# 2011 AIME I Problems/Problem 8

## Problem

In triangle $ABC$, $BC = 23$, $CA = 27$, and $AB = 30$. Points $V$ and $W$ are on $\overline{AC}$ with $V$ on $\overline{AW}$, points $X$ and $Y$ are on $\overline{BC}$ with $X$ on $\overline{CY}$, and points $Z$ and $U$ are on $\overline{AB}$ with $Z$ on $\overline{BU}$. In addition, the points are positioned so that $\overline{UV}\parallel\overline{BC}$, $\overline{WX}\parallel\overline{AB}$, and $\overline{YZ}\parallel\overline{CA}$. Right angle folds are then made along $\overline{UV}$, $\overline{WX}$, and $\overline{YZ}$. The resulting figure is placed on a level floor to make a table with triangular legs. Let $h$ be the maximum possible height of a table constructed from triangle $ABC$ whose top is parallel to the floor. Then $h$ can be written in the form $\frac{k\sqrt{m}}{n}$, where $k$ and $n$ are relatively prime positive integers and $m$ is a positive integer that is not divisible by the square of any prime. Find $k+m+n$. $[asy] unitsize(1 cm); pair translate; pair[] A, B, C, U, V, W, X, Y, Z; A = (1.5,2.8); B = (3.2,0); C = (0,0); U = (0.69*A + 0.31*B); V = (0.69*A + 0.31*C); W = (0.69*C + 0.31*A); X = (0.69*C + 0.31*B); Y = (0.69*B + 0.31*C); Z = (0.69*B + 0.31*A); translate = (7,0); A = (1.3,1.1) + translate; B = (2.4,-0.7) + translate; C = (0.6,-0.7) + translate; U = U + translate; V = V + translate; W = W + translate; X = X + translate; Y = Y + translate; Z = Z + translate; draw (A--B--C--cycle); draw (U--V,dashed); draw (W--X,dashed); draw (Y--Z,dashed); draw (U--V--W--X--Y--Z--cycle); draw (U--A--V,dashed); draw (W--C--X); draw (Y--B--Z); dot("A",A,N); dot("B",B,SE); dot("C",C,SW); dot("U",U,NE); dot("V",V,NW); dot("W",W,NW); dot("X",X,S); dot("Y",Y,S); dot("Z",Z,NE); dot(A); dot(B); dot(C); dot("U",U,NE); dot("V",V,NW); dot("W",W,NW); dot("X",X,dir(-70)); dot("Y",Y,dir(250)); dot("Z",Z,NE);[/asy]$

## Solution

Note that the area is given by Heron's formula and it is $20\sqrt{221}$. Let $h_i$ denote the length of the altitude dropped from vertex i. It follows that $h_b = \frac{40\sqrt{221}}{27}, h_c = \frac{40\sqrt{221}}{30}, h_a = \frac{40\sqrt{221}}{23}$. From similar triangles we can see that $\frac{27h}{h_a}+\frac{27h}{h_c} \le 27 \rightarrow h \le \frac{h_ah_c}{h_a+h_c}$. We can see this is true for any combination of a,b,c and thus the minimum of the upper bounds for h yields $h = \frac{40\sqrt{221}}{57} \rightarrow \boxed{318}$.

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