2015 AMC 10A Problem 19

The figure has a mirror line, so cut along it and work with a right triangle.

1. The sides. The area is $\frac{1}{2} \times \text{leg} \times \text{leg} = 12.5$, so each leg squared is $25$ and each leg is $5$. The long side is $AB = 5\sqrt{2}$.
2. The mirror line. Let $M$ be the middle of $AB$. The segment $CM$ is the height from $C$, it meets $AB$ at a right angle, and it is half of $AB$ (the middle of the long side of a right triangle is the same distance from all three corners):
$$CM = \frac{5\sqrt{2}}{2}.$$
3. The angles. Trisecting the $90^\circ$ angle gives three $30^\circ$ angles. $CM$ cuts the $90^\circ$ angle in half, at $45^\circ$. So the middle $30^\circ$ angle, $\angle DCE$, is split by $CM$ into two angles of $15^\circ$. By the mirror symmetry, $M$ is also the middle of $DE$.
4. One half. Triangle $CMD$ has a right angle at $M$ and a $15^\circ$ angle at $C$. So
$$DM = CM \times \tan 15^\circ.$$
5. Find $\tan 15^\circ$. Draw a $30$-$60$-$90$ triangle with short leg $1$, long leg $\sqrt{3}$ and long side $2$. Extend the long leg past the $30^\circ$ corner by $2$ and join the new point to the far end of the short leg. That makes an isosceles triangle, so the new angle is half of $30^\circ$, which is $15^\circ$. In the big right triangle, the side across from the $15^\circ$ angle is $1$ and the side next to it is $2 + \sqrt{3}$, so
$$\tan 15^\circ = \frac{1}{2 + \sqrt{3}} = 2 - \sqrt{3}.$$
6. The area. Triangle $CDE$ has base $DE = 2 \times DM$ and height $CM$:
$$\text{Area} = \frac{1}{2} \times 2\,DM \times CM = CM^2 \tan 15^\circ = \frac{25}{2}\,(2 - \sqrt{3}) = \frac{50 - 25\sqrt{3}}{2}.$$

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

When a figure has a mirror line, cutting along it usually leaves a right triangle, which is the easiest shape to calculate with. Here that right triangle has a $15^\circ$ angle, so one value, $\tan 15^\circ = 2 - \sqrt{3}$, finishes the problem. It is worth knowing that value, or the quick $30$-$60$-$90$ picture that produces it.

Use coordinates. Put $C = (0,0)$, $A = (5,0)$ and $B = (0,5)$, so $AB$ lies on $x + y = 5$. The ray at $30^\circ$ goes in the direction $(\frac{\sqrt{3}}{2}, \frac{1}{2})$ and reaches the line $x + y = 5$ after a distance of
$$CD = \frac{5}{\frac{\sqrt{3}}{2} + \frac{1}{2}} = \frac{10}{\sqrt{3} + 1} = 5(\sqrt{3} - 1).$$
By symmetry $CE$ is the same. The angle between them is $30^\circ$, so
$$\text{Area} = \frac{1}{2} \cdot CD \cdot CE \cdot \sin 30^\circ = \frac{25(\sqrt{3} - 1)^2}{4} = \frac{25(4 - 2\sqrt{3})}{4} = \frac{50 - 25\sqrt{3}}{2}.$$

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