Difference between revisions of "2016 AMC 12B Problems/Problem 17"
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\textbf{(E)}\ \frac{6}{5}</math> | \textbf{(E)}\ \frac{6}{5}</math> | ||
− | ==Solution== | + | ==Solution 1== |
+ | Get the area of the triangle by [https://artofproblemsolving.com/wiki/index.php/Heron%27s_Formula Heron's Formula]: | ||
+ | <cmath>\sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(12)(3)(4)(5)} = 12\sqrt{5}</cmath> | ||
+ | Use the area to find the height <math>AH</math> with known base <math>BC</math>: | ||
+ | <cmath>Area = 12\sqrt{5} = \frac{1}{2}bh = \frac{1}{2}(8)(AH)</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 the [https://artofproblemsolving.com/wiki/index.php/Angle_Bisector_Theorem Angle Bisector Theorem] on <math> \triangle ACH</math> and <math>\triangle ABH</math>, we get <math>AP:PH = 9:6</math> and <math>AQ:QH = 7:2</math>, respectively. | ||
+ | To find <math>AP</math>, <math>PH</math>, <math>AQ</math>, and <math>QH</math>, 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 = 7:2</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 = \frac{7\sqrt{5}}{3} - \frac{9\sqrt{5}}{5}</math> = <cmath>\boxed{\textbf{(D)}\frac{8}{15}\sqrt{5}}</cmath> | ||
+ | |||
+ | ==Solution 2== | ||
+ | Let the intersection of <math>BD</math> and <math>CE</math> be the point <math>I</math>. Then let the foot of the altitude from <math>I</math> to <math>BC</math> be <math>I'</math>. Note that <math>II'</math> is an inradius and that <math>II' \cdot s = [ABC]</math>, where <math>s</math> is the semiperimeter of the triangle. | ||
+ | |||
+ | Using Heron's Formula, we see that <math>II' \cdot 12 = \sqrt{12 \cdot 3 \cdot 4 \cdot 5} = 12\sqrt{5}</math>, so <math>II' = \sqrt{5}</math>. | ||
+ | |||
+ | Then since <math>II'</math> and <math>AH</math> are parallel, <math>\triangle CI'I \sim \triangle CHP</math> and <math>\triangle BHQ \sim \triangle BI'I</math>. | ||
+ | |||
+ | Thus, <math>\frac{II'}{PQ + QH} = \frac{CI'}{CH}</math> and <math>\frac{II'}{QH} = \frac{BI'}{BH}</math>, so | ||
+ | <math>PQ = \frac{II' \cdot CH}{CI'} - \frac{II' \cdot BH}{BI'}</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> | ||
+ | |||
+ | ==Solution 3 (FAST)== | ||
+ | <math>PQ</math> 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>\boxed{\textbf{(D)}} </math>. | ||
+ | |||
+ | ==Solution 4 == | ||
+ | |||
+ | Let <math>h=AH</math> and <math>BH=x</math>. Then, <math>CH=8-x</math>. By the Pythagorean Theorem on right triangles <math>ABH</math> and <math>ACH</math>, we have <cmath>h^2+x^2=49</cmath> <cmath>x^2+(8-x)^2=81.</cmath> Subtracting the prior from the latter yields <math>-16x+64=32\implies x=2</math>. So, <math>BH=2</math>, <math>CH=6</math>, and <math>AH=3\sqrt{5}</math>. Continue with Solution 1. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=ccB-z4_OHqw | ||
==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}} |
Latest revision as of 19:12, 23 October 2023
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 with known base : Apply the Angle Bisector Theorem on and , we get and , respectively. To find , , , and , 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 (FAST)
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 .
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.
Video Solution
https://www.youtube.com/watch?v=ccB-z4_OHqw
See Also
2016 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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All AMC 12 Problems and Solutions |
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