12. The answer is (A). Let $I$ be the incenter of triangle $ABC$; then triangles $AIB$ and $X I Y$ are similar due to the parallelism between lines $AB$ and $XY$. Let $h$ be the height of trapezoid $ABXY$, which is also the height of triangle $ABC$ relative to base $AB$; the height of triangle $AIB$ relative to $AB$ is the radius of the inscribed circle in $ABC$, whose length is $S_{ABC} / p_{ABC} = S_{ABC} / 34$. On the other hand, $h = 2 S_{ABC} / 17$. Thanks to the similarity mentioned above, $X_{Y}: AB = XY: 17 = (2 S_{ABC} / 17 - S_{ABC} / 34): S_{ABC} / 34$; hence $XY = 3 \cdot 17 = 51$. The area of the trapezoid is therefore $\frac{1}{2}(17 + 51) h = 34 h$.

The height $h$ can be calculated as $\frac{2}{AB} \sqrt{p(p-AB)(p-BC)(p-AC)}$, using Heron's formula, where $p$ is the semiperimeter of $ABC$; substituting the lengths of the sides gives $h = 24$, and thus $S_{ABXY} = 816$. Alternatively, $h$ can be calculated using the Pythagorean theorem: let $x$ and $y$ be the projections of the height onto $AB$, we know that $x + y = 17$, $x^2 + h^2 = 25^2$, and $y^2 + h^2 = 26^2$, from which $y^2 - x^2 = (x + y)(y - x) = 17(x - y) = 51$. Solving the system for $x$ and $y$ gives, for example, $y = 10$, and thus $h = \sqrt{26^2 - 10^2} = 24$.
12. Let $A B C$ be a triangle with sides $A B, C A$, and $B C$ of lengths 17, 25, and 26, respectively. Let $X$ and $Y$ be the intersections of the line parallel to $A B$ passing through $C$ with the angle bisectors of $C \widehat{A} B$ and $A \widehat{B} C$, respectively. What is the area of the trapezoid $A B X Y$?
(A) 816
(B) $338(1+\sqrt{2})$
(C) 784
(D) 408
(E) It cannot be determined with the given data.

## Numerical Answer Problems - 5 points