# Difference between revisions of "2007 AIME I Problems/Problem 12"

## Problem

In isosceles triangle $\triangle ABC$, $A$ is located at the origin and $B$ is located at (20,0). Point $C$ is in the first quadrant with $\displaystyle AC = BC$ and angle $BAC = 75^{\circ}$. If triangle $ABC$ is rotated counterclockwise about point $A$ until the image of $C$ lies on the positive $y$-axis, the area of the region common to the original and the rotated triangle is in the form $p\sqrt{2} + q\sqrt{3} + r\sqrt{6} + s$, where $\displaystyle p,q,r,s$ are integers. Find $\frac{p-q+r-s}2$.

## Solution

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Call the vertices of the new triangle $AB'C'$ ( $A$, the origin, is a vertex of both triangles). $B'C'$ and $AB$ intersect at a single point, $D$. $BC$ intersect at two points; the one with the higher y-coordinate will be $E$, and the other $F$. The intersection of the two triangles is a quadrilateral $ADEF$. Notice that we can find this area by subtracting $[\triangle ADB'] - [\triangle EFB']$.

Since $\displaystyle \angle B'AC'$ and $\displaystyle \angle BAC$ both have measures $75^{\circ}$, both of their complements are $15^{\circ}$, and $\angle DAC' = 90 - 2(15) = 60^{\circ}$. We know that $C'B'A = 75^{\circ}$, and since the angles of a triangle add up to $180^{\circ}$, we find that $ADB' = 180 - 60 - 75 = 45^{\circ}$.

So $ADB'$ is a $45 - 60 - 75 \triangle$. It can be solved by drawing an altitude splitting the $75^{\circ}$ angle into $30^{\circ}$ and $45^{\circ}$ angles – this forms a $\displaystyle 30-60-90$ right triangle and a $\displaystyle 45-45-90$ isosceles right triangle. Since we know that $DB' = 20$, the base of the $\displaystyle 30-60-90$ triangle is $10$, the height is $10\sqrt{3}$, and the base of the $\displaystyle 45-45-90$ is $10\sqrt{3}$. Thus, the total area of $[\triangle ADB'] = \frac{1}{2}(10\sqrt{3})(10\sqrt{3} + 10) = 150 + 50\sqrt{3}$.

Now, we need to find $[\triangle EFB']$, which is a $\displaystyle 15-75-90$ right triangle. We can find its base by subtracting $AF$ from $20$. $\triangle AFB$ is also a $\displaystyle 15-75-90$ triangle, so we find that $AF = 20\sin 75 = 20 \sin (30 + 45) = 20\frac{\sqrt{2} + \sqrt{6}}4 = 5\sqrt{2} + 5\sqrt{6}$. $FB' = 20 - AF = 20 - 5\sqrt{2} - 5\sqrt{6}$.

To solve $[\triangle EFB']$, note that $\displaystyle [\triangle EFB'] = \frac{1}{2} FB' \cdot EF = \frac{1}{2} FB' \cdot (\tan 75 FB')$. Through algebra, we can calculate $(FB')^2 \cdot \tan 75$: $\displaystyle \frac{1}{2}\tan (30 + 45) \cdot (20 - 5\sqrt{2} - 5\sqrt{6})^2$ $\displaystyle = \frac{1}{2} \left(\frac{\frac{1}{\sqrt{3}} + 1}{1 - \frac{1}{\sqrt{3}}}\right) [400 - 100\sqrt{2} - 100\sqrt{6} - 100\sqrt{2} + 50 + 50\sqrt{3} - 100\sqrt{6} + 50\sqrt{3} + 150]$ $= \frac{1}{2}(2 + \sqrt{3})[600 - 200\sqrt{2} - 200\sqrt{6} + 100\sqrt{3}]$ $=- 500\sqrt{2} + 400\sqrt{3} - 300\sqrt{6} +750$

To finish, find $[ADEF] = [\triangle ADB'] - [\triangle EFB']$ $= (150 + 50\sqrt{3}) - (-500\sqrt{2} + 400\sqrt{3} - 300\sqrt{6} + 750)$ $=500\sqrt{2} - 350\sqrt{3} + 300\sqrt{6} - 600$. The solution is $\frac{500 + 350 + 300 + 600}2 = \frac{1750}2 = 875$.