2014 AMC 10A Problems/Problem 22
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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
. By the angle bisector theorem,
. Since
is a
right triangle,
and
. Additionally,
. Now, plugging in the obtained values, we get
and
. Plugging in
for
in the second equation 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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