2005 AMC 10A Problem 23

P21-25~5 minPrint

Circles Triangles: Area & Pythagorean Cut the figure into known shapes (triangles, rectangles, sectors)Exploit symmetry to reduce work or pair up objects

Let $\overline{AB}$ be a diameter of a circle and $C$ be a point on $\overline{AB}$ with $2 \cdot AC = BC$ . Let $D$ and $E$ be points on the circle such that $\overline{DC} \perp \overline{AB}$ and $\overline{DE}$ is a second diameter. What is the ratio of the area of $\triangle DCE$ to the area of $\triangle ABD$ ?

Show solution without answering

Key insight

O is the midpoint of DE, so triangle DCE is twice triangle DCO. Triangles DCO and ABD share the height DC, so compare their bases CO and AB.

Solution

Pick easy numbers. Since $BC$ is twice $AC$ , let $AC = 2$ and $BC = 4$ . Then the diameter $AB = 6$ , the radius is $3$ , and the center $O$ is $3$ from $A$ . So $C$ is between $A$ and $O$ , with $CO = 3 - 2 = 1$ .

1. Split triangle $DCE$ at $O$ . Both diameters pass through the center, so $O$ is the midpoint of $DE$ . The segment $CO$ splits triangle $DCE$ into two triangles, $DCO$ and $ECO$ . They have equal bases ( $DO = OE$ ) and the same height from $C$ , so they have equal areas. Therefore
$\text{area of } DCE = 2 \times \text{area of } DCO.$
2. Compare triangle $DCO$ with triangle $ABD$ . Both have their base on line $AB$ : base $CO = 1$ for one and base $AB = 6$ for the other. Both have their top at $D$ . Since $DC$ is perpendicular to $AB$ , both have the same height, $DC$ . Triangles with the same height compare by their bases:
$\frac{\text{area of } DCO}{\text{area of } ABD} = \frac{1}{6}.$
3. Put the two steps together.
$\frac{\text{area of } DCE}{\text{area of } ABD} = \frac{2 \times \text{area of } DCO}{\text{area of } ABD} = 2 \times \frac{1}{6} = \frac{1}{3}.$

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

Why this works

Area ratios rarely need actual lengths. Two tools did all the work: a line from a corner to the middle of the opposite side cuts a triangle into two equal halves, and triangles with the same height compare by their bases. The length of $DC$ was never needed because it appears in both areas and cancels.

Alternative approach

Compute every area. With $AC = 2$ and $BC = 4$ , in right triangle $DCO$ the hypotenuse $DO$ is the radius $3$ and one leg is $CO = 1$ , so $DC = \sqrt{9 - 1} = 2\sqrt{2}$ . Then the area of $ABD$ is $\frac{1}{2} \cdot 6 \cdot 2\sqrt{2} = 6\sqrt{2}$ . The area of $DCO$ is $\frac{1}{2} \cdot 1 \cdot 2\sqrt{2} = \sqrt{2}$ , so the area of $DCE$ is $2\sqrt{2}$ . The ratio is $\frac{2\sqrt{2}}{6\sqrt{2}} = \frac{1}{3}$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Geometry, Chapter 11: Circles
Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Algebra, Chapter 22: Special Manipulations (§22.3)
Introduction to Geometry, Chapter 12: Circles and Angles