63rd Czech and Slovak Mathematical Olympiad

First we show in two different ways that the line $MI$, a perpendicular to $CI$ through $I$, is tangent to the circle $ABI$. The first way is based on the known fact that $\angle AIC$ and $\angle BIC$ are obtuse angles of measures $90^{\circ }+\frac{1}{2}\beta$ and $90^{\circ }+\frac{1}{2}\alpha$, respectively (in common notation for interior angles of $△ABC$). This fact implies that the line $MI$ forms acute angles $\frac{1}{2}\beta$ and $\frac{1}{2}\alpha$ with the segments $AI$ and $BI$, respectively\textsuperscript{1}, hence the angles congruent with angles $IBA$ and $IAB$ in circle $ABI$ (Fig. 1). Well known properties of inscribed and subtended angles lead to the conclusion that the line $MI$ is tangent to the circle $ABI$. The second reason for this conclusion is based on the known fact that the centre of $ABI$ is the midpoint of the arc $AB$ of circle $ABC$ which lies on the ray $CI$ bisecting $\angle ACB$.



Fig. 1

From the proved tangency of $MI$ to $ABI$ it follows that the point $M$ lies on the line $AB$ outside of the segment $AB$. Moreover, the power $m$ of $M$ with respect to $ABI$ is positive and given by $m=|MI{|}^2=|MA|\cdot |MB|$. Hence $M$ lies in the exterior of the circle $ABC$ (as $AB$ is its chord) and the power of $M$ with respect to $ABC$ is the same $m=|MA|\cdot |MB|$. Since $m=|MI{|}^2<|MC{|}^2$ from the right-angled triangle $CMI$, it holds that $|MA|\cdot |MB|<|MC{|}^2$. This means that the circle $ABC$ intersects the segment $MC$ in an interior point $N$, because $|MN|\cdot |MC|=|MA|\cdot |MB|$ implies that $|MN|<|MC|$ for the second point $N$ of intersection of the ray $MC$ with $ABC$. This proves the first conclusion of the problem.

To show that $\angle CNI$ is a right angle, we use the proved equality $|MC|\cdot |MN|=|MI{|}^2$ and apply a familiar theorem to the leg $MI$ of the right-angled triangle $CMI$: Its altitude from the vertex $I$ meets the hypotenuse $CM$ in such a point $X$ which is determined by equation $|MC|\cdot |MX|=|MI{|}^2$. Thus we have $X=N$ in our case and the solution is complete.

\textsuperscript{1} As a consequence we can see that the assumed existence of the intersection point $M$ is equivalent to the inequality $\frac{1}{2}\alpha \ne \frac{1}{2}\beta$ or $\alpha \ne \beta$. Due to the symmetry we can assume that $\alpha >\beta$ as in our figure; the point $M$ then lies on the ray opposite to ray $AB$ and satisfies $|\angle IMA|=\frac{1}{2}\alpha -\frac{1}{2}\beta$.