of integers k between 1 and 1007 inclusive such that 1007 + k is divisible by k. In other words, (1007 + k) mod k = 0. Simplifying the expression inside the modulus:(1007 + k) mod k = (1007 mod k + k mod k) = 1007 mod k + 0 = 1007 mod k. So the condition is that 1007 mod k equals 0. Which means that k must be a divisor of 1007. Therefore, the number of such k is equal to the number of divisors of 1007.So now the problem reduces to finding the number of positive divisors of 1007. To find the number of divisors, we first need to factorize 1007 into its prime factors. Let's check if 1007 is prime or composite.Starting with small primes:2: 1007 is odd, so not divisible by 2.3: Sum of digits is 1 + 0 + 0 + 7 = 8, which is not divisible by 3, so 1007 isn't divisible by 3.5: Ends with 7, so not divisible by 5.7: Let's divide 1007 by 7. 7*143 = 1001. 1007 - 1001 = 6. So remainder 6, not divisible by 7.11: 11*91 = 1001, same as above. 1007 - 1001 = 6, so remainder 6. Not divisible by 11.13: Let's try 13*77 = 1001. Again, 1007 - 1001 = 6. Not divisible by 13.17: 17*59 = 1003. 1007 - 1003 = 4. Not divisible by 17.19: 19*52 = 988. 1007 - 988 = 19. So 19*53 = 1007. Let me check: 19*50 = 950, 19*3=57, so 950 + 57 = 1007. Yes! So 1007 = 19*53. Are 19 and 53 primes?19 is a prime number. 53: let's check divisibility. 53 is not divisible by 2, 3, 5, 7, 11. Since 7^2 = 49 and 11^2 = 121, which is greater than 53. So 53 is prime. Therefore, the prime factorization of 1007 is 19*53.The number of divisors of a number is found by adding one to each of the exponents in the prime factorization and multiplying them. Here, both 19 and 53 are raised to the first power, so exponents are 1 each. So the number of divisors is (1+1)*(1+1) = 4. Therefore, there are 4 divisors of 1007, which are 1, 19, 53, and 1007.Therefore, the values of k that satisfy the condition are k = 1, 19, 53, and 1007. However, we need to check if k is within the range 1 to 1007. Since 1007 is part of the range, all four divisors are valid. Therefore, there are 4 such pairs where the number in the second row is a multiple of the one above it. Let me verify this conclusion. For each divisor:- When k=1: The pair is (1, 1008). 1008 divided by 1 is 1008, which is an integer. Correct.- When k=19: The pair is (19, 1007 + 19 = 1026). 1026 divided by 19: 19*54 = 1026. Correct.- When k=53: The pair is (53, 1007 + 53 = 1060). 1060 divided by 53: 53*20 = 1060. Correct.- When k=1007: The pair is (1007, 1007 + 1007 = 2014). 2014 divided by 1007 is 2. Correct.So all four pairs are valid. Hence, the answer should be 4, which corresponds to option C. Wait, looking back at the options given:(A) 2(B)(C) 4(D)(E) 6The options are labeled (A), (B), (C), (D), (E) with (C) being 4. So the answer is (C) 4.Final Answerboxed{C}</think>Given a table with 2 rows and 1007 columns, we need to determine how many vertical pairs have the number in the second row as a multiple of the number in the first row. The first row contains numbers from 1 to 1007, and the second row contains numbers from 1008 to 2014. Each pair is ((k, 1007 + k)) for (k) from 1 to 1007. We need to find the number of integers (k) such that (1007 + k) is divisible by (k).This condition simplifies to (1007 mod k = 0), meaning (k) must be a divisor of 1007. The prime factorization of 1007 is (19 times 53), both of which are prime numbers. The divisors of 1007 are 1, 19, 53, and 1007. All these divisors are within the range 1 to 1007.Verifying each divisor:- For (k = 1), the pair is ((1, 1008)) and (1008 div 1 = 1008).- For (k = 19), the pair is ((19, 1026)) and (1026 div 19 = 54).- For (k = 53), the pair is ((53, 1060)) and (1060 div 53 = 20).- For (k = 1007), the pair is ((1007, 2014)) and (2014 div 1007 = 2).All four pairs are valid, leading to the conclusion that there are 4 such pairs. Therefore, the answer is (boxed{C}).,<think>Okay, let's try to figure out how to solve this problem. We have a table with 2 rows and 1007 columns. The first row has numbers from 1 to 1007 in ascending order, and the second row has numbers from 1008 to 2014, also in ascending order. We need to find how many vertical pairs have the number in the second row as a multiple of the number in the first row.So, each column has two numbers: the top one is n (from 1 to 1007), and the bottom one is m = n + 1007 (since 1008 is 1 + 1007, 1009 is 2 + 1007, and so on up to 2014 = 1007 + 1007). Therefore, for each pair, m = n + 1007. The question is asking for how many such pairs m is a multiple of n. In other words, we need to find the number of integers n between 1 and 1007 such that (n + 1007) is divisible by n.Let me rephrase the divisibility condition: m is a multiple of n means that m = k * n for some integer k. Since m = n + 1007, substituting gives:n + 1007 = k * nSubtracting n from both sides:1007 = (k - 1) * nSo, this equation tells us that (k - 1) * n = 1007. Therefore, n must be a divisor of 1007, and (k - 1) is the corresponding quotient. Since n and k are positive integers (n is at least 1, and k must be at least 2 because m = n + 1007 is greater than n, so k must be at least 2 to make k * n > n).Therefore, the number of valid pairs is equal to the number of positive divisors of 1007. Each divisor d of 1007 will correspond to a pair where n = d, and k - 1 = 1007 / d, so k = (1007 / d) + 1. However, we have to ensure that n = d is within the range 1 to 1007. Since 1007 is the number we're factoring, all its divisors are already less than or equal to 1007, so that condition is automatically satisfied.Thus, the problem reduces to finding the number of positive divisors of 1007. To find the number of divisors, we need to factorize 1007 into its prime factors and then apply the formula for the number of divisors, which is (e1 + 1)(e2 + 1)...(en + 1) where e1, e2, etc., are the exponents in the prime factorization.So, let's factorize 1007. Let's check divisibility starting from small primes:1007 ÷ 2: 503.5, not integer.1007 ÷ 3: 335.666..., no.1007 ÷ 5: Ends with 7, so not divisible by 5.1007 ÷ 7: 1007 ÷ 7 = 143.857..., not integer.1007 ÷ 11: 1007 ÷ 11 ≈ 91.545..., not integer.1007 ÷ 13: 1007 ÷ 13 ≈ 77.461, not integer.Next prime is 17: 1007 ÷ 17 ≈ 59.235, nope.19: 1007 ÷ 19. Let's do 19*50 = 950, 1007 - 950 = 57, 57 ÷ 19 = 3. So 19*53 = 19*(50 + 3) = 950 + 57 = 1007. Yes! So 1007 factors into 19 * 53.Therefore, the prime factors of 1007 are 19 and 53, both to the first power. Hence, the number of divisors is (1 + 1)(1 + 1) = 4. Therefore, there are 4 divisors. These divisors are 1, 19, 53, and 1007. So, the possible values of n are 1, 19, 53, and 1007. Let's verify each case:1. For n = 1: m = 1 + 1007 = 1008. 1008 is indeed a multiple of 1 (since every number is a multiple of 1).2. For n = 19: m = 19 + 1007 = 1026. 1026 ÷ 19 = 54, which is an integer.3. For n = 53: m = 53 + 1007 = 1060. 1060 ÷ 53 = 20, which is an integer.4. For n = 1007: m = 1007 + 1007 = 2014. 2014 ÷ 1007 = 2, which is an integer.All four cases check out. Therefore, there are 4 such pairs. Looking back at the answer choices, option (C) is 4. So the answer should be (C).But wait, let me just make sure I didn't make a mistake in the prime factorization. Let me confirm 19*53 again. 19*50=950, 19*3=57, 950+57=1007. Correct. So 19 and 53 are primes? 19 is a prime number. 53: Let's check if 53 is prime. The primes less than sqrt(53) ≈7.28 are 2,3,5,7. 53 is not divisible by any of them, so yes, 53 is prime. Therefore, the prime factorization is correct.Therefore, the number of divisors is indeed (1+1)(1+1)=4, so the answer is 4, which is option (C).Final Answerboxed{C}</think>Given a table with 2 rows and 1007 columns, the first row contains numbers from 1 to 1007, and the second row contains numbers from 1008 to 2014. We need to determine how many vertical pairs exist where the number in the second row is a multiple of the number in the first row.Each pair consists of a number ( n ) in the first row and ( m = n + 1007 ) in the second row. We need ( m ) to be a multiple of ( n ), which translates to the equation ( n + 1007 = k cdot n ) for some integer ( k ). This simplifies to ( 1007 = (k - 1) cdot n ), indicating that ( n ) must be a divisor of 1007.To find the number of such divisors, we factorize 1007. The prime factorization of 1007 is ( 19 times 53 ). The divisors of 1007 are 1, 19, 53, and 1007. Each of these divisors corresponds to a valid pair:- For ( n = 1 ), ( m = 1008 ) (1008 is a multiple of 1)- For ( n = 19 ), ( m = 1026 ) (1026 is a multiple of 19)- For ( n = 53 ), ( m = 1060 ) (1060 is a multiple of 53)- For ( n = 1007 ), ( m = 2014 ) (2014 is a multiple of 1007)Thus, there are 4 such pairs. The answer is (boxed{C}).]

🔑:7. The answer is mathbf{( C )}.The pairs are of the form (k, k+1007). We know that k divides k+1007 if and only if k divides 1007. Since 1007=19 cdot 53, and since the numbers 19 and 53 are prime, the only possible k values are k=1,19,53,1007.[Problem proposed by G. Barbarino]

🧠:[<think>Okay, so I need to find the value of sin(θ + π/4), where θ is an angle whose terminal side passes through the point P(-3, 4). Let me start by recalling some trigonometry basics. First, since the terminal side of angle θ passes through P(-3, 4), that means θ is in a certain quadrant. The coordinates are (-3, 4), so x is negative and y is positive. That places θ in the second quadrant. Right? Because in the second quadrant, x is negative and y is positive. So θ is in Quadrant II.Now, to find sin(θ + π/4), I think I need to use the sine addition formula. The formula for sin(A + B) is sinA cosB + cosA sinB. So applying that here, sin(θ + π/4) would be sinθ cos(π/4) + cosθ sin(π/4). Since cos(π/4) and sin(π/4) are both √2/2, this simplifies to (sinθ + cosθ) * √2/2. So I need to find sinθ and cosθ first.To find sinθ and cosθ, I can use the coordinates of point P. Since the terminal side passes through (-3, 4), we can consider a right triangle formed by the x-axis, the terminal side, and a perpendicular line from P to the x-axis. The coordinates (-3, 4) give the legs of this right triangle. The length of the adjacent side (on the x-axis) is -3, but since we're dealing with lengths for trigonometric functions, we need the hypotenuse. Wait, actually, in trigonometric terms, the radius r is the hypotenuse, calculated by sqrt(x² + y²). So here, x is -3 and y is 4. So r would be sqrt((-3)^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5.So r is 5. Then, sinθ is y/r = 4/5, and cosθ is x/r = -3/5. Wait, but since θ is in the second quadrant, cosine should be negative and sine positive, which matches here. So that's correct.So sinθ = 4/5, cosθ = -3/5. Now plug these into the expression (sinθ + cosθ) * √2/2. Let's compute sinθ + cosθ first. That's 4/5 + (-3/5) = (4 - 3)/5 = 1/5. Then multiply by √2/2. So (1/5) * (√2/2) = √2/10. So sin(θ + π/4) = √2/10. Wait, let me double-check that. Hmm. So sin(θ + π/4) = sinθ cos(π/4) + cosθ sin(π/4). We know sinθ = 4/5, cosθ = -3/5, and cos(π/4) = sin(π/4) = √2/2. So substituting in, that's (4/5)(√2/2) + (-3/5)(√2/2). Let's compute each term separately.First term: (4/5)(√2/2) = (4√2)/10 = (2√2)/5.Second term: (-3/5)(√2/2) = (-3√2)/10.Adding these two terms: (2√2)/5 + (-3√2)/10. To add these, they need a common denominator. The first term can be written as (4√2)/10, so (4√2 - 3√2)/10 = (√2)/10. So that's the same result as before. So √2/10. Hmm, so that seems right. Let me make sure I didn't make a mistake in the calculations. Let's check step by step.Calculating r: sqrt((-3)^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5. Correct.So sinθ = y/r = 4/5, cosθ = x/r = -3/5. Correct.Sine addition formula: sin(a + b) = sin a cos b + cos a sin b. So sinθ cos(π/4) + cosθ sin(π/4). Both cos(π/4) and sin(π/4) are √2/2. So plugging in:sinθ * √2/2 + cosθ * √2/2 = (sinθ + cosθ) * √2/2. Then sinθ + cosθ is 4/5 + (-3/5) = 1/5. Then 1/5 * √2/2 = √2/10. So yes, that's correct.Alternatively, if I compute each term separately:First term: (4/5)(√2/2) = (4√2)/10 = (2√2)/5. Second term: (-3/5)(√2/2) = (-3√2)/10. Then adding (2√2)/5 is equivalent to (4√2)/10, so 4√2/10 - 3√2/10 = √2/10. Same result.Therefore, the value of sin(θ + π/4) is √2/10. But wait, let me check if the angle θ is correctly determined. Since the point is (-3,4), θ is in the second quadrant. So when we compute θ, it's the angle whose reference angle is arctan(4/3), but in the second quadrant. So θ is π - arctan(4/3). But when we use the sine and cosine values, we already accounted for the quadrant by the signs. So since we have sinθ positive and cosθ negative, we have the right values. Therefore, adding π/4 to θ and taking the sine should give √2/10. Let me think if there's another way to approach this. Maybe by converting θ to a different form or using the unit circle, but I think the method I used is correct.Alternatively, if I wanted to compute θ + π/4, maybe using some angle addition, but since θ is arctan(-4/3) adjusted to the second quadrant, but perhaps that complicates things. Alternatively, maybe using vectors. The coordinates (-3,4) can be considered as a vector, and adding π/4 to the angle is equivalent to rotating the vector by π/4. But I think that's more complex. Alternatively, using complex numbers: the point P(-3,4) can be represented as a complex number -3 + 4i. Multiplying by e^{iπ/4} would rotate it by π/4, and then taking the imaginary part would give the sine of the new angle. But I think that's overcomplicating. Let me see.Wait, the sine of θ + π/4 is the same as the imaginary part of [ ( -3 + 4i ) * e^{iπ/4} ] divided by the magnitude. Wait, but maybe not. Because θ is the angle of the point (-3,4), whose magnitude is 5. So the complex number is 5 e^{iθ}. Then θ + π/4 would be 5 e^{i(θ + π/4)}, and sin(θ + π/4) is the imaginary part of e^{i(θ + π/4)}, which is sinθ cosπ/4 + cosθ sinπ/4, which is the same as before. So we still get the same result. So this confirms it.Alternatively, maybe using coordinates. If we rotate the point (-3,4) by -π/4 (since we're adding π/4 to the angle), but that might be a more involved process. Probably the method I used is the most straightforward.Thus, after verifying all steps, I believe the answer is √2/10.Final AnswerThe value of sin left(θ+ dfrac {π}{4}right) is boxed{dfrac{sqrt{2}}{10}}.</think>Given that the initial side of angle ( theta ) coincides with the non-negative semi-axis of the ( x )-axis and its terminal side passes through the point ( P(-3, 4) ), we need to find the value of ( sin left( theta + dfrac{pi}{4} right) ).1. Determine the quadrant of ( theta ): - The coordinates ( P(-3, 4) ) indicate that ( theta ) is in the second quadrant (where ( x ) is negative and ( y ) is positive).2. Calculate the radius ( r ): - The radius ( r ) is the distance from the origin to the point ( P(-3, 4) ): [ r = sqrt{(-3)^2 + 4^2} = sqrt{9 + 16} = sqrt{25} = 5 ]3. Find ( sin theta ) and ( cos theta ): - Using the coordinates ( (-3, 4) ): [ sin theta = frac{y}{r} = frac{4}{5}, quad cos theta = frac{x}{r} = frac{-3}{5} ]4. Apply the sine addition formula: - The formula for ( sin(a + b) ) is ( sin a cos b + cos a sin b ): [ sin left( theta + frac{pi}{4} right) = sin theta cos frac{pi}{4} + cos theta sin frac{pi}{4} ] - Since ( cos frac{pi}{4} = sin frac{pi}{4} = frac{sqrt{2}}{2} ): [ sin left( theta + frac{pi}{4} right) = left( sin theta + cos theta right) cdot frac{sqrt{2}}{2} ] - Substitute ( sin theta = frac{4}{5} ) and ( cos theta = -frac{3}{5} ): [ sin left( theta + frac{pi}{4} right) = left( frac{4}{5} + left( -frac{3}{5} right) right) cdot frac{sqrt{2}}{2} ] [ = left( frac{1}{5} right) cdot frac{sqrt{2}}{2} = frac{sqrt{2}}{10} ]The value of ( sin left( theta + dfrac{pi}{4} right) ) is boxed{dfrac{sqrt{2}}{10}}.,<think>Okay, so I need to find the value of sin(θ + π/4) where the angle θ starts from the positive x-axis and its terminal side passes through the point P(-3, 4). Hmm, let's think about how to approach this step by step.First, I remember that when dealing with trigonometric functions and angles in standard position (where the initial side is on the positive x-axis), we can use the coordinates of a point on the terminal side to find the sine and cosine of the angle. Since the terminal side passes through (-3, 4), that means we can use these coordinates to determine the values of sin θ and cos θ.But wait, the coordinates are (-3, 4), which is in the second quadrant. So θ must be an angle in the second quadrant. That makes sense because the x-coordinate is negative and the y-coordinate is positive there. So θ is between π/2 and π radians.To find sin θ and cos θ, I need to find the hypotenuse of the right triangle formed by the point (-3, 4). The hypotenuse is the distance from the origin to the point P. The formula for that is sqrt(x² + y²). Let me compute that.So, x is -3, y is 4. Then the radius r (which is the hypotenuse) is sqrt((-3)^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5. Okay, so r is 5. That means:sin θ = y/r = 4/5cos θ = x/r = -3/5Right? Because sine is opposite over hypotenuse (which is y/r) and cosine is adjacent over hypotenuse (x/r). Since x is negative here, the cosine will be negative. That checks out because we are in the second quadrant where cosine is negative and sine is positive.Now, the question is asking for sin(θ + π/4). Hmm, that's a sine of a sum of angles. I recall the sine addition formula: sin(a + b) = sin a cos b + cos a sin b. So applying that here, we have:sin(θ + π/4) = sin θ cos(π/4) + cos θ sin(π/4)We already know sin θ and cos θ, and cos(π/4) and sin(π/4) are both sqrt(2)/2. Let me write that out:sin(θ + π/4) = (4/5)(sqrt(2)/2) + (-3/5)(sqrt(2)/2)Let me compute each term separately. First term: (4/5)(sqrt(2)/2) = (4 * sqrt(2)) / 10 = (2 * sqrt(2)) / 5Second term: (-3/5)(sqrt(2)/2) = (-3 * sqrt(2)) / 10So combining these two terms:(2 * sqrt(2)/5) + (-3 * sqrt(2)/10) = To combine them, I need a common denominator. The first term is 2/5, which is 4/10. So:(4 * sqrt(2)/10) - (3 * sqrt(2)/10) = (4 - 3) * sqrt(2)/10 = sqrt(2)/10Wait, that gives me sqrt(2)/10. But is that correct? Let me check my steps again.Wait, original terms:First term: (4/5)(sqrt(2)/2) = (4 * sqrt(2))/10 = 2*sqrt(2)/5Second term: (-3/5)(sqrt(2)/2) = (-3*sqrt(2))/10So 2*sqrt(2)/5 is equal to 4*sqrt(2)/10. Then 4*sqrt(2)/10 - 3*sqrt(2)/10 = (4 - 3)*sqrt(2)/10 = sqrt(2)/10Yes, that's correct. So sin(θ + π/4) is sqrt(2)/10. Hmm, but let me make sure I didn't make a mistake in the calculation.Wait, sqrt(2)/10 is approximately 0.1414, which seems a bit small. Let me check if that makes sense. Let's think about θ. Since the point is (-3,4), θ is in the second quadrant. π/4 is 45 degrees, so θ + π/4 would be an angle that's θ plus 45 degrees. Depending on θ, the sine could be positive or negative. But θ is in the second quadrant. For example, if θ is 135 degrees (3π/4), then θ + π/4 is 180 degrees (π), whose sine is 0. But in this case, θ is not 135 degrees. Let's compute θ's reference angle.The reference angle α is arctan(|y/x|) = arctan(4/3). So θ is π - arctan(4/3). Then θ + π/4 is π - arctan(4/3) + π/4 = (5π/4) - arctan(4/3). Hmm, not sure if that helps.Alternatively, maybe using decimal approximations. Let's compute θ in radians. The reference angle is arctan(4/3). Let's compute that. arctan(4/3) is approximately 0.9273 radians. So θ is π - 0.9273 ≈ 2.2143 radians. Then θ + π/4 ≈ 2.2143 + 0.7854 ≈ 3.0 radians. The sine of 3 radians is sin(3) ≈ 0.1411, which is approximately sqrt(2)/10 ≈ 0.1414. So that checks out. So the answer sqrt(2)/10 is correct.Wait, but just to make sure. Let's compute it step by step again:sin θ = 4/5, cos θ = -3/5sin(θ + π/4) = sin θ cos π/4 + cos θ sin π/4= (4/5)(sqrt(2)/2) + (-3/5)(sqrt(2)/2)= (4/5 - 3/5)(sqrt(2)/2)= (1/5)(sqrt(2)/2)Wait, wait, hold on. Wait, no:Wait, the two terms are (4/5)(sqrt(2)/2) and (-3/5)(sqrt(2)/2). So factoring out sqrt(2)/2, we have [4/5 - 3/5] * sqrt(2)/2 = (1/5)(sqrt(2)/2) = sqrt(2)/10. Yes, that's correct. So that's the same as before.Alternatively, maybe factor sqrt(2)/2 first:sqrt(2)/2 [4/5 - 3/5] = sqrt(2)/2 * 1/5 = sqrt(2)/10.Yes, same result.Alternatively, maybe I can rationalize why it's positive. Since θ is in the second quadrant, θ + π/4 would be in the second or third quadrant? Let's see. θ is between π/2 and π. Adding π/4 (which is 45 degrees) to θ would make it between π/2 + π/4 = 3π/4 and π + π/4 = 5π/4. So θ + π/4 is between 3π/4 and 5π/4. So between 135 degrees and 225 degrees. So depending on θ, it could be in the second or third quadrant.But in our case, θ was calculated as approximately 2.2143 radians, which is about 126.87 degrees (since π is 3.1416, so 2.2143 is about 126.87 degrees). Adding 45 degrees, that's 171.87 degrees, which is still in the second quadrant. Therefore, the sine of θ + π/4 should still be positive, which aligns with our result sqrt(2)/10 being positive. So that makes sense.Alternatively, if θ were exactly 135 degrees (3π/4), then θ + π/4 would be 180 degrees (π), whose sine is 0. But our θ is a bit less than 135 degrees, so θ + π/4 is a bit less than 180 degrees, so sine is positive but approaching zero. Our result is sqrt(2)/10 ≈ 0.141, which is a small positive number, so that makes sense.Alternatively, maybe another approach: converting to rectangular coordinates.Wait, sin(θ + π/4) can also be represented using rotation matrices or something. But maybe that's overcomplicating. Alternatively, since we have coordinates of the point P(-3,4), maybe we can parametrize θ and then find θ + π/4.But no, probably the way I did it is correct.Alternatively, verify using the coordinates. If θ is the angle to point P(-3,4), then θ + π/4 is another angle. If we can find the coordinates of a point Q such that Q is at angle θ + π/4 and the same radius 5, then the y-coordinate of Q would be sin(θ + π/4). But since we are rotating by π/4, perhaps we can use rotation matrices.The coordinates of Q would be (r cos(θ + π/4), r sin(θ + π/4)). But since we know r is 5, and we can compute cos(θ + π/4) and sin(θ + π/4). But this might not be necessary because we already computed sin(θ + π/4) as sqrt(2)/10. However, let's try this method to verify.Rotation by π/4 radians counterclockwise would transform the original point P(-3,4) to a new point Q. The rotation matrix is:[cos π/4, -sin π/4][sin π/4, cos π/4]So multiplying this matrix by the vector [-3, 4]:x' = (-3)cos π/4 - 4 sin π/4y' = (-3)sin π/4 + 4 cos π/4But sin π/4 and cos π/4 are both sqrt(2)/2. So:x' = (-3)(sqrt(2)/2) - 4(sqrt(2)/2) = (-3 -4)(sqrt(2)/2) = (-7 sqrt(2))/2y' = (-3)(sqrt(2)/2) + 4(sqrt(2)/2) = (-3 + 4)(sqrt(2)/2) = (1)(sqrt(2)/2) = sqrt(2)/2Therefore, the coordinates of Q after rotating P by π/4 are (-7 sqrt(2)/2, sqrt(2)/2). The radius is still 5, right? Because rotating a point doesn't change its distance from the origin. Let's check the distance:sqrt( [(-7 sqrt(2)/2)^2 + (sqrt(2)/2)^2 ] ) = sqrt( (49 * 2 /4 ) + (2 /4 ) ) = sqrt( (98/4) + (2/4) ) = sqrt(100/4) = sqrt(25) = 5. Correct.Therefore, the y-coordinate of Q is sqrt(2)/2, and since the radius is 5, sin(θ + π/4) = y'/r = (sqrt(2)/2)/5 = sqrt(2)/10. Which matches the previous result. So that's another way to verify that the answer is indeed sqrt(2)/10.Therefore, after two different methods, I get the same answer. So I'm confident that the value of sin(θ + π/4) is sqrt(2)/10.Final AnswerThe value of sin left(θ+ dfrac {π}{4}right) is boxed{dfrac{sqrt{2}}{10}}.</think>Given that the initial side of angle ( theta ) coincides with the non-negative semi-axis of the ( x )-axis and its terminal side passes through the point ( P(-3, 4) ), we need to find the value of ( sin left( theta + dfrac{pi}{4} right) ).1. Determine the hypotenuse (radius) ( r ): [ r = sqrt{(-3)^2 + 4^2} = sqrt{9 + 16} = sqrt{25} = 5 ]2. Find ( sin theta ) and ( cos theta ): [ sin theta = frac{4}{5}, quad cos theta = frac{-3}{5} ]3. Apply the sine addition formula: [ sin left( theta + dfrac{pi}{4} right) = sin theta cos dfrac{pi}{4} + cos theta sin dfrac{pi}{4} ] Since ( cos dfrac{pi}{4} = sin dfrac{pi}{4} = dfrac{sqrt{2}}{2} ): [ sin left( theta + dfrac{pi}{4} right) = left( frac{4}{5} right) left( frac{sqrt{2}}{2} right) + left( frac{-3}{5} right) left( frac{sqrt{2}}{2} right) ]4. Simplify the expression: [ = frac{4sqrt{2}}{10} - frac{3sqrt{2}}{10} = frac{(4 - 3)sqrt{2}}{10} = frac{sqrt{2}}{10} ]Thus, the value of ( sin left( theta + dfrac{pi}{4} right) ) is (boxed{dfrac{sqrt{2}}{10}}).]

🔑:From the given information, we have x=-3, y=4. We can find the magnitude of the vector, r=sqrt{x^2+y^2}=sqrt{(-3)^2+4^2}=5.Then, we calculate sin θ= dfrac {y}{r}= dfrac {4}{5} and cos θ= dfrac {x}{r}= dfrac {-3}{5}.Now, using the trigonometric identity for the sine of a sum of angles, we get sin (θ+ dfrac {π}{4})= dfrac { sqrt {2}}{2}sin θ+ dfrac { sqrt {2}}{2}cos θ= dfrac { sqrt {2}}{2}( dfrac {4}{5}- dfrac {3}{5})= boxed{dfrac { sqrt {2}}{10}}.

🧠:[<think>Okay, let me try to figure out this problem. So, we have a right triangle with legs a and b, and the equation given is log(a + b) = (1/2) log b + (1/2) log(a + 3b). We need to find the measure of the angle opposite the leg of length a. First, let's recall that in a right triangle, the angles opposite the legs are acute angles, and their measures depend on the ratio of the legs. The tangent of the angle opposite side a is a/b, so if we can find the ratio a/b, we can find the angle using the arctangent function. That seems like a good plan. So, the key is to find a in terms of b or vice versa from the given logarithmic equation.Let me start by simplifying the given equation. The equation is:log(a + b) = (1/2) log b + (1/2) log(a + 3b)Hmm, logarithms can be tricky, but maybe if I combine the terms on the right-hand side. Remember that (1/2) log x + (1/2) log y = (1/2)(log x + log y) = (1/2) log(xy). So, applying that here:Right-hand side (RHS) becomes (1/2) log [b(a + 3b)]So, the equation simplifies to:log(a + b) = (1/2) log [b(a + 3b)]Now, to eliminate the logarithm, maybe we can exponentiate both sides. If I take 10 to the power of both sides (assuming log is base 10, though actually the base doesn't matter here because we're just trying to equate the arguments eventually). So:10^{log(a + b)} = 10^{(1/2) log [b(a + 3b)]}Which simplifies to:a + b = [b(a + 3b)]^{1/2}Wait, because 10^{log x} = x, and 10^{(1/2) log x} = x^{1/2}. So, that's right. So now we have:a + b = sqrt[b(a + 3b)]To get rid of the square root, let's square both sides:(a + b)^2 = b(a + 3b)Let me expand both sides:Left-hand side (LHS): (a + b)^2 = a^2 + 2ab + b^2Right-hand side (RHS): b(a + 3b) = ab + 3b^2So, setting them equal:a^2 + 2ab + b^2 = ab + 3b^2Now, subtract ab + 3b^2 from both sides to bring all terms to the left:a^2 + 2ab + b^2 - ab - 3b^2 = 0Simplify:a^2 + (2ab - ab) + (b^2 - 3b^2) = 0Which is:a^2 + ab - 2b^2 = 0Hmm, okay. So, this is a quadratic equation in terms of a. Let's see. Maybe we can factor this quadratic equation.Looking at a^2 + ab - 2b^2. Let's try to factor it. Let me think of two numbers that multiply to (-2b^2) and add to (1b). So, coefficients: 1 for a^2, 1 for ab, -2 for b^2.Looking for factors of (a + mb)(a + nb) = a^2 + (m + n)ab + mnb^2. So, need m + n = 1 and mn = -2. Solving m + n = 1 and mn = -2. The numbers that satisfy this would be 2 and -1, because 2 + (-1) = 1 and 2*(-1) = -2. Therefore, the factors are (a + 2b)(a - b). Let me check:(a + 2b)(a - b) = a^2 - ab + 2ab - 2b^2 = a^2 + ab - 2b^2. Yes, that's correct.So, the equation factors to:(a + 2b)(a - b) = 0Setting each factor equal to zero:a + 2b = 0 --> a = -2bOr a - b = 0 --> a = bBut since a and b are lengths of legs in a triangle, they must be positive. Therefore, a = -2b is invalid. So, the only solution is a = b.Wait, but if a = b, then the triangle is an isosceles right triangle, so the angle opposite leg a would be 45 degrees. But let me check if this is correct. Let's verify if a = b satisfies the original logarithmic equation.Let me substitute a = b into the original equation:log(a + b) = (1/2) log b + (1/2) log(a + 3b)Substituting a = b:log(b + b) = (1/2) log b + (1/2) log(b + 3b)Simplify:log(2b) = (1/2) log b + (1/2) log(4b)Let's compute each term:Left-hand side: log(2b) = log 2 + log bRight-hand side: (1/2) log b + (1/2) log(4b) = (1/2) log b + (1/2)(log 4 + log b) = (1/2) log b + (1/2) log 4 + (1/2) log b = [ (1/2 + 1/2) log b ] + (1/2) log 4 = log b + (1/2) log 4So, RHS = log b + (1/2) log 4Compare to LHS: log 2 + log bTherefore, LHS vs RHS: log 2 + log b vs log b + (1/2) log 4But log 4 is 2 log 2, so (1/2) log 4 = (1/2)(2 log 2) = log 2. Therefore, RHS = log b + log 2 = log 2 + log b = LHS. So, yes, it checks out. Therefore, a = b is indeed a solution.But wait, let's recall that in the quadratic equation, we had a^2 + ab - 2b^2 = 0, which factored into (a + 2b)(a - b) = 0. So, the only valid solution is a = b. Therefore, the legs are equal, so the triangle is a 45-45-90 triangle, so each acute angle is 45 degrees. Therefore, the angle opposite leg a is 45 degrees. So, the answer is 45 degrees, which is π/4 radians, but since the question asks for the measure, it's probably expecting degrees unless specified otherwise. But wait, let me double-check to ensure that there are no other possible solutions. Since we squared both sides of the equation during our process, sometimes squaring can introduce extraneous solutions. But in this case, we only got a = -2b and a = b. The a = -2b was rejected because lengths can't be negative. So, the only valid solution is a = b. Therefore, the angle is 45 degrees. Therefore, the measure of the angle opposite the leg of length a is 45 degrees, which is π/4 radians. But since the problem doesn't specify radians or degrees, and given the context is a right triangle, likely degrees. So, the answer is 45 degrees. Wait, but let me think again. The problem statement is in a math context, and angles can be in degrees or radians. But unless told otherwise, in such geometry problems, degrees are standard. However, sometimes radians are also possible. But the question says "measure of the angle", which could be either, but in the context of triangles, degrees are more common. But in higher mathematics, radians are standard. Hmm. Wait, the original equation uses log, which is more of an algebra/precalculus problem, so maybe degrees. But let me check. Alternatively, maybe the angle is not 45 degrees. Wait, but if a = b, then the legs are equal, so the triangle is isosceles right-angled, so the angles are 45-45-90. Therefore, the angle opposite a is 45 degrees. That seems correct. Wait a second, but let's suppose that the problem is tricking me. Maybe when we squared both sides, we introduced an extraneous solution, but we already saw that a = b works. Alternatively, maybe there is another solution. Wait, the quadratic equation gave us only a = b and a = -2b, but perhaps we missed something. Let me check. Let's suppose that a ≠ b, but perhaps another solution. Wait, when we squared the equation, could there be another solution?Original equation after simplifying the logs:log(a + b) = (1/2) log [b(a + 3b)]We exponentiated both sides to get:a + b = sqrt[b(a + 3b)]Then squared both sides to get:(a + b)^2 = b(a + 3b)Which led us to the quadratic equation. So, the steps are correct. Since we squared both sides, we have to check that the solutions satisfy the original equation. But since a = -2b would make the left-hand side log(a + b) = log(-2b + b) = log(-b), which is undefined. So, the only solution is a = b, which works. So, the conclusion is correct.Therefore, the angle opposite leg a is 45 degrees. Thus, the answer is 45 degrees. Wait, but maybe the problem expects the answer in radians? The problem says "measure of the angle", and in mathematics, measure can be in degrees or radians. But in a triangle context, unless specified, degrees are more common. However, if the problem is in a calculus or advanced context, radians might be expected. Since this problem involves logarithms and algebra, it's probably safe to assume degrees. But to confirm, let me see if 45 degrees makes sense. Alternatively, perhaps there's a mistake in the process. Let me check again. Let's re-express the original equation step by step.Original equation:log(a + b) = (1/2) log b + (1/2) log(a + 3b)Combine the RHS:(1/2)[log b + log(a + 3b)] = (1/2) log [b(a + 3b)]So:log(a + b) = (1/2) log [b(a + 3b)]Exponentiate both sides:10^{log(a + b)} = 10^{(1/2) log [b(a + 3b)]}Which simplifies to:(a + b) = [b(a + 3b)]^{1/2}Square both sides:(a + b)^2 = b(a + 3b)Expand:a^2 + 2ab + b^2 = ab + 3b^2Subtract ab + 3b^2:a^2 + ab - 2b^2 = 0Factor:(a + 2b)(a - b) = 0Solutions: a = -2b, a = b. Only a = b is valid. Therefore, the triangle has legs a = b, so it's a 45-45-90 triangle. Therefore, angle opposite a is 45 degrees. So, I think that is correct.Therefore, the answer should be 45 degrees. Wait, but in the problem statement, is there any indication that the angle isn't 45 degrees? Maybe not. So, that must be the answer. Therefore, the measure of the angle opposite leg a is 45 degrees, which is π/4 radians. Since the problem says "measure", which can be either, but unless specified, probably degrees. So, I think the answer is 45 degrees. But just to make sure, let's suppose that a = b, then the legs are equal, so the hypotenuse is a√2, and the angles are both 45 degrees. Yep, that's correct. So, everything checks out. Therefore, the angle is 45 degrees.Final AnswerThe measure of the angle opposite the leg of length ( a ) is (boxed{45^circ}).</think>Given a right triangle with legs (a) and (b), we start with the equation:[log (a + b) = frac{1}{2} cdot log b + frac{1}{2} cdot log (a + 3b)]First, we combine the logarithms on the right-hand side:[log (a + b) = frac{1}{2} cdot log [b(a + 3b)]]Exponentiating both sides to eliminate the logarithms, we get:[a + b = sqrt{b(a + 3b)}]Squaring both sides to remove the square root:[(a + b)^2 = b(a + 3b)]Expanding both sides:[a^2 + 2ab + b^2 = ab + 3b^2]Simplifying by bringing all terms to one side:[a^2 + ab - 2b^2 = 0]Factoring the quadratic equation:[(a + 2b)(a - b) = 0]Since (a) and (b) are positive lengths, the solution (a = -2b) is invalid. Thus, we have:[a = b]In a right triangle with legs (a) and (b), if (a = b), the triangle is isosceles, and the angles opposite the legs are each (45^circ). Therefore, the measure of the angle opposite the leg of length (a) is:[boxed{45^circ}],<think>Okay, so I need to solve this problem where we have a right triangle with legs a and b. The equation given is log(a + b) = (1/2) log b + (1/2) log(a + 3b). I need to find the measure of the angle opposite the leg of length a. Hmm, let's start by understanding what the equation is telling me.First, since it's a right triangle, by the Pythagorean theorem, we know that a² + b² = c², where c is the hypotenuse. But the problem doesn't mention c, so maybe that comes into play later. The equation given involves logarithms, so maybe I can manipulate that equation to find a relationship between a and b.Let me rewrite the equation:log(a + b) = (1/2) log b + (1/2) log(a + 3b)I remember that (1/2) log x is the same as log x^(1/2), so maybe I can combine the right-hand side into a single logarithm. Let's try that.Right-hand side: (1/2) log b + (1/2) log(a + 3b) = log b^(1/2) + log(a + 3b)^(1/2)Since log x + log y = log(xy), this becomes log [b^(1/2) * (a + 3b)^(1/2)]Which is log [sqrt(b) * sqrt(a + 3b)] = log [sqrt(b(a + 3b))]So the equation simplifies to:log(a + b) = log sqrt(b(a + 3b))Since if log A = log B, then A = B, so we can remove the logarithms:a + b = sqrt(b(a + 3b))Now, let's square both sides to eliminate the square root:(a + b)² = [sqrt(b(a + 3b))]²Which gives:a² + 2ab + b² = b(a + 3b)Let me expand the right-hand side:b(a + 3b) = a*b + 3b² = ab + 3b²So now the equation is:a² + 2ab + b² = ab + 3b²Let me bring all terms to the left-hand side:a² + 2ab + b² - ab - 3b² = 0Simplify:a² + (2ab - ab) + (b² - 3b²) = 0Which becomes:a² + ab - 2b² = 0So we have a quadratic equation in terms of a and b. Let me factor this:a² + ab - 2b² = 0Looking for factors of the form (a + mb)(a + nb), where m and n are numbers such that mn = -2b² and m + n = 1 (coefficient of ab). Hmm, let's see.Wait, perhaps factor it as:(a + 2b)(a - b) = 0Let me check: (a + 2b)(a - b) = a² - ab + 2ab - 2b² = a² + ab - 2b². Yes, that's correct.So the equation factors into (a + 2b)(a - b) = 0Setting each factor equal to zero:a + 2b = 0 => a = -2bBut a and b are lengths of legs of a triangle, so they must be positive. So this solution is invalid.Alternatively, a - b = 0 => a = bSo a = b. Wait, if a = b, then the triangle is an isosceles right triangle, so the angles are 45° and 45°. But the problem asks for the measure of the angle opposite the leg of length a. If a = b, then both legs are equal, so both angles opposite them are 45°. But wait, let me confirm.Wait, in a right triangle, the legs are the two sides adjacent to the right angle. So the hypotenuse is opposite the right angle. If a and b are the legs, then the angles opposite to them are the other two angles. If a = b, then those angles are both 45°, which makes sense. But the problem is presented as if the answer isn't 45°, so maybe I made a mistake here.Wait, let me check my steps again. Starting from the equation:log(a + b) = (1/2) log b + (1/2) log(a + 3b)Then converted RHS to log sqrt(b(a + 3b))So a + b = sqrt(b(a + 3b))Squared both sides: a² + 2ab + b² = b(a + 3b) = ab + 3b²Then moved all terms left:a² + 2ab + b² - ab - 3b² = a² + ab - 2b² = 0Factored as (a + 2b)(a - b) = 0, leading to a = b or a = -2bSince a and b are positive, a = b. Therefore, the triangle has legs a and a, so the angle opposite leg a (which is one of the acute angles) is 45°. But the problem seems to expect a different answer, so maybe there's a mistake here.Wait, perhaps I miscalculated when factoring. Let me check:a² + ab - 2b². Let's use the quadratic formula for a in terms of b.Treat this as a quadratic equation in a:a² + (b)a + (-2b²) = 0Using quadratic formula:a = [-b ± sqrt(b² - 4*1*(-2b²))]/2= [-b ± sqrt(b² + 8b²)]/2= [-b ± sqrt(9b²)]/2= [-b ± 3b]/2So two solutions:a = (-b + 3b)/2 = (2b)/2 = ba = (-b - 3b)/2 = (-4b)/2 = -2bSo same as before. So only valid solution is a = b. So angle is 45°, which is π/4 radians. But maybe the problem is designed so that even though the algebra gives a = b, perhaps there's a step I missed?Wait, let me confirm by plugging a = b into the original equation.Original equation:log(a + b) = (1/2) log b + (1/2) log(a + 3b)If a = b, then LHS is log(b + b) = log(2b)RHS: (1/2) log b + (1/2) log(b + 3b) = (1/2) log b + (1/2) log(4b) = (1/2)(log b + log 4b) = (1/2) log(4b²) = (1/2)(log 4 + log b²) = (1/2)(log 4 + 2 log b) = (1/2)(2 log 2 + 2 log b) = log 2 + log b = log(2b)Which equals LHS. So that checks out. Therefore, a = b is indeed a solution.But then, the angle opposite leg a is 45°, right? Because in a right triangle with legs equal, the angles are both 45°. So the answer should be 45°, or π/4 radians.But the problem might be expecting a different answer, which makes me think maybe I made a wrong assumption. Wait, perhaps there's a different solution. Let me check again.Wait, if a = b, then in the right triangle, tan(theta) = opposite/adjacent = a/b = 1, so theta = 45°, which is correct.Alternatively, maybe the problem is not in the algebra but in the application. Wait, but the steps all check out. Let me try another approach.Suppose we let t = a/b, so a = tb. Then we can express everything in terms of t.Original equation:log(a + b) = (1/2) log b + (1/2) log(a + 3b)Substitute a = tb:log(tb + b) = (1/2) log b + (1/2) log(tb + 3b)Factor b inside the logs:log(b(t + 1)) = (1/2) log b + (1/2) log(b(t + 3))Using log rules:log b + log(t + 1) = (1/2) log b + (1/2)[log b + log(t + 3)]Simplify RHS:(1/2) log b + (1/2) log b + (1/2) log(t + 3) = log b + (1/2) log(t + 3)So equation becomes:log b + log(t + 1) = log b + (1/2) log(t + 3)Subtract log b from both sides:log(t + 1) = (1/2) log(t + 3)Multiply both sides by 2:2 log(t + 1) = log(t + 3)Which is equivalent to:log(t + 1)^2 = log(t + 3)Therefore:(t + 1)^2 = t + 3Expand left side:t² + 2t + 1 = t + 3Bring all terms to left:t² + 2t + 1 - t - 3 = t² + t - 2 = 0Factor:(t + 2)(t - 1) = 0Solutions t = -2 or t = 1Since t = a/b and a, b are positive, t = 1, so a/b = 1 => a = bSame result as before. Therefore, angle opposite a is 45 degrees. So the answer is 45°, which is π/4 radians.But the problem says "Calculate the measure of the angle opposite the leg of length a." If a = b, then the angle opposite a is 45 degrees. So maybe the answer is 45 degrees. But maybe I need to verify once again.Alternatively, is there a possibility that the equation allows another solution? Let me check with t = -2. If t = -2, then a = -2b, which is invalid because lengths can't be negative. So the only valid solution is t = 1, meaning a = b. Therefore, the angle is 45 degrees. Therefore, the answer should be 45 degrees.But maybe the problem expects the answer in radians? The question doesn't specify, but since it's a trigonometry problem, both degrees and radians are possible. But in most cases, unless specified, degrees are used for angles in such contexts. However, since the user hasn't specified, perhaps both are acceptable, but the standard answer is π/4 radians or 45°. Let me check the problem statement again.The original problem says: "Calculate the measure of the angle opposite the leg of length a." The word "measure" can be in degrees or radians. But since it's a right triangle, commonly angles are given in degrees unless stated otherwise. So 45 degrees. But maybe I should present it as π/4? Let me check.Alternatively, perhaps the answer is 60 degrees? But that would be if the triangle were 30-60-90, but that's not the case here. Wait, no, if a = b, it's definitely 45-45-90. So 45 degrees is correct. Therefore, I think the answer is 45 degrees, or π/4 radians. Since the problem may expect a box around the answer, but I need to check if there's any mistake in my steps.Wait, another way to check: if a = b, then the original equation holds. Therefore, the angle is 45 degrees. Therefore, the answer is 45°, which is π/4. Let me confirm once again.Alternatively, maybe the problem is designed to trick us, but given the steps all check out, I think 45 degrees is correct. Therefore, the measure of the angle opposite leg a is 45 degrees.Final AnswerThe measure of the angle opposite the leg of length (a) is boxed{45^circ}.</think>Given a right triangle with legs (a) and (b), we start with the equation:[log (a + b) = frac{1}{2} cdot log b + frac{1}{2} cdot log (a + 3b)]First, we combine the right-hand side using logarithm properties:[log (a + b) = log left( sqrt{b} cdot sqrt{a + 3b} right) = log sqrt{b(a + 3b)}]Equating the arguments of the logarithms:[a + b = sqrt{b(a + 3b)}]Squaring both sides to eliminate the square root:[(a + b)^2 = b(a + 3b)]Expanding both sides:[a^2 + 2ab + b^2 = ab + 3b^2]Rearranging terms:[a^2 + ab - 2b^2 = 0]Factoring the quadratic equation:[(a + 2b)(a - b) = 0]Since (a) and (b) are positive, the solution (a = -2b) is invalid, leaving:[a = b]In a right triangle with legs (a) and (b), if (a = b), the triangle is isosceles, and the angles opposite the legs are each (45^circ).Thus, the measure of the angle opposite the leg of length (a) is (boxed{45^circ}).]

🔑:## Solution.Using the rules for logarithms, we getbegin{array}{ll}log (a+b)=frac{1}{2} cdot log [b cdot(a+3 b)] & log (a+b)^{2}=log [b cdot(a+3 b)] & 1 text { point } (a+b)^{2}=b cdot(a+3 b) . & 1 text { point }end{array}From this, we have a^{2}+a b-2 b^{2}=0.After dividing by b^{2}, we getbegin{array}{r}left(frac{a}{b}right)^{2}+frac{a}{b}-2=0 text { or } operatorname{tg}^{2} alpha+operatorname{tg} alpha-2=0 operatorname{tg} alpha=1 text { or } operatorname{tg} alpha=-2end{array}In a right-angled triangle, the solution can only be operatorname{tg} alpha=1, i.e., alpha=45^{circ}.1 point

🧠:[<think>Okay, so I have this problem here with vectors a and b. Let me take a deep breath and tackle each part step by step. First, part (1) asks for the coordinates of 2a + b. Hmm, vector addition and scalar multiplication. Alright, vectors a and b are given in component form, which should make this straightforward. Vector a is (1, 0). If I need to compute 2a, that should just be multiplying each component by 2. So 2 times 1 is 2, and 2 times 0 is 0. So 2a is (2, 0). Then, vector b is (-1, 2). Now, adding 2a and b together, I just add their corresponding components. So the x-component is 2 + (-1), which is 1. The y-component is 0 + 2, which is 2. So 2a + b should be (1, 2). Let me double-check that. Starting with a = (1,0), scaling by 2 gives (2,0). Then adding b, which is (-1,2), so 2 + (-1) = 1 and 0 + 2 = 2. Yep, that seems right. So part (1) answer is (1,2). Moving on to part (2), which is to find the magnitude of vector (a - b). So first, I need to compute a - b, then find its length. Let's start with subtracting the vectors. Vector a is (1, 0) and vector b is (-1, 2). So subtracting component-wise: 1 - (-1) for the x-component, and 0 - 2 for the y-component. That gives 1 + 1 = 2 for x, and 0 - 2 = -2 for y. Therefore, a - b is (2, -2). Now, the magnitude of this vector is calculated by the square root of the sum of the squares of its components. So the formula is |v| = sqrt(x^2 + y^2). Applying that here: x is 2, so 2 squared is 4; y is -2, so (-2)^2 is 4. Adding those together: 4 + 4 = 8. Then the square root of 8. Hmm, sqrt(8) can be simplified. Since 8 is 4 times 2, sqrt(4*2) = sqrt(4)*sqrt(2) = 2*sqrt(2). So |a - b| is 2 times the square root of 2. Let me verify these steps again. For part (2): a - b. Original vectors: a = (1,0), b = (-1,2). Subtracting: 1 - (-1) = 2, 0 - 2 = -2. So vector (2, -2). The magnitude is sqrt(2^2 + (-2)^2) = sqrt(4 + 4) = sqrt(8) = 2*sqrt(2). That seems correct. Wait, just making sure I didn't mix up the components. For a - b, yes, it's (a_x - b_x, a_y - b_y). So 1 - (-1) is 2, and 0 - 2 is -2. Yes, that's right. Then squaring each component: 2 squared is 4, (-2) squared is 4. Sum is 8. Square root of 8 is 2.828... but in exact terms, that's 2*sqrt(2). So both parts look good. Let me recap:1) 2a + b: 2*(1,0) + (-1,2) = (2,0) + (-1,2) = (1,2). 2) |a - b|: a - b = (2, -2), magnitude is sqrt(4 + 4) = sqrt(8) = 2√2. I think that's all. I don't see any mistakes here. The operations are straightforward component-wise addition and scalar multiplication, followed by applying the magnitude formula. Final Answer(1) boxed{(1, 2)}(2) boxed{2sqrt{2}}</think>Given vectors (overrightarrow{a} = (1, 0)) and (overrightarrow{b} = (-1, 2)).(1) To find the coordinates of (2overrightarrow{a} + overrightarrow{b}):- Calculate (2overrightarrow{a}) by scaling each component of (overrightarrow{a}) by 2: (2overrightarrow{a} = 2(1, 0) = (2, 0)).- Add the vectors (2overrightarrow{a}) and (overrightarrow{b}) component-wise: (2overrightarrow{a} + overrightarrow{b} = (2, 0) + (-1, 2) = (2 + (-1), 0 + 2) = (1, 2)).(2) To find (|overrightarrow{a} - overrightarrow{b}|):- Subtract the vectors (overrightarrow{a}) and (overrightarrow{b}) component-wise: (overrightarrow{a} - overrightarrow{b} = (1, 0) - (-1, 2) = (1 - (-1), 0 - 2) = (2, -2)).- Calculate the magnitude of the resulting vector: (|overrightarrow{a} - overrightarrow{b}| = sqrt{(2)^2 + (-2)^2} = sqrt{4 + 4} = sqrt{8} = 2sqrt{2}). Final Answer(1) (boxed{(1, 2)})(2) (boxed{2sqrt{2}}),<think>Okay, so I have these two vectors here, vector a is (1, 0) and vector b is (-1, 2). The problem has two parts. Let me start with the first one: finding the coordinates of 2a + b. Alright, vectors, right? They can be scaled and added component-wise. Let me recall how that works. If I have a vector, say (x, y), and I multiply it by a scalar, like 2, then each component gets multiplied by 2. So 2*(x, y) would be (2x, 2y). Similarly, adding two vectors is done by adding their corresponding components. So (x1, y1) + (x2, y2) = (x1+x2, y1+y2). Got it.So for part (1), I need to compute 2 times vector a plus vector b. Let's break it down step by step. First, calculate 2a. Vector a is (1, 0), so multiplying by 2 would give (2*1, 2*0) which is (2, 0). Then I need to add vector b, which is (-1, 2), to this result. Adding the components: the x-component is 2 + (-1) and the y-component is 0 + 2. Let me write that out:2a + b = (2*1 + (-1), 2*0 + 2) = (2 - 1, 0 + 2) = (1, 2).Wait, that seems straightforward. Let me double-check. 2 times a is (2, 0). Then adding (-1, 2) gives (2 + (-1), 0 + 2) which is indeed (1, 2). Yep, that looks right. So part (1) should be (1, 2).Moving on to part (2): Find the magnitude of vector a minus vector b, which is written as |a - b|. Hmm. First, I need to compute the vector a - b, and then find its magnitude. Let me recall how vector subtraction works. Similar to addition, you subtract component-wise. So (x1, y1) - (x2, y2) = (x1 - x2, y1 - y2). Then, the magnitude of a vector (x, y) is the square root of (x squared plus y squared), right? So |(x, y)| = sqrt(x² + y²).Let me compute a - b first. Vector a is (1, 0) and vector b is (-1, 2). So subtracting the components: x-component is 1 - (-1) and y-component is 0 - 2. That gives:a - b = (1 - (-1), 0 - 2) = (1 + 1, -2) = (2, -2).Now, the magnitude of this resulting vector (2, -2) is sqrt(2² + (-2)²) = sqrt(4 + 4) = sqrt(8). Hmm, sqrt(8) can be simplified further. Since 8 is 4*2, sqrt(4*2) = sqrt(4)*sqrt(2) = 2*sqrt(2). So |a - b| is 2√2.Wait, let me check that again. The components after subtraction are 2 and -2. Squaring both gives 4 and 4, summing to 8, square root of 8 is indeed 2√2. That seems correct. Let me verify the subtraction again: a is (1,0), b is (-1,2). So a - b is (1 - (-1), 0 - 2). 1 - (-1) is 1 + 1 = 2. 0 - 2 is -2. So yes, (2, -2). Then the magnitude is sqrt(4 + 4) = sqrt(8) = 2√2. Perfect.So part (1) answer is (1, 2) and part (2) answer is 2√2. Let me just recap to make sure I didn't mix up any components. First part: scaling and adding. Second part: subtracting and then magnitude. All steps seem correct. I don't think I made any arithmetic errors here, but let me check once more.For part (1):2a = 2*(1, 0) = (2, 0)Adding b: (2, 0) + (-1, 2) = (2 + (-1), 0 + 2) = (1, 2). Correct.Part (2):a - b = (1 - (-1), 0 - 2) = (2, -2)| (2, -2) | = sqrt(2² + (-2)^2) = sqrt(4 + 4) = sqrt(8) = 2√2. Correct.Yep, looks like everything checks out. I think I'm confident with these answers.Final Answer(1) boxed{(1, 2)} (2) boxed{2sqrt{2}}</think>Given vectors (overrightarrow{a} = (1, 0)) and (overrightarrow{b} = (-1, 2)).(1) To find the coordinates of (2overrightarrow{a} + overrightarrow{b}):First, compute (2overrightarrow{a}):[2overrightarrow{a} = 2(1, 0) = (2, 0)]Then add (overrightarrow{b}):[2overrightarrow{a} + overrightarrow{b} = (2, 0) + (-1, 2) = (2 - 1, 0 + 2) = (1, 2)](2) To find (|overrightarrow{a} - overrightarrow{b}|):First, compute (overrightarrow{a} - overrightarrow{b}):[overrightarrow{a} - overrightarrow{b} = (1, 0) - (-1, 2) = (1 + 1, 0 - 2) = (2, -2)]Then find the magnitude of the resulting vector:[|overrightarrow{a} - overrightarrow{b}| = sqrt{2^2 + (-2)^2} = sqrt{4 + 4} = sqrt{8} = 2sqrt{2}] Final Answer(1) (boxed{(1, 2)}) (2) (boxed{2sqrt{2}})]

🔑: Solution:# Part (1): Finding the coordinates of 2overrightarrow{a}+overrightarrow{b}Given vectors overrightarrow{a}=(1,0) and overrightarrow{b}=(-1,2), we calculate 2overrightarrow{a}+overrightarrow{b} as follows:- First, we find 2overrightarrow{a} by multiplying overrightarrow{a} by 2: 2overrightarrow{a} = 2 cdot (1,0) = (2cdot1, 2cdot0) = (2,0).- Then, we add overrightarrow{b} to 2overrightarrow{a}: 2overrightarrow{a}+overrightarrow{b} = (2,0) + (-1,2) = (2+(-1), 0+2) = (1,2).Therefore, the coordinates of 2overrightarrow{a}+overrightarrow{b} are boxed{(1,2)}.# Part (2): Finding |overrightarrow{a}-overrightarrow{b}|Given overrightarrow{a}=(1,0) and overrightarrow{b}=(-1,2), we calculate overrightarrow{a}-overrightarrow{b} as follows:- Subtract overrightarrow{b} from overrightarrow{a}: overrightarrow{a}-overrightarrow{b} = (1,0) - (-1,2) = (1-(-1), 0-2) = (2,-2).- To find the magnitude |overrightarrow{a}-overrightarrow{b}|, we use the formula for the magnitude of a vector in mathbb{R}^2, which is sqrt{x^2+y^2}, where (x,y) are the coordinates of the vector:|overrightarrow{a}-overrightarrow{b}| = sqrt{(2)^2+(-2)^2} = sqrt{4+4} = sqrt{8} = 2sqrt{2}.Thus, the magnitude of overrightarrow{a}-overrightarrow{b} is boxed{2sqrt{2}}.

🧠:[<think>Okay, so I have this problem here: The external angles of a triangle are proportional to the numbers 5:7:8. I need to find the angle between the altitudes of this triangle drawn from the vertices of its smaller angles. Hmm, let me try to break this down step by step. First, let's recall what external angles are. Wait, in a triangle, the external angle at a vertex is equal to the sum of the two internal angles at the other vertices. Right? So, each external angle is supplementary to its corresponding internal angle. That means if the external angles are proportional to 5:7:8, then each external angle is 5x, 7x, 8x for some x. But the sum of the external angles of any polygon is 360 degrees, right? So, for a triangle, the sum of all external angles is 360 degrees. Therefore, 5x + 7x + 8x = 360. Let me compute that. 5x + 7x is 12x, plus 8x is 20x. So, 20x = 360. Then, x = 360 / 20 = 18. Therefore, the external angles are 5*18=90, 7*18=126, and 8*18=144 degrees. But these are the external angles. Since each external angle is supplementary to the internal angle, the internal angles would be 180 - external angle. So, the internal angles are 180 - 90 = 90 degrees, 180 - 126 = 54 degrees, and 180 - 144 = 36 degrees. Let me verify that. 90 + 54 + 36 = 180, which is correct for a triangle's internal angles. So, the triangle has angles of 90°, 54°, and 36°. So, that's a right-angled triangle with angles 90°, 54°, and 36°. Now, the problem asks for the angle between the altitudes of this triangle drawn from the vertices of its smaller angles. Let me make sure I understand this. The triangle has three angles: 90°, 54°, and 36°. The smaller angles would be the ones with the smaller measures, which are 36° and 54°, right? Since 36 is smaller than 54, which is smaller than 90. So, the vertices opposite these angles would be the sides opposite to these angles. Wait, no, the altitudes are drawn from the vertices of the smaller angles. So, vertices with the smaller angles. So, the angles at these vertices are 36° and 54°. So, the triangle has three vertices: let's say vertex A, B, and C. Let me assign the angles. Let's say angle at A is 36°, angle at B is 54°, and angle at C is 90°. So, right-angled at C. Then, the altitudes from the vertices of the smaller angles would be the altitudes from A and B. So, the altitude from A and the altitude from B. Then, we need to find the angle between these two altitudes. Wait, but how exactly? Let me visualize this. If the triangle is right-angled at C, then the sides opposite to the angles would be: side opposite angle A (36°) is BC, opposite angle B (54°) is AC, and opposite angle C (90°) is AB. So, AB is the hypotenuse. Then, the altitude from vertex A (which is angle 36°) would be a perpendicular line from A to the opposite side BC. Similarly, the altitude from vertex B (angle 54°) would be a perpendicular line from B to the opposite side AC. Then, these two altitudes intersect each other at some point inside the triangle, and the angle between them is what we need to find. Alternatively, maybe the problem is referring to the angle formed at the point where these two altitudes meet. Hmm. Wait, but in a triangle, the three altitudes intersect at the orthocenter. So, if we have two altitudes, they intersect at the orthocenter, and the angle between them at that point is the angle we need. But in a right-angled triangle, the orthocenter is at the vertex of the right angle. Wait, yes! In a right-angled triangle, the altitudes are the two legs and the altitude from the right angle to the hypotenuse. Wait, but let's recall: in any triangle, the altitudes are the perpendicular segments from each vertex to the opposite side. In a right-angled triangle, say right-angled at C, then the altitude from C is the same as the vertex C, because the legs themselves are the altitudes. The altitude from A would be the perpendicular from A to BC, and the altitude from B would be the perpendicular from B to AC. But in the right-angled triangle, the orthocenter is at the right-angled vertex C, because the two legs are the altitudes from A and B, intersecting at C. Wait, maybe I'm getting confused here.Wait, let's clarify. Let me denote the triangle as ABC, right-angled at C. So, angle C is 90°, angle A is 36°, angle B is 54°. Then, sides: AB is hypotenuse, AC and BC are legs. The altitude from vertex A is a line from A perpendicular to BC. Similarly, the altitude from vertex B is a line from B perpendicular to AC. The orthocenter of a right-angled triangle is at the right-angle vertex, which is C. So, the three altitudes of the triangle are: from A to BC (which meets BC at some point, say D), from B to AC (meeting AC at some point, say E), and from C to AB (which is the same as point C, since it's already a right angle). Wait, no, the altitude from C would be the same as the vertex C only if we consider the altitude to AB, but in reality, the altitude from C is the same as the vertex C in this case? Wait, no. Wait, in a triangle, the altitude from a vertex is the perpendicular segment from that vertex to the opposite side. In the case of vertex C in triangle ABC, the altitude from C is the same as the vertex C because the side opposite to C is AB, and the altitude from C to AB is the same as the vertex C if ABC is right-angled at C. Wait, no. Wait, in a right-angled triangle, the altitude from the right angle to the hypotenuse is a different segment. For example, in triangle ABC right-angled at C, the altitude from C to AB is a segment from C to the hypotenuse AB, meeting it at a right angle. That's different from the vertex C itself. So, actually, in a right-angled triangle, the three altitudes are: the two legs (AC and BC) and the altitude from C to AB. Therefore, the orthocenter is at point C, because that's where the three altitudes intersect. Because the altitude from A is the leg BC, the altitude from B is the leg AC, and the altitude from C is the altitude to AB. But since BC and AC meet at point C, which is also where the altitude from C (the orthocenter) is located. Wait, actually, in a right-angled triangle, the orthocenter coincides with the vertex of the right angle. So, point C is the orthocenter. But in this problem, we are asked about the angle between the altitudes drawn from the vertices of the smaller angles. The smaller angles are at vertices A (36°) and B (54°). So, the altitudes from A and B. Wait, the altitude from A is BC (since in a right-angled triangle, the legs are the altitudes). Wait, no, wait. If the triangle is right-angled at C, then the sides AC and BC are the legs. The altitude from A to BC would be a different segment. Wait, maybe I need to draw this mentally. Let me consider triangle ABC with right angle at C. So, AC and BC are perpendicular. Then, the altitude from A to BC: since BC is already a leg, and A is one endpoint of AC. If I draw a perpendicular from A to BC, but BC is already a horizontal line (for example), and A is above BC. Wait, but in this case, point A is connected to point C, and BC is the other leg. So, to draw an altitude from A to BC, which is a perpendicular line from A to BC. But since AC is perpendicular to BC, the altitude from A to BC is just the segment AC itself. Wait, because AC is perpendicular to BC. Therefore, the altitude from A to BC is AC. Similarly, the altitude from B to AC is BC. Therefore, the altitudes from A and B are the legs AC and BC themselves, which meet at point C, forming a right angle. Therefore, the angle between these two altitudes is 90 degrees. But that seems too straightforward, and the problem probably expects a different answer. Maybe I misunderstood the problem. Wait, the problem says "the angle between the altitudes of this triangle drawn from the vertices of its smaller angles". So, in the triangle, the smaller angles are 36° and 54°, so their vertices are A and B. So, the altitudes from A and B. But in the right-angled triangle, the altitudes from A and B are the legs BC and AC, respectively, which meet at point C, forming a right angle. Therefore, the angle between them is 90°. But that seems contradictory because if the triangle is right-angled, then the legs are perpendicular, so the angle between them is 90°, which would be the answer. But maybe the problem is not a right-angled triangle? Wait, no, according to the given ratios of external angles, we derived the internal angles as 36°, 54°, and 90°, so it is a right-angled triangle. Wait, maybe I made a mistake here. Let me check again. The external angles are proportional to 5:7:8. So, 5x + 7x + 8x = 360, which gives x=18, so external angles are 90°, 126°, and 144°. Then internal angles are 180 - 90=90°, 180 - 126=54°, 180 - 144=36°, which adds up to 180. So, that's correct. Therefore, the triangle is right-angled. Therefore, the altitudes from the smaller angles (36° and 54°) are the legs, and their angle is 90°. Therefore, the answer should be 90°, which is π/2 radians. But that seems too easy, and maybe I misrepresented the problem. Wait, perhaps I'm confusing internal and external angles. Wait, external angles are the angles formed by one side and the extension of the other side at each vertex. For a triangle, each external angle is equal to the sum of the two non-adjacent internal angles. So, in that case, if external angles are proportional to 5:7:8, then each external angle is 5x,7x,8x. Then, as before, 5x+7x+8x=360, x=18, external angles 90°,126°,144°, so internal angles 90°,54°,36°. So, triangle is right-angled at C. Then, the altitudes from A and B are the legs BC and AC, which meet at 90°, so the angle between the altitudes is 90°. But maybe the problem is expecting something different. Wait, maybe the triangle is not right-angled. Wait, no, 90° is one of the internal angles. So, unless there's a miscalculation here. Let me check again. Sum of external angles is always 360° for any polygon. For a triangle, each external angle is equal to the sum of the two internal angles at the other vertices. Wait, is that correct? Let's recall. For a triangle, the measure of an external angle is equal to the sum of the measures of the two non-adjacent internal angles. So, for example, the external angle at vertex A is equal to the sum of the internal angles at B and C. So, if the external angles are given, then each external angle equals the sum of the two opposite internal angles. So, if we denote the internal angles as α, β, γ, then the external angles are α' = β + γ, β' = α + γ, γ' = α + β. Then, the given ratios are α':β':γ' =5:7:8. So, α' =5k, β'=7k, γ'=8k. Then, since α' + β' + γ' = (β + γ) + (α + γ) + (α + β) = 2(α + β + γ) = 2*180=360°, which matches. Therefore, 5k +7k +8k=20k=360°, so k=18°, so external angles are 90°,126°,144°, and internal angles are 180 - external angles. Wait, but hold on: internal angle at A is 180 - α', where α' is the external angle at A. But α' = β + γ, so internal angle at A is 180 - (β + γ) = α. Because α + β + γ =180. So, yes, internal angle α = 180 - α', where α' is the external angle at A. Therefore, internal angles are 180 - 90=90°, 180 -126=54°, 180 -144=36°. So, correct. Therefore, the triangle is right-angled. Therefore, the triangle is right-angled at the vertex with 90°, which is the largest angle. Therefore, the two smaller angles are 36° and 54°, which are at vertices A and B. The altitudes from these vertices would be the legs BC and AC, which are perpendicular to each other, hence the angle between them is 90°. Therefore, the answer is 90°, which is π/2 radians, so the answer is 90°, boxed as boxed{90}. Wait, but maybe I need to verify this with coordinates. Let me assign coordinates to the triangle. Let’s place the right angle at C (0,0), vertex B along the x-axis (c,0), and vertex A along the y-axis (0,b). Then, angle at A is 36°, angle at B is 54°, angle at C is 90°. Let’s compute the coordinates. In a right-angled triangle, the legs are AC and BC. Let’s denote AC = b, BC = a, and AB = c. Then, the angles at A and B can be determined by tan(36°) = opposite/adjacent = BC/AC = a/b, and tan(54°) = AC/BC = b/a. Since tan(54°) = cot(36°), which is 1/tan(36°), which checks out. So, let's say AC = 1, then BC = tan(36°) ≈ 0.7265. Then, the coordinates would be C(0,0), B(0.7265, 0), A(0,1). Then, the altitude from A to BC. Wait, but in this coordinate system, BC is along the x-axis from (0,0) to (0.7265,0). The altitude from A(0,1) to BC is the perpendicular from A to BC. But since BC is horizontal, the altitude from A is vertical, which is the line x=0, which is AC itself. Therefore, the altitude from A is AC, which is the same as the leg AC. Similarly, the altitude from B to AC is the horizontal line y=0, which is BC. Therefore, these two altitudes are AC and BC, which meet at point C(0,0) at a right angle. Therefore, the angle between them is 90°, which confirms the previous conclusion. Therefore, the answer is 90 degrees. But let me think again. The problem says "the angle between the altitudes of this triangle drawn from the vertices of its smaller angles". So, since the triangle is right-angled, the two smaller angles are 36° and 54°, which are at vertices A and B. The altitudes from A and B are the legs AC and BC, which are perpendicular. So, the angle between them is 90°, so the answer is 90°. Alternatively, maybe I misread the problem. Maybe it's not a right-angled triangle? Wait, but according to the given ratio of external angles, the internal angles are 90°,54°,36°, so it is a right-angled triangle. Alternatively, maybe the problem is not a right-angled triangle, and I miscalculated the internal angles. Wait, let's check again. External angles are 5:7:8. Sum of external angles is 360°, so 5x +7x +8x=20x=360°, x=18. Therefore, external angles are 90°,126°,144°. Therefore, internal angles are 180-90=90°, 180-126=54°, 180-144=36°, which sums to 180. So, that's correct. So, it's a right-angled triangle. Therefore, my initial conclusion seems correct. Therefore, the angle between the two altitudes from the smaller angles is 90°, so the answer is boxed{90} degrees. But wait, maybe the problem is intended to be more complex. Let me think again. Maybe in a non-right-angled triangle, but in this case, according to the given ratios, it is right-angled. Alternatively, perhaps the problem is referring to the angle between the two altitudes when they are not the legs of the triangle. Wait, but in a right-angled triangle, the altitudes from the acute angles are the legs themselves. Wait, maybe in a different configuration. Suppose the triangle is not right-angled, but according to our calculation, it is. So perhaps the answer is indeed 90°. Alternatively, let's suppose that I made a mistake in considering the altitudes. Let me consider another approach. Suppose we have triangle ABC with angles at A:36°, B:54°, C:90°. Let's find the coordinates of the orthocenter. But in a right-angled triangle, the orthocenter is at the right-angle vertex, which is C. The altitudes from A and B intersect at C. Therefore, the angle between the altitudes from A and B is the angle at C, which is 90°, so the angle between them is 90°. Therefore, the answer is 90°, so I think that is correct. Alternatively, perhaps the problem is expecting a different interpretation. If the triangle were not right-angled, the computation would be more involved, but in this case, given the ratios of the external angles lead to a right-angled triangle. Therefore, I think the correct answer is 90 degrees, and the boxed answer is boxed{90}.Final Answerboxed{90}</think>The external angles of a triangle are proportional to the numbers 5:7:8. To find the angle between the altitudes drawn from the vertices of its smaller angles, we start by determining the internal angles of the triangle.1. Calculate the external angles: - The sum of the external angles of any polygon is 360 degrees. Given the ratio 5:7:8, we have: [ 5x + 7x + 8x = 360^circ implies 20x = 360^circ implies x = 18^circ ] - The external angles are (5 times 18^circ = 90^circ), (7 times 18^circ = 126^circ), and (8 times 18^circ = 144^circ).2. Determine the internal angles: - Each internal angle is supplementary to its corresponding external angle: [ 180^circ - 90^circ = 90^circ, quad 180^circ - 126^circ = 54^circ, quad 180^circ - 144^circ = 36^circ ] - The internal angles are (90^circ), (54^circ), and (36^circ), forming a right-angled triangle with the right angle at the vertex opposite the largest external angle.3. Identify the vertices and altitudes: - The triangle is right-angled at the vertex with the (90^circ) angle. The smaller angles are (36^circ) and (54^circ) at the other two vertices. - The altitudes from the vertices of the smaller angles (36° and 54°) are the legs of the right-angled triangle. These legs are perpendicular to each other.4. Conclusion: - The angle between the altitudes (which are the legs of the right-angled triangle) is (90^circ).Thus, the angle between the altitudes of the triangle drawn from the vertices of its smaller angles is (boxed{90}).,<think>Alright, so I have this problem here: The external angles of a triangle are proportional to the numbers 5:7:8. I need to find the angle between the altitudes of this triangle drawn from the vertices of its smaller angles. Hmm, okay. Let me start by recalling some basic triangle properties.First, external angles. Wait, the external angles of a triangle? But in a triangle, each external angle is equal to the sum of the two non-adjacent internal angles. Also, the sum of all external angles in any polygon is 360 degrees. So, for a triangle, each external angle is 180 degrees minus the corresponding internal angle. So, if the external angles are proportional to 5:7:8, maybe I can find each external angle's measure and then find the internal angles?Let me write that down. Let the external angles be 5x, 7x, and 8x. Since the sum of external angles in a polygon is 360 degrees, their sum should be 360. So, 5x + 7x + 8x = 360. That's 20x = 360, so x = 18. Therefore, the external angles are 5*18=90°, 7*18=126°, and 8*18=144°. Now, since each external angle is 180° minus the internal angle, the internal angles should be 180° - 90°=90°, 180° - 126°=54°, and 180° - 144°=36°. So the internal angles of the triangle are 90°, 54°, and 36°. Wait, but let me check: the sum of internal angles in a triangle is 180°, so 90 + 54 + 36 is indeed 180. Correct.So, this is a right-angled triangle with angles 90°, 54°, and 36°. The sides opposite these angles would be in the ratio of sine of these angles. But maybe I don't need the sides yet. The problem asks for the angle between the altitudes drawn from the vertices of its smaller angles. The smaller angles would be the ones with the smaller measures, which are 36° and 54°, since 90° is the largest. So, the vertices at the 36° and 54° angles each have an altitude drawn from them, and we need the angle between these two altitudes.Hmm. So, first, perhaps I need to visualize or sketch the triangle. Let me imagine a right-angled triangle where the right angle is at vertex C, angle A is 54°, and angle B is 36°. So, sides: opposite angle A (54°) is BC, opposite angle B (36°) is AC, and the hypotenuse AB is opposite the right angle at C.Wait, maybe labeling the triangle would help. Let me assign labels:Let’s say triangle ABC, right-angled at C. So angle C is 90°, angle A is 54°, angle B is 36°. Then, sides: AB is the hypotenuse, BC is opposite angle A (54°), AC is opposite angle B (36°).Now, the altitudes from the vertices of the smaller angles. The smaller angles are 36° and 54°, which are at vertices B and A, respectively. So, we need to find the altitudes from vertex B and vertex A, and then find the angle between these two altitudes.Wait, in a triangle, the altitude from a vertex is the perpendicular line from that vertex to the opposite side. So, altitude from B would be perpendicular to AC, and altitude from A would be perpendicular to BC.But in a right-angled triangle, some altitudes coincide with the legs. Wait, in triangle ABC, right-angled at C, the altitude from C is the same as the vertex C itself, but altitudes from A and B would be different.Wait, altitude from A: since A is one of the vertices, and the triangle is right-angled at C, the altitude from A to BC. Let me think. In triangle ABC, side BC is opposite angle A (54°), so BC is of length, let's say, if we take hypotenuse AB as 1 (for simplicity), then BC = sin(54°), and AC = sin(36°). Wait, but maybe instead of assigning lengths, I can compute coordinates.Alternatively, maybe coordinate geometry would help here. Let me place the triangle in coordinate system. Let’s place point C at the origin (0,0). Since it's right-angled at C, let’s put point B along the x-axis and point A along the y-axis. So, coordinates: C(0,0), B(b, 0), A(0, a). Then, angle at B is 36°, angle at A is 54°, right?Wait, angle at B is 36°, angle at A is 54°, angle at C is 90°. So, the sides: AB is hypotenuse, AC is adjacent to angle B, BC is adjacent to angle A.Wait, in the coordinate system, side AC is from (0,0) to (0,a), so length AC = a. Side BC is from (0,0) to (b,0), length BC = b. Hypotenuse AB is from (b,0) to (0,a), length sqrt(a² + b²).The angles at A and B can be calculated using trigonometry. For angle at B (36°), tan(angle B) = opposite/adjacent = AC / BC = a / b. So, tan(36°) = a / b. Therefore, a = b tan(36°). Similarly, angle at A is 54°, tan(angle A) = BC / AC = b / a. But since a = b tan(36°), then tan(54°) = b / (b tan(36°)) = 1 / tan(36°). But tan(54°) is indeed equal to cot(36°), since tan(54°) = tan(90° - 36°) = cot(36°). So that's consistent.So, given that, maybe we can assign specific lengths. Let’s choose b = 1 for simplicity. Then, a = tan(36°). Let me calculate tan(36°). Approximately, tan(36°) ≈ 0.7265. So, coordinates would be C(0,0), B(1, 0), A(0, 0.7265). Hypotenuse AB connects (1,0) to (0, 0.7265).Now, we need to find the altitudes from vertices A and B. Let's start with the altitude from A to BC. But BC is the base from (0,0) to (1,0), which is along the x-axis. The altitude from A to BC is the perpendicular line from A to BC. Since BC is horizontal, the altitude from A to BC is vertical. But point A is already on the y-axis, so the altitude from A to BC is just the line from A(0, 0.7265) straight down to BC at point C(0,0). Wait, but that's just the side AC itself. But in a right-angled triangle, the legs are the altitudes corresponding to the other vertices. Wait, yes, in a right-angled triangle, the two legs are the altitudes from the acute angles. So, altitude from A is AC, and altitude from B is BC. But then, the angle between these two altitudes would be the angle between AC and BC, which is the right angle, 90°. But that can't be, because the problem is asking for the angle between the altitudes from the smaller angles, which are 36° and 54°, but in this case, the altitudes are the legs themselves, and the angle between them is 90°, but maybe I made a mistake here.Wait, no, hold on. Wait, in a right-angled triangle, the altitude from the right angle is the same as the vertex, but the altitudes from the other two vertices (the acute angles) are different. Wait, no, in a right-angled triangle, the two legs are the altitudes corresponding to the opposite vertices. Let me think. For triangle ABC, right-angled at C. The altitude from A to BC: since BC is the side opposite A, the altitude from A would be the same as the leg AC? Wait, no. Wait, the altitude from a vertex is a line segment from that vertex perpendicular to the opposite side. So, in triangle ABC, the altitude from A to BC: BC is the side opposite A. So, to draw the altitude from A to BC, we need a line from A perpendicular to BC. But since the triangle is right-angled at C, BC is the horizontal side, and AC is vertical. So, point A is already vertically above point C, so the altitude from A to BC is indeed AC, which is a leg of the triangle. Similarly, the altitude from B to AC is BC, the other leg. Therefore, the two altitudes from A and B are the legs AC and BC, and the angle between them is the angle at C, which is 90°. But that seems too straightforward, and the answer would be 90°, but the problem says "the angle between the altitudes of this triangle drawn from the vertices of its smaller angles." Wait, but in this case, the smaller angles are 36° and 54°, so the vertices at B (36°) and A (54°). The altitudes from these vertices are BC and AC, which meet at angle C, which is 90°, so the angle between the altitudes is 90°. But the problem might be more complicated if the triangle wasn't right-angled. Maybe I made a wrong assumption here.Wait, but let's double-check the problem statement: "The external angles of a triangle are proportional to the numbers 5:7:8." So we found the internal angles are 90°, 54°, and 36°, which makes it a right-angled triangle. Then the altitudes from the smaller angles (36° and 54°) are the legs, and their angle is 90°. But is that the case? Let me check with an example.Alternatively, maybe I have a misunderstanding about external angles. Let me confirm. The external angle at each vertex is equal to the sum of the two internal angles at the other vertices. So, for a triangle, each external angle is 180° minus the internal angle. So, if the external angles are in proportion 5:7:8, then as I did before, 5x + 7x + 8x = 360°, leading to x=18°, so external angles 90°, 126°, 144°, internal angles 90°, 54°, 36°. So that seems correct.Therefore, the triangle is right-angled, with angles 90°, 54°, 36°, so sides in ratio opposite these angles: hypotenuse (opposite 90°), side opposite 54°, and side opposite 36°. Therefore, the triangle is similar to a triangle with sides in ratio sin(90°) : sin(54°) : sin(36°), which is 1 : ~0.8090 : ~0.5878. So, maybe 1, 0.8090, 0.5878 as sides.But if we place this triangle in coordinate system with right angle at C(0,0), then the legs would be AC and BC, with lengths corresponding to 0.5878 and 0.8090, and hypotenuse AB of length 1. Then, the altitudes from A and B would be the legs themselves, so the angle between them is 90°, which is the right angle. Therefore, is the answer 90°? But the problem seems more complex, so maybe I messed up.Wait, wait a second. Maybe in the problem statement, the external angles are proportional to 5:7:8, but perhaps they are considered in a different order. For example, the external angles correspond to the vertices, but depending on the order, maybe the internal angles could be different. Wait, no, the external angles are in the ratio 5:7:8, but the sum is 360°, so regardless of order, each external angle is 5x,7x,8x, and the internal angles are 180° - 5x, 180° -7x, 180° -8x. But wait, that would be incorrect. Wait, each external angle is equal to 180° minus the corresponding internal angle. Therefore, if the external angles are in ratio 5:7:8, then the internal angles would be 180° -5x, 180° -7x, 180° -8x. But the sum of internal angles is 180°, so:(180 -5x) + (180 -7x) + (180 -8x) = 180540 -20x = 180540 -180 =20x360=20x => x=18Therefore, the external angles are 5*18=90°, 7*18=126°, 8*18=144°, as before, and internal angles are 180-90=90°, 180-126=54°, 180-144=36°, as before. So, the triangle is right-angled. Therefore, the altitudes from the smaller angles (36° and 54°) are the legs of the triangle, and the angle between them is 90°. Therefore, the answer is 90°, which is π/2 radians. But the problem might expect a different approach, maybe considering non-right-angled triangles? Wait, but according to the given ratio of external angles, the triangle must be right-angled. So maybe the answer is 90°, but let me verify once again.Alternatively, perhaps I made a mistake in assuming which angles the external angles correspond to. Wait, in a triangle, each vertex has an external angle. The external angle is equal to the sum of the two non-adjacent internal angles. So, if the external angles are in ratio 5:7:8, then the corresponding internal angles would be 180 -5x, 180 -7x, 180 -8x. But since the sum of internal angles is 180, as above. So the calculation seems correct.Alternatively, maybe the problem is not right-angled? Wait, but according to the calculation, one of the internal angles is 90°, so it is right-angled. Therefore, the altitudes from the smaller angles are the legs, and the angle between them is 90°. Therefore, the answer should be 90°, which is 90 degrees. But maybe the problem is tricking me? Let me check again.Wait, suppose the triangle is not right-angled. How could that be? If the external angles are in proportion 5:7:8, but the calculation leads to one internal angle being 90°, so it must be right-angled. Therefore, the answer should be 90°, but since the problem is given in a math competition or something, maybe expecting a different answer. Alternatively, maybe the triangle is not right-angled, and I made a wrong calculation. Let me check the steps again.Sum of external angles: 360 degrees. Ratios 5:7:8. Therefore, 5x +7x +8x=20x=360, so x=18. Therefore, external angles: 90°, 126°, 144°, internal angles: 90°, 54°, 36°. Correct. So triangle is right-angled. Then altitudes from the smaller angles (36° and 54°) are the legs, meeting at the right angle. So the angle between them is 90°. So answer is 90°. Hmm.Alternatively, maybe the problem is referring to the angle between the two altitudes outside the triangle? But in a right-angled triangle, the altitudes from the acute angles are the legs themselves, so they meet at the right angle vertex. So the angle between them is 90°. Therefore, unless there's a misunderstanding in the problem statement, the answer should be 90°.Wait, but maybe I misread the problem. It says "the angle between the altitudes of this triangle drawn from the vertices of its smaller angles." The smaller angles are 36° and 54°, so the vertices at those angles. The altitudes from those vertices. So in the right-angled triangle, the altitude from 54° (vertex A) is the leg AC, and the altitude from 36° (vertex B) is the leg BC. The angle between AC and BC is 90°, which is the right angle. Therefore, the angle between the altitudes is 90°. So maybe the answer is 90°, which is π/2 radians. But maybe the problem expects a more complicated answer? Let me check with another approach.Alternatively, suppose the triangle is not right-angled. Wait, but according to our calculation, one angle is 90°, so it must be right-angled. Therefore, the triangle is right-angled. Therefore, the angle between the altitudes is 90°. Therefore, the answer is 90 degrees. But let me check again with coordinates.Let me assign coordinates again. Let’s place the right angle at C(0,0), B(1,0), A(0, tan36°). So, A is (0, tan36°). Then, altitude from A is the same as AC, which is the vertical line from A(0, tan36°) to C(0,0). Altitude from B is BC, the horizontal line from B(1,0) to C(0,0). These two altitudes meet at C, forming a right angle. So the angle between them is 90°, so the answer is 90°. Therefore, the angle between the two altitudes is 90°, so boxed answer is boxed{90^circ}.But wait, maybe the problem is referring to another angle? Like, not the angle at the vertex where they meet, but the angle between the two altitude lines somewhere else? For example, if the two altitudes are extended, do they form another angle? But in this case, the altitudes are AC and BC, which are perpendicular, so anywhere they intersect, the angle between them is 90°. So, regardless, the angle between the two altitudes is 90°, so answer is 90°.Wait, but maybe in another configuration. Wait, suppose the triangle is not right-angled. Wait, but according to the given ratio of external angles, one of the internal angles is 90°, so it must be right-angled. Therefore, the answer is 90°. Therefore, I think the answer is 90 degrees. Maybe the problem is straightforward, but I overcomplicated it. Let me see.Alternatively, perhaps the problem is in Russian or another language, and "external angles" might refer to something else. Wait, in some contexts, external angles can mean angles outside the triangle, but in standard terms, external angles at each vertex are the angles formed by one side and the extension of the other side. Their sum is 360° for any polygon, which holds here. So, given that, the calculation is correct.Therefore, I think the answer is 90°, boxed as boxed{90^circ}.Final Answerboxed{90^circ}</think>Given that the external angles of a triangle are proportional to the numbers 5:7:8, we need to find the angle between the altitudes drawn from the vertices of its smaller angles.1. Determine the external angles: - Let the external angles be (5x), (7x), and (8x). - The sum of the external angles of any polygon is (360^circ). - Therefore, (5x + 7x + 8x = 360^circ). - Solving for (x), we get (20x = 360^circ implies x = 18^circ). - The external angles are (5 times 18^circ = 90^circ), (7 times 18^circ = 126^circ), and (8 times 18^circ = 144^circ).2. Find the internal angles: - Each internal angle is (180^circ - text{external angle}). - Internal angles are (180^circ - 90^circ = 90^circ), (180^circ - 126^circ = 54^circ), and (180^circ - 144^circ = 36^circ). - Verify the sum of internal angles: (90^circ + 54^circ + 36^circ = 180^circ), which is correct.3. Identify the triangle type and vertices: - The triangle is right-angled with angles (90^circ), (54^circ), and (36^circ). - The smaller angles are (36^circ) and (54^circ).4. Determine the altitudes: - In a right-angled triangle, the legs are the altitudes from the vertices of the smaller angles. - The altitude from the vertex with (36^circ) (vertex B) is one leg (BC). - The altitude from the vertex with (54^circ) (vertex A) is the other leg (AC).5. Find the angle between the altitudes: - The angle between the legs (altitudes) BC and AC in a right-angled triangle is the right angle at vertex C.Thus, the angle between the altitudes of the triangle drawn from the vertices of its smaller angles is (boxed{90^circ}).]

🔑:Answer: 90^{0}.Solution. The sum of the exterior angles of a triangle is180^{circ} cdot 3-180^{circ}=360^{circ}From this, we get that one of the exterior angles of the triangle is frac{360^{circ}}{5+7+8} cdot 5=90^{0}, and therefore the adjacent interior angle of the triangle is 90^{0}. Consequently, the other two angles are the smaller angles of the triangle, and the heights drawn from them coincide with the legs of the triangle, the angle between which is 90^{circ}.Comment. An answer without justification - 0 points.

🧠:[<think>Okay, so I have this cube, right? Let me visualize it. The cube is named ABCD-A₁B₁C₁D₁, and each edge has length 1. The problem says that there's a line segment MN where M is on the ray AA₁ and N is on the ray BC. Also, this segment MN intersects the edge C₁D₁ at point L. I need to find the minimum length of MN. Hmm, okay. Let me try to break this down.First, let me sketch the cube mentally. Cube ABCD-A₁B₁C₁D₁. So, the base is ABCD, and the top face is A₁B₁C₁D₁. Edges like AA₁, BB₁, etc., connect the base to the top face. All edges are length 1. Now, point M is on the ray AA₁. That means M can be anywhere along the line extending from A through A₁, even beyond A₁. Similarly, point N is on the ray BC, which is the line from B through C, possibly beyond C. The segment MN must intersect the edge C₁D₁ at some point L. The goal is to find the minimal possible length of MN.Alright, coordinates might help here. Let me assign coordinates to the cube. Let's place point A at the origin (0,0,0). Then, since the cube has edge length 1, the coordinates of the other points can be determined:- A: (0,0,0)- B: (1,0,0)- C: (1,1,0)- D: (0,1,0)- A₁: (0,0,1)- B₁: (1,0,1)- C₁: (1,1,1)- D₁: (0,1,1)So, ray AA₁ is the line from (0,0,0) through (0,0,1) and beyond. Similarly, ray BC is from (1,0,0) through (1,1,0) and beyond. Edge C₁D₁ is from (1,1,1) to (0,1,1).Since M is on ray AA₁, let's parametrize its position. Let me use a parameter t for M. If t=0, M is at A; t=1, M is at A₁; t>1, M is above A₁. Similarly, point N is on ray BC. Let's parametrize that with a parameter s. Starting at B (s=0), moving towards C (s=1), and beyond (s>1). So:Coordinates of M: (0, 0, t), where t ≥ 0.Coordinates of N: (1, s, 0), where s ≥ 0. Wait, actually, BC goes from B(1,0,0) to C(1,1,0), so the direction is along the y-axis. So, parametrizing N as (1, s, 0) where s ≥ 0. Correct.Now, the line segment MN connects (0,0,t) to (1,s,0). We need this segment to intersect the edge C₁D₁, which is from (1,1,1) to (0,1,1). Wait, edge C₁D₁ is actually along the x-axis at y=1, z=1. So any point on C₁D₁ is (x,1,1) where x goes from 1 to 0. So the coordinates for L would be (x,1,1) for some x between 0 and 1.So the problem reduces to: find t and s such that the line connecting (0,0,t) and (1,s,0) passes through some point (x,1,1) on C₁D₁, and then find the minimum length of MN over all such possible t and s.Hmm, okay. Let me try to model this. Let's parametrize the line MN. The line from M(0,0,t) to N(1,s,0). Let's write parametric equations for MN. Let parameter be u, going from 0 to 1.So, the parametric equations:x(u) = 0 + u*(1 - 0) = uy(u) = 0 + u*(s - 0) = u*sz(u) = t + u*(0 - t) = t - u*tSo, the line is (u, u*s, t - u*t) for u ∈ [0,1]We need this line to pass through a point L on C₁D₁, which is (x,1,1). So, we need there to exist some u and x such that:x = u1 = u*s1 = t - u*tAdditionally, since L is on C₁D₁, x must be between 0 and 1 (since edge C₁D₁ goes from x=1 to x=0 at y=1, z=1). So x ∈ [0,1], which means u ∈ [0,1].So, substituting x = u into the other equations:From the y-coordinate: 1 = u*s => s = 1/uFrom the z-coordinate: 1 = t - u*t => 1 = t*(1 - u) => t = 1/(1 - u)So, we have s = 1/u and t = 1/(1 - u). But u must be in (0,1), because if u=0, s would be undefined, and if u=1, t would be undefined. So u ∈ (0,1). Also, since x = u must be in [0,1], which it is because u is between 0 and 1.Therefore, for each u ∈ (0,1), we can define s and t such that the line MN passes through L = (u, 1, 1). Then, MN has endpoints M(0,0,t) = (0,0,1/(1 - u)) and N(1,s,0) = (1,1/u,0). Then, the length of MN can be computed as the distance between these two points.So, compute the distance between (0,0,1/(1 - u)) and (1,1/u,0). Let's do that.Distance MN squared is:(1 - 0)^2 + (1/u - 0)^2 + (0 - 1/(1 - u))^2= 1 + (1/u²) + (1/(1 - u))²So, MN = sqrt(1 + 1/u² + 1/(1 - u)^2)Therefore, to find the minimum MN, we need to minimize the function f(u) = 1 + 1/u² + 1/(1 - u)^2 over u ∈ (0,1)So, the problem reduces to minimizing f(u) = 1 + 1/u² + 1/(1 - u)^2 for 0 < u < 1.This seems like a calculus problem. Let's find the derivative of f(u) and set it to zero.First, write f(u) as:f(u) = 1 + u^(-2) + (1 - u)^(-2)Compute derivative f’(u):f’(u) = -2u^(-3) + 2(1 - u)^(-3)Set derivative to zero:-2/u³ + 2/(1 - u)³ = 0Multiply both sides by u³(1 - u)³ / 2:- (1 - u)³ + u³ = 0So,u³ - (1 - u)³ = 0Expand (1 - u)^3:1 - 3u + 3u² - u³So,u³ - (1 - 3u + 3u² - u³) = 0Simplify:u³ -1 + 3u - 3u² + u³ = 0Combine like terms:2u³ - 3u² + 3u -1 = 0So, 2u³ -3u² +3u -1=0We need to solve this cubic equation for u in (0,1). Let me check if u=1 is a solution: 2 -3 +3 -1=1≠0. u=0: -1≠0. Let's try rational roots. Possible rational roots are ±1, 1/2.Test u=1/2:2*(1/8) -3*(1/4) +3*(1/2) -1 = 1/4 - 3/4 + 3/2 -1 = (1/4 -3/4) + (3/2 -1) = (-1/2) + (1/2) = 0. Yes! u=1/2 is a root.Therefore, (u - 1/2) is a factor. Let's perform polynomial division or factor it out.Divide 2u³ -3u² +3u -1 by (u - 1/2). Let's use synthetic division:Coefficients: 2 | -3 | 3 | -1Root at u=1/2:Bring down 2Multiply by 1/2: 1Add to next coefficient: -3 +1= -2Multiply by 1/2: -1Add to next coefficient: 3 + (-1)=2Multiply by1/2:1Add to last coefficient: -1 +1=0So, the cubic factors as (u -1/2)(2u² -2u +2). Let's verify:(u -1/2)(2u² -2u +2) = 2u³ -2u² +2u -u² +u -1 = 2u³ -3u² +3u -1. Correct.Now, set 2u² -2u +2 =0. Discriminant: 4 -16= -12 <0, so no real roots. Thus, the only real root is u=1/2.Therefore, critical point at u=1/2. Now, check if this is a minimum.Second derivative test? Or check values around u=1/2.Let me compute f(u) at u=1/2:f(1/2) =1 +1/(1/2)^2 +1/(1 -1/2)^2 =1 +4 +4=9. So f(1/2)=9, so MN=3.Check the behavior as u approaches 0+:1/u² tends to infinity, so f(u) tends to infinity.As u approaches 1-:1/(1 - u)^2 tends to infinity, so f(u) tends to infinity.Thus, the function f(u) has a unique critical point at u=1/2, which must be the minimum. Hence, the minimal MN is 3.Wait, but let me verify this with another value. For example, take u=1/3:f(1/3)=1 +1/(1/9) +1/(2/3)^2=1 +9 + (9/4)=10 + 2.25=12.25, which is 12.25, so sqrt(12.25)=3.5, which is greater than 3.Similarly, u=1/4:f(1/4)=1 +16 +1/(9/16)=1 +16 +16/9≈17 +1.777≈18.777, sqrt≈4.33, which is larger than 3.Alternatively, at u=2/3:f(2/3)=1 +1/(4/9) +1/(1 -2/3)^2=1 +9/4 +9=10 + 9/4=12.25, same as before. So yes, 3.5.So seems like the minimal value is indeed at u=1/2, giving MN=3. Therefore, the minimal MN is 3.But wait, let me check the coordinates again. When u=1/2, then:s=1/u=2, t=1/(1 -u)=2.Therefore, point M is (0,0,2) and point N is (1,2,0). The line connecting these points passes through L=(1/2,1,1). Wait, let me check.Wait, according to parametrization:x(u)=u, y(u)=u*s=u*(1/u)=1. When u=1/2, y(u)=1/2 * s. Wait, but s=1/u, so y(u)=1/2*(2)=1. So z(u)=t - u*t=2 - (1/2)*2=2 -1=1. So when u=1/2, the point is (1/2,1,1), which is on C₁D₁ (since x=1/2 is between 0 and1, y=1, z=1). So yes, correct.So, the line MN from (0,0,2) to (1,2,0) passes through (1/2,1,1). The distance between (0,0,2) and (1,2,0) is sqrt((1)^2 + (2)^2 + (-2)^2)=sqrt(1 +4 +4)=sqrt(9)=3. Correct.So, the minimal MN is 3. Therefore, the answer is 3. So boxed, boxed{3}.Final AnswerThe minimum value of ( MN ) is (boxed{3}).</think>Given a cube (ABCD-A_1B_1C_1D_1) with edge length 1, we need to find the minimum length of the line segment (MN) where (M) is on the ray (AA_1) and (N) is on the ray (BC), and (MN) intersects the edge (C_1D_1) at point (L).1. Coordinates Assignment: - (A = (0,0,0)), (B = (1,0,0)), (C = (1,1,0)), (D = (0,1,0)) - (A_1 = (0,0,1)), (B_1 = (1,0,1)), (C_1 = (1,1,1)), (D_1 = (0,1,1))2. Parametrization: - Point (M) on ray (AA_1): ((0, 0, t)) where (t geq 0) - Point (N) on ray (BC): ((1, s, 0)) where (s geq 0)3. Line Segment (MN): - Parametric equations for line (MN): ((u, us, t - ut)) for (u in [0, 1]) - Intersection with (C_1D_1) (coordinates ((x, 1, 1))): - (x = u) - (1 = us implies s = frac{1}{u}) - (1 = t - ut implies t = frac{1}{1 - u})4. Distance Calculation: - Distance (MN) squared: (1 + left(frac{1}{u}right)^2 + left(frac{1}{1 - u}right)^2) - Minimize (f(u) = 1 + frac{1}{u^2} + frac{1}{(1 - u)^2}) for (0 < u < 1)5. Critical Point Analysis: - Derivative (f'(u) = -frac{2}{u^3} + frac{2}{(1 - u)^3}) - Solve (-frac{2}{u^3} + frac{2}{(1 - u)^3} = 0 implies u = frac{1}{2}) - Check (fleft(frac{1}{2}right) = 1 + 4 + 4 = 9 implies sqrt{9} = 3)Thus, the minimum value of (MN) is (boxed{3}).,<think>Okay, so I need to find the minimum length of the line segment MN in this cube. Let me try to visualize the cube first. The cube is labeled ABCD-A1B1C1D1, and the edge length is 1. Points M and N are on the rays AA1 and BC respectively. Also, the segment MN intersects the edge C1D1 at point L. Hmm, okay.First, let me sketch the cube in my mind. Let me recall that in a cube labeled like this, ABCD is the base square, and A1B1C1D1 is the top square. So AA1, BB1, CC1, DD1 are the vertical edges. The edges like AB, BC, CD, DA are the base edges, and similarly for the top.Now, point M is on the ray AA1. That means it can be anywhere along the line extending from A through A1, right? Similarly, point N is on the ray BC, which is the line extending from B through C. So BC is a horizontal edge on the base, from B to C, but the ray BC would go beyond point C, right?And the segment MN must intersect the edge C1D1 at point L. Edge C1D1 is the top edge connecting C1 to D1. So, for MN to intersect C1D1, the line segment must pass through that edge. So MN starts at M on AA1 ray, goes through L on C1D1, and ends at N on BC ray.Our goal is to find the minimal length of MN. So we need to find positions of M and N such that when we connect them, the line passes through C1D1, and the length of MN is as short as possible.Let me think about coordinates. Maybe assigning coordinates to the cube will make this easier. Let's place the cube in a 3D coordinate system. Let me set point A at the origin (0,0,0). Then:- A: (0,0,0)- B: (1,0,0)- C: (1,1,0)- D: (0,1,0)- A1: (0,0,1)- B1: (1,0,1)- C1: (1,1,1)- D1: (0,1,1)So, the edge AA1 goes from (0,0,0) to (0,0,1) and beyond. So any point M on ray AA1 can be represented as (0,0, t) where t ≥ 0. Similarly, the ray BC starts at B (1,0,0) and goes through C (1,1,0) and beyond. So any point N on ray BC can be represented as (1,1 + s, 0) where s ≥ 0. Wait, hold on. The direction from B to C is along the y-axis. So moving from B (1,0,0) to C (1,1,0), the ray BC would have direction vector (0,1,0). So parametrizing point N on ray BC: starting at B (1,0,0), moving in the direction of C, so point N can be written as (1, t, 0) where t ≥ 0. Wait, but if t=0, it's at B, t=1 at C, and t>1 beyond C. That makes sense. So maybe I should parameterize M and N as follows:Point M is on ray AA1. Since AA1 goes from A (0,0,0) upwards to A1 (0,0,1) and beyond. Let's let parameter h be the height above A, so M is at (0,0,h) where h ≥ 0.Point N is on ray BC. Since BC goes from B (1,0,0) through C (1,1,0) and beyond. Let's let parameter k be the distance along BC from B, so N is at (1, k, 0) where k ≥ 0. Wait, but BC has length 1, so if k=1, N is at C, and k>1 would be beyond C.Now, we need the line segment MN to intersect edge C1D1. Edge C1D1 is from C1 (1,1,1) to D1 (0,1,1). So points on C1D1 have coordinates (x,1,1) where x goes from 1 to 0. So the intersection point L is some point on C1D1: (x,1,1) with 0 ≤ x ≤ 1.So, given points M (0,0,h) and N (1,k,0), the line MN must pass through some point L (x,1,1) on C1D1. Our goal is to find h and k such that this condition is satisfied, and then find the minimal length of MN.Alternatively, since L is on both MN and C1D1, we can parametrize the line MN and find where it intersects C1D1, then impose that the intersection point is on C1D1.Let me parametrize the line MN. Let's write parametric equations for MN. Starting at M (0,0,h) and going to N (1,k,0). The vector from M to N is (1 - 0, k - 0, 0 - h) = (1, k, -h). So parametric equations can be written as:x = 0 + t*1y = 0 + t*kz = h + t*(-h)Where t ranges from 0 to 1 to cover the segment MN. So:x = ty = t*kz = h - h*tWe need this line to pass through a point L on C1D1, which is (x,1,1). So there must be some parameter t such that:t = x (from x-coordinate)t*k = 1 (from y-coordinate)h - h*t = 1 (from z-coordinate)But also, since L is on C1D1, its x-coordinate x must be between 0 and 1, so t must be between 0 and 1. Wait, but t is the parameter along MN. If MN is going from M (0,0,h) to N (1,k,0), then t ranges from 0 to 1. But for the intersection point L to be on edge C1D1, which is at z=1, we need z-coordinate to be 1. Let me write down the equations:From z-coordinate: h - h*t = 1From y-coordinate: t*k = 1From x-coordinate: t = x (but x is between 0 and 1 because L is on C1D1)So let's solve for t from the z-coordinate equation:h(1 - t) = 1 => 1 - t = 1/h => t = 1 - 1/hWait, but h must be greater than or equal to 1, right? Because t must be less than or equal to 1 (since t is between 0 and 1 for the segment MN). So 1 - 1/h ≤ 1 => -1/h ≤ 0, which is true because h > 0. But also, t must be ≥ 0 because it's along the segment from M to N. So:t = 1 - 1/h ≥ 0 => 1 - 1/h ≥ 0 => 1/h ≤ 1 => h ≥ 1Therefore, h must be at least 1. So point M is at least at height 1 on AA1, which is at point A1 and beyond.Now, from the y-coordinate equation: t*k = 1. Since t = 1 - 1/h, we have:(1 - 1/h)*k = 1 => k = 1 / (1 - 1/h) = h / (h - 1)Since h ≥ 1, the denominator h - 1 ≥ 0, so h / (h - 1) is positive. Also, as h approaches 1 from above, k approaches infinity. As h increases, k approaches 1.So now, we have expressions for k in terms of h: k = h / (h - 1)Now, the coordinates of M are (0,0,h) and N are (1, k, 0) = (1, h/(h - 1), 0)We can compute the distance MN using the distance formula:MN = sqrt[(1 - 0)^2 + (h/(h - 1) - 0)^2 + (0 - h)^2]Simplify:MN = sqrt[1 + (h^2)/(h - 1)^2 + h^2]Let me write that as:MN = sqrt[1 + h^2 + (h^2)/(h - 1)^2]Hmm, that looks complicated. Maybe we can simplify this expression.First, note that the expression inside the square root is:1 + h^2 + h^2/(h - 1)^2Let me factor h^2:1 + h^2 [1 + 1/(h - 1)^2]But not sure if that helps. Alternatively, maybe combine terms:Let me write all terms over (h - 1)^2:= [ (h - 1)^2 + h^2(h - 1)^2 + h^2 ] / (h - 1)^2Wait, that might not be helpful. Let's compute the expression step by step.First, compute (h^2)/(h - 1)^2:= h^2 / (h^2 - 2h + 1)So the total expression inside sqrt:1 + h^2 + h^2/(h - 1)^2 = 1 + h^2 + h^2/(h^2 - 2h + 1)Alternatively, perhaps we can write this as:Let me denote t = h - 1. Since h ≥ 1, t ≥ 0. Then h = t + 1.Substitute into k:k = h/(h - 1) = (t + 1)/t = 1 + 1/tSo k = 1 + 1/tNow, MN distance becomes:sqrt[1 + (t + 1)^2 + (t + 1)^2/(t^2)]Let's compute each term:1) The first term is 1.2) The second term is (t + 1)^2 = t^2 + 2t + 13) The third term is (t + 1)^2 / t^2 = (t^2 + 2t + 1)/t^2 = 1 + 2/t + 1/t^2So total expression inside sqrt:1 + (t^2 + 2t + 1) + (1 + 2/t + 1/t^2) =1 + t^2 + 2t + 1 + 1 + 2/t + 1/t^2 =t^2 + 2t + 3 + 2/t + 1/t^2Hmm, that's:t^2 + 1/t^2 + 2t + 2/t + 3Not sure if that's helpful. Alternatively, maybe factor terms:t^2 + 1/t^2 + 2(t + 1/t) + 3Let me note that t + 1/t is a common substitution. Let u = t + 1/t. Then t^2 + 1/t^2 = u^2 - 2.So the expression becomes:(u^2 - 2) + 2u + 3 = u^2 + 2u + 1 = (u + 1)^2Wow, that's clever! So:t^2 + 1/t^2 + 2t + 2/t + 3 = (u + 1)^2 where u = t + 1/tTherefore, MN = sqrt[(u + 1)^2] = |u + 1| = u + 1 since u is positive (because t > 0).But u = t + 1/t, so MN = t + 1/t + 1But t = h - 1, and h ≥ 1, so t ≥ 0. However, t cannot be zero because h = t + 1 would be 1, but h must be greater than 1 as per earlier. Wait, no. Wait, when h approaches 1 from above, t approaches 0+. So t > 0.So MN = t + 1/t + 1Now, to find the minimum of MN, we need to minimize the expression t + 1/t + 1 for t > 0.But t + 1/t is a standard expression. We know that t + 1/t ≥ 2, with equality when t = 1. So MN ≥ 2 + 1 = 3. But wait, that can't be right, because the cube has edge length 1, and MN is a diagonal through the cube. Wait, but maybe in 3D, but MN is passing through C1D1, so maybe it's longer? Wait, but let's check.Wait, hold on. If t = 1, then MN = 1 + 1/1 + 1 = 3. So MN would be 3. But if t approaches 0, MN approaches infinity, and as t approaches infinity, MN approaches infinity as well. So the minimum is 3? But that seems too large. Let me verify.Wait, but maybe there's a mistake here. Let's check when t = 1.t = 1, which implies h = t + 1 = 2. Then k = 1 + 1/t = 2. So point M is (0,0,2), point N is (1,2,0). Then the distance MN would be sqrt[(1-0)^2 + (2 - 0)^2 + (0 - 2)^2] = sqrt[1 + 4 + 4] = sqrt[9] = 3. That's correct.But is there a way to have MN shorter than 3?Wait, let's think. Maybe the minimal MN is not achieved at t = 1. Maybe there's another parameterization. Wait, but according to the substitution, we have MN expressed as t + 1/t + 1, which has its minimum at t = 1. Hmm. So the minimal MN is 3. But is that possible?Wait, but in the cube with edge length 1, how can there be a segment of length 3? From (0,0,2) to (1,2,0), which is outside the cube. But the problem states that M is on the ray AA1 and N is on the ray BC. So they can be outside the cube.But the segment MN must intersect edge C1D1, which is on the cube. So even if M and N are outside the cube, as long as the segment passes through C1D1, it's allowed.But the minimal distance is 3? Let me see if there's a shorter MN.Wait, maybe there's a different approach. Let's think geometrically.Since MN has to pass through C1D1, maybe we can use reflections to find the shortest path.In some cases, reflecting points across surfaces can help find the shortest path that intersects a certain line.But here, it's a 3D problem. Maybe reflecting in 3D?Alternatively, think of L as a fixed point on C1D1, and then find the shortest MN passing through L, then minimize over L.But since L can be anywhere on C1D1, perhaps we can parametrize L as (x,1,1) where 0 ≤ x ≤ 1.For a fixed L, the shortest MN passing through L would be the straight line from M on AA1 to N on BC passing through L. So for fixed L, the minimal MN is the sum of distances from L to M and L to N, but since M and N are on rays, perhaps the minimal MN is the straight line through L.Wait, but since M is on AA1 and N is on BC, the line MN must pass through L. So for each L, there is a unique line passing through L, intersecting AA1 and BC. The question is to find the minimal length of such lines MN as L varies along C1D1.Alternatively, parametrize L as (x,1,1) with x ∈ [0,1], find MN length in terms of x, then minimize over x.Let me try that.Let L be (x,1,1). Then the line MN passes through L, M is on AA1 (0,0,t) where t ≥ 0, and N is on BC (1,s,0) where s ≥ 0.Wait, but we need the line MN to pass through L. So let's write parametric equations for the line MN.Suppose the line MN passes through M (0,0,t), L (x,1,1), and N (1,s,0). Let's find the parametric equations.From M to N, the direction vector is (1 - 0, s - 0, 0 - t) = (1, s, -t). The line can be parametrized as:(0,0,t) + k*(1, s, -t), where k is a parameter.This line must pass through L (x,1,1). So there exists some k such that:0 + k*1 = x0 + k*s = 1t + k*(-t) = 1So:k = xk*s = 1 => x*s = 1 => s = 1/xt - x*t = 1 => t(1 - x) = 1 => t = 1/(1 - x)Therefore, for a given x ∈ (0,1), since x cannot be 0 or 1 (as that would make t or s undefined), we have:s = 1/xt = 1/(1 - x)Therefore, the coordinates are:M: (0,0,t) = (0,0,1/(1 - x))N: (1,s,0) = (1,1/x,0)So the length MN is the distance between (0,0,1/(1 - x)) and (1,1/x,0):MN = sqrt[(1 - 0)^2 + (1/x - 0)^2 + (0 - 1/(1 - x))^2]Simplify:MN = sqrt[1 + (1/x^2) + (1/(1 - x))^2]So now, the problem reduces to minimizing the function f(x) = 1 + 1/x² + 1/(1 - x)² for x ∈ (0,1)Find the minimum of f(x) = 1 + 1/x² + 1/(1 - x)²This seems like a calculus problem. Take derivative and set to zero.Let me compute f'(x):f'(x) = d/dx [1 + x^{-2} + (1 - x)^{-2}]= -2x^{-3} + 2(1 - x)^{-3}Set derivative to zero:-2/x³ + 2/(1 - x)³ = 0=> -2/x³ = -2/(1 - x)³Multiply both sides by -1:2/x³ = 2/(1 - x)³Divide both sides by 2:1/x³ = 1/(1 - x)³Take reciprocals:x³ = (1 - x)³Take cube roots:x = 1 - x=> 2x = 1=> x = 1/2So critical point at x = 1/2. Now, check second derivative or test intervals to confirm it's a minimum.Compute f''(x):First, f'(x) = -2/x³ + 2/(1 - x)³f''(x) = 6/x⁴ + 6/(1 - x)^4Which is always positive for x ∈ (0,1). Therefore, the function is convex, and x=1/2 is the point of minimum.Therefore, minimal value occurs at x=1/2.Compute f(1/2):f(1/2) = 1 + 1/( (1/2)^2 ) + 1/( (1 - 1/2)^2 ) = 1 + 1/(1/4) + 1/(1/4) = 1 + 4 + 4 = 9Therefore, MN = sqrt(9) = 3Wait, same result as before. So the minimal MN is 3.But let me cross-verify. When x=1/2,t = 1/(1 - 1/2) = 2s = 1/(1/2) = 2So point M is (0,0,2), point N is (1,2,0). The distance between them is sqrt(1^2 + 2^2 + (-2)^2) = sqrt(1 + 4 + 4) = sqrt(9) = 3.Alternatively, if we use the previous substitution where t = h -1 = 2 -1 =1, so t=1, then MN = t +1/t +1=1+1+1=3. Same result.So according to both parameterizations, the minimal MN is 3.But wait, intuitively, in a cube of edge length 1, having a segment of length 3 seems quite long, but given that the points are outside the cube, maybe it's possible.Alternatively, maybe there's a shorter path that goes through the cube. Wait, but according to the calculations, 3 is the minimum. Let me check with another point.Suppose x = 1/3,Then t = 1/(1 - 1/3) = 1/(2/3) = 3/2s = 1/(1/3) = 3Distance MN = sqrt(1 + 9 + (3/2)^2) = sqrt(1 +9 + 9/4) = sqrt(10 + 2.25) = sqrt(12.25)=3.5>3.Another point x=1/4,t=1/(1 -1/4)=4/3,s=4,distance= sqrt(1 + 16 + (16/9))≈sqrt(17 +1.777)=sqrt(18.777)≈4.33>3.So indeed, the minimal MN is 3. So the answer should be 3.But let me check for x=1/2,The line MN goes from (0,0,2) to (1,2,0) passing through (1/2,1,1). Let's see if that's correct.Parametrize the line from (0,0,2) to (1,2,0). The direction vector is (1,2,-2). Parametric equations:x = 0 + t*1y = 0 + t*2z = 2 + t*(-2)We need to see if this passes through (1/2,1,1).Set z=1: 2 - 2t =1 => t=0.5Then x=0.5*1=0.5, y=0.5*2=1. So yes, at t=0.5, the point is (0.5,1,1), which is on C1D1 edge (since x=0.5, y=1, z=1). So correct.Therefore, MN length is indeed 3, passing through (0.5,1,1).Is there a way to get a shorter MN?Suppose we take x closer to 0 or 1, but as seen earlier, the length increases.Alternatively, perhaps considering points inside the cube?Wait, but if M is on AA1 ray and N is on BC ray, even if they are inside the cube, but the line MN has to intersect C1D1, which is on the top face.But in that case, M would be between A and A1 (h between 0 and1), but earlier we saw that h must be at least 1. Wait, if M is between A and A1, h ∈ [0,1), then from z-coordinate equation:h(1 - t) =1.If h <1, then 1 - t =1/h >1, so t =1 -1/h <0, which is not allowed since t must be between 0 and1 for the segment MN. Therefore, M cannot be below A1. Therefore, M must be on AA1 beyond A1 (h ≥1) and N must be on BC beyond C (k ≥1).Therefore, the minimal MN is indeed 3.But wait, the problem says "the endpoints of the line segment MN be M on the ray AA1, and point N on the ray BC", and "MN intersects the edge C1D1 at point L". So even if we could have MN passing through C1D1 with M and N inside the cube, but according to the earlier analysis, M must be above A1 and N beyond C. Therefore, the minimal MN is 3.Therefore, the answer is 3. So boxed answer is boxed{3}Final Answerboxed{3}</think>Given a cube (ABCD-A_1B_1C_1D_1) with edge length 1, we need to find the minimum length of the line segment (MN) where (M) is on the ray (AA_1) and (N) is on the ray (BC), and (MN) intersects the edge (C_1D_1) at point (L).1. Coordinates Setup: - The cube is placed in a 3D coordinate system with (A) at the origin ((0,0,0)). - Points on the rays are parameterized: - (M) on (AA_1) as ((0, 0, h)) where (h geq 1). - (N) on (BC) as ((1, k, 0)) where (k geq 1).2. Intersection Condition: - The line segment (MN) must intersect the edge (C_1D_1) at point (L) with coordinates ((x, 1, 1)) where (0 leq x leq 1).3. Parametric Equations: - The parametric equations for the line (MN) are derived and solved for the intersection with (C_1D_1): - (x = t) - (y = tk) - (z = h - ht) - Solving for (t) when (z = 1) gives (t = 1 - frac{1}{h}). - Substituting (t) into the (y)-coordinate equation gives (k = frac{h}{h - 1}).4. Distance Calculation: - The distance (MN) is calculated using the coordinates of (M) and (N): [ MN = sqrt{(1 - 0)^2 + left(frac{h}{h - 1} - 0right)^2 + (0 - h)^2} ] - Simplifying the expression and substituting (h = t + 1) leads to the expression: [ MN = sqrt{1 + h^2 + left(frac{h^2}{(h - 1)^2}right)} ] - Further simplification using (t = h - 1) transforms the expression into: [ MN = sqrt{(t + frac{1}{t} + 1)^2} ] - Minimizing (t + frac{1}{t} + 1) gives the minimum value when (t = 1), leading to (MN = 3).5. Verification: - The critical point analysis confirms that the minimum value occurs at (x = frac{1}{2}), leading to the shortest distance (MN = 3).Thus, the minimum value of (MN) is (boxed{3}).]

🔑:7.3.Let A M=x, B N=y. ThenM N=sqrt{x^{2}+y^{2}+1} text {. }Draw L L_{1} perp D C at point L_{1}. Thus,begin{array}{l}frac{1}{x-1}=frac{A A_{1}}{M A_{1}}=frac{L M}{L N}=frac{N C}{C B}=frac{y-1}{1} Rightarrow x y=x+y Rightarrow x^{2} y^{2} geqslant 4 x y Rightarrow x y geqslant 4 .end{array}Therefore, M N=sqrt{x^{2}+y^{2}+1} geqslant sqrt{2 x y+1} geqslant 3,with equality holding if and only if x=y=2.