Difference between revisions of "2017 AMC 12A Problems/Problem 24"
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==Solution 3== | ==Solution 3== | ||
− | Denote <math>P</math> to be the intersection between line <math>AE</math> and circle <math>O</math>. Note that <math>\angle GFE = \angle ACG = \angle APG = 180 - \angle GPE</math>, making <math>\angle GFE + \angle GDE = 180</math>. Thus, <math>PEFG</math> is a cyclic quadrilateral. | + | Denote <math>P</math> to be the intersection between line <math>AE</math> and circle <math>O</math>. Note that <math>\angle GFE = \angle ACG = \angle APG = 180 - \angle GPE</math>, making <math>\angle GFE + \angle GDE = 180</math>. Thus, <math>PEFG</math> is a cyclic quadrilateral. Using [[Power of a Point]] on <math>X</math> gives <math>XP \cdot XE = XG \cdot XF</math>. |
− | + | Since <math>\triangle ADX \sim \triangle EYX</math> and <math>\triangle ACX \sim \triangle EFX</math>, <math>AX/XE = XD/YX = 9/16</math>. Using Power of a Point on <math>X</math> again, <math>(AX)(PX) = (BX)(DX)</math>. Plugging in <math>AX=9/16 XE</math> gives: | |
<cmath>\dfrac{9}{16}(XE)(PX) = (BX)(DX) = \dfrac{9}{16}(FX)(GX)</cmath> | <cmath>\dfrac{9}{16}(XE)(PX) = (BX)(DX) = \dfrac{9}{16}(FX)(GX)</cmath> | ||
By [[Law of Cosines]], we can find <math>BD = \sqrt{51}</math>, as in Solution 1. Now, <math>BX = 3/4 (\sqrt{51})</math> and <math>DX = 1/4 (\sqrt{51})</math>, making <math>\dfrac{9}{16}(FX)(GX) = \left( \dfrac{\sqrt{51}}{4}\right)\left( \dfrac{3\sqrt{51}}{4}\right) = \dfrac{3(51)}{16}</math>. This gives us <math>FX \cdot GX = \boxed{\textbf{(A)}\ 17}</math> as a result. | By [[Law of Cosines]], we can find <math>BD = \sqrt{51}</math>, as in Solution 1. Now, <math>BX = 3/4 (\sqrt{51})</math> and <math>DX = 1/4 (\sqrt{51})</math>, making <math>\dfrac{9}{16}(FX)(GX) = \left( \dfrac{\sqrt{51}}{4}\right)\left( \dfrac{3\sqrt{51}}{4}\right) = \dfrac{3(51)}{16}</math>. This gives us <math>FX \cdot GX = \boxed{\textbf{(A)}\ 17}</math> as a result. | ||
-Solution by sml1809 | -Solution by sml1809 | ||
+ | |||
+ | ===Note=== | ||
+ | |||
+ | You could have also got the relation <math>XP \cdot XE = XG \cdot XF</math> as follows: From the similarities, <math>AX/XE = CX/XF = 9/16</math>. PoP on <math>X</math> gives <math>(AX)(PX) = (CX)(GX)</math>. Plugging in <math>AX = 9/16 XE</math> and <math>CX=9/16 FX</math> gives | ||
+ | <cmath>9/16 (XE)(PX) = 9/16(FX)(GX),</cmath> | ||
+ | implying that <math>(XP)(XE) = (XG)(XP)</math>. | ||
+ | |||
+ | ~sml1809 | ||
==Video Solution by Richard Rusczyk== | ==Video Solution by Richard Rusczyk== |
Revision as of 19:39, 7 September 2023
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
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
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.