Serbian Mathematical Olympiad

Solution:

Recall that points $P$ and $Q$ inside $\triangle ABC$ are called isogonally conjugate if $\text{PAB = QAC}$ and $\text{PBC = QBA}$. Then it also holds that $\text{PCA = QCB}$.

Lemma. The feet of the perpendiculars from points $P$ and $Q$ to the lines $BC, CA$ and $AB$ lie on the same circle.

Proof. Let $P_{a}$ and $Q_{a}$ be, respectively, the feet of the perpendiculars from $P$ and $Q$ to $BC$; similarly we denote $P_{b}, Q_{b}, P_{c}, Q_{c}$. From $\text{AP b P c = APP c = AQQ b = AQ c Q b}$ it follows that $\triangle AP_{b}P_{c} \sim \triangle AQ_{c}Q_{b}$, so the points $P_{b}, P_{c}, Q_{b}, Q_{c}$ lie on the same circle $k$, whose center is the intersection of the perpendicular bisectors of the segments $P_{b}Q_{b}$ and $P_{c}Q_{c}$, which is precisely the midpoint $U$ of the segment $PQ$. Similarly, the points $P_{c}, P_{a}, Q_{c}, Q_{a}$ are also equidistant from the point $U$, so both $P_{a}$ and $Q_{a}$ also lie on the circle $k$.

Suppose that the perpendiculars from $X, Y$ and $Z$ to $BC, CA$ and $AB$, respectively, intersect at the point $P$. If the point $Q$ is isogonally conjugate to the point $P$ in $\triangle ABC$, then by the Lemma the feet of the perpendiculars from $Q$ to $BC, CA$ and $AB$ are precisely the points $A_{1}, B_{1}$ and $C_{1}$.

Let us denote by $A_{0}, B_{0}$ and $C_{0}$, respectively, the intersections of the internal angle bisectors at $A, B$ and $C$ with the opposite sides. As usual, $BC = a, CA = b$ and $AB = c$. From the ratio $BA_{0} : A_{0}C = c : b$ we find $BA_{1} = A_{0}C = \frac{ab}{b+c}$ and, similarly, $A_{1}C = \frac{ac}{b+c}, CB_{1} = \frac{bc}{c+a}, B_{1}A = \frac{ba}{c+a}, AC_{1} = \frac{ca}{a+b}$ and $C_{1}B = \frac{cb}{a+b}$. We now have

$$\begin{aligned} 0 & = \left(BA_{1}^{2} - A_{1}C^{2}\right) + \left(CB_{1}^{2} - B_{1}A^{2}\right) + \left(AC_{1}^{2} - C_{1}B^{2}\right) \\ & = \frac{a^{2}(b-c)}{b+c} + \frac{b^{2}(c-a)}{c+a} + \frac{c^{2}(a-b)}{a+b} \\ & = \frac{a^{4}(b-c) + b^{4}(c-a) + c^{4}(a-b) - (b-c)(c-a)(a-b)(ab+bc+ca)}{(b+c)(c+a)(a+b)} \\ & = -\frac{(b-c)(c-a)(a-b)(a+b+c)^{2}}{(b+c)(c+a)(a+b)} \end{aligned}$$

from which it follows that $a = b$ or $a = c$ or $b = c$.

Second solution. As in the first solution, $BA_{1} = A_{0}C = \frac{ab}{b+c}, A_{1}C = \frac{ac}{b+c}, CB_{1} = \frac{bc}{c+a}$, $B_{1}A = \frac{ba}{c+a}, AC_{1} = \frac{ca}{a+b}$ and $C_{1}B = \frac{cb}{a+b}$. Let us denote $BX = x, CY = y$ and $AZ = z$. The power of the point $A$ gives $AB_{1} \cdot AY = AC_{1} \cdot AZ$, i.e. $\frac{by}{c+a} + \frac{cz}{a+b} = \frac{b^{2}}{c+a}$. Similarly we obtain $\frac{cz}{a+b} + \frac{ax}{b+c} = \frac{c^{2}}{a+b}$ and $\frac{ax}{b+c} + \frac{by}{c+a} = \frac{a^{2}}{b+c}$. From this it follows that $\frac{2ax}{b+c} = \frac{a^{2}}{b+c} + \frac{c^{2}}{a+b} - \frac{b^{2}}{c+a}$ which reduces to $x = \frac{1}{2}a - \frac{(b+c)(b-c)(b^{2} + c^{2} + ab + ac + bc)}{2a(a+b)(a+c)}$, etc. The condition that the three perpendiculars are concurrent is

$$\begin{aligned} 0 & = (a+b)(a+c)(b+c)\left[x^{2} - (a-x)^{2} + y^{2} - (b-y)^{2} + z^{2} - (c-z)^{2}\right] \\ & = (b+c)^{2}(c-b)(T - a^{2}) + (c+a)^{2}(a-c)(T - b^{2}) + (a+b)^{2}(b-a)(T - c^{2}) \\ & = a^{2}(b+c)^{2}(b-c) + b^{2}(c+a)^{2}(c-a) + c^{2}(a+b)^{2}(a-b) - (a-b)(b-c)(c-a)T \\ & = -(a-b)(b-c)(c-a)(a+b+c)^{2} \end{aligned}$$

where $T = a^{2} + b^{2} + c^{2} + ab + bc + ca$.