Difference between revisions of "2005 AIME II Problems/Problem 12"
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too and by SAS we know that <math>\triangle{FOE}=\triangle{FOG}</math> so FG=400. Now we have a right <math>\triangle{BFG}</math> with legs x and 500-x and hypotenuse 400. Now by [http://www.artofproblemsolving.com/Wiki/index.php/Pythagorean_theorem Pythagorean Theorem], | too and by SAS we know that <math>\triangle{FOE}=\triangle{FOG}</math> so FG=400. Now we have a right <math>\triangle{BFG}</math> with legs x and 500-x and hypotenuse 400. Now by [http://www.artofproblemsolving.com/Wiki/index.php/Pythagorean_theorem Pythagorean Theorem], | ||
− | <math>(500-x)^2+x^2=400^2</math> | + | <math>\displaystyle(500-x)^2+x^2=400^2</math> |
− | <math>250000-1000x+2x^2=16000</math> | + | <math> 250000-1000x+2x^2=16000</math> |
− | <math>90000-1000x+2x^2=0</math> | + | <math>\displaystyle 90000-1000x+2x^2=0</math> |
applying the quadratic formula we get that | applying the quadratic formula we get that |
Revision as of 19:20, 26 July 2007
Contents
Problem
Square has center
and
are on
with
and
between
and
and
Given that
where
and
are positive integers and
is not divisible by the square of any prime, find
Solution
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
Draw the perpendicular from , with the intersection at
. Denote
and
, and
(since
and
). The tangent of
, and of
.
By the tangent addition rule , we see that
. Since
,
. We know that
, so we can substitute this to find that
.
A second equation can be set up using . To solve for
,
. This is a quadratic with roots
. Since
, use the smaller root,
.
Now, . The answer is
.
Solution 2
Since we are given that EF=400 label BF=x and EA=500-x. Rotate about O until EF lies on BC. Now we know that
therefore
+
also since
is the center of the square. Label the new triangle that we created
.Now we know that rotation preserves angles and side lenghts so
and
. Now draw
and
. Notice that
=
since rotations preserve the same angles so
=
too and by SAS we know that so FG=400. Now we have a right
with legs x and 500-x and hypotenuse 400. Now by Pythagorean Theorem,
applying the quadratic formula we get that
=
and since we want p+q+r we get that is 307.
See also
2005 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |