# 2007 AMC 10B Problems/Problem 11

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## Problem

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

$\mathrm{(A)}\ 2\pi \qquad\mathrm{(B)}\ 5\pi/2 \qquad\mathrm{(C)}\ 81\pi/32 \qquad\mathrm{(D)}\ 3\pi \qquad\mathrm{(E)}\ 7\pi/2$

## Solution

### Solution 1

Let $\triangle ABC$ have vertex $A$ and center $O$, with foot of altitude from $A$ 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 Pythagorean Theorem (with radius $r$, height $OD = h$) on $\triangle OBD, ABD$ \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 $r^2 \pi = \left(\frac {9}{4\sqrt {2}}\right)^2 \pi = \frac {81}{32} \pi \Rightarrow \mathrm{(C)}$.

### Solution 2

By $A = \frac {1}{2}Bh = \frac {abc}{4R}$ (or we could use $s = 4$ and Heron's formula), $$R = \frac {abc}{2Bh} = \frac {3 \cdot 3 \cdot 2}{2(2)(2\sqrt {2})} = \frac {9}{4\sqrt {2}}$$ and the answer is $R^2 \pi = \mathrm{(C)}$

Alternatively, by the Extended Law of Sines, $$2R = \frac {AC}{\sin \angle ABC} = \frac {3}{\frac {2\sqrt {2}}{3}} \Longrightarrow R = \frac {9}{4\sqrt {2}}$$ Answer follows as above.