Difference between revisions of "2016 AMC 12B Problems/Problem 17"
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By the Dual Principle, <math>CI' = 5</math> and <math>BI' = 3</math>. With the same method as Solution 1, <math>CH = 6</math> and <math>BH = 2</math>. | By the Dual Principle, <math>CI' = 5</math> and <math>BI' = 3</math>. With the same method as Solution 1, <math>CH = 6</math> and <math>BH = 2</math>. | ||
Then <math>PQ = \frac{8}{15} II' = </math> <cmath>\boxed{\textbf{(D)}\frac{8}{15}\sqrt{5}}</cmath> | Then <math>PQ = \frac{8}{15} II' = </math> <cmath>\boxed{\textbf{(D)}\frac{8}{15}\sqrt{5}}</cmath> | ||
+ | |||
+ | ==Solution 3 (if you're running short on time)== | ||
+ | PQ lies on altitude <math>AH</math>, which we find to have a length of <math>3\sqrt{5}</math> by Heron's Formula and dividing twice the area by <math>BC</math>. From H we can construct a segment <math>HX</math> with <math>X</math> on <math>CE</math> such that <math>HX</math> is parallel to <math>EB</math>. A similar construction gives <math>Y</math> on <math>BD</math> such that <math>HY</math> is parallel to <math>DC</math>. We can hence generate a system of ratios that will allow us to find <math>PQ/AH</math>. Note that such a system will generate a rational number for the ratio <math>PQ/AH</math>. Thus, we choose the only answer that has a <math>\sqrt{5}</math> term in it, giving us <math>\textbf{D} </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 16:29, 11 February 2018
Contents
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)
PQ 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 .
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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