Problem:
Circles ω1​ and ω2​ intersect at points X and Y. Line â„“ is tangent to ω1​ and ω2​ at A and B, respectively, with line AB closer to point X than to Y. Circle ω passes through A and B intersecting ω1​ again at Dî€ =A and intersecting ω2​ again at Cî€ =B. The three points C,Y, and D are collinear, XC=67,XY=47, and XD=37. Find AB2.
Solution:
Note that lines AD,XY,BC are the radical axes of pairs of circles ω and ω1​, ω1​ and ω2​,ω2​ and ω, respectively. Therefore lines AD,XY, and BC are either parallel to each other or concurrent. In the former case, ω1​ and ω2​ would be the same size, and by symmetry, CX=DX, contradicting the given condition. Hence it must be that lines AD,XY, and BC are concurrent at a point Z. Denote by M the intersection of segments ZY and AB. By the Power of a Point Theorem it follows that MA2=MX⋅MY=MB2. In particular,
AB2​=4MA2=4MX⋅MY=4MX(MX+XY)=(2MX+XY)2−XY2=(MY+MX)2−XY2​
Claim: (MX+MY)2=CX⋅DX. The claim implies that AB2=CX⋅DX− XY2=37⋅67−472=270​.
The proof of the claim is based on the following three observations: ZAXB is cyclic; â–³XZC is similar to â–³XDZ; and ZAYB is a parallelogram.
Because BCYX is cyclic, ∠XBZ=∠XYC. Because ADYX is cyclic, ∠XAZ= ∠XYD. Because C,Y, and D are collinear, ∠XAZ+∠XBZ=∠XYD+ ∠XYC=180∘, from which it follows that ZAXB is cyclic, establishing the first observation. This is also a direct consequence of Miquel's Theorem.
Because BCYX and ZAXB are cyclic, ∠XCB=∠XYB and ∠ABX= ∠AZX. Because AB is tangent to ω2​ at B,∠ABX=∠XYB. Combining the three equations yields ∠XCZ=∠XCB=∠XYB=∠ABX=∠AZX= ∠DZX. Likewise, ∠XZC=∠XDZ. Hence △XZC is similar to △XDZ, establishing the second observation.
As in the previous paragraph, ∠XYB=∠AZX or BY∥AZ. Similarly, AY∥ BZ. Thus ZAYB is a parallelogram, establishing the third observation.
Because ZAYB is a parallelogram, MY=MZ and MX+MY=XM+MZ= XZ. Because △XZC and △XDZ are similar, XCXZ​=XZXD​ or XZ2=XC⋅XD. Combining the last two equations yields (MX+MY)2=XZ2=XC⋅XD, establishing the claim.
The problems on this page are the property of the MAA's American Mathematics Competitions
