MathDB

In the adjoining figure, the circle meets the sides of an equilateral triangle at six points. If

AG=2

GF=13

FC=1

, and

HJ=7

, then

DE

equals
[asy] size(200); defaultpen(fontsize(10)); real r=sqrt(22); pair B=origin, A=16*dir(60), C=(16,0), D=(10-r,0), E=(10+r,0), F=C+1*dir(120), G=C+14*dir(120), H=13*dir(60), J=6*dir(60), O=circumcenter(G,H,J); dot(A^^B^^C^^D^^E^^F^^G^^H^^J); draw(Circle(O, abs(O-D))^^A--B--C--cycle, linewidth(0.7)); label("

A

", A, N); label("

B

", B, dir(210)); label("

C

", C, dir(330)); label("

D

", D, SW); label("

E

", E, SE); label("

F

", F, dir(170)); label("

G

", G, dir(250)); label("

H

", H, SE); label("

J

", J, dir(0)); label("2", A--G, dir(30)); label("13", F--G, dir(180+30)); label("1", F--C, dir(30)); label("7", H--J, dir(-30));[/asy]

\textbf {(A) } 2\sqrt{22} \qquad \textbf {(B) } 7\sqrt{3} \qquad \textbf {(C) } 9 \qquad \textbf {(D) } 10 \qquad \textbf {(E) } 13

Problem Statement