Difference between revisions of "2017 AMC 12A Problems/Problem 24"
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First we note that <math>\triangle AXD \sim \triangle EXY</math> with a ratio of <math>\frac{DX}{XY} = \frac{9}{16}.</math> Then <math>\triangle ACX \sim \triangle EFX</math> with a ratio of <math>\frac{AX}{XE} = \frac{DX}{XY} = \frac{9}{16}</math>, so <math>\frac{XF}{CX} = \frac{16}{9}.</math> | First we note that <math>\triangle AXD \sim \triangle EXY</math> with a ratio of <math>\frac{DX}{XY} = \frac{9}{16}.</math> Then <math>\triangle ACX \sim \triangle EFX</math> with a ratio of <math>\frac{AX}{XE} = \frac{DX}{XY} = \frac{9}{16}</math>, so <math>\frac{XF}{CX} = \frac{16}{9}.</math> | ||
Now we find the length of <math>BD</math>. Because the quadrilateral is cyclic, we can simply use the Law of Cosines. <cmath>BD^2=3^2+8^2-48\cos\angle BAD=2^2+6^2-24\cos (180-\angle BAD)=2^2+6^2+24\cos\angle BAD</cmath><cmath>\rightarrow \cos\angle BAD = \frac{11}{24}</cmath><cmath>\rightarrow BD=\sqrt{51}</cmath> | Now we find the length of <math>BD</math>. Because the quadrilateral is cyclic, we can simply use the Law of Cosines. <cmath>BD^2=3^2+8^2-48\cos\angle BAD=2^2+6^2-24\cos (180-\angle BAD)=2^2+6^2+24\cos\angle BAD</cmath><cmath>\rightarrow \cos\angle BAD = \frac{11}{24}</cmath><cmath>\rightarrow BD=\sqrt{51}</cmath> | ||
− | By Power of a Point, <math>CX\cdot XG = DX\cdot XB = \frac{\sqrt{51}}{4} \frac{3\sqrt{51}}{4}</math>. Thus <math>XF\cdot XG = \frac{XF}{CX} CX\cdot XG = \frac{51}{3} = \boxed{(A) 17}.</math> | + | By Power of a Point, <math>CX\cdot XG = DX\cdot XB = \frac{\sqrt{51}}{4} \frac{3\sqrt{51}}{4}</math>. Thus <math>XF\cdot XG = \frac{XF}{CX} CX\cdot XG = \frac{51}{3} = \boxed{\textbf{(A)}\ 17}.</math> |
-solution by FRaelya | -solution by FRaelya |
Revision as of 21:45, 9 February 2017
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
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
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.