2015 AMC 10A Problem 19

From $\frac{1}{2}\ell^2 = 12.5$ the legs are $\ell = 5$, so the hypotenuse is $AB = 5\sqrt{2}$. Let $M$ be the midpoint of $AB$; then $CM \perp AB$ and $CM = \frac{1}{2}AB = \frac{5\sqrt{2}}{2}$.

The trisecting rays make angles of $30^\circ$ and $60^\circ$ with $CA$. The line $CM$ is the $45^\circ$ line, so it bisects $\angle DCE$ and, by the symmetry of the isosceles triangle, $M$ is the midpoint of $DE$. In right triangle $CMD$, $\angle MCD = 45^\circ - 30^\circ = 15^\circ$, so
$$DM = CM \tan 15^\circ.$$
Recall $\tan 15^\circ = \dfrac{1 - \cos 30^\circ}{\sin 30^\circ} = \dfrac{1 - \frac{\sqrt{3}}{2}}{\frac{1}{2}} = 2 - \sqrt{3}$.

The area of $\triangle CDE$ is $\frac{1}{2} \cdot DE \cdot CM = DM \cdot CM = CM^2 \tan 15^\circ$:
$$\left(\frac{5\sqrt{2}}{2}\right)^2 (2 - \sqrt{3}) = \frac{25}{2}(2 - \sqrt{3}) = \frac{50 - 25\sqrt{3}}{2}.$$

The answer is $\boxed{\textbf{(D)}\ \dfrac{50-25\sqrt{3}}{2}}$.

The altitude to the hypotenuse is the axis of symmetry of an isosceles right triangle, so it splits the angle-trisector triangle into two congruent right triangles with a $15^\circ$ angle. A single trig value, $\tan 15^\circ = 2 - \sqrt{3}$, then finishes the job. Whenever a figure has a mirror symmetry, cut along it: half the figure is usually a right triangle.

Place $C = (0,0)$, $A = (5,0)$, $B = (0,5)$. Ray $CD$ has direction $(\cos 30^\circ, \sin 30^\circ)$ and meets $x + y = 5$ at distance $CD = \frac{5}{\cos 30^\circ + \sin 30^\circ} = \frac{10}{\sqrt{3} + 1} = 5(\sqrt{3} - 1)$; by symmetry $CE = CD$. Then $[CDE] = \frac{1}{2} CD \cdot CE \sin 30^\circ = \frac{25(\sqrt{3}-1)^2}{4} = \frac{25(4 - 2\sqrt{3})}{4} = \frac{50 - 25\sqrt{3}}{2}$.

• Splitting $AB$ into three equal parts as if angle trisectors trisected the opposite side, giving area $\frac{25}{6}$ (choice E).
• Using $\tan 30^\circ$ instead of $\tan 15^\circ$ for the half-angle at $C$, or computing $CM$ as $5$ rather than $\frac{5\sqrt{2}}{2}$.