2005 AMC 10A Problem 23

Let $O$ be the center. Scale so that $AC = 2$ and $BC = 4$; then $AB = 6$, the radius is $3$, and $CO = AO - AC = 3 - 2 = 1$.

Step 1: Both diameters pass through $O$, so $O$ is the midpoint of $DE$. In $\triangle DCE$ the segment $CO$ is a median, and a median splits a triangle into two pieces of equal area. Hence
$$[DCE] = 2\,[DCO].$$

Step 2: Triangles $DCO$ and $ABD$ both have a base on line $AB$ ($CO$ and $AB$ respectively) and share the same apex $D$, so both have height $DC$. Their areas are in the ratio of their bases:
$$\frac{[DCO]}{[ABD]} = \frac{CO}{AB} = \frac{1}{6}.$$

Combining,
$$\frac{[DCE]}{[ABD]} = \frac{2\,[DCO]}{[ABD]} = \frac{2}{6} = \frac{1}{3}.$$

The answer is $\boxed{\textbf{(C)}\ \frac{1}{3}}$.

Area ratios rarely need actual lengths. Two tools do the work: a median halves area, and triangles sharing an altitude compare by base. The right angle at $C$ and the value of $DC$ are red herrings; only the positions $C$ and $O$ along the diameter matter, which is why the answer is a clean $\frac{1}{3}$.

Compute: $DC^2 = AC \cdot CB = 8$, so $DC = 2\sqrt{2}$ and $[ABD] = \frac{1}{2}\cdot 6 \cdot 2\sqrt{2} = 6\sqrt{2}$. In right triangle $DCO$ (legs $1$ and $2\sqrt{2}$, hypotenuse $3$) the altitude from $C$ to $DO$ is $\frac{1 \cdot 2\sqrt{2}}{3}$, so $[DCE] = \frac{1}{2} \cdot 6 \cdot \frac{2\sqrt{2}}{3} = 2\sqrt{2}$. Ratio $\frac{2\sqrt{2}}{6\sqrt{2}} = \frac{1}{3}$.

• Computing DC via power of a point and then working out the altitude from C to DE with messy radicals, instead of comparing triangles that share the altitude DC.
• Placing $C$ at $AC = \frac{1}{3}$ of the radius rather than of the diameter, which changes $CO$ and produces $\frac{1}{6}$ or $\frac{1}{2}$.