Difference between revisions of "2014 AIME II Problems/Problem 11"
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==Problem 11== | ==Problem 11== | ||
− | In <math>\triangle RED</math>, <math>\measuredangle DRE=75^{\circ}</math> and <math>\measuredangle RED=45^{\circ}</math>. <math> | + | In <math>\triangle RED</math>, <math>\measuredangle DRE=75^{\circ}</math> and <math>\measuredangle RED=45^{\circ}</math>. <math>|RD|=1</math>. Let <math>M</math> be the midpoint of segment <math>\overline{RD}</math>. Point <math>C</math> lies on side <math>\overline{ED}</math> such that <math>\overline{RC}\perp\overline{EM}</math>. Extend segment <math>\overline{DE}</math> through <math>E</math> to point <math>A</math> such that <math>CA=AR</math>. Then <math>AE=\frac{a-\sqrt{b}}{c}</math>, where <math>a</math> and <math>c</math> are relatively prime positive integers, and <math>b</math> is a positive integer. Find <math>a+b+c</math>. |
==Solution== | ==Solution== |
Revision as of 21:42, 28 February 2015
Problem 11
In ,
and
.
. Let
be the midpoint of segment
. Point
lies on side
such that
. Extend segment
through
to point
such that
. Then
, where
and
are relatively prime positive integers, and
is a positive integer. Find
.
Solution
Let be the foot of the perpendicular from
to
, so
. Since triangle
is isosceles,
is the midpoint of
, and
. Thus,
is a parallelogram and
. We can then use coordinates. Let
be the foot of altitude
and set
as the origin. Now we notice special right triangles! In particular,
and
, so
,
, and
midpoint
and the slope of
, so the slope of
Instead of finding the equation of the line, we use the definition of slope: for every
to the left, we go
up. Thus,
, and
, so the answer is
.
See also
2014 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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.