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Balti Tee 2011 · Valikvooru ülesanne

Geomeetria

Let AA and BB be two circles, external to each other. Let ℓ\ell be a line not meeting the circles. For any point XX on ℓ\ell, let EE be a point of contact of a tangent to AA through XX, and FF a point of contact of a tangent to BB through XX. Find the position of XX on ℓ\ell such that EX+FXEX + FX is minimized.

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Lahendus

Denote the centres of AA and BB by AA and BB, respectively. Let CC and DD be the feet of the perpendiculars from AA and BB to ℓ\ell. Let GG be a point of contact of AA and a tangent to AA from CC. Define HH similarly on BB. Diagram for the mathnet 018u 1 of bw-cand-2011-mn-018u. Now, by Pythagoras,

XE2=AX2−AE2=XC2+CA2−AE2=XC2+AG2+GC2−AE2=XC2+CG2.XE^2 = AX^2 - AE^2 = XC^2 + CA^2 - AE^2 = XC^2 + AG^2 + GC^2 - AE^2 = XC^2 + CG^2.

Now choose a point MM on CACA such that CM=CGCM = CG. (MM is a point of intersection of CACA and the circle with center CC through GG). Then XE=XMXE = XM. Similarly, if NN is the point on DBDB such that DN=DHDN = DH and MM, NN lie on different sides of ℓ\ell, then XF=XNXF = XN. So minimizing EX+FXEX + FX is equivalent to minimizing MX+XNMX + XN. Clearly, if PP is the point of intersection of the lines ℓ\ell and MNMN, then PP solves the problem.