Difference between revisions of "2016 AMC 12B Problems/Problem 17"
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<cmath>CH = BC - BH = 8 - 2 = 6</cmath> | <cmath>CH = BC - BH = 8 - 2 = 6</cmath> | ||
Apply angle bisector theorem on triangle <math>ACH</math> and triangle <math>ABH</math>, we get <math>AP:PH = 9:6</math> and <math>AQ:QH = 7:2</math>, respectively. | Apply angle bisector theorem on triangle <math>ACH</math> and triangle <math>ABH</math>, we get <math>AP:PH = 9:6</math> and <math>AQ:QH = 7:2</math>, respectively. | ||
− | To find AP, PH, AQ, and QH, apply variables, such that <math>AP:PH = 9:6</math> is <math>\frac{3\sqrt{5} - x}{x} = \frac{9}{6}</math> and <math>AQ:QH = 9:6</math> is <math>\frac{3\sqrt{5} - y}{y} = \frac{7}{2}</math>. Solving them out, you will get <math>AP = \frac{9\sqrt{5}}{5}</math>, <math>PH = \frac{6\sqrt{5}}{5}</math>, <math>AQ = \frac{7\sqrt{5}}{3}</math>, and <math>QH = \frac{2\sqrt{5}}{3}</math>. Then, since <math>AP + PQ = AQ</math> according to the Segment Addition Postulate, and thus manipulating, you get | + | To find AP, PH, AQ, and QH, apply variables, such that <math>AP:PH = 9:6</math> is <math>\frac{3\sqrt{5} - x}{x} = \frac{9}{6}</math> and <math>AQ:QH = 9:6</math> is <math>\frac{3\sqrt{5} - y}{y} = \frac{7}{2}</math>. Solving them out, you will get <math>AP = \frac{9\sqrt{5}}{5}</math>, <math>PH = \frac{6\sqrt{5}}{5}</math>, <math>AQ = \frac{7\sqrt{5}}{3}</math>, and <math>QH = \frac{2\sqrt{5}}{3}</math>. Then, since <math>AP + PQ = AQ</math> according to the Segment Addition Postulate, and thus manipulating, you get <math>PQ = AQ - AP = </math>\frac{7\sqrt{5}}{3}<math> - </math>\frac{9\sqrt{5}}{5}<cmath> = </cmath>\boxed{\textbf{(D)}\frac{8}{15}\sqrt{5}}<math></math> |
==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 08:07, 3 March 2017
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.
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
\frac{7\sqrt{5}}{3}
\frac{9\sqrt{5}}{5}
\boxed{\textbf{(D)}\frac{8}{15}\sqrt{5}}$$ (Error compiling LaTeX. Unknown error_msg)
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.