Difference between revisions of "2011 AIME I Problems/Problem 2"
(Created page with '== Problem == In rectangle <math>ABCD</math>, <math>AB=12</math> and <math>BC=10</math>. Points <math>E</math> and <math>F</math> lie inside rectangle <math>ABCD</math> so that …') |
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== Solution == | == Solution == | ||
− | Let us call the point where <math>\overline{EF}</math> intersects <math>\overline{AD}</math> point <math>G</math>, and the point where <math>\overline{EF}</math> intersects <math>\overline{BC}</math> point <math>H</math>. Since angles <math>FHB</math> and <math>EGA</math> are both right angles, and angles <math>BEF</math> and <math>DFE</math> are congruent due to parallelism, right triangles <math>BHE</math> and <math> | + | Let us call the point where <math>\overline{EF}</math> intersects <math>\overline{AD}</math> point <math>G</math>, and the point where <math>\overline{EF}</math> intersects <math>\overline{BC}</math> point <math>H</math>. Since angles <math>FHB</math> and <math>EGA</math> are both right angles, and angles <math>BEF</math> and <math>DFE</math> are congruent due to parallelism, right triangles <math>BHE</math> and <math>DGF</math> are similar. This implies that <math>\frac{BH}{GD} = \frac{9}{8}</math>. Since <math>BC=10</math>, <math>BH+GD=BH+HC=BC=10</math>. (<math>HC</math> is the same as <math>GD</math> because they are opposite sides of a rectangle.) Now, we have a system: |
<math>\frac{BH}{GD}=\frac{9}8</math> | <math>\frac{BH}{GD}=\frac{9}8</math> |
Revision as of 02:56, 29 March 2011
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
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: