Difference between revisions of "1992 AIME Problems/Problem 14"
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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>. | 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>. | ||
− | == Solution == | + | == Solution 1== |
+ | 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 \begin{align*} | ||
+ | \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}. | ||
+ | \end{align*} 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 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> | ||
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+ | == 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. | 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. | ||
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<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>\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>. | + | <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>. |
== See also == | == See also == |
Revision as of 22:01, 5 December 2018
Contents
[hide]Problem
In triangle , , , and are on the sides , , and , respectively. Given that , , and are concurrent at the point , and that , find .
Solution 1
Let and Due to triangles and having the same base, Therefore, we have \begin{align*}
\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}.
\end{align*} Thus, we are given Combining and expanding gives We desire Expanding gives
Solution 2
Using mass points, let the weights of , , and be , , and respectively.
Then, the weights of , , and are , , and respectively.
Thus, , , and .
Therefore:
.
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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