Difference between revisions of "2017 AMC 12A Problems/Problem 24"
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==Problem== | ==Problem== | ||
− | Quadrilateral <math>ABCD</math> is inscribed in circle <math>O</math> and has side lengths <math>AB=3, BC=2, CD=6</math>, and <math>DA=8</math>. Let <math>X</math> and <math>Y</math> be points on | + | Quadrilateral <math>ABCD</math> is inscribed in circle <math>O</math> and has side lengths <math>AB=3, BC=2, CD=6</math>, and <math>DA=8</math>. Let <math>X</math> and <math>Y</math> be points on <math>\overline{BD}</math> such that <math>\frac{DX}{BD} = \frac{1}{4}</math> and <math>\frac{BY}{BD} = \frac{11}{36}</math>. |
− | |||
− | <math>\overline{BD}</math> such that <math>\frac{DX}{BD} = \frac{1}{4}</math> and <math>\frac{BY}{BD} = \frac{11}{36}</math>. | ||
Let <math>E</math> be the intersection of line <math>AX</math> and the line through <math>Y</math> parallel to <math>\overline{AD}</math>. Let <math>F</math> be the intersection of line <math>CX</math> and the line through <math>E</math> parallel to <math>\overline{AC}</math>. Let <math>G</math> be the point on circle <math>O</math> other than <math>C</math> that lies on line <math>CX</math>. What is <math>XF\cdot XG</math>? | Let <math>E</math> be the intersection of line <math>AX</math> and the line through <math>Y</math> parallel to <math>\overline{AD}</math>. Let <math>F</math> be the intersection of line <math>CX</math> and the line through <math>E</math> parallel to <math>\overline{AC}</math>. Let <math>G</math> be the point on circle <math>O</math> other than <math>C</math> that lies on line <math>CX</math>. What is <math>XF\cdot XG</math>? | ||
<math>\textbf{(A) }17\qquad\textbf{(B) }\frac{59 - 5\sqrt{2}}{3}\qquad\textbf{(C) }\frac{91 - 12\sqrt{3}}{4}\qquad\textbf{(D) }\frac{67 - 10\sqrt{2}}{3}\qquad\textbf{(E) }18</math> | <math>\textbf{(A) }17\qquad\textbf{(B) }\frac{59 - 5\sqrt{2}}{3}\qquad\textbf{(C) }\frac{91 - 12\sqrt{3}}{4}\qquad\textbf{(D) }\frac{67 - 10\sqrt{2}}{3}\qquad\textbf{(E) }18</math> | ||
+ | |||
+ | ==Diagram== | ||
+ | |||
+ | <asy> | ||
+ | size(8cm); | ||
+ | real r = 4.01754; | ||
+ | draw(circle((0,0), r)); | ||
+ | pair C = r * dir(-30), B = r * dir(28.83-30), A = r * dir(72.68-30), D = r * dir(241.98-30); | ||
+ | draw(A--B--C--D--cycle); draw(B--D); | ||
+ | pair X = B * 1/4 + D * 3/4, Y = B * 25/36 + D * 11/36; | ||
+ | label("A", A, N); label("B", B, NE); label ("C", C, E); label("D", D, S); | ||
+ | label("Y", Y, N); label("X", X, N); | ||
+ | pair G = X * 1.445 - C*0.445; | ||
+ | label("G", G, NW); | ||
+ | pair E = Y + (D - A) * 1.48; | ||
+ | draw(Y--E); | ||
+ | draw(A--E); | ||
+ | label("E", E, S); | ||
+ | pair F = E + (A - C) * 1.45; | ||
+ | draw(C--F--E); | ||
+ | label("F",F,NW); | ||
+ | </asy> | ||
+ | ~raxu, put in by fuzimiao2013 | ||
==Solution 1== | ==Solution 1== | ||
− | + | Using the given ratios, note that <math>\frac{XY}{BD} = 1 - \frac{1}{4} - \frac{11}{36} = \frac{4}{9}.</math> | |
− | + | ||
+ | By AA Similarity, <math>\triangle AXD \sim \triangle EXY</math> with a ratio of <math>\frac{DX}{XY} = \frac{9}{16}</math> and <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{\textbf{(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> | ||
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We shall make use of the pairs of similar triangles present in the problem, Ptolemy's Theorem, and Power of a Point. | We shall make use of the pairs of similar triangles present in the problem, Ptolemy's Theorem, and Power of a Point. | ||
Let <math>Z</math> be the intersection of <math>AC</math> and <math>BD</math>. First, from <math>ABCD</math> being a cyclic quadrilateral, we have that <math>\triangle BCZ \sim \triangle AZD</math>, <math>\triangle BZA \sim \triangle CDZ</math>. Therefore, <math>\frac{2}{BZ} = \frac{8}{AZ}</math>, <math>\frac{6}{CZ} = \frac{3}{BZ}</math>, and <math>\frac{2}{CZ} = \frac{8}{DZ}</math>, so we have <math>BZ = \frac{1}{2}CZ</math>, <math>AZ = 2CZ</math>, and <math>DZ = 4CZ</math>. By Ptolemy's Theorem, <cmath>(AB)(CD) + (BC)(DA) = (AC)(BD) = (AZ + ZC)(BZ + ZD)</cmath> <cmath>\rightarrow 3 \cdot 6 + 2 \cdot 8 = 34 = \left(2CZ + ZC\right)\left(\frac{1}{2}CZ + 4CZ\right) = \frac{27}{2}CZ^2.</cmath> Thus, <math>CZ^2 = \frac{68}{27}</math>. Then, by Power of a Point, <math>GX \cdot XC = BX \cdot XD = \frac{3}{4} \cdot \frac{1}{4}BD^2 = \frac{3}{16} \cdot \left(\frac{9}{2}CZ\right)^2 = \frac{9 \cdot 17}{16}</math>. So, <math>XG = \frac{9 \cdot 17}{16XC}</math>. | Let <math>Z</math> be the intersection of <math>AC</math> and <math>BD</math>. First, from <math>ABCD</math> being a cyclic quadrilateral, we have that <math>\triangle BCZ \sim \triangle AZD</math>, <math>\triangle BZA \sim \triangle CDZ</math>. Therefore, <math>\frac{2}{BZ} = \frac{8}{AZ}</math>, <math>\frac{6}{CZ} = \frac{3}{BZ}</math>, and <math>\frac{2}{CZ} = \frac{8}{DZ}</math>, so we have <math>BZ = \frac{1}{2}CZ</math>, <math>AZ = 2CZ</math>, and <math>DZ = 4CZ</math>. By Ptolemy's Theorem, <cmath>(AB)(CD) + (BC)(DA) = (AC)(BD) = (AZ + ZC)(BZ + ZD)</cmath> <cmath>\rightarrow 3 \cdot 6 + 2 \cdot 8 = 34 = \left(2CZ + ZC\right)\left(\frac{1}{2}CZ + 4CZ\right) = \frac{27}{2}CZ^2.</cmath> Thus, <math>CZ^2 = \frac{68}{27}</math>. Then, by Power of a Point, <math>GX \cdot XC = BX \cdot XD = \frac{3}{4} \cdot \frac{1}{4}BD^2 = \frac{3}{16} \cdot \left(\frac{9}{2}CZ\right)^2 = \frac{9 \cdot 17}{16}</math>. So, <math>XG = \frac{9 \cdot 17}{16XC}</math>. | ||
− | Next, observe that <math>\triangle ACX \sim \triangle EFX</math>, so <math>\frac{XE}{XF} = \frac{AX}{XC}</math>. Also, <math>\triangle{AXD} \sim \triangle{EXY}</math>, so <math>\frac{8}{AX} = \frac{EY}{ | + | Next, observe that <math>\triangle ACX \sim \triangle EFX</math>, so <math>\frac{XE}{XF} = \frac{AX}{XC}</math>. Also, <math>\triangle{AXD} \sim \triangle{EXY}</math>, so <math>\frac{8}{AX} = \frac{EY}{XE}</math>. We can compute <math>EY = \frac{128}{9}</math> after noticing that <math>XY = BD - BY - DX = BD - \frac{11}{36}BD - \frac{1}{4}BD = \frac{4}{9}BD</math> and that <math>\frac{8}{DX} = \frac{32}{BD} = \frac{EY}{XY} = \frac{EY}{\frac{4}{9}BD}</math>. So, <math>\frac{8}{AX} = \frac{128}{9XE}</math>. Then, <math>\frac{XE}{AX} = \frac{XF}{XC} = \frac{16}{9} \rightarrow XF = \frac{16}{9}XC</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> | 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== | ||
+ | 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}} |
Latest revision as of 03:24, 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.