Problem 2

Let

A

be one of the two distinct intersection points of unequal coplanar circles

C_1,C_2

with centers

O_1,O_2

. One common tangent touches them at

P_1,P_2

, and the other at

Q_1,Q_2

. Let

M_1,M_2

be the midpoints of

P_1Q_1,P_2Q_2

. Prove that

\angle O_1AO_2=\angle M_1AM_2

Step 2 of 5: Use the tangent-point chord

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O_iM_i\perp P_iQ_i,\qquad O_iM_i\cdot O_iS=O_iP_i^2=O_iA^2

Detailed analysis

Reflection in $O_1O_2$ exchanges the two common tangents, hence it exchanges $P_i,Q_i$ . Thus $O_iO_j$ is the perpendicular bisector of $P_iQ_i$ , so $M_i$ lies on $O_1O_2$ . In the right triangle formed by $S,O_i,P_i$ , the altitude to the hypotenuse gives $O_iM_i\cdot O_iS=O_iP_i^2$ . Since $A,P_i$ lie on $C_i$ , $O_iP_i=O_iA$ .