AMC10 2007 10B, Problem 11

A circle passes through the three vertices of an isosceles triangle that has two sides of length $3$ and a base of length $2$. What is the area of this circle?

Solution: Let $\triangle ABC$ have vertex $A$ and center $O$, with foot of altitude from $A$ intersecting $BC$ at $D$. [asy] import olympiad; pair B=(0,0), C=(2,0), A=(1,3), D=(1,0); pair O=circumcenter(A,B,C); draw(A--B--C--A--D); draw(B--O--C); draw(circumcircle(A,B,C)); dot(O); label("$A$",A,N); label("$B$",B,S); label("$C$",C,S); label("$D$",D,S); label("$O$",O,W); label("$r$",(O+A)/2,SE); label("$r$",(O+B)/2,N); label("$h$",(O+D)/2,SE); label("$3$",(A+B)/2,NW); label("$1$",(B+D)/2,N); [/asy] Then by the Pythagorean Theorem (with radius $r$ and height $OD = h$) on $\triangle OBD$ and $\triangle ABD$ \begin{align} h^2 + 1 & = r^2 \\ (h + r)^2 + 1 & = 9 \end{align} Substituting and solving gives $r = \frac {9}{4\sqrt {2}}$. Then the area of the circle is $r^2 \pi = \left(\frac {9}{4\sqrt {2}}\right)^2 \pi = \fbox{\frac{81}{32}\pi}$.

Answer: \frac{81}{32}\pi