Difference between revisions of "1992 AIME Problems/Problem 14"
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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> | 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 == | + | == Solution 2 (Mass Points) == |
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. | ||
Latest revision as of 17:17, 20 January 2025
Contents
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 (Mass Points)
Using mass points, let the weights of , , and be , , and respectively.
Then, the weights of , , and are , , and respectively.
Thus, , , and .
Therefore:
.
Solution 3
As in above solutions, find (where in barycentric coordinates). Now letting we get , and so .
~Lcz
Solution 4 (Ceva's Theorem)
A consequence of Ceva's theorem sometimes attributed to Gergonne is that , and similarly for cevians and . Now we apply Gergonne several times and do algebra:
~ 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.