63rd Czech and Slovak Mathematical Olympiad

For any point $X$ of the plane $ABC$, let $X_a,X_b$ and $X_c$ denote the reflections of $X$ through the lines $BC$, $CA$ and $AB$, respectively (Fig. 5). First we prove that the distances between any two of the points $X_a,X_b$ and $X_c$ are given in general by formulæ $$|X_a{X}_b|=2|XC|\sin \gamma ,|X_a{X}_c|=2|XB|\sin \beta ,|X_b{X}_c|=2|XA|\sin \alpha ,(1)$$ in which $\alpha ,\beta ,\gamma$ denote the interior angles of the triangle $ABC$ as usual. Fig. 5 It suffices to prove the first equality (1) which is obvious if $X=C$, because then $X_a=X_b(=X)$. If $X\ne C$, then the segment $XC$ is a diameter of a circle (see Fig. 5) which passes through the marked orthogonal projections $P_a$ and $P_b$ of $X$ onto $BC$ and $CA$, respectively (Thales' theorem). Since the chord $P_a{P}_b$ subtends inscribed angles $\gamma$ and $180^{\circ }-\gamma$, Law of Sines implies that $|P_a{P}_b|=|XC|\sin \gamma$. Using the homothety with centre $X$ and ratio $2$, we conclude that $|X_a{X}_b|=2|P_a{P}_b|$, and hence the equalities (1) are established for any point $X$.

The proved formulæ (1) imply that our task is to find exactly such points $X$ in the plane $ABC$ that satisfy $$2|XA|\sin \alpha =2|XB|\sin \beta =2|XC|\sin \gamma >0$$ (recall that the triangle $X_a{X}_b{X}_c$ has to be equilateral). Otherwise speaking, we look for all points $X$ whose distances to $A$, $B$ and $C$ are positive and proportional as follows: $$|XA|:|XB|:|XC|=\frac{1}{\sin \alpha }:\frac{1}{\sin \beta }:\frac{1}{\sin \gamma }=\frac{1}{|BC|}:\frac{1}{|AC|}:\frac{1}{|AB|}$$ (we have turned from angles to sides of $△ABC$ using Law of Sines again). Such points $X$ are determined as common points of the following three circles of Apollonius (i.e. sets of points in the plane which have a specified ratio of distances to two fixed points): $$k_a:\frac{|XB|}{|XC|}=\frac{|AB|}{|AC|},k_b:\frac{|XA|}{|XC|}=\frac{|AB|}{|BC|},k_c:\frac{|XA|}{|XB|}=\frac{|AC|}{|BC|}(2)$$ It is clear that any point shared by two of the circles lies on the third circle as well. It follows from (2) that $A\in k_a$, $B\in k_b$ and $C\in k_c$, which simplifies the construction of the three circles in practice: If the bisectors of interior angles in $△ABC$ cut its interior in segments $AK$, $BL$ and $CM$ (Fig. 6), then $K\in k_a$, $L\in k_b$ and $M\in k_c$ (an immediate consequence of the well known proportions such as $|KB|:|KC|=|AB|:|AC|$). Hence the centre of $k_a$ can be constructed as the intersection point of the line $BC$ and the perpendicular bisector of the segment $AK$ (excluding the case $|AB|=|AC|$, when $k_a$ becomes simply the perpendicular bisector of $BC$). Similarly, using the perpendicular bisectors of $BL$ and $CM$ we get centres of $k_b$ and $k_c$, respectively.



Fig. 6

Despite of the fact that the requested locus of points $X$ is determined (by an Euclidean construction), we have to discuss how the number of solutions depends on

a) If the triangle $ABC$ is equilateral, the "circles" $k_a,k_b,k_c$ are in fact perpendicular bisectors of the sides of $△ABC$. Consequently, the problem has a unique solution — a point $X$ which coincides with the incentre of $△ABC$.

b) If the triangle $ABC$ is isosceles (but not equilateral), say if $|AB|\ne |AC|=|BC|$, then the circle $k_c$ is a perpendicular bisector of the base $AB$ which meets the circle $k_a$ in two points, because $k_c$ meets the interior of the chord $AK$, and hence the both arcs $AK$ of the circle $k_a$ as well. Consequently, the problem has two solutions.

c) Suppose that the triangle $ABC$ is scalene, with the largest side, say $AB$ (as in Fig. 6). Then the ratio $|XB|/|XC|$ for points $X\in k_a$ is larger than $1$, because of $A\in k_a$. Hence $B$ lies in the interior $k_a$, while $C$ lies in its exterior. The last together with $A\in k_a$ implies that $L$, an interior point of $AC$, lies in the exterior of $k_a$. Thus $k_a$ intersects the chord $BL$ of $k_b$ which means that $k_a$ and $k_b$ meet in two points. Consequently, the problem has two solutions.