2016 AMC 12B Problems/Problem 17
Contents
[hide]Problem
In shown in the figure,
,
,
, and
is an altitude. Points
and
lie on sides
and
, respectively, so that
and
are angle bisectors, intersecting
at
and
, respectively. What is
?
Solution 1
Get the area of the triangle by heron's formula:
Use the area to find the height AH with known base BC:
Apply angle bisector theorem on triangle
and triangle
, we get
and
, respectively.
To find AP, PH, AQ, and QH, apply variables, such that
is
and
is
. Solving them out, you will get
,
,
, and
. Then, since
according to the Segment Addition Postulate, and thus manipulating, you get
=
Solution 2
Let the intersection of and
be the point
. Then let the foot of the altitude from
to
be
. Note that
is an inradius and that
, where
is the semiperimeter of the triangle.
Using Heron's Formula, we see that , so
.
Then since and
are parallel,
and
.
Thus, and
, so
.
By the Dual Principle, and
. With the same method as Solution 1,
and
.
Then
Solution 3 (if you're running short on time)
lies on altitude
, which we find to have a length of
by Heron's Formula and dividing twice the area by
. From H we can construct a segment
with
on
such that
is parallel to
. A similar construction gives
on
such that
is parallel to
. We can hence generate a system of ratios that will allow us to find
. Note that such a system will generate a rational number for the ratio
. Thus, we choose the only answer that has a
term in it, giving us $\boxed\textbf{(D)}}$ (Error compiling LaTeX. Unknown error_msg).
Solution 4
Let and
. Then,
. By the Pythagorean Theorem on right triangles
and
, we have
Subtracting the prior from the latter yields
. So,
,
, and
. Continue with Solution 1.
See Also
2016 AMC 12B (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 |
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