17. The measures of the sides of triangle $ABC$ are: $\overline{AC}=\overline{BC}=10 \text{~cm}, \overline{AB}=12 \text{~cm}$. How many cm does the part of the perimeter of $ABC$ formed by the points for which the distance from $A$ is less than the distance from $C$ measure?
(A) $27 / 2$
(B) 16
(C) 12
(D) $25 / 2$
(E) $40 / 3$
Problem 17. The answer is (E).

Let $r$ be the perpendicular bisector of segment $A C$: the part of the perimeter we are interested in is precisely the points on the perimeter belonging to the half-plane determined by $r$ containing $A$. Let then $R$ and $S$ be the points where this perpendicular bisector intersects the perimeter of the triangle, with $R$ belonging to $A C$; it is evident that $S$ belongs to the side $A B$, since point $B$ is $10 \, \text{cm}$ from $C$ and $12 \, \text{cm}$ from $A$. Let $H$ be the projection of $C$ onto $A B$ and $x = S H$, so we have $R C = 5$ and $C S = A S = 6 + x$. Applying the Pythagorean theorem to triangle $A H C$ and triangle $C H B$, we find: $C H = \sqrt{10^{2} - 6^{2}} = 8$ and

$$
(6 + x)^{2} = 8^{2} + x^{2}
$$

from which $x = \frac{7}{3}$. The part of the perimeter in question is therefore $R A + A H + H S = 5 + 6 + \frac{7}{3} = \frac{40}{3}$. (Problem proposed by P. Negrini)