HMMT February

Geometry Difficulty 4.6 AIME Find the answer United States

Problem:

Right triangle $\triangle DEF$ with $\angle D = 90^{\circ}$ and $\angle F = 30^{\circ}$ is inscribed in equilateral triangle $\triangle ABC$ such that $D$ , $E$ , and $F$ lie on segments $\overline{BC}$ , $\overline{CA}$ , and $\overline{AB}$ , respectively. Given that $BD = 7$ and $DC = 4$ , compute $DE$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solutions — 2

Solution 1

Solution:

From $\angle E = 60^{\circ}$ , we get that $\angle AEF = 120^{\circ} - \angle CED = \angle CDE$ . Therefore, $\triangle AEF \sim \triangle CDE$ . Since $EF:DE = 2:1$ , the ratio of similarity must be $2:1$ , so $AE = 2CD = 8$ . Recall $ABC$ has side length $7 + 4 = 11$ , so $EC = 11 - 8 = 3$ . Law of Cosines on $\triangle CDE$ gives $DE^2 = \sqrt{3^2 + 4^2 - 3\cdot 4} = \sqrt{13}$ .

Solution 2

Solution:

Let $\odot (DEF)$ meet $AC$ again at point $X$ . Then, $\angle FXA = 180^{\circ} - \angle FXE = \angle FDE = 90^{\circ}$ and $\angle XDC = 180^{\circ} - \angle DCX - \angle DXC = 120^{\circ} - \angle DXE = 120^{\circ} - \angle DFE = 90^{\circ}$ . It follows that $CX = 2CD = 8$ , so $AX = 11 - CX = 3$ , and $AF = 2AX = 6$ . Thus, Law of Cosines on $\triangle AEF$ gives $EF = \sqrt{8^2 + 6^2 - 8\cdot 6} = 2\sqrt{13}$ , implying that $DE = \sqrt{13}$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.