2007 AMC 10B Problems/Problem 11

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.

2007 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

Page

Toolbox

Search

2007 AMC 10B Problems/Problem 11

Problem

Contents

Solution

Solution 1

Solution 2

See also