Difference between revisions of "2011 AIME I Problems/Problem 2"
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− | == Solution == | + | == Solution 1 == |
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: | 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: | ||
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Since <math>21</math> isn't divisible by any perfect square, our answer is: | Since <math>21</math> isn't divisible by any perfect square, our answer is: | ||
+ | <math>3+21+12=\boxed{36}</math> | ||
+ | |||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | Extend lines <math>BE</math> and <math>CD</math> to meet at point <math>G</math>. | ||
+ | Draw the altitude <math>GH</math> from point <math>G</math> to line <math>BA</math> extended. | ||
+ | |||
+ | <math>GE=DF=8</math> | ||
+ | <math>GB=17</math> | ||
+ | |||
+ | In right <math>\bigtriangleup GHB</math>, <math>GH=10</math>, <math>GB=17</math>, thus by Pythagoras Theorem we have: | ||
+ | <math>HB=\sqrt{17^2-10^2}=3\sqrt{21}</math> | ||
+ | |||
+ | <math>EF=GD=3\sqrt{21}-12</math> | ||
+ | |||
+ | Thus our answer is: | ||
<math>3+21+12=\boxed{36}</math> | <math>3+21+12=\boxed{36}</math> |
Revision as of 03:20, 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 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: