Question

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 $\frac{AO}{OA'}\cdot\frac{BO}{OB'}\cdot\frac{CO}{OC'}.$

Answer

Answer: 94

Concepts:mass points ratio and proportion algebraic manipulation

Difficulty rating: 2350

Small Hint:

Let $(\alpha,\beta,\gamma)$ be normalized barycentric coordinates of $O$

Big Hint:

Write the three ratios as $\frac{1-\alpha}{\alpha}$ , $\frac{1-\beta}{\beta}$ , and $\frac{1-\gamma}{\gamma}$

Solution:

Let $\alpha+\beta+\gamma=1$ be the barycentric coordinates of $O.$ Then $\begin{aligned}x=\frac{AO}{OA'}&=\frac{\beta+\gamma}{\alpha},\\y=\frac{BO}{OB'}&=\frac{\gamma+\alpha}{\beta},\\z=\frac{CO}{OC'}&=\frac{\alpha+\beta}{\gamma}.\end{aligned}$ Expanding both sides using $\alpha+\beta+\gamma=1$ gives the standard identity $xyz=x+y+z+2.$ Since $x+y+z=92,$ the requested product is $94.$

1992 AIME Problem 14

Answer: 94

Difficulty rating: 2350

Small Hint:

Big Hint:

Solution:

Problem 14 in Other Years