# 2008 iTest Problems/Problem 99

## Problem

Given a convex, $n$-sided polygon $P$, form a $2n$-sided polygon $\text{clip}(P)$ by cutting off each corner of $P$ at the edges’ trisection points. In other words, $\text{clip}(P)$ is the polygon whose vertices are the $2n$ edge trisection points of $P$, connected in order around the boundary of $P$. Let $P_1$ be an isosceles trapezoid with side lengths $13, 13, 13$, and $3$, and for each $i \geq 2$, let $P_i = \text{clip}(P_{i-1})$. This iterative clipping process approaches a limiting shape $P_\infty = \lim_{i \rightarrow \infty} P_i$. If the difference of the areas of $P_{10}$ and $P_{\infty}$ is written as a fraction $\frac{x}{y}$ in lowest terms, calculate the number of positive integer factors of $x \cdot y$.

## Solution

Let $D_n$ be the difference in the areas between $P_n$ and $P_{n+1}$. Let our trapezoid be $P_1 = ABCD$ (and $[ABCD] = \frac{12(3+13)}{2} = 96$); then without loss of generality construct diagonal $BD$.

$[asy] pathpen = linewidth(0.7);pen d = linetype("4 4")+linewidth(0.7); pair A=(0,0),B=(5,12),C=(8,12),D=(13,0),A1=B/3,A2=D/3; D(MP("A",A)--MP("B",B,N)--MP("C",C,N)--MP("D",D)--cycle); D(B--D); D(MP("A_1",A1,NW) -- MP("A_2",A2)); D(D+2(C-D)/3 -- B+2(C-B)/3); D(4B/9 -- (A+D)/3+2*((A+B)/3 - (A+D)/3)/3, d); D(4D/9 -- (A+D)/3 + ((A+B)/3 - (A+D)/3)/3, d); [/asy]$

Let $A_1, A_2$ be the trisection points on $\overline{AB},\overline{AD}$, respectively, that are closest to $A$. Then the operation $\text{clip}(P)$ deletes $\triangle A_1AA_2$. Since $A_1A/AB = 1/3, A_2A/AD = 1/3$, and $\triangle A_1AA_2, \triangle BAD$ share common $\angle A$, we have $\triangle A_1AA_2 \sim \triangle BAD$ by side ratio $1/3$. Their areas are in the ratio $(1/3)^2 = 1/9$.

Similarly, $[C_1CC_2] = \frac{1}{9}[BCD]$, and $[A_1AA_2] + [C_1CC_2] = \frac{1}{9}[ABCD]$. Cutting along diagonal $AC$, we get the same result, so $D_1 = \frac{2}{9}P_1$.

We now consider the effects of the second clipping. Without loss of generality consider what happens along the vertex $A_1$ of $P_2$. Let $A_{11}$ be the trisection point along $\overline{AB}$ (again closest to $A_1$), and $A_{12}$ be the trisection point along $\overline{A_1A_2}$. Now $\frac{A_{1}A_{11}}{AA_{1}} = \frac{(AB/3)/3}{AB/3} = \frac{1}{3}$ and $\frac{A_{1}A_{12}}{A_{1}A_{2}} = \frac{1}{3}$, and $\angle AA_1A_2 = 180 - \angle A_{11}A_1A_{12} \Longrightarrow \sin(\angle AA_1A_2) = \sin(\angle A_{11}A_1A_{12})$. Using the $\frac{1}{2}ab\sin C$ definition of the area of a triangle, we see that $[A_1A_{11}A_{12}] = \frac{1}{9}[AA_1A_2]$. A similar clipping about $A_2$ gives $[A_2A_{21}A_{22}] = \frac{1}{9}[AA_1A_2]$; around each clipped region in $D_1$, we clip a new area $2/9D_1$. Generalizing, we have the recursion $D_n = \frac 29 \cdot D_{n-1}$.

Then, $P_n = P_1 - D_1 - D_2 - \cdots - D_{n-1} = 96 - 96\left(\left(\frac 29\right) + \left(\frac 29\right)^2 + \cdots + \left(\frac 29\right)^{n-1}\right)$. Hence,

$P_{10} - P_{\infty} = 96\left(\left(\frac 29\right)^{10} + \left(\frac 29\right)^{11} + \cdots \right) = \left(\frac 29\right)^{10}\left(\frac{1}{1-2/9}\right) = \frac{2^{15}}{3^{17} \cdot 7}.$

Then $xy = 2^{15} \cdot 3^{17} \cdot 7$ has $16 \cdot 18 \cdot 2 = \boxed{576}$ factors.