Difference between revisions of "2016 AMC 12B Problems/Problem 17"
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<cmath>Area = 12\sqrt{5} = \frac{1}{2}bh = \frac{1}{2}(8)(AH)</cmath> | <cmath>Area = 12\sqrt{5} = \frac{1}{2}bh = \frac{1}{2}(8)(AH)</cmath> | ||
<cmath>AH = 3\sqrt{5}</cmath> | <cmath>AH = 3\sqrt{5}</cmath> | ||
+ | <cmath>BH = \sqrt{AB^2 - AH^2} = \sqrt{7^2 - (3\sqrt{5})^2} = 2</cmath> | ||
+ | <cmath>CH = BC - BH = 8 - 2 = 6</cmath> | ||
+ | Apply angle bisector theorem on triangle ACH and triangle ABH, we get AP:PH = 9:6 and AQ:QH = 7:2, respectively. | ||
+ | From now, you can simply use the answer choices because only choice D has \sqrt{5} in it and we know that AH = 3\sqrt{5} the segments on it all have integral lengths, so that \sqrt{5} will remain there. | ||
+ | However, by scaling up the length ratio: | ||
+ | <math>AH:AP:PH = 45:27:18 and AQ:QH =45:35:10</math> | ||
+ | we get <math>AH:PQ = 45:(18 - 10) = 45 : 8</math> | ||
+ | <math>PQ = 3\sqrt{5} * \frac{8}{45} = \boxed{frac{8}{15}\sqrt{5}}</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 22:23, 21 February 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 ACH and triangle ABH, we get AP:PH = 9:6 and AQ:QH = 7:2, respectively.
From now, you can simply use the answer choices because only choice D has \sqrt{5} in it and we know that AH = 3\sqrt{5} the segments on it all have integral lengths, so that \sqrt{5} will remain there.
However, by scaling up the length ratio:
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.