2014 AIME II Problems/Problem 11
Contents
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 .
Solution
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 .
Solution 2
Let Meanwhile, because 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
Finally, what is ? It comes out to .
We got the three sides. Now all that is left is using the Law of Cosines. There we can equate and solve for it.
Taking and using , of course, we find out (after some calculation) that . The step before? .
See also
2014 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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