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Ukrainian National Mathematical Olympiad

Geometry Difficulty 6.7 National Olympiad Prove it Ukraine

Two circles $w_1, w_2$ are externally tangent at a point $Q$ . A common external tangent line to these circles (that doesn't pass through $Q$ ) is tangent to $w_1$ at a point $B$ , and $BA$ is a diameter of this circle. The point $A$ belongs to the line tangent to the circle $w_2$ at a point $C$ such that $B$ and $C$ are in the same half-plane with respect to the line $AQ$ . Prove that the circle $w_1$ bisects the segment $BC$ .

Fig. 6.

Solution

Let $K$ be the point where the common tangent touches the circle $w_2$ (Fig. 6). Consider the common tangent line to the two circles that passes through the point $Q$ . Suppose it intersects the line $BK$ at a point $P$ . By the properties of lines tangent to circles,
$PB = PQ = PK.$
Therefore, $\angle BQK = 90^\circ$ . Since $AB$ is a diameter of $w_1$ , we have that $\angle AQB = 90^\circ$ , and so the points $A, Q, K$ are collinear. It is given in the problem statement that $AB \perp BK$ , which implies that
$AB^2 = AQ \cdot AK \text{ and } AC^2 = AQ \cdot AK$
(from the properties of the right triangle $\triangle ABK$ and the properties of secant and tangent lines to the circle $w_2$ ). It then follows that $AB = AC$ . If $w_1$ intersects $BC$ at a point $M$ , then $\angle BMA = 90^\circ$ , and so $AM$ is the altitude of the isosceles triangle $\triangle ABC$ . This proves that $BM = MC$ , as required.

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