1992 AIME Problem 14

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

Let

(\alpha,\beta,\gamma)

be normalized barycentric coordinates of

O

Write the three ratios as

\frac{1-\alpha}{\alpha}

,

\frac{1-\beta}{\beta}

, and

\frac{1-\gamma}{\gamma}

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.

Problem 14 in other years