2007 AMC 10B Problem 11

Call the top corner $A$ and the base corners $B$ and $C$, so $AB = AC = 3$ and $BC = 2$. We need the radius $R$ of the circle through all three corners.

1. Find the height. Drop a line from $A$ straight down to the middle $M$ of the base. Then $BM = 1$, and triangle $ABM$ has a right angle at $M$:
$$AM = \sqrt{3^2 - 1^2} = \sqrt{8} = 2\sqrt{2}.$$
2. Place the center. The triangle is symmetric about line $AM$, so the circle's center $O$ is on that line. It is $R$ away from $A$, so it is $2\sqrt{2} - R$ above $M$.
3. Use the corner $B$. The center is also $R$ away from $B$. Triangle $OMB$ has a right angle at $M$, legs $OM = 2\sqrt2 - R$ and $MB = 1$, and long side $OB = R$:
$$R^2 = (2\sqrt{2} - R)^2 + 1^2.$$
4. Solve. Multiply out the square: $(2\sqrt2 - R)^2 = 8 - 4\sqrt2\,R + R^2$. So
$$R^2 = 8 - 4\sqrt{2}\,R + R^2 + 1.$$
The $R^2$ on each side cancels, leaving $4\sqrt{2}\,R = 9$, so $R = \dfrac{9}{4\sqrt2}$.
5. Area. We only need $R^2$:
$$R^2 = \frac{81}{16 \times 2} = \frac{81}{32}, \qquad \text{area} = \pi R^2 = \frac{81}{32}\pi.$$

The answer is $\boxed{\textbf{(C)}\ \frac{81}{32}\pi}$.

The center of the circle through a triangle's corners is the same distance from every corner. For an isosceles triangle it must lie on the line of symmetry, so there is only one unknown. Writing "distance to the top corner equals distance to a bottom corner" as a Pythagorean equation locates it, and the $R^2$ terms always cancel in this kind of setup.

There is a formula for this radius: $R = \dfrac{abc}{4K}$, where $a, b, c$ are the sides and $K$ is the area. Here $K = \frac12 \times 2 \times 2\sqrt2 = 2\sqrt2$, so
$$R = \frac{3 \times 3 \times 2}{4 \times 2\sqrt2} = \frac{9}{4\sqrt2},$$
the same as before, and the area is again $\frac{81}{32}\pi$.