15th Czech-Polish-Slovak Mathematics Competition

Let $E$ be the intersection point of the circle $l$ and the line $AC$ ($E\ne A$). Since $k$ lies in the half-plane $ACB$ and $l$ lies in the half-plane $ABC$, the point $X$ lies inside of the angle $BAC$ (Fig. 4). Using the well known fact about the angle between a tangent and a chord of a given circle we get $\angle XAE=\angle XBA$ and $\angle XAB=\angle XEA$. Therefore the triangles $ABX$ and $EAX$ are similar.

Denote by $\gamma$ the measure of $\angle ACB$. We have $\angle AOB=2\gamma$, as $\angle AOB$ is the central angle corresponding to the inscribed angle of measure $\gamma$ in the circumcircle of the triangle $ABC$.

The line $BH$ passes through the centre of $l$ and is perpendicular to its chord $AE$, so it is in fact the axis of symmetry of the chord $AE$. Therefore $AH=HE$, and

from the isosceles triangle $EAH$ with $AH\perp BC$, we obtain $\angle EAH=90^{\circ }-\gamma$, hence $\angle AHE=2\gamma$.



Fig. 4

The triangles $ABO$ and $EAH$ are both isosceles, and have the vertex angle of the same measure, so they are similar. We have two pairs of similar triangles with the same ratio of similitude $AB:EA$. Since $O$ and $X$ lie on the same side of $AB$, and $H$ and $X$ lie on the same side of $EA$, the quadrilaterals${}^1$ $ABXO$ and $EAXH$ are similar.

Consider the rotation with centre $X$ which maps the ray $XB$ onto the ray $XA$. With respect to the derived similarity, the ray $XO$ is mapped onto the ray $XH$ under this rotation. Therefore we have

$$\angle HXO=\angle AXB=180^{\circ }-\angle BAC.$$

The last identity follows from the fact that $\angle AXB$ is the inscribed angle corresponding to the chord $AB$ of $k$, while $\angle CAB$ is the angle between this chord and its tangent, and both $X$ and $C$ lie in the same half-plane determined by this chord.

${}^1$ These quadrilaterals may be degenerate – three of the four vertices may be collinear.

Remark. Instead of rotation, one may consider the spiral similarity which maps $ABXO$ onto $EAXH$. Since it maps $BA$ onto $AC$, it is clear that the rotating part of this map rotates all the lines by angle $180^{\circ }-\angle BAC$.

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Alternative solution.

(Outline.) It is possible to solve the problem using coordinates. Let $A$ be the origin and $B$ lies on the $x$-axis. Denote by $b$ the first coordinate of $B$ and let $(c,v)$ be the coordinates of $C$. Routine calculations give

$$\begin{gathered} A=(0,0),B=(b,0),C=(c,v),H=\left(c,\frac{c(b-c)}{v}\right),O=\left(\frac{1}{2}b,\frac{c^2+v^2-bc}{2v}\right), \\ {S}_k=\left(\frac{1}{2}b,-\frac{bc}{2v}\right),S_l=\left(0,\frac{bc}{v}\right),X=\left(\frac{6b{c}^2}{9c^2+v^2},\frac{2bcv}{9c^2+v^2}\right),Y=\left(\frac{1}{2}b,\frac{bc}{2v}\right). \end{gathered}$$

Here, $S_k$ and $S_l$ are the centres of $k$ and $l$, respectively, and $Y$ is the intersection point of the line $BH$ and the line $O{S}_k$ (these two lines are perpendicular to $AC$ and $AB$, respectively, hence they form the angle of the same measure as $\angle BAC$).

Instead of expressing the measure of $\angle HXO$, we will show that $X,O,H$, and $Y$ are concyclic, that is, the determinant

$$\left|\begin{array}{cccc}\frac{(6b{c}^2{)}^2+(2bcv{)}^2}{(9c^2+v^2{)}^2}& \frac{6b{c}^2}{9c^2+v^2}& \frac{2bcv}{9c^2+v^2}& 1\\ \frac{b^2}{4}+\frac{(c^2+v^2-bc{)}^2}{4v^2}& \frac{b}{2}& \frac{c^2+v^2-bc}{2v}& 1\\ c^2+\frac{c^2(b-c{)}^2}{v^2}& c& \frac{c(b-c)}{v}& 1\\ \frac{b^2}{4}+\frac{b^2{c}^2}{4v^2}& \frac{b}{2}& \frac{bc}{2v}& 1\end{array}\right|$$

evaluates to zero. In fact, this is an easy exercise, provided we know basic tricks from linear algebra (adding a scalar multiple of one row to another row does not change the value of the determinant; the same is valid for columns; when checking only zero value, we can multiply any row/column by a nonzero scalar).

It remains to show that among the two possible values, $\angle BAC$ and $180^{\circ }-\angle BAC$, of an inscribed angle corresponding to the chord $HO$ of the circumcircle of the triangle $HOY$, the latter always apply for $\angle HXO$. Also, the case $Y=O$ or $Y=H$ should be handled separately. One may use some kind of continuity arguments to show that we cannot "jump" from one value to another.