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 $\displaystyle AB'C'$ ( $A$ , the origin, is a vertex of both triangles). $\displaystyle B'C'$ and $\displaystyle AB$ intersect at a single point, $D$ . $\displaystyle BC$ 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 $\displaystyle ADEF$ . 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 $\displaystyle ADB'$ 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 $\displaystyle DB' = 20$ , 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 $(FB')^2 \cdot \tan 75$ :

$\displaystyle \frac{1}{2}\tan (30 + 45) \cdot (20 - 5\sqrt{2} - 5\sqrt{6})^2$

$\displaystyle = \frac{1}{2} \left(\frac{\frac{1}{\sqrt{3}} + 1}{1 - \frac{1}{\sqrt{3}}}\right) \left[400 - 100\sqrt{2} - 100\sqrt{6} - 100\sqrt{2} + 50 + 50\sqrt{3} - 100\sqrt{6} + 50\sqrt{3} + 150\right]$

$= \frac{1}{2}(2 + \sqrt{3})[600 - 200\sqrt{2} - 200\sqrt{6} + 100\sqrt{3}]$

$=- 500\sqrt{2} + 400\sqrt{3} - 300\sqrt{6} +750$

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

2007 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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2007 AIME I Problems/Problem 12

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