Difference between revisions of "2014 AMC 10A Problems/Problem 22"
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==Problem== | ==Problem== | ||
− | In rectangle <math>ABCD</math>, <math>AB=20</math> and <math>BC=10</math>. Let <math>E</math> be a point on <math>\overline{CD}</math> such that <math>\angle CBE=15^\circ</math>. What is <math>AE</math>? | + | In rectangle <math>ABCD</math>, <math>\overline{AB}=20</math> and <math>\overline{BC}=10</math>. Let <math>E</math> be a point on <math>\overline{CD}</math> such that <math>\angle CBE=15^\circ</math>. What is <math>\overline{AE}</math>? |
<math> \textbf{(A)}\ \dfrac{20\sqrt3}3\qquad\textbf{(B)}\ 10\sqrt3\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 11\sqrt3\qquad\textbf{(E)}\ 20 </math> | <math> \textbf{(A)}\ \dfrac{20\sqrt3}3\qquad\textbf{(B)}\ 10\sqrt3\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 11\sqrt3\qquad\textbf{(E)}\ 20 </math> |
Revision as of 12:03, 28 December 2015
Contents
[hide]Problem
In rectangle ,
and
. Let
be a point on
such that
. What is
?
Solution
Note that . (If you do not know the tangent half-angle formula, it is
). Therefore, we have
. Since
is a
triangle,
Solution 2 (non-trig)
Let be a point on line
such that points
and
are distinct and that
. By the angle bisector theorem,
. Since
is a
right triangle,
and
. Additionally,
. Now, plugging in the obtained values, we get
and
. Plugging the first equation into the second yields
, so
. Because
is a
triangle,
.
See Also
2014 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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 10 Problems and Solutions |
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