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.

### Page 3

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 $$x_4 = 20$$.

### Page 4

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).

### Page 5

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.