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

## 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

Note that there are only two possible locations for points $D_1$ and $D_2$, as they are both $\sqrt{111}$ from point $A$ and $\sqrt{11}$ from point $B$, so they are the two points where a circle centered at $A$ with radius $\sqrt{111}$ and a circle centered at $B$ with radius $\sqrt{11}$ intersect. Let $D_1$ be the point on the opposite side of $\overline{AB}$ from $C$, and $D_2$ the point on the same side of $\overline{AB}$ as $C$.

Let $\theta$ be the measure of angle $BAD_1$ (which is also the measure of angle $BAD_2$); by the Law of Cosines, $$\sqrt{11}^2 = \sqrt{111}^2 + \sqrt{111}^2 - 2 \cdot \sqrt{111} \cdot \sqrt{111} \cdot cos\;\theta$$ $$11 = 222\;(1 - cos\;\theta)$$

There are two equilateral triangles with $\overline{AD_1}$ as a side; let $E_1$ be the third vertex that is farthest from $C$, and $E_2$ be the third vertex that is nearest to $C$.

Angle $E_1AC = E_1AD_1 + D_1AB + BAC = 60 + \theta + 60 = 120 + \theta$; by the Law of Cosines, $$(E_1C)^2 = (E_1A)^2 + (AC)^2 - 2 (E_1A) (E_1C)\;cos\;(120 + \theta)$$ $$= 111 + 111 - 222\;cos\;(120 + \theta)$$ Angle $E_2AC = \theta$; by the Law of Cosines, $$(E_2C)^2 = (E_2A)^2 + (AC)^2 - 2 (E_2A) (E_2C)\;cos\;\theta = 111 + 111 - 222\,cos\;\theta$$

There are two equilateral triangles with $\overline{AD_2}$ as a side; let $E_3$ be the third vertex that is farthest from $C$, and $E_4$ be the third vertex that is nearest to $C$.

Angle $E_3AC = E_3AB + BAC = (60 - \theta) + 60 = 120 - \theta$; by the Law of Cosines, $$(E_3C)^2 = (E_3A)^2 + (AC)^2 - 2 (E_3A) (E_3C)\;cos\;(120 - \theta)$$ $$= 111 + 111 - 222\;cos\;(120 - \theta)$$ Angle $E_4AC = \theta$; by the Law of Cosines, $$(E_4C)^2 = (E_4A)^2 + (AC)^2 - 2 (E_4A) (E_4C)\;cos\;\theta = 111 + 111 - 222\;cos\;\theta$$

The solution is: $$(E_1C)^2 + (E_3C)^2 + (E_2C)^2 + (E_4C)^2$$ $$= 222\;(1 - cos\;(120 + \theta)) + 222\;(1 - cos\;(120 - \theta)) + 222\;(1 - cos\;\theta) + 222\;(1 - cos\;\theta)$$ $$= 222\;((1 - (cos\;120\;cos\;\theta - sin\;120\;sin\;\theta)) + (1 - (cos\;120\;cos\;\theta + sin\;120\;sin\;\theta)) + 2\;(1 -\;cos\;\theta))$$ $$= 222\;(1 - cos\;120\;cos\;\theta + sin\;120\;sin\;\theta + 1 - cos\;120\;cos\;\theta - sin\;120\;sin\;\theta + 2 - 2\;cos\;\theta)$$ $$= 222\;(1 + \frac{1}{2}\;cos\;\theta + 1 + \frac{1}{2}\;cos\;\theta + 2 - 2\;cos\;\theta)$$ $$= 222\;(4 - cos\;\theta)$$ $$= 666 + 222\;(1 - cos\;\theta)$$ Substituting $11$ for $222\;(1 - cos\;\theta)$ gives the solution $666 + 11 = \framebox{677}.$