2007 AMC 10B Problem 11

Let the triangle have apex $A$ and base $BC = 2$ with $AB = AC = 3$. Drop the altitude from $A$ to the midpoint $M$ of $BC$. Then $BM = 1$ and
$$AM = \sqrt{3^2 - 1^2} = 2\sqrt{2}.$$

By symmetry the center $O$ of the circle lies on line $AM$, at distance $R$ from $A$, so $OM = 2\sqrt{2} - R$. The center is also at distance $R$ from $B$, and triangle $OMB$ is right-angled at $M$:
$$R^2 = (2\sqrt{2} - R)^2 + 1^2 = 8 - 4\sqrt{2}\,R + R^2 + 1.$$
The $R^2$ terms cancel, leaving $4\sqrt{2}\,R = 9$, so
$$R = \frac{9}{4\sqrt{2}}, \qquad R^2 = \frac{81}{32}.$$

The circle's area is $\pi R^2 = \frac{81}{32}\pi$.

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

An isosceles triangle's circumcenter lies on its axis of symmetry, so one unknown $R$ and one right triangle suffice. Writing "distance to apex equals distance to base vertex" is the general way to locate a circumcenter without memorized formulas, and the $R^2$ cancellation is typical.

Use $R = \dfrac{abc}{4K}$. The area is $K = \frac12 \cdot 2 \cdot 2\sqrt{2} = 2\sqrt{2}$, so
$$R = \frac{3 \cdot 3 \cdot 2}{4 \cdot 2\sqrt{2}} = \frac{9}{4\sqrt{2}},$$
and $\pi R^2 = \frac{81}{32}\pi$ as before.

• Assuming the circumcenter is the midpoint of the base or the foot of the altitude, or confusing the circumradius with the inradius.
• Reporting $R^2 = 81/32$ or $R$ itself rather than the area, or dropping the $\pi$ and mismatching the choices.