Difference between revisions of "2011 AIME I Problems/Problem 2"
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<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 16:30, 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. 
