answer：To find the value of ( m ) such that the vectors ( overrightarrow{a} = (-2, 3) ) and ( overrightarrow{b} = (3, m) ) are perpendicular, I start by recalling that two vectors are perpendicular if their dot product is zero.The dot product of ( overrightarrow{a} ) and ( overrightarrow{b} ) is calculated as:[(-2) times 3 + 3 times m = -6 + 3m]Setting this equal to zero gives:[-6 + 3m = 0]Solving for ( m ):[3m = 6 m = 2]

answer：Okay, so I have this problem here: the seventh and tenth terms of a geometric sequence are 7 and 21, respectively. I need to find the 13th term of this progression. Hmm, let me think about how to approach this.First, I remember that in a geometric sequence, each term is found by multiplying the previous term by a common ratio, which is usually denoted as 'r'. So, if I denote the first term as 'a', then the nth term of the sequence can be written as:a_n = a * r^(n-1)So, for the seventh term, which is given as 7, that would be:a_7 = a * r^(7-1) = a * r^6 = 7Similarly, the tenth term is 21, so:a_10 = a * r^(10-1) = a * r^9 = 21Alright, so I have two equations here:1) a * r^6 = 72) a * r^9 = 21I need to find the 13th term, which would be:a_13 = a * r^(13-1) = a * r^12So, I need to find 'a' and 'r' first, right? Let me see how I can do that.Looking at the two equations I have, maybe I can divide the second equation by the first to eliminate 'a'. Let's try that.Divide equation 2 by equation 1:(a * r^9) / (a * r^6) = 21 / 7Simplify the left side: The 'a's cancel out, and r^9 divided by r^6 is r^(9-6) = r^3.So, r^3 = 21 / 7 = 3Therefore, r^3 = 3. So, to find 'r', I can take the cube root of both sides:r = 3^(1/3)Hmm, 3^(1/3) is the cube root of 3. I can leave it like that for now, or maybe express it as 3^(1/3). Either way, it's a real number, so that's fine.Now that I have 'r', I can plug it back into one of the original equations to find 'a'. Let's use equation 1:a * r^6 = 7We know that r^3 = 3, so r^6 is (r^3)^2 = 3^2 = 9.So, substituting back:a * 9 = 7Therefore, a = 7 / 9Okay, so now I have both 'a' and 'r'. Let me write them down:a = 7/9r = 3^(1/3)Now, I need to find the 13th term, which is a * r^12.Let me compute that:a_13 = a * r^12 = (7/9) * (3^(1/3))^12Simplify the exponent on 'r':(3^(1/3))^12 = 3^(12/3) = 3^4 = 81So, substituting back:a_13 = (7/9) * 81Now, let's compute that:First, 81 divided by 9 is 9, so (7/9) * 81 = 7 * 9 = 63So, the 13th term is 63.Wait, let me double-check my steps to make sure I didn't make a mistake.1. I wrote the general formula for the nth term of a geometric sequence: a_n = a * r^(n-1). That seems correct.2. Plugged in n=7 and n=10 to get two equations: a * r^6 = 7 a * r^9 = 213. Divided the second equation by the first to eliminate 'a': (a * r^9) / (a * r^6) = 21 / 7 => r^3 = 3, so r = 3^(1/3). That seems right.4. Then, substituted r^3 = 3 into equation 1 to find 'a': a * (r^3)^2 = 7 => a * 9 = 7 => a = 7/9. Correct.5. Then, for the 13th term: a_13 = a * r^12 = (7/9) * (3^(1/3))^12 Simplify exponent: (3^(1/3))^12 = 3^(12/3) = 3^4 = 81. Correct.6. Then, (7/9) * 81 = 7 * 9 = 63. Yes, that's correct.So, all the steps check out. Therefore, the 13th term is 63.Alternatively, I can think about the terms in terms of their positions. Since the seventh term is 7 and the tenth term is 21, that's three terms apart. So, from term 7 to term 10, it's multiplied by r three times, which is why r^3 = 3. Then, from term 10 to term 13, it's another three terms, so multiplying by r^3 again. So, term 13 would be term 10 multiplied by r^3, which is 21 * 3 = 63. That's another way to see it without computing 'a'.Wait, that's a good point. Maybe I can solve it that way too, just to confirm.So, term 7 is 7, term 10 is 21. The ratio between term 10 and term 7 is 21 / 7 = 3. Since term 10 is three terms after term 7, that ratio is r^3 = 3. So, r = 3^(1/3). Then, term 13 is three terms after term 10, so term 13 = term 10 * r^3 = 21 * 3 = 63. Yep, same result.So, that's another way to think about it without having to find 'a'. So, that's reassuring.I think I'm confident that 63 is the correct answer.**Final Answer**The 13th term of the geometric sequence is boxed{63}.

answer：First, I need to find the modulus of the complex number (1 + i). The modulus of a complex number (a + bi) is given by (sqrt{a^2 + b^2}). For (1 + i), this is (sqrt{1^2 + 1^2} = sqrt{2}).Next, I'll use the property of moduli that states (|z^n| = |z|^n). Applying this to ((1 + i)^6), the modulus is ((sqrt{2})^6).Calculating ((sqrt{2})^6) involves raising (sqrt{2}) to the power of 6. Since ((sqrt{2})^2 = 2), raising this result to the power of 3 gives (2^3 = 8).Therefore, the modulus of ((1 + i)^6) is 8.

answer：Alright, so I have this problem: Among all the triangles inscribed in a given circle, find the one with the maximum sum of the squares of the lengths of its sides. Hmm, okay. Let me try to figure this out step by step.First, I know that all triangles inscribed in a circle are called cyclic triangles, right? So, we're dealing with cyclic triangles here. The circle is given, so the radius is fixed. Let's denote the radius as ( R ). The goal is to maximize the sum ( a^2 + b^2 + c^2 ), where ( a ), ( b ), and ( c ) are the lengths of the sides of the triangle.I remember that in a circle, the length of a chord is related to the central angle subtended by the chord. The formula for the length of a chord is ( 2R sin theta ), where ( theta ) is half the central angle. Wait, actually, let me recall that correctly. The chord length formula is ( 2R sin left( frac{alpha}{2} right) ), where ( alpha ) is the central angle corresponding to that chord.So, if we have a triangle inscribed in a circle, each side corresponds to a central angle. Let me denote the central angles corresponding to sides ( a ), ( b ), and ( c ) as ( alpha ), ( beta ), and ( gamma ) respectively. Since the triangle is inscribed in the circle, the sum of these central angles should be ( 2pi ) radians, right? Because the total angle around a circle is ( 2pi ).So, ( alpha + beta + gamma = 2pi ).Now, the lengths of the sides can be expressed in terms of these central angles:( a = 2R sin left( frac{alpha}{2} right) )( b = 2R sin left( frac{beta}{2} right) )( c = 2R sin left( frac{gamma}{2} right) )Therefore, the sum of the squares of the sides is:( a^2 + b^2 + c^2 = 4R^2 left[ sin^2 left( frac{alpha}{2} right) + sin^2 left( frac{beta}{2} right) + sin^2 left( frac{gamma}{2} right) right] )So, our goal is to maximize this expression:( S = sin^2 left( frac{alpha}{2} right) + sin^2 left( frac{beta}{2} right) + sin^2 left( frac{gamma}{2} right) )subject to the constraint ( alpha + beta + gamma = 2pi ).Hmm, okay. So, this is an optimization problem with a constraint. I think I can use Lagrange multipliers here. Let me set up the Lagrangian.Let me denote ( x = frac{alpha}{2} ), ( y = frac{beta}{2} ), and ( z = frac{gamma}{2} ). Then, ( 2x + 2y + 2z = 2pi ), so ( x + y + z = pi ).So, now, the expression to maximize becomes:( S = sin^2 x + sin^2 y + sin^2 z )with the constraint ( x + y + z = pi ).Alright, so I can use Lagrange multipliers here. Let me define the function:( f(x, y, z) = sin^2 x + sin^2 y + sin^2 z )and the constraint:( g(x, y, z) = x + y + z - pi = 0 )The Lagrangian is:( mathcal{L}(x, y, z, lambda) = sin^2 x + sin^2 y + sin^2 z - lambda (x + y + z - pi) )Taking partial derivatives with respect to ( x ), ( y ), ( z ), and ( lambda ), and setting them equal to zero.First, partial derivative with respect to ( x ):( frac{partial mathcal{L}}{partial x} = 2 sin x cos x - lambda = 0 )Similarly, for ( y ):( frac{partial mathcal{L}}{partial y} = 2 sin y cos y - lambda = 0 )And for ( z ):( frac{partial mathcal{L}}{partial z} = 2 sin z cos z - lambda = 0 )And the constraint:( x + y + z = pi )So, from the first three equations, we have:( 2 sin x cos x = lambda )( 2 sin y cos y = lambda )( 2 sin z cos z = lambda )Therefore, all three expressions are equal:( 2 sin x cos x = 2 sin y cos y = 2 sin z cos z )Which simplifies to:( sin x cos x = sin y cos y = sin z cos z )Hmm, so ( sin x cos x ) is equal for all three variables. Let me denote ( k = sin x cos x ). So, ( sin x cos x = sin y cos y = sin z cos z = k ).I know that ( sin 2theta = 2 sin theta cos theta ), so ( sin x cos x = frac{1}{2} sin 2x ). Therefore, ( frac{1}{2} sin 2x = frac{1}{2} sin 2y = frac{1}{2} sin 2z = k ).So, ( sin 2x = sin 2y = sin 2z = 2k ).Therefore, all three angles ( 2x ), ( 2y ), and ( 2z ) have the same sine value.Now, the sine function is positive in the first and second quadrants, so ( 2x ), ( 2y ), ( 2z ) must be in ( (0, pi) ) since ( x, y, z ) are each less than ( pi ) (since ( x + y + z = pi )).So, the possible solutions for ( 2x = 2y = 2z ) or ( 2x = pi - 2y ), etc. But since all three sines are equal, the angles could either be equal or supplementary.But if ( 2x = pi - 2y ), then ( x + y = frac{pi}{2} ). Similarly, if ( 2y = pi - 2z ), then ( y + z = frac{pi}{2} ). But then, adding these, ( x + y + y + z = pi ), but ( x + y + z = pi ), so this would imply ( y = 0 ), which is not possible because all angles must be positive in a triangle.Therefore, the only feasible solution is that all three angles are equal, i.e., ( 2x = 2y = 2z ), which implies ( x = y = z ).So, ( x = y = z = frac{pi}{3} ).Therefore, each central angle is ( 2x = frac{2pi}{3} ).So, each side of the triangle corresponds to a central angle of ( frac{2pi}{3} ).Therefore, the triangle is equilateral, since all central angles are equal, which would make all sides equal as well.Wait, but hold on. If the central angles are all ( frac{2pi}{3} ), then the triangle is equilateral. Is that correct?Yes, because in a circle, if all central angles are equal, the chords are equal, so the triangle is equilateral.So, does that mean that the equilateral triangle inscribed in the circle has the maximum sum of the squares of its sides?But let me verify this because sometimes intuition can be misleading.Alternatively, maybe another triangle could give a higher sum.Wait, let's compute the sum for the equilateral triangle.Each side is ( 2R sin left( frac{pi}{3} right) = 2R times frac{sqrt{3}}{2} = R sqrt{3} ).Therefore, each side squared is ( 3R^2 ), so the sum is ( 3 times 3R^2 = 9R^2 ).Wait, hold on, that can't be right. Wait, no, each side is ( R sqrt{3} ), so each squared is ( 3R^2 ). Therefore, the sum is ( 3 times 3R^2 = 9R^2 ). Hmm, okay.But let me check another triangle, say, a right-angled triangle. Let's take a right-angled triangle inscribed in the circle. In that case, the hypotenuse is the diameter, so length ( 2R ). The other two sides can be calculated using Pythagoras.Let me suppose the triangle is right-angled, so one angle is ( frac{pi}{2} ), and the other two angles sum to ( frac{pi}{2} ).Wait, but in terms of central angles, the central angle corresponding to the hypotenuse is ( pi ), since it's the diameter. The other two sides correspond to central angles ( alpha ) and ( beta ), such that ( alpha + beta = pi ).So, the lengths of the sides would be:Hypotenuse: ( 2R ).Other sides: ( 2R sin left( frac{alpha}{2} right) ) and ( 2R sin left( frac{beta}{2} right) ).Since ( alpha + beta = pi ), let me set ( alpha = theta ), so ( beta = pi - theta ).Therefore, the sides are ( 2R sin left( frac{theta}{2} right) ) and ( 2R sin left( frac{pi - theta}{2} right) = 2R sin left( frac{pi}{2} - frac{theta}{2} right) = 2R cos left( frac{theta}{2} right) ).So, the sum of the squares is:( (2R)^2 + left( 2R sin left( frac{theta}{2} right) right)^2 + left( 2R cos left( frac{theta}{2} right) right)^2 )Simplify:( 4R^2 + 4R^2 sin^2 left( frac{theta}{2} right) + 4R^2 cos^2 left( frac{theta}{2} right) )Factor out ( 4R^2 ):( 4R^2 left[ 1 + sin^2 left( frac{theta}{2} right) + cos^2 left( frac{theta}{2} right) right] )But ( sin^2 x + cos^2 x = 1 ), so:( 4R^2 [1 + 1] = 8R^2 )So, the sum of the squares is ( 8R^2 ) for a right-angled triangle.Wait, earlier, for the equilateral triangle, the sum was ( 9R^2 ). So, ( 9R^2 ) is greater than ( 8R^2 ), so the equilateral triangle gives a higher sum.Hmm, that's interesting. So, the equilateral triangle seems to give a higher sum than the right-angled triangle.But let me test another triangle. Maybe an isosceles triangle with two sides equal but not equilateral.Suppose we have a triangle with two sides corresponding to central angles ( alpha ) and the third side corresponding to central angle ( 2pi - 2alpha ).Wait, no, because the sum of the central angles must be ( 2pi ). So, if two central angles are ( alpha ), the third is ( 2pi - 2alpha ).So, the sides would be:( a = 2R sin left( frac{alpha}{2} right) )( b = 2R sin left( frac{alpha}{2} right) )( c = 2R sin left( frac{2pi - 2alpha}{2} right) = 2R sin (pi - alpha) = 2R sin alpha )Therefore, the sum of the squares is:( 2 times [2R sin (frac{alpha}{2})]^2 + [2R sin alpha]^2 )Simplify:( 2 times 4R^2 sin^2 left( frac{alpha}{2} right) + 4R^2 sin^2 alpha )Which is:( 8R^2 sin^2 left( frac{alpha}{2} right) + 4R^2 sin^2 alpha )Let me express ( sin^2 alpha ) in terms of ( sin^2 left( frac{alpha}{2} right) ).Recall that ( sin alpha = 2 sin left( frac{alpha}{2} right) cos left( frac{alpha}{2} right) ), so ( sin^2 alpha = 4 sin^2 left( frac{alpha}{2} right) cos^2 left( frac{alpha}{2} right) ).Therefore, the sum becomes:( 8R^2 sin^2 left( frac{alpha}{2} right) + 4R^2 times 4 sin^2 left( frac{alpha}{2} right) cos^2 left( frac{alpha}{2} right) )Simplify:( 8R^2 sin^2 left( frac{alpha}{2} right) + 16R^2 sin^2 left( frac{alpha}{2} right) cos^2 left( frac{alpha}{2} right) )Factor out ( 8R^2 sin^2 left( frac{alpha}{2} right) ):( 8R^2 sin^2 left( frac{alpha}{2} right) left[ 1 + 2 cos^2 left( frac{alpha}{2} right) right] )Hmm, let me compute this expression for some specific values of ( alpha ).First, let me take ( alpha = frac{2pi}{3} ), which would make the triangle equilateral.Then, ( frac{alpha}{2} = frac{pi}{3} ), so ( sin left( frac{pi}{3} right) = frac{sqrt{3}}{2} ), and ( cos left( frac{pi}{3} right) = frac{1}{2} ).Plugging in:( 8R^2 times left( frac{3}{4} right) times left[ 1 + 2 times left( frac{1}{4} right) right] = 8R^2 times frac{3}{4} times left( 1 + frac{1}{2} right) = 8R^2 times frac{3}{4} times frac{3}{2} = 8R^2 times frac{9}{8} = 9R^2 ).Which matches our previous result.Now, let me take ( alpha = frac{pi}{2} ), so the triangle is right-angled.Then, ( frac{alpha}{2} = frac{pi}{4} ), so ( sin left( frac{pi}{4} right) = frac{sqrt{2}}{2} ), and ( cos left( frac{pi}{4} right) = frac{sqrt{2}}{2} ).Plugging in:( 8R^2 times left( frac{1}{2} right) times left[ 1 + 2 times left( frac{1}{2} right) right] = 8R^2 times frac{1}{2} times left( 1 + 1 right) = 8R^2 times frac{1}{2} times 2 = 8R^2 ).Which again matches our previous result.Now, let's try ( alpha = frac{pi}{3} ). Then, ( frac{alpha}{2} = frac{pi}{6} ), so ( sin left( frac{pi}{6} right) = frac{1}{2} ), ( cos left( frac{pi}{6} right) = frac{sqrt{3}}{2} ).Plugging in:( 8R^2 times left( frac{1}{4} right) times left[ 1 + 2 times left( frac{3}{4} right) right] = 8R^2 times frac{1}{4} times left( 1 + frac{3}{2} right) = 8R^2 times frac{1}{4} times frac{5}{2} = 8R^2 times frac{5}{8} = 5R^2 ).So, the sum is ( 5R^2 ), which is less than both ( 8R^2 ) and ( 9R^2 ).Hmm, so when ( alpha ) is smaller, the sum is smaller. What about when ( alpha ) is larger?Let me try ( alpha = frac{3pi}{4} ), so ( frac{alpha}{2} = frac{3pi}{8} ). Then, ( sin left( frac{3pi}{8} right) ) is approximately ( sin(67.5^circ) approx 0.9239 ), and ( cos left( frac{3pi}{8} right) approx 0.3827 ).So, computing:( 8R^2 times (0.9239)^2 times left[ 1 + 2 times (0.3827)^2 right] )First, ( (0.9239)^2 approx 0.8536 ).Then, ( (0.3827)^2 approx 0.1464 ), so ( 2 times 0.1464 approx 0.2928 ).Therefore, the bracket becomes ( 1 + 0.2928 = 1.2928 ).So, the total is ( 8R^2 times 0.8536 times 1.2928 approx 8R^2 times 1.092 approx 8.736R^2 ).Which is less than ( 9R^2 ), but more than ( 8R^2 ).Hmm, so it's higher than the right-angled triangle but still less than the equilateral triangle.So, it seems that as ( alpha ) increases from ( frac{pi}{2} ) to ( frac{2pi}{3} ), the sum increases, reaching a maximum at ( alpha = frac{2pi}{3} ), which is the equilateral triangle.Therefore, this suggests that the equilateral triangle gives the maximum sum.But just to be thorough, let me consider another case where the triangle is very "flat", approaching a degenerate triangle.Suppose one central angle approaches ( 2pi ), and the other two approach 0. Then, the corresponding sides would be almost ( 2R ) and two sides approaching 0.So, the sum of the squares would approach ( (2R)^2 + 0 + 0 = 4R^2 ), which is much less than ( 9R^2 ). So, that's worse.Alternatively, suppose two central angles approach ( pi ), and the third approaches 0. Then, the sides would approach ( 2R ) each, and the third side approaches 0. So, the sum would approach ( 4R^2 + 4R^2 + 0 = 8R^2 ), which is still less than ( 9R^2 ).Therefore, it seems that the maximum occurs when all central angles are equal, i.e., the triangle is equilateral.Alternatively, another way to approach this problem is by using vector algebra.Let me consider the circle centered at the origin with radius ( R ). Let the triangle have vertices at points ( A ), ( B ), and ( C ) on the circumference. So, the position vectors of these points can be represented as ( vec{A} ), ( vec{B} ), and ( vec{C} ), each with magnitude ( R ).The lengths of the sides can be expressed using the dot product:( a = |vec{B} - vec{C}| )( b = |vec{A} - vec{C}| )( c = |vec{A} - vec{B}| )So, ( a^2 = |vec{B} - vec{C}|^2 = (vec{B} - vec{C}) cdot (vec{B} - vec{C}) = |vec{B}|^2 + |vec{C}|^2 - 2 vec{B} cdot vec{C} = 2R^2 - 2 vec{B} cdot vec{C} )Similarly,( b^2 = 2R^2 - 2 vec{A} cdot vec{C} )( c^2 = 2R^2 - 2 vec{A} cdot vec{B} )Therefore, the sum ( a^2 + b^2 + c^2 ) is:( 6R^2 - 2 ( vec{A} cdot vec{B} + vec{A} cdot vec{C} + vec{B} cdot vec{C} ) )So, to maximize ( a^2 + b^2 + c^2 ), we need to minimize ( vec{A} cdot vec{B} + vec{A} cdot vec{C} + vec{B} cdot vec{C} ).Hmm, okay. So, let me denote ( S = vec{A} cdot vec{B} + vec{A} cdot vec{C} + vec{B} cdot vec{C} ). We need to minimize ( S ).Alternatively, perhaps we can express this in terms of angles between the vectors.Let me denote ( theta_{AB} ) as the angle between ( vec{A} ) and ( vec{B} ), ( theta_{AC} ) as the angle between ( vec{A} ) and ( vec{C} ), and ( theta_{BC} ) as the angle between ( vec{B} ) and ( vec{C} ).Since the points are on a circle, these angles correspond to the central angles between the points.So, ( vec{A} cdot vec{B} = R^2 cos theta_{AB} ), similarly for the others.Therefore, ( S = R^2 ( cos theta_{AB} + cos theta_{AC} + cos theta_{BC} ) )So, we need to minimize ( cos theta_{AB} + cos theta_{AC} + cos theta_{BC} ).But we know that in a triangle inscribed in a circle, the sum of the central angles is ( 2pi ). So, ( theta_{AB} + theta_{AC} + theta_{BC} = 2pi ).Wait, actually, no. Wait, the central angles correspond to the arcs between the points. So, if we have three points on a circle, the central angles between them add up to ( 2pi ). So, yes, ( theta_{AB} + theta_{AC} + theta_{BC} = 2pi ).But in reality, the central angles are the angles between each pair of points, so they are the angles at the center between each pair of vertices. So, for three points, the three central angles add up to ( 2pi ).So, we have ( theta_{AB} + theta_{AC} + theta_{BC} = 2pi ).Therefore, we need to minimize ( cos theta_{AB} + cos theta_{AC} + cos theta_{BC} ) with the constraint ( theta_{AB} + theta_{AC} + theta_{BC} = 2pi ).Hmm, okay. So, perhaps using Lagrange multipliers again.Let me set ( x = theta_{AB} ), ( y = theta_{AC} ), ( z = theta_{BC} ). Then, ( x + y + z = 2pi ).We need to minimize ( f(x, y, z) = cos x + cos y + cos z ).So, the Lagrangian is:( mathcal{L}(x, y, z, lambda) = cos x + cos y + cos z - lambda (x + y + z - 2pi) )Taking partial derivatives:( frac{partial mathcal{L}}{partial x} = -sin x - lambda = 0 )Similarly,( frac{partial mathcal{L}}{partial y} = -sin y - lambda = 0 )( frac{partial mathcal{L}}{partial z} = -sin z - lambda = 0 )And the constraint:( x + y + z = 2pi )From the first three equations:( -sin x = lambda )( -sin y = lambda )( -sin z = lambda )Therefore, ( sin x = sin y = sin z ).So, ( sin x = sin y = sin z = k ), say.So, each of ( x ), ( y ), ( z ) must satisfy ( sin theta = k ).Since ( x, y, z ) are angles between ( 0 ) and ( 2pi ), but in our case, since they are central angles between points on a circle, each angle is between ( 0 ) and ( 2pi ), but more specifically, between ( 0 ) and ( pi ) because beyond ( pi ), the chord length starts decreasing.Wait, actually, chord length is ( 2R sin left( frac{theta}{2} right) ), so as ( theta ) increases from 0 to ( pi ), chord length increases, and then decreases from ( pi ) to ( 2pi ). So, chord lengths are symmetric around ( pi ).But in our case, the central angles ( x ), ( y ), ( z ) must each be less than ( pi ) because if one central angle is greater than ( pi ), the triangle would be "reflected" on the other side of the circle, but the chord lengths would be the same as if the angle were less than ( pi ).Wait, actually, no. The chord length formula is ( 2R sin left( frac{theta}{2} right) ), which is symmetric around ( pi ), because ( sin left( frac{theta}{2} right) = sin left( frac{2pi - theta}{2} right) ).Therefore, chord lengths depend only on the smaller arc between two points, so effectively, each central angle is considered as the smaller angle between two points, so each central angle is between ( 0 ) and ( pi ).Therefore, in our case, ( x, y, z in (0, pi) ).Therefore, the equation ( sin x = sin y = sin z = k ) implies that either ( x = y = z ), or some of them are ( pi - x ), but since all are less than ( pi ), the only solution is ( x = y = z ).Therefore, ( x = y = z = frac{2pi}{3} ).Therefore, each central angle is ( frac{2pi}{3} ), which again implies that the triangle is equilateral.Therefore, this method also leads us to the conclusion that the equilateral triangle maximizes the sum ( a^2 + b^2 + c^2 ).Alternatively, another approach is to use the formula for the sum of squares of the sides in terms of the circumradius.I recall that in any triangle, the formula ( a^2 + b^2 + c^2 = 2(s^2 - r^2 - 4Rr) ) exists, but I might be misremembering. Alternatively, perhaps using the formula involving the circumradius.Wait, actually, in any triangle, the formula ( a^2 + b^2 + c^2 = 2(p^2 + q^2 + r^2) ) where ( p ), ( q ), ( r ) are the medians, but that might not be helpful here.Alternatively, perhaps using the identity that relates the sum of squares of the sides to the circumradius and the angles.Wait, in any triangle, ( a = 2R sin A ), ( b = 2R sin B ), ( c = 2R sin C ), where ( A ), ( B ), ( C ) are the angles of the triangle.Therefore, ( a^2 + b^2 + c^2 = 4R^2 ( sin^2 A + sin^2 B + sin^2 C ) ).So, we need to maximize ( sin^2 A + sin^2 B + sin^2 C ) given that ( A + B + C = pi ).Hmm, so this is similar to the earlier problem but in terms of the angles of the triangle rather than the central angles.So, let me consider this function ( f(A, B, C) = sin^2 A + sin^2 B + sin^2 C ) with the constraint ( A + B + C = pi ).Again, using Lagrange multipliers.Define the Lagrangian:( mathcal{L}(A, B, C, lambda) = sin^2 A + sin^2 B + sin^2 C - lambda (A + B + C - pi) )Taking partial derivatives:( frac{partial mathcal{L}}{partial A} = 2 sin A cos A - lambda = 0 )Similarly,( frac{partial mathcal{L}}{partial B} = 2 sin B cos B - lambda = 0 )( frac{partial mathcal{L}}{partial C} = 2 sin C cos C - lambda = 0 )And the constraint:( A + B + C = pi )So, from the partial derivatives, we have:( 2 sin A cos A = lambda )( 2 sin B cos B = lambda )( 2 sin C cos C = lambda )Therefore, ( sin A cos A = sin B cos B = sin C cos C )Again, using the identity ( sin 2theta = 2 sin theta cos theta ), so ( sin A cos A = frac{1}{2} sin 2A ), etc.Therefore, ( frac{1}{2} sin 2A = frac{1}{2} sin 2B = frac{1}{2} sin 2C )Thus, ( sin 2A = sin 2B = sin 2C )So, similar to before, the sines of the doubled angles are equal.Now, since ( A + B + C = pi ), each angle is less than ( pi ), so ( 2A ), ( 2B ), ( 2C ) are less than ( 2pi ).Therefore, the possible solutions are either all equal or some are supplementary.But if ( 2A = pi - 2B ), then ( A + B = frac{pi}{2} ). Similarly, ( 2B = pi - 2C ) implies ( B + C = frac{pi}{2} ). Then, adding these, ( A + 2B + C = pi ). But since ( A + B + C = pi ), this implies ( B = 0 ), which is impossible.Therefore, the only solution is ( 2A = 2B = 2C ), which implies ( A = B = C = frac{pi}{3} ).Therefore, the triangle is equilateral.Thus, this method also confirms that the equilateral triangle maximizes the sum ( a^2 + b^2 + c^2 ).Alternatively, another approach is to use Jensen's inequality, since the function ( sin^2 x ) is convex or concave in the interval ( (0, pi) ).Wait, let me check the second derivative of ( sin^2 x ).First derivative: ( 2 sin x cos x = sin 2x ).Second derivative: ( 2 cos 2x ).So, the second derivative is ( 2 cos 2x ).In the interval ( x in (0, pi) ), ( 2x in (0, 2pi) ), so ( cos 2x ) varies between ( -1 ) and ( 1 ). Therefore, the second derivative can be positive or negative, so ( sin^2 x ) is neither convex nor concave over the entire interval ( (0, pi) ). Therefore, Jensen's inequality may not be directly applicable.Alternatively, perhaps considering the function ( f(x) = sin^2 x ) over ( x in (0, pi) ), and given that we have three variables ( A, B, C ) with ( A + B + C = pi ), we can analyze whether the maximum occurs at the symmetric point.Given that in all the methods above, the maximum occurs when all angles are equal, it's likely that the function ( sin^2 x + sin^2 y + sin^2 z ) is maximized when ( x = y = z ).Therefore, this further supports the conclusion that the equilateral triangle gives the maximum sum.Alternatively, let me think about the physical interpretation. If we imagine the triangle being pulled outward on the circle, the configuration that maximizes the sum of squares of the sides would be the most "spread out" one, which is the equilateral triangle.Alternatively, considering that the sum of squares is related to the moment of inertia. If we think of the triangle's vertices as masses on the circle, the moment of inertia about the center would be proportional to ( a^2 + b^2 + c^2 ). The configuration that maximizes the moment of inertia is when the masses are as far apart as possible, which again would be the equilateral triangle.Therefore, putting all this together, it seems that the equilateral triangle inscribed in the given circle has the maximum sum of the squares of its sides.Thus, the answer is that the triangle is equilateral, and the maximum sum is ( 9R^2 ), where ( R ) is the radius of the given circle.**Final Answer**The triangle with the maximum sum of the squares of its sides is an equilateral triangle, and the maximum sum is boxed{9R^2}.

answer：Alright, so I have this problem: I need to find the largest five-digit positive integer that isn't a multiple of 11, and also, any number you get by deleting some of its digits isn't divisible by 11 either. Hmm, okay, let's break this down.First, I know that a number is divisible by 11 if the alternating sum of its digits is a multiple of 11. So, for a five-digit number ABCDE, the test would be (A + C + E) - (B + D) is divisible by 11. If that's the case, then the number is divisible by 11. So, I need to make sure that ABCDE isn't divisible by 11.But also, any number obtained by deleting some digits shouldn't be divisible by 11 either. That means if I delete one digit, making it a four-digit number, or delete two digits, making it a three-digit number, and so on, none of those should be divisible by 11. So, all possible shorter numbers formed by the digits of ABCDE must not be divisible by 11.Since we're looking for the largest five-digit number, I should start from 99999 and work my way down until I find a number that satisfies both conditions. But 99999 is a multiple of 11, right? Let me check: (9 + 9 + 9) - (9 + 9) = 27 - 18 = 9, which isn't a multiple of 11. Wait, so 99999 isn't divisible by 11? Hmm, maybe I was wrong.Wait, let me recalculate: For 99999, the alternating sum is (9 + 9 + 9) - (9 + 9) = 27 - 18 = 9. Since 9 isn't a multiple of 11, 99999 isn't divisible by 11. So, maybe 99999 is a candidate? But I need to check if all its subnumbers are not divisible by 11.But wait, 99999 has subnumbers like 9999, 999, 99, etc. Let's check 9999: (9 + 9) - (9 + 9) = 18 - 18 = 0, which is a multiple of 11. So, 9999 is divisible by 11. Therefore, 99999 is out because one of its subnumbers is divisible by 11.So, I need to go down to the next number: 99998. Let's check if 99998 is divisible by 11. The alternating sum is (9 + 9 + 8) - (9 + 9) = 26 - 18 = 8, which isn't a multiple of 11. So, 99998 isn't divisible by 11. Now, check its subnumbers.First, the four-digit subnumbers: 9999, 9998, 9998, 9989, etc. Wait, 9999 is still a subnumber, which is divisible by 11 as we saw earlier. So, 99998 is also out because it contains 9999 as a subnumber.Hmm, so maybe I need to find a five-digit number that doesn't have any four-digit subnumbers divisible by 11. That means all four-digit numbers formed by deleting one digit from ABCDE shouldn't be divisible by 11.Similarly, all three-digit subnumbers shouldn't be divisible by 11, and all two-digit subnumbers shouldn't be divisible by 11 either.So, perhaps the strategy is to construct a five-digit number where every possible subnumber (of any length) isn't divisible by 11. That sounds complicated, but maybe there's a pattern or a way to ensure this.Let me think about the two-digit case first. For a two-digit number AB, it's divisible by 11 if A = B. Because 11, 22, ..., 99 are the two-digit multiples of 11. So, to prevent any two-digit subnumber from being divisible by 11, all adjacent digits in ABCDE must be different. Wait, no, actually, any two digits, not just adjacent ones. So, if any two digits in the number are the same, then deleting the digits in between would result in a two-digit number with equal digits, which is divisible by 11.Wait, is that true? For example, take 99999. If I delete the middle three digits, I get 99, which is divisible by 11. Similarly, in 99998, deleting the middle three digits gives 98, which isn't divisible by 11, but deleting other digits might give 99, which is. So, actually, if any two digits in the number are the same, regardless of their position, deleting the digits in between would result in a two-digit number with equal digits, which is divisible by 11.Therefore, to prevent any two-digit subnumber from being divisible by 11, all digits in ABCDE must be distinct. Because if any two digits are the same, you can delete the digits in between to get a two-digit number with equal digits, which is bad.Wait, but in the number 99998, the digits aren't all distinct. So, that's why it's problematic. So, if I have all digits distinct, then no two digits are the same, so no two-digit subnumber can be divisible by 11. That seems like a good start.So, the five-digit number must have all distinct digits. That reduces the problem to finding the largest five-digit number with all distinct digits, such that the number itself isn't divisible by 11, and none of its four-digit, three-digit, etc., subnumbers are divisible by 11.But wait, even if all digits are distinct, some four-digit subnumbers might still be divisible by 11. For example, take 98765. Let's check if 9876 is divisible by 11. The alternating sum is (9 + 7 + 6) - (8 + 6) = 22 - 14 = 8, which isn't divisible by 11. What about 9875? (9 + 7 + 5) - (8 + 7) = 21 - 15 = 6, not divisible by 11. Hmm, maybe 98765 is okay? But wait, let's check all four-digit subnumbers.Wait, 98765 has four four-digit subnumbers: 9876, 9875, 9865, 8765. Let's check each:- 9876: (9 + 7 + 6) - (8 + 6) = 22 - 14 = 8, not divisible by 11.- 9875: (9 + 7 + 5) - (8 + 7) = 21 - 15 = 6, not divisible by 11.- 9865: (9 + 6 + 5) - (8 + 6) = 20 - 14 = 6, not divisible by 11.- 8765: (8 + 6 + 5) - (7 + 6) = 19 - 13 = 6, not divisible by 11.Okay, so 98765 seems okay in terms of four-digit subnumbers. But let's check the three-digit subnumbers. For example, 987: (9 + 7) - 8 = 16 - 8 = 8, not divisible by 11. 986: (9 + 6) - 8 = 15 - 8 = 7, not divisible by 11. 985: (9 + 5) - 8 = 14 - 8 = 6, not divisible by 11. 976: (9 + 6) - 7 = 15 - 7 = 8, not divisible by 11. 975: (9 + 5) - 7 = 14 - 7 = 7, not divisible by 11. 965: (9 + 5) - 6 = 14 - 6 = 8, not divisible by 11. 876: (8 + 6) - 7 = 14 - 7 = 7, not divisible by 11. 875: (8 + 5) - 7 = 13 - 7 = 6, not divisible by 11. 865: (8 + 5) - 6 = 13 - 6 = 7, not divisible by 11. 765: (7 + 5) - 6 = 12 - 6 = 6, not divisible by 11.So, all three-digit subnumbers are okay. Now, two-digit subnumbers: since all digits are distinct, none of them can be like 11, 22, etc., so they are all safe.So, 98765 seems to satisfy the conditions. But wait, is 98765 itself divisible by 11? Let's check: (9 + 7 + 5) - (8 + 6) = 21 - 14 = 7, which isn't a multiple of 11. So, 98765 isn't divisible by 11.But is 98765 the largest such number? Because 98765 is less than 99999, but maybe there's a larger number with all distinct digits that also satisfies the conditions.Wait, 98765 is the largest five-digit number with all distinct digits, right? Because it goes from 9 down to 5. So, 98765 is the largest five-digit number with all distinct digits. So, if 98765 satisfies the conditions, then it's the answer.But let me just confirm: is there a larger number than 98765 that also satisfies the conditions? For example, 98766 is larger, but it has repeating digits, so it's out. 98767 is also repeating, same with 98768, 98769, 98770, etc., all have repeating digits or end with 0, which might not be necessary, but the point is, 98765 is the largest with all distinct digits.Wait, but 98765 is 98765, but maybe 98764 is larger? No, 98765 is larger. Wait, no, 98765 is the largest with distinct digits.Wait, but let me think again. If I try 98765, it's the largest five-digit number with all distinct digits. So, if it satisfies the conditions, it's the answer. But let me check if any of its subnumbers are divisible by 11.Wait, I already checked the four-digit, three-digit, and two-digit subnumbers, and none were divisible by 11. So, 98765 seems to satisfy all the conditions.But wait, let me think about the number 98765. Is there any subnumber of length four or five that is divisible by 11? We checked four-digit subnumbers, but let me check the five-digit number itself: 98765. As we saw, (9 + 7 + 5) - (8 + 6) = 21 - 14 = 7, not divisible by 11. So, it's safe.But wait, let me think about another angle. Maybe there's a number larger than 98765 that doesn't have all distinct digits but still satisfies the condition that no subnumber is divisible by 11. Because 98765 is the largest with all distinct digits, but maybe a number like 99765 is larger and still satisfies the conditions.Wait, 99765 has two 9s. So, deleting the middle digits would give 99, which is divisible by 11. So, that's bad. So, 99765 is out.Similarly, 98875 has two 8s. Deleting the middle digits would give 88, which is divisible by 11. So, that's also bad.So, any number with repeating digits would have a two-digit subnumber that's divisible by 11. Therefore, to satisfy the condition, the number must have all distinct digits. So, 98765 is indeed the largest such number.Wait, but let me think again. Is 98765 the largest five-digit number with all distinct digits? Or is there a larger one? For example, 98765 is 98765, but 98765 is less than 98766, but 98766 has repeating digits. Similarly, 98767, 98768, 98769, 98770, etc., all have repeating digits or end with 0, which might not be necessary, but the point is, 98765 is the largest with all distinct digits.Wait, but actually, 98765 is less than 98765? Wait, no, 98765 is 98765, and the next number with all distinct digits would be 98765 + 1 = 98766, which has repeating digits. So, yes, 98765 is the largest five-digit number with all distinct digits.But hold on, let me check 98765 again. Is there any subnumber that's divisible by 11? For example, the four-digit number 9876: (9 + 7 + 6) - (8 + 6) = 22 - 14 = 8, not divisible by 11. 9875: (9 + 7 + 5) - (8 + 7) = 21 - 15 = 6, not divisible by 11. 9865: (9 + 6 + 5) - (8 + 6) = 20 - 14 = 6, not divisible by 11. 8765: (8 + 6 + 5) - (7 + 6) = 19 - 13 = 6, not divisible by 11.Three-digit subnumbers: 987: (9 + 7) - 8 = 16 - 8 = 8, not divisible by 11. 986: (9 + 6) - 8 = 15 - 8 = 7, not divisible by 11. 985: (9 + 5) - 8 = 14 - 8 = 6, not divisible by 11. 976: (9 + 6) - 7 = 15 - 7 = 8, not divisible by 11. 975: (9 + 5) - 7 = 14 - 7 = 7, not divisible by 11. 965: (9 + 5) - 6 = 14 - 6 = 8, not divisible by 11. 876: (8 + 6) - 7 = 14 - 7 = 7, not divisible by 11. 875: (8 + 5) - 7 = 13 - 7 = 6, not divisible by 11. 865: (8 + 5) - 6 = 13 - 6 = 7, not divisible by 11. 765: (7 + 5) - 6 = 12 - 6 = 6, not divisible by 11.Two-digit subnumbers: All are distinct, so none are like 11, 22, etc. So, none are divisible by 11.Therefore, 98765 seems to satisfy all the conditions. But wait, is 98765 the largest? Because 98765 is less than 99999, but maybe there's a larger number that doesn't have all distinct digits but still satisfies the condition that no subnumber is divisible by 11.Wait, but as I thought earlier, any number with repeating digits would have a two-digit subnumber that's divisible by 11. So, to avoid that, the number must have all distinct digits. Therefore, 98765 is indeed the largest such number.But let me think again. Maybe there's a number larger than 98765 that doesn't have all distinct digits but still doesn't have any subnumber divisible by 11. For example, maybe 98764 is larger? No, 98765 is larger. Wait, 98765 is 98765, and 98764 is smaller.Wait, no, 98765 is larger than 98764. So, 98765 is the largest five-digit number with all distinct digits.But wait, let me think about the number 98765. Is there a way to make a larger number with all distinct digits? For example, 98765 is 98765, but 98765 is less than 98765? Wait, no, that's the same number.Wait, maybe I can rearrange the digits to get a larger number. For example, 98765 is already the largest possible with digits 9,8,7,6,5. So, that's the maximum.Therefore, I think 98765 is the answer.But wait, let me check 98765 again. Is there any subnumber that's divisible by 11? For example, the four-digit number 9876: (9 + 7 + 6) - (8 + 6) = 22 - 14 = 8, not divisible by 11. 9875: (9 + 7 + 5) - (8 + 7) = 21 - 15 = 6, not divisible by 11. 9865: (9 + 6 + 5) - (8 + 6) = 20 - 14 = 6, not divisible by 11. 8765: (8 + 6 + 5) - (7 + 6) = 19 - 13 = 6, not divisible by 11.Three-digit subnumbers: 987: (9 + 7) - 8 = 16 - 8 = 8, not divisible by 11. 986: (9 + 6) - 8 = 15 - 8 = 7, not divisible by 11. 985: (9 + 5) - 8 = 14 - 8 = 6, not divisible by 11. 976: (9 + 6) - 7 = 15 - 7 = 8, not divisible by 11. 975: (9 + 5) - 7 = 14 - 7 = 7, not divisible by 11. 965: (9 + 5) - 6 = 14 - 6 = 8, not divisible by 11. 876: (8 + 6) - 7 = 14 - 7 = 7, not divisible by 11. 875: (8 + 5) - 7 = 13 - 7 = 6, not divisible by 11. 865: (8 + 5) - 6 = 13 - 6 = 7, not divisible by 11. 765: (7 + 5) - 6 = 12 - 6 = 6, not divisible by 11.Two-digit subnumbers: All are distinct, so none are divisible by 11.So, 98765 seems to satisfy all the conditions. Therefore, I think the answer is 98765.But wait, let me think again. Is there a larger number than 98765 that doesn't have all distinct digits but still doesn't have any subnumber divisible by 11? For example, maybe 98764 is larger? No, 98765 is larger. Wait, 98765 is 98765, and 98764 is smaller.Wait, no, 98765 is larger than 98764. So, 98765 is the largest five-digit number with all distinct digits. Therefore, it's the answer.But just to be thorough, let me check 98765 one more time. The number itself: (9 + 7 + 5) - (8 + 6) = 21 - 14 = 7, not divisible by 11. All four-digit subnumbers: none are divisible by 11. All three-digit subnumbers: none are divisible by 11. All two-digit subnumbers: none are divisible by 11. So, yes, 98765 is the answer.Wait, but hold on. Let me think about the number 98765. Is there any way that a subnumber could be divisible by 11? For example, the number 98765. If I delete the first digit, I get 8765, which we've checked. If I delete the second digit, I get 9765. Let's check 9765: (9 + 6 + 5) - (7 + 6) = 20 - 13 = 7, not divisible by 11. If I delete the third digit, I get 9865, which we've checked. If I delete the fourth digit, I get 9875, which we've checked. If I delete the fifth digit, I get 9876, which we've checked.Similarly, for three-digit subnumbers, all are safe. So, yes, 98765 is the answer.But wait, let me think about another angle. Maybe the number 98765 isn't the largest possible. For example, maybe 98765 is less than 98765? Wait, no, that's the same number. So, I think I've confirmed enough that 98765 is the answer.But just to be 100% sure, let me think about the number 98765. Is there any subnumber that's divisible by 11? Let's list all possible subnumbers:Four-digit subnumbers:- 9876: 9+7+6 - (8+6) = 22-14=8- 9875: 9+7+5 - (8+7)=21-15=6- 9865: 9+6+5 - (8+6)=20-14=6- 8765: 8+6+5 - (7+6)=19-13=6Three-digit subnumbers:- 987: 9+7 -8=16-8=8- 986: 9+6 -8=15-8=7- 985: 9+5 -8=14-8=6- 976: 9+6 -7=15-7=8- 975: 9+5 -7=14-7=7- 965: 9+5 -6=14-6=8- 876: 8+6 -7=14-7=7- 875: 8+5 -7=13-7=6- 865: 8+5 -6=13-6=7- 765: 7+5 -6=12-6=6Two-digit subnumbers:- 98: 9-8=1- 97: 9-7=2- 96: 9-6=3- 95: 9-5=4- 87: 8-7=1- 86: 8-6=2- 85: 8-5=3- 76: 7-6=1- 75: 7-5=2- 65: 6-5=1None of these are multiples of 11, so all subnumbers are safe.Therefore, I can confidently say that 98765 is the largest five-digit number that isn't divisible by 11, and none of its subnumbers are divisible by 11 either.

answer：Alright, so I have this train problem to solve. Let me try to parse it step by step. The problem says:"A train starts its journey, then stops after 1 hour due to an incident and remains halted for half an hour. After that, it continues at 3/4 of its original speed, resulting in a delay of 3 1/2 hours upon reaching its destination. If the incident had occurred 90 miles further ahead, the train would have arrived only 3 hours late. What is the total distance of the whole journey in miles?"The options are (A) 400, (B) 465, (C) 600, (D) 640, (E) 550.Hmm, okay. So, let me try to break this down.First, the train starts its journey. It goes for 1 hour, then stops for half an hour due to an incident. After that, it continues at 3/4 of its original speed, which causes it to be 3.5 hours late. If the incident had happened 90 miles further ahead, the delay would have been only 3 hours. We need to find the total distance of the journey.Let me denote some variables to model this.Let’s say the original speed of the train is v miles per hour. Let the total distance of the journey be D miles.In the first scenario:1. The train travels for 1 hour at speed v, covering a distance of v miles.2. Then it stops for 0.5 hours.3. After that, it continues the remaining distance at 3/4 v speed.Because of the stoppage and the reduced speed, the train arrives 3.5 hours late.In the second scenario:1. The incident occurs 90 miles further ahead. So, instead of stopping after 1 hour (which is v miles), it stops after traveling (v + 90) miles.2. It stops for 0.5 hours again.3. Then it continues the remaining distance at 3/4 v speed.4. In this case, the delay is only 3 hours.So, we have two different scenarios with different delays, and we can set up equations based on that.First, let's figure out the total time taken in each scenario and compare it to the original time without any incident.Original time without any incident would be D / v hours.In the first scenario:- Time taken before stopping: 1 hour.- Stoppage time: 0.5 hours.- Remaining distance after stopping: D - v miles.- Speed after stopping: (3/4)v.- Time taken to cover remaining distance: (D - v) / ( (3/4)v ) = (4(D - v)) / (3v) hours.Total time taken in first scenario: 1 + 0.5 + (4(D - v))/(3v) = 1.5 + (4(D - v))/(3v).This total time is equal to the original time plus 3.5 hours delay.So,1.5 + (4(D - v))/(3v) = D / v + 3.5.Similarly, in the second scenario:- The incident occurs 90 miles further ahead, so the train travels v + 90 miles before stopping.- Time taken to travel v + 90 miles: Since speed is v, time is (v + 90)/v hours.- Stoppage time: 0.5 hours.- Remaining distance: D - (v + 90) miles.- Speed after stopping: (3/4)v.- Time taken to cover remaining distance: (D - v - 90) / ( (3/4)v ) = (4(D - v - 90)) / (3v) hours.Total time taken in second scenario: (v + 90)/v + 0.5 + (4(D - v - 90))/(3v).This total time is equal to the original time plus 3 hours delay.So,(v + 90)/v + 0.5 + (4(D - v - 90))/(3v) = D / v + 3.Now, we have two equations:1. 1.5 + (4(D - v))/(3v) = D / v + 3.5.2. (v + 90)/v + 0.5 + (4(D - v - 90))/(3v) = D / v + 3.Let me write these equations more clearly.Equation 1:1.5 + (4(D - v))/(3v) = D / v + 3.5.Equation 2:(v + 90)/v + 0.5 + (4(D - v - 90))/(3v) = D / v + 3.Let me simplify both equations.Starting with Equation 1:1.5 + (4(D - v))/(3v) = D / v + 3.5.Let me subtract D / v from both sides:1.5 + (4(D - v))/(3v) - D / v = 3.5.Simplify the left side:1.5 + [4(D - v) - 3D] / (3v) = 3.5.Compute numerator:4D - 4v - 3D = D - 4v.So,1.5 + (D - 4v)/(3v) = 3.5.Subtract 1.5:(D - 4v)/(3v) = 2.Multiply both sides by 3v:D - 4v = 6v.So,D = 10v.Okay, so from Equation 1, we get D = 10v.Now, let's move to Equation 2:(v + 90)/v + 0.5 + (4(D - v - 90))/(3v) = D / v + 3.First, note that D = 10v, so we can substitute D with 10v in Equation 2.So, let's do that.First, compute each term:(v + 90)/v = 1 + 90/v.0.5 remains as is.(4(D - v - 90))/(3v) = 4(10v - v - 90)/(3v) = 4(9v - 90)/(3v) = (36v - 360)/(3v) = 12 - 120/v.So, putting it all together:1 + 90/v + 0.5 + 12 - 120/v = 10v / v + 3.Simplify the left side:1 + 0.5 + 12 + (90/v - 120/v) = 13.5 - 30/v.Right side:10 + 3 = 13.So, equation becomes:13.5 - 30/v = 13.Subtract 13:0.5 - 30/v = 0.So,0.5 = 30/v.Multiply both sides by v:0.5v = 30.So,v = 30 / 0.5 = 60.So, the original speed v is 60 mph.Then, since D = 10v, D = 10 * 60 = 600 miles.Wait, so the total distance is 600 miles. Let me check if this is one of the options. Option C is 600, so that's likely the answer.But let me verify the calculations again to make sure.From Equation 1:We had D = 10v.From Equation 2, substituting D = 10v, we found v = 60, so D = 600.Let me verify with the first scenario:Original time without incident: D / v = 600 / 60 = 10 hours.In the first scenario:- Traveled 1 hour at 60 mph: 60 miles.- Stopped for 0.5 hours.- Remaining distance: 600 - 60 = 540 miles.- Speed reduced to 3/4 * 60 = 45 mph.- Time to cover 540 miles: 540 / 45 = 12 hours.- Total time: 1 + 0.5 + 12 = 13.5 hours.- Original time is 10 hours, so delay is 13.5 - 10 = 3.5 hours. Correct.Second scenario:Incident occurs 90 miles further ahead, so 60 + 90 = 150 miles from start.Time to reach incident point: 150 / 60 = 2.5 hours.Stopped for 0.5 hours.Remaining distance: 600 - 150 = 450 miles.Speed reduced to 45 mph.Time to cover 450 miles: 450 / 45 = 10 hours.Total time: 2.5 + 0.5 + 10 = 13 hours.Original time is 10 hours, so delay is 13 - 10 = 3 hours. Correct.So, both scenarios check out. Therefore, the total distance is 600 miles.**Final Answer**boxed{C}