Difference between revisions of "2017 AMC 12A Problems/Problem 24"
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<math>\textbf{(A) }17\qquad\textbf{(B) }\frac{59 - 5\sqrt{2}}{3}\qquad\textbf{(C) }\frac{91 - 12\sqrt{3}}{4}\qquad\textbf{(D) }\frac{67 - 10\sqrt{2}}{3}\qquad\textbf{(E) }18</math> | <math>\textbf{(A) }17\qquad\textbf{(B) }\frac{59 - 5\sqrt{2}}{3}\qquad\textbf{(C) }\frac{91 - 12\sqrt{3}}{4}\qquad\textbf{(D) }\frac{67 - 10\sqrt{2}}{3}\qquad\textbf{(E) }18</math> | ||
− | ==Solution== | + | ==Solution 1== |
It is easy to see that <math>\frac{XY}{BD} = 1 - \frac{1}{4} - \frac{11}{36} = \frac{4}{9}.</math> | It is easy to see that <math>\frac{XY}{BD} = 1 - \frac{1}{4} - \frac{11}{36} = \frac{4}{9}.</math> | ||
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-solution by FRaelya | -solution by FRaelya | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | We shall make use of the pairs of similar triangles present in the problem, Ptolemy's Theorem, and Power of a Point. | ||
+ | Let <math>Z</math> be the intersection of <math>AC</math> and <math>BD</math>. First, from <math>ABCD</math> being a cyclic quadrilateral, we have that <math>\triangle BCZ \sim \triangle AZD</math>, <math>\triangle BZA \sim \triangle CDZ</math>. Therefore, <math>\frac{2}{BZ} = \frac{8}{AZ}</math>, <math>\frac{6}{CZ} = \frac{3}{BZ}</math>, and <math>\frac{2}{CZ} = \frac{8}{DZ}</math>, so we have <math>BZ = \frac{1}{2}CZ</math>, <math>AZ = 2CZ</math>, and <math>DZ = 4CZ</math>. By Ptolemy's Theorem, <cmath>(AB)(CD) + (BC)(DA) = (AC)(BD) = (AZ + ZC)(BZ + ZD)</cmath> <cmath>\rightarrow 3 \cdot 6 + 2 \cdot 8 = 34 = \left(2CZ + ZC\right)\left(\frac{1}{2}CZ + 4CZ\right) = \frac{27}{2}CZ^2.</cmath> Thus, <math>CZ^2 = \frac{68}{27}</math>. Then, by Power of a Point, <math>GX \cdot XC = BX \cdot XD = \frac{3}{4} \cdot \frac{1}{4}BD^2 = \frac{3}{16} \cdot \left(\frac{9}{2}CZ\right)^2 = \frac{9 \cdot 17}{16}</math>. So, <math>XG = \frac{9 \cdot 17}{16XC}</math>. | ||
+ | Next, observe that <math>\triangle ACX \sim \triangle EFX</math>, so <math>\frac{XE}{XF} = \frac{AX}{XC}</math>. Also, <math>\triangle{AXD} \sim \triangle{EXY}</math>, so <math>\frac{8}{AX} = \frac{EY}{XY}</math>. We can compute <math>EY = \frac{128}{9}</math> after noticing that <math>XY = BD - BY - DX = BD - \frac{11}{36}BD - \frac{1}{4}BD = \frac{4}{9}BD</math> and that <math>\frac{8}{DX} = \frac{32}{BD} = \frac{EY}{XY} = \frac{EY}{\frac{4}{9}BD}</math>. So, <math>\frac{8}{AX} = \frac{128}{9XE}</math>. Then, <math>\frac{XE}{AX} = \frac{XF}{XC} = \frac{16}{9} \rightarrow XF = \frac{16}{9}XC</math>. | ||
+ | |||
+ | Multiplying our equations for <math>XF</math> and <math>XG</math> yields that <math>XF \cdot XG = \frac{9 \cdot 17}{16XC} \cdot \frac{16}{9}XC = \boxed{\textbf{(A)}\ 17}.</math> | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2017|ab=A|num-b=23|num-a=25}} | {{AMC12 box|year=2017|ab=A|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 00:56, 12 July 2017
Contents
[hide]Problem
Quadrilateral is inscribed in circle and has side lengths , and . Let and be points on
such that and . Let be the intersection of line and the line through parallel to . Let be the intersection of line and the line through parallel to . Let be the point on circle other than that lies on line . What is ?
Solution 1
It is easy to see that First we note that with a ratio of Then with a ratio of , so Now we find the length of . Because the quadrilateral is cyclic, we can simply use the Law of Cosines. By Power of a Point, . Thus
-solution by FRaelya
Solution 2
We shall make use of the pairs of similar triangles present in the problem, Ptolemy's Theorem, and Power of a Point. Let be the intersection of and . First, from being a cyclic quadrilateral, we have that , . Therefore, , , and , so we have , , and . By Ptolemy's Theorem, Thus, . Then, by Power of a Point, . So, . Next, observe that , so . Also, , so . We can compute after noticing that and that . So, . Then, .
Multiplying our equations for and yields that
See Also
2017 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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.