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2017 AMC 12A Problem 24

Problem 24 of 25HarderGeometry

Quadrilateral ABCDABCD is inscribed in circle OO and has sides AB=3,AB=3, BC=2,BC=2, CD=6,CD=6, and DA=8.DA=8. Let XX and YY be points on BDBD such that DXBD=14\dfrac{DX}{BD}=\dfrac{1}{4} and BYBD=1136.\dfrac{BY}{BD}=\dfrac{11}{36}. Let EE be the intersection of line AXAX and the line through YY parallel to AD.AD. Let FF be the intersection of line CXCX and the line through EE parallel to AC.AC. Let GG be the point on circle OO other than CC that lies on line CX.CX. What is XF⋅XG?XF\cdot XG?

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Solution

Because YE∥ADYE\parallel AD and EF∥AC,EF\parallel AC, we get △XEY∼△XAD\triangle XEY\sim\triangle XAD and △XEF∼△XAC,\triangle XEF\sim\triangle XAC, giving XYXE=XDXA\dfrac{XY}{XE}=\dfrac{XD}{XA} and XFXE=XCXA.\dfrac{XF}{XE}=\dfrac{XC}{XA}. Hence XCXD=XFXY,\dfrac{XC}{XD}=\dfrac{XF}{XY}, so XF⋅XD=XC⋅XY.XF\cdot XD=XC\cdot XY. Power of a Point at XX gives XC⋅XG=XD⋅XB,XC\cdot XG=XD\cdot XB, and combining yields XF⋅XG=XB⋅XY.XF\cdot XG=XB\cdot XY. With d=BD,d=BD, DX=14dDX=\dfrac14 d and BY=1136d,BY=\dfrac{11}{36}d, so XF⋅XG=(d−14d)⋅(d−14d−1136d)=34d⋅49d=d23. \begin{aligned} XF\cdot XG &=\left(d-\tfrac14 d\right) \\ &\quad {}\cdot\left(d-\tfrac14 d-\tfrac{11}{36}d\right) \\ &=\dfrac34 d\cdot\dfrac49 d=\dfrac{d^2}{3}. \end{aligned} Since ABCDABCD is cyclic, ∠BAD\angle BAD and ∠BCD\angle BCD are supplementary. The Law of Cosines on △ABD\triangle ABD and △CBD\triangle CBD gives 73−d248=d2−4024,\dfrac{73-d^2}{48}=\dfrac{d^2-40}{24}, so d2=51.d^2=51. Therefore XF⋅XG=513=17.XF\cdot XG=\dfrac{51}{3}=17. Thus, the correct answer is A.
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Tagged: similarity · power of a point · law of cosines

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