Problem 2006

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 $p \neq I$ 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 $PX'Z'$ is similar to $IXZ$, and since $P$ is closer to $AB$ than $I$ is, the perimeter of $PX'Z'$ 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 $A_1B_1C_1$, $A_2B_2C_2$ and $A_3B_3C_3$ (with $P = C_1 = A_2 = B_3$) as shown. The sum of the areas of the triangles is given by the expression
$$\frac{1}{2}a_1b_1 + \frac{1}{2}a_2b_2 + \frac{1}{2}a_3b_3.$$

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
$$a_1 : a_2 : a_3 = b_1 : b_2 : b_3 = t_1 : t_2 : t_3 \quad \text{with} \quad t_1 + t_2 + t_3 = 1.$$
It therefore follows that
$$\begin{aligned} \frac{1}{2}a_1b_1 + \frac{1}{2}a_2b_2 + \frac{1}{2}a_3b_3 &= \frac{1}{2}h_c \cdot c \cdot (t_1^2 + t_2^2 + t_3^2) \\ &\geq h_c \cdot c \cdot \left( \frac{t_1 + t_2 + t_3}{2} \right)^2 \\ &= \frac{h_c \cdot c}{4}, \end{aligned}$$
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