Difference between revisions of "2012 AIME II Problems/Problem 13"

(Created page with "== Problem 13 == Equilateral <math>\triangle ABC</math> has side length <math>\sqrt{111}</math>. There are four distinct triangles <math>AD_1E_1</math>, <math>AD_1E_2</math>, <ma...")
 
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Equilateral <math>\triangle ABC</math> has side length <math>\sqrt{111}</math>. There are four distinct triangles <math>AD_1E_1</math>, <math>AD_1E_2</math>, <math>AD_2E_3</math>, and <math>AD_2E_4</math>, each congruent to <math>\triangle ABC</math>,
 
Equilateral <math>\triangle ABC</math> has side length <math>\sqrt{111}</math>. There are four distinct triangles <math>AD_1E_1</math>, <math>AD_1E_2</math>, <math>AD_2E_3</math>, and <math>AD_2E_4</math>, each congruent to <math>\triangle ABC</math>,
 
with <math>BD_1 = BD_2 = \sqrt{11}</math>. Find <math>\sum_{k=1}^4(CE_k)^2</math>.
 
with <math>BD_1 = BD_2 = \sqrt{11}</math>. Find <math>\sum_{k=1}^4(CE_k)^2</math>.
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== Solution ==
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== See also ==
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{{AIME box|year=2012|n=II|num-b=12|num-a=14}}

Revision as of 17:22, 31 March 2012

Problem 13

Equilateral $\triangle ABC$ has side length $\sqrt{111}$. There are four distinct triangles $AD_1E_1$, $AD_1E_2$, $AD_2E_3$, and $AD_2E_4$, each congruent to $\triangle ABC$, with $BD_1 = BD_2 = \sqrt{11}$. Find $\sum_{k=1}^4(CE_k)^2$.


Solution

See also

2012 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
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