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

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== Problem ==
 
== Problem ==
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In triangle <math>ABC^{}_{}</math>, <math>\displaystyle A'</math>, <math>\displaystyle B'</math>, and <math>\displaystyle C'</math> are on the sides <math>\displaystyle BC</math>, <math>AC^{}_{}</math>, and <math>AB^{}_{}</math>, respectively. Given that <math>\displaystyle AA'</math>, <math>\displaystyle BB'</math>, and <math>\displaystyle CC'</math> are concurrent at the point <math>O^{}_{}</math>, and that <math>\frac{AO^{}_{}}{OA'}+\frac{BO}{OB'}+\frac{CO}{OC'}=92</math>, find <math>\frac{AO}{OA'}\cdot \frac{BO}{OB'}\cdot \frac{CO}{OC'}</math>.
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In triangle <math>ABC^{}_{}</math>, <math>A'</math>, <math>B'</math>, and <math>C'</math> are on the sides <math>BC</math>, <math>AC^{}_{}</math>, and <math>AB^{}_{}</math>, respectively. Given that <math>AA'</math>, <math>BB'</math>, and <math>CC'</math> are concurrent at the point <math>O^{}_{}</math>, and that <math>\frac{AO^{}_{}}{OA'}+\frac{BO}{OB'}+\frac{CO}{OC'}=92</math>, find <math>\frac{AO}{OA'}\cdot \frac{BO}{OB'}\cdot \frac{CO}{OC'}</math>.
  
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== Solution ==
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== Solution 1==
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{{solution}}
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Let <math>K_A=[BOC], K_B=[COA],</math> and <math>K_C=[AOB].</math>  Due to triangles <math>BOC</math> and <math>ABC</math> having the same base,  <cmath>\frac{AO}{OA'}+1=\frac{AA'}{OA'}=\frac{[ABC]}{[BOC]}=\frac{K_A+K_B+K_C}{K_A}.</cmath>  Therefore, we have
 +
<cmath>\frac{AO}{OA'}=\frac{K_B+K_C}{K_A}</cmath> <cmath>\frac{BO}{OB'}=\frac{K_A+K_C}{K_B}</cmath> <cmath>\frac{CO}{OC'}=\frac{K_A+K_B}{K_C}.</cmath>
 +
Thus, we are given <cmath>\frac{K_B+K_C}{K_A}+\frac{K_A+K_C}{K_B}+\frac{K_A+K_B}{K_C}=92.</cmath> Combining and expanding gives <cmath>\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.</cmath> We desire <math>\frac{(K_B+K_C)(K_C+K_A)(K_A+K_B)}{K_AK_BK_C}.</math> Expanding this gives <cmath>\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}.</cmath>
 +
 
 +
== Solution 2 ==
 +
Using [[mass points]], let the weights of <math>A</math>, <math>B</math>, and <math>C</math> be <math>a</math>, <math>b</math>, and <math>c</math> respectively.
 +
 
 +
Then, the weights of <math>A'</math>, <math>B'</math>, and <math>C'</math> are <math>b+c</math>, <math>c+a</math>, and <math>a+b</math> respectively.
 +
 
 +
Thus, <math>\frac{AO^{}_{}}{OA'} = \frac{b+c}{a}</math>, <math>\frac{BO^{}_{}}{OB'} = \frac{c+a}{b}</math>, and <math>\frac{CO^{}_{}}{OC'} = \frac{a+b}{c}</math>.
 +
 
 +
Therefore:
 +
<math>\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}</math> <math>= \frac{2abc+b^2c+bc^2+c^2a+ca^2+a^2b+ab^2}{abc} =</math>
 +
 
 +
<math>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} </math> <math>= 2 + \frac{AO^{}_{}}{OA'}+\frac{BO}{OB'}+\frac{CO}{OC'} = 2+92 = \boxed{094}</math>.
 +
 
 +
== Solution 3 ==
 +
 
 +
As in above solutions, find <math>\sum_{cyc} \frac{y+z}{x}=92</math> (where <math>O=(x:y:z)</math> in barycentric coordinates). Now letting <math>y=z=1</math> we get <math>\frac{2}{x}+2(x+1)=92 \implies x+\frac{1}{x}=45</math>, and so <math>\frac{2}{x}(x+1)^2=2(x+\frac{1}{x}+2)=2 \cdot 47 = 94</math>.
 +
 
 +
~Lcz
 +
 
 +
== 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:
 +
 
 +
<cmath>\begin{align*}
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\frac{AO}{OA'}\frac{BO}{OB'}\frac{CO}{OC'} &=
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\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}}_{\text{Ceva}} +
 +
\underbrace{\frac{AC'\cdot BA'\cdot CB'}{C'B\cdot A'C\cdot B'A}}_{\text{Ceva}} +
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\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*}</cmath>
 +
 
 +
~ proloto
  
 
== See also ==
 
== See also ==
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* [[1992 AIME Problems/Problem 13 | Previous Problem]]
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{{AIME box|year=1992|num-b=13|num-a=15}}
−
 
 
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* [[1992 AIME Problems/Problem 15 | Next Problem]]
 
  
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* [[1992 AIME Problems]]
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[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 00:26, 16 August 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:

\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 (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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All AIME Problems and Solutions

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