The Obligatory PowerPoint Puzzle
Answer: ROOT BEER
Samuel Yeom

This puzzle is about the Power of a Point Theorem in circle geometry. Combined with the constraint that they must be integers, the values of the \( x_i \)'s are uniquely determined. Reading these values in order alphanumerically (A=1, B=2, etc.), we get the final answer.

Page 2

By Power of Point, we have \( x_1 (x_1 + 2) = x_2 (x_2 + 9) \). At this point, we could just let WolframAlpha solve this equation, but here is one way to solve it algebraically.

First, complete the square and multiply both sides by 4 to get \( (2x_1 + 2)^2 - 4 = (2x_2 + 9)^2 - 81 \). Letting \( \alpha = 2x_1 + 2 \) and \( \beta = 2x_2 + 9 \), we can rewrite the equation as \( \beta^2 - \alpha^2 = (\beta+\alpha)(\beta-\alpha) = 77 \). Note that \( \alpha \) and \( \beta \) are both positive integers. Therefore, \( (\beta+\alpha, \beta-\alpha) \) is either (11, 7) or (77, 1). Of these, only (77, 1) results in integer values of \( x_1 \) and \( x_2 \), which are 18 and 15, respectively.

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Power of Point gives us \( 4x_3 = 3x_4 \). In addition, triangle inequality on triangle \( ADX \) forces \( 12 < x_3 < 18 \). The only integer solution to both constraints is \( x_3 = 15 \) and \( x_4 = 20 \).

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Power of Point gives us \( 8(x_5 + 8) = x_6(x_6 + 11) \). The left-hand side of this equation is a multiple of 8, so the right-hand side must also be a multiple of 8, which only happens when \( x_6 \equiv 0 \text{ or } 5 \pmod 8 \). This still leaves infinitely many possibilities for the pair \( (x_5, x_6) \), starting with (2, 5), (11, 8), (31, 13), etc., but the only possibility that makes triangle \( ACX \) valid is (2, 5).

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By Power of Point, we have \( 12^2 = 8x_8 \), which we can solve to get \( x_8 = 18 \). Then, the diameter \( BC \) is 10, so the radius \( x_7 \) is 5.

For this problem, there is an alternative solution that does not use Power of Point. Instead, we can apply the Pythagorean Theorem on triangle \( AOX \) to solve for the radius \( x_7 \) first.