Difference between revisions of "2007 AIME II Problems/Problem 3"
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== Problem == | == Problem == | ||
[[Square]] <math>ABCD</math> has side length <math>13</math>, and [[point]]s <math>E</math> and <math>F</math> are exterior to the square such that <math>BE=DF=5</math> and <math>AE=CF=12</math>. Find <math>EF^{2}</math>. | [[Square]] <math>ABCD</math> has side length <math>13</math>, and [[point]]s <math>E</math> and <math>F</math> are exterior to the square such that <math>BE=DF=5</math> and <math>AE=CF=12</math>. Find <math>EF^{2}</math>. | ||
+ | <asy>unitsize(0.2 cm); | ||
− | + | pair A, B, C, D, E, F; | |
+ | |||
+ | A = (0,13); | ||
+ | B = (13,13); | ||
+ | C = (13,0); | ||
+ | D = (0,0); | ||
+ | E = A + (12*12/13,5*12/13); | ||
+ | F = D + (5*5/13,-5*12/13); | ||
+ | |||
+ | draw(A--B--C--D--cycle); | ||
+ | draw(A--E--B); | ||
+ | draw(C--F--D); | ||
+ | |||
+ | dot("$A$", A, W); | ||
+ | dot("$B$", B, dir(0)); | ||
+ | dot("$C$", C, dir(0)); | ||
+ | dot("$D$", D, W); | ||
+ | dot("$E$", E, N); | ||
+ | dot("$F$", F, S);</asy> | ||
__TOC__ | __TOC__ | ||
+ | |||
== Solution == | == Solution == | ||
Revision as of 14:48, 1 May 2015
Problem
Square has side length
, and points
and
are exterior to the square such that
and
. Find
.
Contents
[hide]Solution
Solution 1
Extend and
to their points of intersection. Since
and are both
right triangles, we can come to the conclusion that the two new triangles are also congruent to these two (use ASA, as we know all the sides are
and the angles are mostly complementary). Thus, we create a square with sides
.
is the diagonal of the square, with length
; the answer is
.
Solution 2
A slightly more analytic/brute-force approach:
Drop perpendiculars from and
to
and
, respectively; construct right triangle
with right angle at K and
. Since
, we have
. Similarly,
. Since
, we have
.
Now, we see that . Also,
. By the Pythagorean Theorem, we have
. Therefore,
.
See also
2007 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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.