Maths Olympiad Prep

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Geometry Difficulty 5.6 AIME, harder Prove it Iran

ω\omega is a circle with diameter ABAB. Points C,DC, D lie on ω\omega such that C,DC, D are on different sides of ABAB. A line passing through CC and parallel to ADAD cuts ABAB at FF, and a line passing through DD and parallel to ACAC cuts ABAB at EE. A,B,CA, B, C and DD are in a way that E,FE, F are inside ω\omega. The line passing through EE and perpendicular to ABAB cuts BDBD at XX and the line passing through FF and perpendicular to ABAB cuts BCBC at YY. Prove that the perimeter of triangle AXY\triangle AXY equals to 2CD2CD.

Solution

Let KK be the intersection point of CDCD, AXAX and LL be the intersection point of CDCD, AYAY.

Figure 1

Then, since DAEXDAEX is a cyclic quadrilateral, we have
KDX=BAC=AED=AXD    DK=KX=AK. \angle KDX = \angle BAC = \angle AED = \angle AXD \implies DK = KX = AK.
Similarly, AL=LY=LCAL = LY = LC. By Thales's theorem, we have XY=2KLXY = 2KL. Therefore
AX+XY+AY=AK+KL+LA=KD+KL+LC=CD AX + XY + AY = AK + KL + LA = KD + KL + LC = CD
Which completes the proof.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.