Difference between revisions of "2006 AMC 12A Problems/Problem 17"
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<cmath>AE^2 = 9 + 5\sqrt{2} + r^2.</cmath> | <cmath>AE^2 = 9 + 5\sqrt{2} + r^2.</cmath> | ||
− | But it can also be seen that <math>\angle BDA = 45^\circ</math> since <math> | + | But it can also be seen that <math>\angle BDA = 45^\circ</math>. Therefore, since <math>D</math> lies on <math>\overline{BE}</math>, <math>\angle ADE = 135^\circ</math>. Using the Law of Cosines on <math>\triangle ADE</math>, we see |
<cmath>AE^2 &= s^2 + r^2 - 2sr\cos(135^\circ)</cmath> | <cmath>AE^2 &= s^2 + r^2 - 2sr\cos(135^\circ)</cmath> |
Revision as of 00:32, 9 December 2013
Contents
[hide]Problem
Square has side length
, a circle centered at
has radius
, and
and
are both rational. The circle passes through
, and
lies on
. Point
lies on the circle, on the same side of
as
. Segment
is tangent to the circle, and
. What is
?

Solution
Solution 1
One possibility is to use the coordinate plane, setting at the origin. Point
will be
and
will be
since
, and
are collinear and contain a diagonal of
. The Pythagorean theorem results in
This implies that and
; dividing gives us
.
Solution 2
First note that angle is right since
is tangent to the circle. Using the Pythagorean Theorem on
, then, we see
But it can also be seen that . Therefore, since
lies on
,
. Using the Law of Cosines on
, we see
\[AE^2 &= s^2 + r^2 - 2sr\cos(135^\circ)\] (Error compiling LaTeX. Unknown error_msg)
Thus, since and
are rational,
and
. So
,
, and
.
See also
2006 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.