Difference between revisions of "1992 AIME Problems/Problem 14"
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== Solution 1== | == 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 | + | 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> | |
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== Solution 2 == | == Solution 2 == |
Revision as of 22:04, 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 Thus, we are given Combining and expanding gives We desire Expanding this 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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