Difference between revisions of "1992 AIME Problems/Problem 14"
(A different solution involving Ceva's Theorem explicitly) |
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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: | 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*} |
+ | \frac{AO}{OA'}\frac{BO}{OB'}\frac{CO}{OC'} &= | ||
\left(\frac{AB'}{B'C}+\frac{AC'}{C'B}\right) | \left(\frac{AB'}{B'C}+\frac{AC'}{C'B}\right) | ||
\left(\frac{BC'}{C'A}+\frac{BA'}{A'C}\right) | \left(\frac{BC'}{C'A}+\frac{BA'}{A'C}\right) | ||
− | \left(\frac{CB'}{B'A}+\frac{CA'}{A'B}\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}}_{Ceva} + | + | &=\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}}_{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}}_{Gergonne} + | + | \underbrace{\frac{AB'}{B'C} + \frac{AC'}{C'B}}_{\text{Gergonne}} + |
− | \underbrace{\frac{BA'}{A'C} + \frac{BC'}{C'A}}_{Gergonne} + | + | \underbrace{\frac{BA'}{A'C} + \frac{BC'}{C'A}}_{\text{Gergonne}} + |
− | \underbrace{\frac{CA'}{A'B} + \frac{CB'}{B'A}}_{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}</ | + | &= 1 + 1 + \underbrace{\frac{AO}{OA'} + \frac{BO}{OB'} + \frac{CO}{OC'}}_{92} = \boxed{94} |
+ | \end{align*}</cmath> | ||
~ proloto | ~ proloto |
Latest revision as of 00:26, 16 August 2023
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
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 |
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