Difference between revisions of "2017 AMC 12A Problems/Problem 24"
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<cmath>[ABCD] = [ABC] + [ACD] = \frac12 \cdot ab \cdot \sin B + \frac12 \cdot cd \cdot \sin B = \frac12 (ab + cd) \sin B</cmath> | <cmath>[ABCD] = [ABC] + [ACD] = \frac12 \cdot ab \cdot \sin B + \frac12 \cdot cd \cdot \sin B = \frac12 (ab + cd) \sin B</cmath> | ||
− | <cmath>\frac{ab + cd}{ad + bc} = \frac{ \sin A }{ \sin B} = \frac{ \frac{q}{2R} }{ \frac{p}{2R} } = \frac{q}{p}</cmath> | + | <cmath>\frac{ab + cd}{ad + bc} = \frac{ \sin A }{ \sin B} = \frac{ \frac{q}{2R} }{ \frac{p}{2R} } = \frac{q}{p}, \quad p = \frac{q(ad + bc)}{ab + cd}</cmath> |
+ | |||
+ | <cmath>\frac{q(ad + bc)}{ab + cd} \cdot q = ac + bd</cmath> | ||
+ | |||
+ | <cmath>BD^2 = q^2 = \frac{ (ac + bd)(ab + cd) }{ad + bc} = \frac{(3 \cdot 6 + 2 \cdot 8)(3 \cdot 2 + 6 \cdot 8)}{3 \cdot 8 + 2 \cdot 6} = 51</cmath> | ||
+ | |||
+ | <cmath>XF \cdot XG = \frac{51}{3} = \boxed{\textbf{(A) } 17}</cmath> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
==Video Solution by Richard Rusczyk== | ==Video Solution by Richard Rusczyk== |
Revision as of 03:12, 7 October 2023
Contents
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
Solution 3
Denote to be the intersection between line
and circle
. Note that
, making
. Thus,
is a cyclic quadrilateral. Using Power of a Point on
gives
.
Since and
,
. Using Power of a Point on
again,
. Plugging in
gives:
By Law of Cosines, we can find
, as in Solution 1. Now,
and
, making
. This gives us
as a result.
-Solution by sml1809
Note
You could have also got the relation as follows: From the similarities,
. PoP on
gives
. Plugging in
and
gives
implying that
.
~sml1809
Solution 4
,
,
By Power of a Point,
By multiplying the equations we get
,
,
By substitution,
Let ,
,
,
,
, and
By Ptolemy's theorem,
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.