2007 AMC 10A Problem 15

P11-15~4 minPrint

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

Four circles of radius $1$ are each tangent to two sides of a square and externally tangent to a circle of radius $2$ , as shown. What is the area of the square?

Show solution without answering

Key insight

Walk along a diagonal from the center to a corner: 2 + 1 = 3 to a small circle's center, then the diagonal of a 1-by-1 square to the corner.

Solution

By symmetry the big circle is at the center of the square, and each small circle sits in a corner, with its center on a diagonal. So measure half of one diagonal, piece by piece.

1. Center of square to center of a small circle. The circles just touch, so the distance between centers is the sum of the radii: $2 + 1 = 3$ .
2. Center of the small circle to the corner. The small circle touches both sides at that corner, so its center is $1$ away from each side. The center and the corner are opposite corners of a $1 \times 1$ square, so this distance is $\sqrt{1^2 + 1^2} = \sqrt2$ .
3. Half the diagonal is $3 + \sqrt2$ , so the whole diagonal is $6 + 2\sqrt2$ .
4. Side of the square. A square's diagonal is its side times $\sqrt2$ , so
$\text{side} = \frac{6 + 2\sqrt2}{\sqrt2} = \frac{6}{\sqrt2} + 2 = 3\sqrt2 + 2.$
5. Area.
$(3\sqrt2 + 2)^2 = (3\sqrt2)^2 + 2 \cdot 3\sqrt2 \cdot 2 + 2^2 = 18 + 12\sqrt2 + 4 = 22 + 12\sqrt2.$

The answer is $\boxed{\textbf{(B)}\ 22 + 12\sqrt {2}}$ .

Why this works

Every "tangent" in the problem turns into a distance: two touching circles have centers apart by the sum of their radii, and a circle touching a side has its center one radius from that side. Lining these pieces up along a line of symmetry, here the diagonal, turns the picture into a single sum.

Alternative approach

Join the four small centers; they form a smaller square with the same center. Its diagonal runs from one small center through the big circle to the opposite small center: $1 + 2 + 2 + 1 = 6$ . So its side is $\frac{6}{\sqrt2} = 3\sqrt2$ . The big square sticks out by one radius ( $1$ ) on each side, so its side is $3\sqrt2 + 2$ and its area is $22 + 12\sqrt2$ .

Concise solution

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

By symmetry the big circle sits at the center of the square and each small circle's center lies on a diagonal. Trace one diagonal from the center to a corner.

From the center to a small circle's center: the circles are externally tangent, so this distance is $2 + 1 = 3$ .

From the small circle's center to the corner: the small circle is tangent to both sides meeting at that corner, so its center is $1$ unit from each side, i.e. at the far vertex of a $1 \times 1$ square in the corner. That distance is $\sqrt{2}$ .

Half the diagonal is $3 + \sqrt{2}$ , so the full diagonal is $6 + 2\sqrt{2}$ and the side is
$s = \frac{6 + 2\sqrt2}{\sqrt2} = 3\sqrt2 + 2.$
The area is
$s^2 = (3\sqrt2 + 2)^2 = 18 + 12\sqrt2 + 4 = 22 + 12\sqrt2.$

The answer is $\boxed{\textbf{(B)}\ 22 + 12\sqrt {2}}$ .

Review in the AoPS books

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