# 1992 AIME Problems/Problem 14

## Problem

In triangle $ABC^{}_{}$, $A'$, $B'$, and $C'$ are on the sides $BC$, $AC^{}_{}$, and $AB^{}_{}$, respectively. Given that $AA'$, $BB'$, and $CC'$ are concurrent at the point $O^{}_{}$, and that $\frac{AO^{}_{}}{OA'}+\frac{BO}{OB'}+\frac{CO}{OC'}=92$, find $\frac{AO}{OA'}\cdot \frac{BO}{OB'}\cdot \frac{CO}{OC'}$.

## Solution 1

Let $K_A=[BOC], K_B=[COA],$ and $K_C=[AOB].$ Due to triangles $BOC$ and $ABC$ having the same base, $$\frac{AO}{OA'}+1=\frac{AA'}{OA'}=\frac{[ABC]}{[BOC]}=\frac{K_A+K_B+K_C}{K_A}.$$ Therefore, we have $$\frac{AO}{OA'}=\frac{K_B+K_C}{K_A}$$ $$\frac{BO}{OB'}=\frac{K_A+K_C}{K_B}$$ $$\frac{CO}{OC'}=\frac{K_A+K_B}{K_C}.$$ Thus, we are given $$\frac{K_B+K_C}{K_A}+\frac{K_A+K_C}{K_B}+\frac{K_A+K_B}{K_C}=92.$$ Combining and expanding gives $$\frac{K_A^2K_B+K_AK_B^2+K_A^2K_C+K_AK_C^2+K_B^2K_C+K_BK_C^2}{K_AK_BK_C}=92.$$ We desire $\frac{(K_B+K_C)(K_C+K_A)(K_A+K_B)}{K_AK_BK_C}.$ Expanding this gives $$\frac{K_A^2K_B+K_AK_B^2+K_A^2K_C+K_AK_C^2+K_B^2K_C+K_BK_C^2}{K_AK_BK_C}+2=\boxed{094}.$$

## Solution 2

Using mass points, let the weights of $A$, $B$, and $C$ be $a$, $b$, and $c$ respectively.

Then, the weights of $A'$, $B'$, and $C'$ are $b+c$, $c+a$, and $a+b$ respectively.

Thus, $\frac{AO^{}_{}}{OA'} = \frac{b+c}{a}$, $\frac{BO^{}_{}}{OB'} = \frac{c+a}{b}$, and $\frac{CO^{}_{}}{OC'} = \frac{a+b}{c}$.

Therefore: $\frac{AO}{OA'}\cdot \frac{BO}{OB'}\cdot \frac{CO}{OC'} = \frac{b+c}{a} \cdot \frac{c+a}{b} \cdot \frac{a+b}{c}$ $= \frac{2abc+b^2c+bc^2+c^2a+ca^2+a^2b+ab^2}{abc} =$ $2+\frac{bc(b+c)}{abc}+\frac{ca(c+a)}{abc}+\frac{ab(a+b)}{abc} = 2 + \frac{b+c}{a} + \frac{c+a}{b} + \frac{a+b}{c}$ $= 2 + \frac{AO^{}_{}}{OA'}+\frac{BO}{OB'}+\frac{CO}{OC'} = 2+92 = \boxed{094}$.

## Solution 3

As in above solutions, find $\sum_{cyc} \frac{y+z}{x}=92$ (where $O=(x:y:z)$ in barycentric coordinates). Now letting $y=z=1$ we get $\frac{2}{x}+2(x+1)=92 \implies x+\frac{1}{x}=45$, and so $\frac{2}{x}(x+1)^2=2(x+\frac{1}{x}+2)=2 \cdot 47 = 94$.

~Lcz

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