Austria 2010

a) Let $I$ be the incenter of $ABC$. Let $X$ be the common point of $AB$ with the line through $I$ parallel to $CA$, and $Y$ be the common point of $CA$ with the line through $I$ parallel to $AB$. $AXIY$ is a parallelogram, and since $I$ is the incenter of $ABC$, we have $\angle IAX=\angle IAY$. Since $\angle IAX=\angle AIX$ must also hold in the parallelogram $AXIY$, we see that $\angle IAX=\angle AIX$ holds. The triangle $AIX$ is therefore isosceles with $XA=XI$. If $Z$ denotes the common point of $AB$ with the line through $I$ parallel to $BC$, we similarly obtain $ZB=ZI$, and it therefore follows that $$XI+XZ+ZI=XA+XZ+ZB=AB$$ holds as claimed.

b) We assume that a point $p\ne I$ with this property exists. Such a point must lie between one of the sides of the triangle and the line parallel to this side through $I$. Without loss of generality, we assume it lies between $AB$ and $YI$. The triangle $P{X}^{\prime }{Z}^{\prime }$ is similar to $IXZ$, and since $P$ is closer to $AB$ than $I$ is, the perimeter of $P{X}^{\prime }{Z}^{\prime }$ is certainly smaller than that of $IXZ$, which is equal to the length of $AB$. $P$ therefore does not fulfill the required condition. We see that $I$ is the only point with this property. qed

c) The point $P$ determines three triangles $A_1{B}_1{C}_1$, $A_2{B}_2{C}_2$ and $A_3{B}_3{C}_3$ (with $P=C_1=A_2=B_3$) as shown. The sum of the areas of the triangles is given by the expression $$\frac{1}{2}{a}_1{b}_1+\frac{1}{2}{a}_2{b}_2+\frac{1}{2}{a}_3{b}_3.$$ Since $b_1+b_2+b_3=c=|AB|$ and $a_1+a_2+a_3=h_c$ obviously hold, and all three triangles are similar to $ABC$, we have $$a_1:a_2:a_3=b_1:b_2:b_3=t_1:t_2:t_3\text{with}{t}_1+t_2+t_3=1.$$ It therefore follows that $$\begin{array}{rl}\frac{1}{2}{a}_1{b}_1+\frac{1}{2}{a}_2{b}_2+\frac{1}{2}{a}_3{b}_3& =\frac{1}{2}{h}_c\cdot c\cdot (t_1^2+t_2^2+t_3^2)\\ & \ge h_c\cdot c\cdot {\left(\frac{t_1+t_2+t_3}{2}\right)}^2\\ & =\frac{h_c\cdot c}{4},\end{array}$$ with equality holding iff $t_1=t_2=t_3=\frac{1}{3}$. The sum of the areas is therefore minimized if the distance of $P$ from each of the sides is equal to one third of each altitude. This is the case for the centroid of $ABC$, and we see that this is the point with the required property. qed