Difference between revisions of "1992 AIME Problems/Problem 14"
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== Solution 3 == | == 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> | + | 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 | ~Lcz |
Revision as of 12:30, 2 December 2020
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:
.
Solution 3
As in above solutions, find (where in barycentric coordinates). Now letting we get , and so .
~Lcz
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.