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. | ||
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~Lcz | ~Lcz | ||
+ | |||
+ | == Solution 4 (Ceva's Theorem) == | ||
+ | |||
+ | 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{BC'}{C'A}+\frac{BA'}{A'C}\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}}_{\text{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}}_{\text{Gergonne}} + | ||
+ | \underbrace{\frac{BA'}{A'C} + \frac{BC'}{C'A}}_{\text{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} | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | ~ proloto | ||
== See also == | == See also == |
Latest revision as of 17:17, 20 January 2025
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 (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 |
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