2014 AIME II Problems/Problem 11
In , and . . Let be the midpoint of segment . Point lies on side such that . Extend segment through to point such that . Then , where and are relatively prime positive integers, and is a positive integer. Find .
Let be the foot of the perpendicular from to , so . Since triangle is isosceles, is the midpoint of , and . Thus, is a parallelogram and . We can then use coordinates. Let be the foot of altitude and set as the origin. Now we notice special right triangles! In particular, and , so , , and midpoint and the slope of , so the slope of Instead of finding the equation of the line, we use the definition of slope: for every to the left, we go up. Thus, , and , so the answer is .
Let Meanwhile, since is similar to (angle, side, and side- and ratio), must be 2. Now, notice that is , because of the parallel segments and .
Now we just have to calculate . Using the Law of Sines, or perhaps using altitude , we get . , which equals
Using Law of Sine in , we find = .
We got the three sides of . Now using the Law of Cosines on . There we can equate and solve for it. We got . Then rationalize the denominator, we get .
Let be the foot of the perpendicular from to , so . Since is isosceles, is the midpoint of , and by midpoint theorem . Thus, is a parallelogram and therefore . We can now use coordinates with as origin and along the -axis.
Let instead of (in the end we will scale down by ). Since , we get , and therefore .
We use sine-law in to find the coordinates : Since slope, and , it follows that slope. If then we have Now .
Scaling down by , we get , so our answer is .
|2014 AIME II (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|
|All AIME Problems and Solutions|
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.