1992 AIME Problems/Problem 14


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 $\frac{2}{x}(x+1)^2=2(x+\frac{1}{x}+2)=2 \cdot 47 = 94$.


Solution 4 (Ceva's Theorem)

A consequence of Ceva's theorem sometimes attributed to Gergonne is that $\frac{AO}{OA'}=\frac{AC'}{C'B}+\frac{AB'}{B'C}$, and similarly for cevians $BB'$ and $CC'$. Now we apply Gergonne several times and do algebra:

\begin{align*} \frac{AO}{OA'}\frac{BO}{OB'}\frac{CO}{OC'} &= \left(\frac{AB'}{B'C}+\frac{AC'}{C'B}\right) \left(\frac{BC'}{C'A}+\frac{BA'}{A'C}\right)  \left(\frac{CB'}{B'A}+\frac{CA'}{A'B}\right)\\ &=\underbrace{\frac{AB'\cdot CA'\cdot BC'}{B'C\cdot A'B\cdot C'A}}_{\text{Ceva}} + \underbrace{\frac{AC'\cdot BA'\cdot CB'}{C'B\cdot A'C\cdot B'A}}_{\text{Ceva}} + \underbrace{\frac{AB'}{B'C} + \frac{AC'}{C'B}}_{\text{Gergonne}} + \underbrace{\frac{BA'}{A'C} + \frac{BC'}{C'A}}_{\text{Gergonne}} + \underbrace{\frac{CA'}{A'B} + \frac{CB'}{B'A}}_{\text{Gergonne}}\\ &= 1 + 1 + \underbrace{\frac{AO}{OA'} + \frac{BO}{OB'} + \frac{CO}{OC'}}_{92} = \boxed{94} \end{align*}

~ proloto

See also

1992 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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