2007 AMC 10A Problem 24

P21-25~5 minPrint

Circles Quadrilaterals & Polygon Areas Add construction lines/points (drop altitudes, extend segments, connect centers)Cut the figure into known shapes (triangles, rectangles, sectors)

Circles centered at $A$ and $B$ each have radius $2$ , as shown. Point $O$ is the midpoint of $\overline{AB}$ , and $OA = 2\sqrt {2}$ . Segments $OC$ and $OD$ are tangent to the circles centered at $A$ and $B$ , respectively, and $EF$ is a common tangent. What is the area of the shaded region $ECODF$ ?

Show solution without answering

Key insight

Radii to E and F make a rectangle. The shaded part is that rectangle minus two right triangles with legs 2 and two 45-degree circle slices.

Solution

The shaded region has curved edges, so build a simple shape around it and subtract the extra pieces.

1. Make a rectangle. Draw radii $AE$ and $BF$ . A radius to a point where a line touches the circle is perpendicular to that line, so $AE$ and $BF$ both stand straight up from $AB$ , with length $2$ . So $ABFE$ is a rectangle.
2. Rectangle area. $AB = 2 \times OA = 4\sqrt2$ and the height is $2$ , so the area is $2 \times 4\sqrt2 = 8\sqrt2$ .
3. What is extra? Near $A$ , the rectangle also contains triangle $ACO$ and the slice of circle $A$ between radii $AC$ and $AE$ . The same two pieces appear near $B$ .
4. Triangle $ACO$ . $OC$ touches circle $A$ at $C$ , so $\angle ACO = 90^\circ$ . With $AC = 2$ and $AO = 2\sqrt2$ , the Pythagorean theorem gives
$OC^2 = (2\sqrt2)^2 - 2^2 = 8 - 4 = 4, \qquad OC = 2.$
Both legs are $2$ , so the area is $\frac12 \times 2 \times 2 = 2$ , and the angle at $A$ is $45^\circ$ .
5. Slice $CAE$ . $AE$ is at $90^\circ$ to $AB$ , and $AC$ is at $45^\circ$ to $AB$ , so the slice between them is $90^\circ - 45^\circ = 45^\circ$ . That is $\frac{45}{360} = \frac18$ of the circle:
$\frac18 \times \pi \times 2^2 = \frac{\pi}{2}.$
6. Subtract both pairs. By symmetry, the pieces near $B$ are the same size:
$8\sqrt2 - 2 \times 2 - 2 \times \frac{\pi}{2} = 8\sqrt2 - 4 - \pi.$

The answer is $\boxed{\textbf{(B)}\ 8\sqrt {2} - 4 - \pi}$ .

Why this works

Areas bounded by arcs and tangent lines are rarely found directly. Draw radii to the touching points to make a rectangle, then take away triangles and circle slices. The right angle between a radius and a tangent line tells you the triangle's shape, and the triangle's angle tells you how big the slice is.

Concise solution

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

Draw radii $AC$ , $AE$ , $BD$ , $BF$ . Since $EF$ is tangent to both circles at their topmost points, $AE$ and $BF$ are perpendicular to $EF$ and to $AB$ , so $ABFE$ is a rectangle with $AB = 2 \cdot 2\sqrt2 = 4\sqrt2$ and height $2$ : area $8\sqrt2$ .

The shaded region is this rectangle with four pieces cut away from the bottom corners and the middle: triangle $ACO$ , sector $CAE$ , triangle $BDO$ , and sector $DBF$ .

Triangle $ACO$ : $OC$ is tangent at $C$ , so $\angle ACO = 90^\circ$ . With $AC = 2$ and $AO = 2\sqrt2$ , the Pythagorean theorem gives $OC = 2$ . So $ACO$ is an isosceles right triangle with area $\frac12 \cdot 2 \cdot 2 = 2$ , and $\angle CAO = 45^\circ$ .

Sector $CAE$ : $\angle EAO = 90^\circ$ and $\angle CAO = 45^\circ$ , so the sector spans $45^\circ$ : area $\frac{45}{360} \cdot \pi \cdot 2^2 = \frac{\pi}{2}$ .

By symmetry the pieces at $B$ are identical. Subtract everything:
$8\sqrt2 - 2 \cdot 2 - 2 \cdot \frac{\pi}{2} = 8\sqrt2 - 4 - \pi.$

The answer is $\boxed{\textbf{(B)}\ 8\sqrt {2} - 4 - \pi}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 11: Circles
Introduction to Geometry, Chapter 8: Quadrilaterals
Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2, §4.3)
Introduction to Geometry, Chapter 19: Problem Solving Strategies in Geometry (§19.1)