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Difference between revisions of "1992 AIME Problems/Problem 14"

(→‎Solution 3)
(A different solution involving Ceva's Theorem explicitly)
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~Lcz
 
~Lcz
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== Solution 4 (Ceva's Theorem) ==
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 +
A consequence of Ceva's theorem sometimes attributed to Gergonne is that <math>\frac{AO}{OA'}=\frac{AC'}{C'B}+\frac{AB'}{B'C}</math>, and similarly for cevians <math>BB'</math> and <math>CC'</math>.  Now we apply Gergonne several times and do algebra:
 +
 +
<math>\frac{AO}{OA'}\frac{BO}{OB'}\frac{CO}{OC'} =
 +
\left(\frac{AB'}{B'C}+\frac{AC'}{C'B}\right)
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\left(\frac{BC'}{C'A}+\frac{BA'}{A'C}\right)
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\left(\frac{CB'}{B'A}+\frac{CA'}{A'B}\right) = \\
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\underbrace{\frac{AB'\cdot CA'\cdot BC'}{B'C\cdot A'B\cdot C'A}}_{Ceva} +
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\underbrace{\frac{AC'\cdot BA'\cdot CB'}{C'B\cdot A'C\cdot B'A}}_{Ceva} +
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\underbrace{\frac{AB'}{B'C} + \frac{AC'}{C'B}}_{Gergonne} +
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\underbrace{\frac{BA'}{A'C} + \frac{BC'}{C'A}}_{Gergonne} +
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\underbrace{\frac{CA'}{A'B} + \frac{CB'}{B'A}}_{Gergonne} = \\
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1 + 1 + \underbrace{\frac{AO}{OA'} + \frac{BO}{OB'} + \frac{CO}{OC'}}_{92} = \boxed{94}</math>
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 +
~ proloto
  
 
== See also ==
 
== See also ==

Revision as of 22:37, 17 July 2023

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

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:

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

~ proloto

See also

1992 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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