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> | ||
+ | |||
+ | ==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. 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> | ||
+ | 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 | ||
+ | |||
+ | ===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 | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | <math>\because</math> <math>AC \parallel EF</math>, <math>\quad \therefore</math> <math>\triangle ACX \sim \triangle EFX</math>, <math>\quad \frac{XF}{XC} = \frac{XE}{XA}</math> | ||
+ | |||
+ | By Power of a Point, <math>XG \cdot XC = XD \cdot XB</math> | ||
+ | |||
+ | By multiplying the <math>2</math> equations we get <math>XF \cdot XG = \frac{XE}{XA} \cdot XD \cdot XB</math> | ||
+ | |||
+ | <math>\because</math> <math>YE \parallel AD</math>, <math>\quad \therefore</math> <math>\triangle EYX \sim \triangle ADX</math>, <math>\quad \frac{XD}{XY} = \frac{XA}{XE}, \quad XD \cdot XE = XA \cdot XY, \quad XD = \frac{XA \cdot XY}{XE}</math> | ||
+ | |||
+ | By substitution, <math>XF \cdot XG = \frac{XE}{XA} \cdot \frac{XA \cdot XY}{XE} \cdot XB = XY \cdot XB = \frac{4}{9} BD \cdot \frac{3}{4} BD = \frac{BD^2}{3}</math> | ||
+ | |||
+ | Let <math>a = AB</math>, <math>b = BC</math>, <math>c = CD</math>, <math>d = AD</math>, <math>p = AC</math>, and <math>q = BD</math> | ||
+ | |||
+ | By Ptolemy's theorem, <math>p \cdot q = a \cdot c + b \cdot d</math> | ||
+ | |||
+ | <cmath>[ABD] = \frac12 \cdot ad \cdot \sin A, \quad [BCD] = \frac12 \cdot bc \cdot \sin C = \frac12 \cdot bc \cdot \sin A</cmath> | ||
+ | |||
+ | <cmath>[ABC] = \frac12 \cdot ab \cdot \sin B, \quad [ACD] = \frac12 \cdot cd \cdot \sin D = \frac12 \cdot cd \cdot \sin B</cmath> | ||
+ | |||
+ | <cmath>[ABCD] = [ABD] + [BCD] = \frac12 \cdot ad \cdot \sin A + \frac12 \cdot bc \cdot \sin A = \frac12 (ad + bc) \sin A</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}, \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== |
Latest revision as of 02:24, 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.