Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2007 AIME I Problems/Problem 12

Revision as of 14:52, 16 March 2007 by Azjps (talk | contribs) (→‎Solution 2: got it, complete solution)

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

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

Solution 1

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 $\frac{500 + 350 + 300 + 600}2 = \frac{1750}2 = 875$.

Solution 2

Redefine the points in the same manner as the last time ($\displaystyle \triangle AB'C'$, intersect at $D$, $E$, and $F$). This time, notice that $[ADEF] = [\triangle AB'C'] - ([\triangle ADC'] + [\triangle EFB'])$.

The area of $[\triangle AB'C'] = [\triangle ABC]$. The altitude of $\triangle ABC$ is clearly $10 \tan 75 = 10 \tan (30 + 45)$. The tangent addition rule yields $10(2 + \sqrt{3})$ (see above). Thus, $\displaystyle [\triangle ABC] = \frac{1}{2} 20 \cdot (20 + 10\sqrt{3}) = 200 + 100\sqrt{3}$.

The area of $[\triangle ADC']$ (with a side on the y-axis) can be found by splitting it into two triangles, $30-60-90$ and $15-75-90$ right triangles. $AC' = AC = \frac{10}{\sin 75}$. The sine subtraction rule shows that $\frac{10}{\sin 15} = \frac{10}{\frac{\sqrt{6} - \sqrt{2}}4} = \frac{40}{\sqrt{6} - \sqrt{2}} = 10(\sqrt{6} + \sqrt{2})$. $AC'$, in terms of the height of $\triangle ADC'$, is equal to $h(\sqrt{3} + \tan 75) = h(\sqrt{3} + 2 + \sqrt{3})$. $[ADC'] = \frac 12 AC' \cdot h = \frac 12 (10\sqrt{6} + 10\sqrt{2})(\frac{10\sqrt{6} + 10\sqrt{2}}{2\sqrt{3} + 2}$$= \frac{(800 + 400\sqrt{3})}{(2 + \sqrt{3})}\cdot\frac{2 - \sqrt{3}}{2-\sqrt{3}}$$= \frac{400\sqrt{3} + 400}8 = 50\sqrt{3} + 50$.

The area of $[\triangle EFB']$ was found in the previous solution to be $- 500\sqrt{2} + 400\sqrt{3} - 300\sqrt{6} +750$.

Therefore, $[ADEF] = (200 + 100\sqrt{3}) - \left((50 + 50\sqrt{3}) + (-500\sqrt{2} + 400\sqrt{3} - 300\sqrt{6} +750)\right) = 500\sqrt{2} - 350\sqrt{3} + 300\sqrt{6} - 600$, and our answer is $875$.

See also

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
All AIME Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_12&oldid=14082"