💡Okay, so I have this problem where I need to prove that the sum of the squares of the distances between all pairs of points on a unit circle is less than or equal to ( k^2 ). The points are labeled ( A_1, A_2, ldots, A_k ), and ( d(A_i, A_j) ) is the distance between points ( A_i ) and ( A_j ). First, I should recall that the unit circle has a radius of 1, so each point ( A_i ) can be represented in the coordinate system as ( (cos x_i, sin x_i) ) where ( x_i ) is the angle made with the positive x-axis. That makes sense because any point on the unit circle can be described using cosine and sine functions based on its angle.Now, the distance between two points ( A_i ) and ( A_j ) on the unit circle can be found using the distance formula. So, if ( A_i ) is ( (cos x_i, sin x_i) ) and ( A_j ) is ( (cos x_j, sin x_j) ), then the distance ( d(A_i, A_j) ) is:[d(A_i, A_j) = sqrt{(cos x_j - cos x_i)^2 + (sin x_j - sin x_i)^2}]I remember there's a trigonometric identity that relates the distance between two points on a unit circle to the cosine of the angle between them. Let me try to recall it. I think it's something like ( d(A_i, A_j) = 2sinleft(frac{theta}{2}right) ) where ( theta ) is the angle between them. But wait, actually, squaring both sides might be more helpful here since the problem involves the square of the distance.Let me square the distance formula:[d(A_i, A_j)^2 = (cos x_j - cos x_i)^2 + (sin x_j - sin x_i)^2]Expanding the squares:[= cos^2 x_j - 2cos x_j cos x_i + cos^2 x_i + sin^2 x_j - 2sin x_j sin x_i + sin^2 x_i]Combine like terms:[= (cos^2 x_j + sin^2 x_j) + (cos^2 x_i + sin^2 x_i) - 2(cos x_j cos x_i + sin x_j sin x_i)]Since ( cos^2 theta + sin^2 theta = 1 ) for any angle ( theta ), this simplifies to:[= 1 + 1 - 2(cos x_j cos x_i + sin x_j sin x_i)][= 2 - 2cos(x_j - x_i)]Oh, that's right! So the square of the distance between two points on the unit circle is ( 2 - 2cos(x_j - x_i) ). That's a useful identity. So, ( d(A_i, A_j)^2 = 2 - 2cos(x_j - x_i) ).Now, the problem asks for the sum of these squared distances over all pairs ( (i, j) ) where ( 1 leq i < j leq k ). So, let's write that sum:[sum_{1 leq i < j leq k} d(A_i, A_j)^2 = sum_{1 leq i < j leq k} [2 - 2cos(x_j - x_i)]]I can split this sum into two separate sums:[= sum_{1 leq i < j leq k} 2 - 2 sum_{1 leq i < j leq k} cos(x_j - x_i)]The first sum is just adding 2 for each pair ( (i, j) ). How many such pairs are there? Well, for ( k ) points, the number of pairs is ( binom{k}{2} = frac{k(k - 1)}{2} ). So, the first sum becomes:[2 times frac{k(k - 1)}{2} = k(k - 1)]So, now the expression is:[k(k - 1) - 2 sum_{1 leq i < j leq k} cos(x_j - x_i)]Now, I need to evaluate or bound the second sum, which is ( sum_{1 leq i < j leq k} cos(x_j - x_i) ). Hmm, this seems tricky. I wonder if there's a way to relate this sum to something else or perhaps find an upper bound for it.I recall that the cosine function has a maximum value of 1 and a minimum value of -1. So, each term ( cos(x_j - x_i) ) is between -1 and 1. Therefore, the entire sum ( sum_{1 leq i < j leq k} cos(x_j - x_i) ) is between ( -binom{k}{2} ) and ( binom{k}{2} ).But I need to find an upper bound for the entire expression ( k(k - 1) - 2 sum cos(x_j - x_i) ). Since the sum of cosines can be as low as ( -binom{k}{2} ), substituting that into the expression would give the maximum possible value of the original sum.Let me try that. If ( sum cos(x_j - x_i) geq -binom{k}{2} ), then:[-2 sum cos(x_j - x_i) leq -2 times (-binom{k}{2}) = 2 times binom{k}{2}]So, substituting back into the expression:[k(k - 1) - 2 sum cos(x_j - x_i) leq k(k - 1) + 2 times binom{k}{2}]But wait, ( binom{k}{2} = frac{k(k - 1)}{2} ), so:[2 times binom{k}{2} = 2 times frac{k(k - 1)}{2} = k(k - 1)]Therefore, the expression becomes:[k(k - 1) + k(k - 1) = 2k(k - 1)]So, the sum ( sum_{1 leq i < j leq k} d(A_i, A_j)^2 leq 2k(k - 1) ).But the problem states that this sum should be less than or equal to ( k^2 ). Hmm, ( 2k(k - 1) ) is actually equal to ( 2k^2 - 2k ), which is less than ( k^2 ) only when ( k^2 - 2k < 0 ), which happens when ( k < 2 ). But for ( k geq 2 ), ( 2k(k - 1) ) is greater than ( k^2 ). For example, when ( k = 3 ), ( 2k(k - 1) = 12 ) and ( k^2 = 9 ), so 12 > 9. So, my upper bound is actually higher than ( k^2 ), which doesn't help me prove the original statement.Hmm, so maybe my approach is missing something. I need a better way to bound the sum ( sum cos(x_j - x_i) ).Wait, perhaps instead of just using the fact that cosine is between -1 and 1, I can find a more precise bound for the sum. Maybe using some properties of the cosine function or some identity.I remember that the sum of cosines can sometimes be expressed in terms of other trigonometric functions or perhaps using complex exponentials. Let me think about that.Alternatively, maybe I can relate this sum to the magnitude squared of some vector. Let me consider the points on the unit circle as complex numbers. Let ( z_i = e^{ix_i} ), so each ( z_i ) is a point on the unit circle in the complex plane.Then, the distance squared between ( z_i ) and ( z_j ) is:[|z_i - z_j|^2 = (z_i - z_j)(overline{z_i} - overline{z_j}) = |z_i|^2 + |z_j|^2 - z_i overline{z_j} - overline{z_i} z_j]Since each ( |z_i| = 1 ), this simplifies to:[1 + 1 - z_i overline{z_j} - overline{z_i} z_j = 2 - (z_i overline{z_j} + overline{z_i} z_j)]But ( z_i overline{z_j} = e^{i(x_i - x_j)} ) and ( overline{z_i} z_j = e^{i(x_j - x_i)} ). So, their sum is:[e^{i(x_i - x_j)} + e^{-i(x_i - x_j)} = 2cos(x_i - x_j)]Therefore, ( |z_i - z_j|^2 = 2 - 2cos(x_i - x_j) ), which matches what I had earlier.So, the sum ( sum_{1 leq i < j leq k} |z_i - z_j|^2 ) is equal to ( sum_{1 leq i < j leq k} [2 - 2cos(x_i - x_j)] ), which is the same as before.But perhaps expressing this in terms of complex numbers can help me find a better bound. Let me think about the sum:[sum_{1 leq i < j leq k} |z_i - z_j|^2]I can write this as:[sum_{1 leq i < j leq k} (2 - 2cos(x_i - x_j)) = 2 binom{k}{2} - 2 sum_{1 leq i < j leq k} cos(x_i - x_j)]Which is the same as:[k(k - 1) - 2 sum_{1 leq i < j leq k} cos(x_i - x_j)]So, again, I need to bound the sum ( S = sum_{1 leq i < j leq k} cos(x_i - x_j) ).I wonder if there's a way to express this sum in terms of the sum of the cosines of individual angles. Let me consider the sum:[sum_{i=1}^k sum_{j=1}^k cos(x_i - x_j)]This is the sum over all pairs, including ( i = j ). For ( i = j ), ( cos(x_i - x_j) = cos(0) = 1 ). So, the total sum is:[sum_{i=1}^k sum_{j=1}^k cos(x_i - x_j) = sum_{i=1}^k sum_{j=1}^k cos(x_i - x_j)]But this can be rewritten using the identity for the sum of cosines:[sum_{i=1}^k sum_{j=1}^k cos(x_i - x_j) = left( sum_{i=1}^k cos x_i right)^2 + left( sum_{i=1}^k sin x_i right)^2]Wait, is that correct? Let me check.I recall that ( cos(a - b) = cos a cos b + sin a sin b ). So, if I expand ( sum_{i=1}^k sum_{j=1}^k cos(x_i - x_j) ), it becomes:[sum_{i=1}^k sum_{j=1}^k [cos x_i cos x_j + sin x_i sin x_j] = left( sum_{i=1}^k cos x_i right)^2 + left( sum_{i=1}^k sin x_i right)^2]Yes, that's correct. So, the double sum is equal to the square of the sum of cosines plus the square of the sum of sines.Therefore, the sum ( S = sum_{1 leq i < j leq k} cos(x_i - x_j) ) can be related to this double sum. Let's see:The double sum ( sum_{i=1}^k sum_{j=1}^k cos(x_i - x_j) ) includes all pairs, including ( i = j ). There are ( k ) terms where ( i = j ), each contributing 1, so the total sum is:[sum_{i=1}^k sum_{j=1}^k cos(x_i - x_j) = left( sum_{i=1}^k cos x_i right)^2 + left( sum_{i=1}^k sin x_i right)^2 = k + 2 sum_{1 leq i < j leq k} cos(x_i - x_j)]Wait, no. Let me clarify. The double sum includes all pairs, including ( i = j ). So, the total sum is:[sum_{i=1}^k sum_{j=1}^k cos(x_i - x_j) = sum_{i=1}^k cos(0) + 2 sum_{1 leq i < j leq k} cos(x_i - x_j)][= k + 2S]But from earlier, we also have:[sum_{i=1}^k sum_{j=1}^k cos(x_i - x_j) = left( sum_{i=1}^k cos x_i right)^2 + left( sum_{i=1}^k sin x_i right)^2]Let me denote ( C = sum_{i=1}^k cos x_i ) and ( S = sum_{i=1}^k sin x_i ). Then, the double sum is ( C^2 + S^2 ).So, we have:[C^2 + S^2 = k + 2S]Wait, no, that would mean ( C^2 + S^2 = k + 2S ), but ( S ) is already used as the sum of sines. Let me use different notation to avoid confusion. Let me denote the sum ( sum_{1 leq i < j leq k} cos(x_i - x_j) ) as ( T ). Then, the double sum is:[sum_{i=1}^k sum_{j=1}^k cos(x_i - x_j) = k + 2T]And also:[sum_{i=1}^k sum_{j=1}^k cos(x_i - x_j) = left( sum_{i=1}^k cos x_i right)^2 + left( sum_{i=1}^k sin x_i right)^2]Let me denote ( C = sum_{i=1}^k cos x_i ) and ( D = sum_{i=1}^k sin x_i ). Then:[C^2 + D^2 = k + 2T]So, solving for ( T ):[2T = C^2 + D^2 - k][T = frac{C^2 + D^2 - k}{2}]Therefore, the sum ( T = sum_{1 leq i < j leq k} cos(x_i - x_j) = frac{C^2 + D^2 - k}{2} ).Now, going back to the original expression for the sum of squared distances:[sum_{1 leq i < j leq k} d(A_i, A_j)^2 = k(k - 1) - 2T][= k(k - 1) - 2 times frac{C^2 + D^2 - k}{2}][= k(k - 1) - (C^2 + D^2 - k)][= k(k - 1) - C^2 - D^2 + k][= k^2 - k - C^2 - D^2 + k][= k^2 - C^2 - D^2]So, the sum of squared distances is ( k^2 - (C^2 + D^2) ).Now, I need to find an upper bound for this expression. Since ( C^2 + D^2 ) is non-negative, the maximum value of ( k^2 - (C^2 + D^2) ) occurs when ( C^2 + D^2 ) is minimized.But wait, ( C ) and ( D ) are the sums of cosines and sines of the angles ( x_i ). What's the minimum value of ( C^2 + D^2 )?I recall that ( C^2 + D^2 ) represents the square of the magnitude of the vector sum of all the points ( A_i ). Since each ( A_i ) is a unit vector, the magnitude of their sum ( |C + iD| ) can vary depending on the configuration of the points.The maximum value of ( C^2 + D^2 ) occurs when all points are the same, i.e., when all ( x_i ) are equal. In that case, ( C = k cos x ) and ( D = k sin x ), so ( C^2 + D^2 = k^2 ).The minimum value of ( C^2 + D^2 ) occurs when the points are evenly distributed around the circle, such that their vector sum is zero. For example, if the points are the vertices of a regular polygon, their vector sum is zero, so ( C = 0 ) and ( D = 0 ), hence ( C^2 + D^2 = 0 ).Wait, but is that always the case? If the points are not symmetrically distributed, the vector sum might not be zero. However, regardless of their distribution, ( C^2 + D^2 ) is always non-negative, so the minimum value is 0.Therefore, ( C^2 + D^2 geq 0 ), which implies that:[k^2 - (C^2 + D^2) leq k^2]Hence, the sum of squared distances is less than or equal to ( k^2 ).Wait, that seems too straightforward. Let me double-check.We have:[sum_{1 leq i < j leq k} d(A_i, A_j)^2 = k^2 - (C^2 + D^2)]Since ( C^2 + D^2 geq 0 ), subtracting it from ( k^2 ) will always give a value less than or equal to ( k^2 ). Therefore, the sum is indeed less than or equal to ( k^2 ).But wait, earlier when I tried to bound the sum using the individual cosines, I ended up with a higher upper bound, ( 2k(k - 1) ), which is larger than ( k^2 ) for ( k geq 2 ). So, this approach gives a tighter bound, which is exactly what we needed.Therefore, the key was to express the sum of squared distances in terms of ( C^2 + D^2 ), which is always non-negative, and thus the sum is bounded above by ( k^2 ).I think this makes sense. By expressing the sum in terms of the vector sums, we can leverage the fact that the magnitude squared of any real vector is non-negative, which gives us the necessary bound.So, to summarize:1. Express each point ( A_i ) as ( (cos x_i, sin x_i) ).2. Use the distance formula to find ( d(A_i, A_j)^2 = 2 - 2cos(x_i - x_j) ).3. Sum this over all pairs ( (i, j) ) to get ( k(k - 1) - 2T ), where ( T ) is the sum of cosines.4. Express ( T ) in terms of ( C^2 + D^2 ), where ( C ) and ( D ) are the sums of cosines and sines of the angles.5. Show that ( C^2 + D^2 geq 0 ), leading to the sum of squared distances being ( k^2 - (C^2 + D^2) leq k^2 ).This completes the proof.

💡Let ( S ) represent the number of sheets of paper and ( E ) the number of envelopes in each box.1. **Ann's Usage:** - Ann used all the envelopes and had 30 sheets of paper left. - Each letter requires 2 sheets of paper. - Therefore, the number of letters Ann wrote is ( frac{S - 30}{2} ). - Since she used all the envelopes, the number of letters is also equal to ( E ). - This gives the equation: [ frac{S - 30}{2} = E ] [ S - 30 = 2E quad text{(Equation 1)} ]2. **Bob's Usage:** - Bob used all the sheets of paper and had 20 envelopes left. - Each letter requires 4 sheets of paper. - Therefore, the number of letters Bob wrote is ( frac{S}{4} ). - Since he had 20 envelopes left, the total number of envelopes is ( frac{S}{4} + 20 ). - This gives the equation: [ E = frac{S}{4} + 20 quad text{(Equation 2)} ]3. **Solving the Equations:** - Substitute Equation 2 into Equation 1: [ S - 30 = 2left(frac{S}{4} + 20right) ] [ S - 30 = frac{S}{2} + 40 ] [ S - frac{S}{2} = 70 ] [ frac{S}{2} = 70 ] [ S = 140 ]4. **Verification:** - If ( S = 140 ), then from Equation 2: [ E = frac{140}{4} + 20 = 35 + 20 = 55 ] - Check Ann's usage: [ S - 30 = 140 - 30 = 110 = 2E = 2 times 55 = 110 quad text{(Correct)} ] - Check Bob's usage: [ E = 55 = frac{140}{4} + 20 = 35 + 20 = 55 quad text{(Correct)} ]**Conclusion:**The number of sheets of paper in each box is (140).

💡Okay, so I have this function f(x) = 2x³ - a x² + 4, and I know that x = 1 is a critical point. I need to find two things: first, the intervals where the function is monotonically increasing, and second, the minimum value of the function when x is between -1 and 2.Alright, let's start by recalling what a critical point is. A critical point occurs where the derivative of the function is zero or undefined. Since our function is a polynomial, its derivative will be defined everywhere, so the critical points are just where the derivative equals zero.So, first step: find the derivative of f(x). The derivative of 2x³ is 6x², the derivative of -a x² is -2a x, and the derivative of 4 is 0. So, f'(x) = 6x² - 2a x.Now, since x = 1 is a critical point, plugging x = 1 into the derivative should give us zero. Let's do that:f'(1) = 6(1)² - 2a(1) = 6 - 2a = 0.Solving for a: 6 - 2a = 0 => 2a = 6 => a = 3.Okay, so a is 3. That means our function is f(x) = 2x³ - 3x² + 4.Now, moving on to part (1): finding the intervals where the function is monotonically increasing. To do this, we need to analyze the sign of the derivative f'(x). If f'(x) is positive, the function is increasing; if it's negative, the function is decreasing.We already have f'(x) = 6x² - 6x (since a = 3). Let's factor that:f'(x) = 6x(x - 1).So, the derivative is zero at x = 0 and x = 1. These are our critical points. To determine the intervals where the derivative is positive or negative, we can test intervals around these critical points.Let's consider the intervals:1. x < 0: Let's pick x = -1. Plugging into f'(x): 6*(-1)*(-1 - 1) = 6*(-1)*(-2) = 12, which is positive. So, f'(x) > 0 for x < 0.2. 0 < x < 1: Let's pick x = 0.5. Plugging into f'(x): 6*(0.5)*(0.5 - 1) = 6*(0.5)*(-0.5) = -1.5, which is negative. So, f'(x) < 0 for 0 < x < 1.3. x > 1: Let's pick x = 2. Plugging into f'(x): 6*2*(2 - 1) = 12*1 = 12, which is positive. So, f'(x) > 0 for x > 1.Therefore, the function is increasing on (-∞, 0) and (1, ∞), and decreasing on (0, 1).For part (2), we need to find the minimum value of f(x) when x is in the interval [-1, 2]. To find the minimum, we should evaluate f(x) at the critical points within the interval and at the endpoints.First, let's identify the critical points in [-1, 2]. From earlier, we have critical points at x = 0 and x = 1, both of which are within [-1, 2].So, we need to evaluate f(x) at x = -1, x = 0, x = 1, and x = 2.Let's compute each:1. f(-1) = 2*(-1)³ - 3*(-1)² + 4 = 2*(-1) - 3*(1) + 4 = -2 - 3 + 4 = -1.2. f(0) = 2*(0)³ - 3*(0)² + 4 = 0 - 0 + 4 = 4.3. f(1) = 2*(1)³ - 3*(1)² + 4 = 2 - 3 + 4 = 3.4. f(2) = 2*(2)³ - 3*(2)² + 4 = 2*8 - 3*4 + 4 = 16 - 12 + 4 = 8.Now, comparing these values: f(-1) = -1, f(0) = 4, f(1) = 3, f(2) = 8. The smallest value is -1 at x = -1.So, the minimum value of f(x) on the interval [-1, 2] is -1.Wait, just to make sure I didn't make a calculation mistake. Let me double-check f(-1):f(-1) = 2*(-1)^3 - 3*(-1)^2 + 4 = 2*(-1) - 3*(1) + 4 = -2 - 3 + 4 = (-2 - 3) + 4 = -5 + 4 = -1. Yep, that's correct.And f(0) is definitely 4, f(1) is 3, and f(2) is 8. So, no issues there.Therefore, I'm confident that the function is increasing on (-∞, 0) and (1, ∞), and the minimum value on [-1, 2] is -1.

💡First, calculate the total cost of the red notebooks: 3 notebooks multiplied by 4 each equals 12.Next, calculate the total cost of the green notebooks: 2 notebooks multiplied by 2 each equals 4.Add the costs of the red and green notebooks to find the total spent on these two types: 12 plus 4 equals 16.Subtract this amount from the total spending to find out how much was spent on blue notebooks: 37 minus 16 equals 21.Determine the number of blue notebooks by dividing the amount spent on blue notebooks by the cost per blue notebook: 21 divided by 3 equals 7 blue notebooks.Finally, add up all the notebooks purchased: 3 red notebooks plus 2 green notebooks plus 7 blue notebooks equals a total of 12 notebooks.

💡Okay, so I have this problem here where I'm given a complex number z = -3 + 2i, and I need to find out which quadrant its conjugate, denoted as (overline{z}), is located in the complex plane. The options are the first, second, third, or fourth quadrant.Alright, let's start by recalling what a complex number is. A complex number is generally written as z = a + bi, where 'a' is the real part and 'b' is the imaginary part. In this case, z = -3 + 2i, so the real part is -3 and the imaginary part is 2.Now, the question is about the conjugate of z, which is (overline{z}). I remember that the conjugate of a complex number is found by changing the sign of the imaginary part. So, if z = a + bi, then (overline{z}) = a - bi. Applying that to our given z, which is -3 + 2i, the conjugate should be -3 - 2i. Let me write that down: (overline{z}) = -3 - 2i.Next, I need to figure out where this conjugate, -3 - 2i, is located in the complex plane. The complex plane is similar to the Cartesian coordinate system, where the horizontal axis represents the real part and the vertical axis represents the imaginary part. So, each complex number corresponds to a point (a, b) where 'a' is the real part and 'b' is the imaginary part.For (overline{z}) = -3 - 2i, the real part is -3 and the imaginary part is -2. So, plotting this point on the complex plane, it would be at (-3, -2). Now, I need to determine which quadrant this point is in.Quadrants in the complex plane are labeled similarly to the Cartesian coordinate system:- The first quadrant has points where both the real and imaginary parts are positive.- The second quadrant has points where the real part is negative and the imaginary part is positive.- The third quadrant has points where both the real and imaginary parts are negative.- The fourth quadrant has points where the real part is positive and the imaginary part is negative.Looking at the point (-3, -2), both the real part (-3) and the imaginary part (-2) are negative. That means this point is in the third quadrant.Wait, let me double-check to make sure I didn't mix up the quadrants. Sometimes it's easy to get confused between the second and third quadrants because both involve negative values. But no, in the second quadrant, the real part is negative and the imaginary part is positive, whereas in the third quadrant, both are negative. Since both -3 and -2 are negative, it's definitely the third quadrant.Just to visualize it, if I imagine the complex plane, the origin is at (0,0). Moving to the left along the real axis, I reach -3, and moving down along the imaginary axis, I reach -2. So, the point (-3, -2) is in the lower-left area of the plane, which is the third quadrant.I think that's solid. I don't see any mistakes in my reasoning. The conjugate of z is -3 - 2i, which corresponds to the point (-3, -2) in the complex plane, and that point is in the third quadrant.**Final Answer**The point corresponding to (overline{z}) is located in the boxed{C} quadrant.

💡First, I need to determine how many people are in each of the four equal groups. Since the total population is 480, I divide this by 4 to find the size of each group.Next, I know that two of these groups represent males. To find the total number of males, I multiply the number of people in one group by 2.This calculation gives me the total number of males in the town.