Difference between revisions of "2011 AIME I Problems/Problem 2"
(→Solution 2) |
(→Solution 2) |
||
Line 63: | Line 63: | ||
<math>HB=\sqrt{17^2-10^2}=3\sqrt{21}</math> | <math>HB=\sqrt{17^2-10^2}=3\sqrt{21}</math> | ||
− | <math>EF | + | <math>HA=EF=3\sqrt{21}-12</math> |
Thus our answer is: | Thus our answer is: |
Revision as of 17:30, 15 March 2015
Contents
Problem
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 .
Solution 1
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:
and
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:
Solution 2
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:
See also
2011 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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.