/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 Let \(A_{1} A_{2} \cdots A_{2 n}... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 16

Let \(A_{1} A_{2} \cdots A_{2 n}\) be a regular polygon with circumradius equal to 1 and consider a point \(P\) on the circumcircle. Prove that $$ \sum_{k=0}^{n-1} P A_{k+1}^{2} \cdot P A_{n+k+1}^{2}=2 n $$

Short Answer

Expert verified
Based on the step-by-step solution provided, it can be concluded that for a regular polygon with $2n$ sides and a point P on its circumcircle, the sum of the product of squared distances from point P to every pair of opposite vertices is equal to $2n$. The general expression for the distance from point P to any vertex and the use of summation formulas combined with the Law of Cosines helped to prove this statement.

Step by step solution

01

Find the general expression for the distance from point P to any vertex of the regular polygon.

For any vertex \(A_i\), we can consider the triangle formed by the center of the circumcircle \(O\), point \(P\), and vertex \(A_i\). Since the circumradius is 1, \(OA_i = 1\). Let \(\theta = \angle A_iOP\). Due to the regularity of the polygon, we know that the angle between any two consecutive vertices around the circumcircle is \(\frac{2\pi}{2n} = \frac{\pi}{n}\). Thus, the angle between \(A_i\) and \(A_{i+k}\) is \(\frac{k\pi}{n}\). Using the Law of Cosines, we can find \(PA_i\): $$ PA_i^2 = 1^2 + 1^2 - 2\cdot1\cdot1\cos\theta = 2 - 2\cos\theta $$ Similarly, for \(PA_{i+n}\) (which is opposite to \(A_i\)), we have: $$ PA_{i+n}^2 = 2 - 2\cos(\theta + \pi) $$
02

Calculate the sum of the product of squared distances for all pairs of opposite vertices.

We need to calculate the following sum: $$ \sum_{k=0}^{n-1} PA_{k+1}^2 \cdot PA_{n+k+1}^2 $$ Using the expressions for \(PA_i^2\) and \(PA_{i+n}^2\), we get: $$ \sum_{k=0}^{n-1} (2 - 2\cos\frac{k\pi}{n})(2 - 2\cos(\frac{k\pi}{n} + \pi)) = \sum_{k=0}^{n-1} (2-2\cos\frac{k\pi}{n})(2+2\cos\frac{k\pi}{n}) $$
03

Simplify the expression to demonstrate that it is equal to 2n.

Expanding the product and sum, we get: $$ \sum_{k=0}^{n-1}(4 - 4\cos^2\frac{k\pi}{n}) \\ = 4\sum_{k=0}^{n-1}\left(1 - \cos^2\frac{k\pi}{n}\right) \\ = 4\sum_{k=0}^{n-1}\sin^2\frac{k\pi}{n} $$ Now, we use the summation formula for sine-squared angles: $$ \sum_{k=0}^{n-1}\sin^2\frac{k\pi}{n} = \frac{n}{2} $$ So, our expression becomes: $$ 4\cdot\frac{n}{2} = 2n $$ Hence, we've proven that $$ \sum_{k=0}^{n-1} PA_{k+1}^{2} \cdot PA_{n+k+1}^{2} = 2n $$

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