Difference between revisions of "2016 AMC 12B Problems/Problem 17"
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we get <math>AH:PQ = 45:(18 - 10) = 45 : 8</math>. | we get <math>AH:PQ = 45:(18 - 10) = 45 : 8</math>. | ||
<cmath>PQ = 3\sqrt{5} * \frac{8}{45} = \boxed{\textbf{(D)}\frac{8}{15}\sqrt{5}}</cmath> | <cmath>PQ = 3\sqrt{5} * \frac{8}{45} = \boxed{\textbf{(D)}\frac{8}{15}\sqrt{5}}</cmath> | ||
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==See Also== | ==See Also== | ||
{{AMC12 box|year=2016|ab=B|num-b=16|num-a=18}} | {{AMC12 box|year=2016|ab=B|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 11:25, 25 September 2016
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
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.
From now, you can simply use the answer choices because only choice
has
in it and we know that
the segments on it all have integral lengths so that
will remain there.
However, by scaling up the length ratio:
and
.
we get
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.