2011 AIME I Problems/Problem 2
In rectangle , and . Points and lie inside rectangle so that ,,,, and line intersects segment . The length can be expressed in the form , where ,, and are positive integers and is not divisible by the square of any prime. Find .
Let us call the point where intersects point , and the point where intersects point . Since angles and are both right angles, and angles and are congruent due to parallelism, right triangles and are similar. This implies that . Since , . ( is the same as because they are opposite sides of a rectangle.) Now, we have a system:
Solving this system (easiest by substitution), we get that:
Using the Pythagorean Theorem, we can solve for the remaining sides of the two right triangles:
Notice that adding these two sides would give us twelve plus the overlap . This means that:
Since isn't divisible by any perfect square, our answer is:
Extend lines and to meet at point . Draw the altitude from point to line extended.
In right , , , thus by Pythagoras Theorem we have:
Thus our answer is:
|2011 AIME I (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|
|All AIME Problems and Solutions|