Difference between revisions of "2017 AMC 12A Problems/Problem 24"
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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> | 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> | ||
+ | |||
+ | ==Video Solution by Richard Rusczyk== | ||
+ | https://www.youtube.com/watch?v=JdERP0d0W64&list=PLyhPcpM8aMvLZmuDnM-0vrFniLpo7Orbp&index=4 | ||
+ | - AMBRIGGS | ||
==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 16:00, 30 July 2022
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
?
Diagram
~raxu, put in by fuzimiao2013
Solution 1
Using the given ratios, note that
By AA Similarity, with a ratio of
and
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
Video Solution by Richard Rusczyk
https://www.youtube.com/watch?v=JdERP0d0W64&list=PLyhPcpM8aMvLZmuDnM-0vrFniLpo7Orbp&index=4 - AMBRIGGS
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.