# Difference between revisions of "2007 AMC 10B Problems/Problem 11"

## Problem

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?

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

## Solutions

### 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 $r$ and 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 $r^2 \pi = \left(\frac {9}{4\sqrt {2}}\right)^2 \pi = \boxed{\mathrm{(C) \ } \frac {81}{32} \pi}$.

### 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 $R^2 \pi = \boxed{\mathrm{(C) \ } \frac {81}{32} \pi}$

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.

### Solution 3

Extend segment $AD$ to $R$ on Circle $O$.

$[asy] import olympiad; pair B=(0,0), C=(2,0), A=(1,3), D=(1,0), R=(1,-0.35); pair O=circumcenter(A,B,C); draw(A--B--C--A--D--R--C); 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("$$R$$",R,S); label("$$r$$",(O+A)/2,SE); label("$$r$$",(O+R)/2,SE); label("$$3$$",(A+C)/2,NE); label("$$1$$",(C+D)/2,N); [/asy]$

By the Pythagorean Theorem $$AD^2 = 3^2 - 1^2$$ $$AD = 2\sqrt{2}$$

$\triangle ADC$ is similar to $\triangle ACR$, so $$\frac {2\sqrt{2}}{3} = \frac {3}{2r}$$ which gives us $$2r = \frac {9}{2\sqrt{2}} = \frac {9\sqrt{2}}{4}$$ therefore $$r = \frac {9\sqrt{2}}{8}$$

The area of the circle is therefore $\pi r^2 = \left(\frac {9\sqrt{2}}{8}\right)^2 \pi = \boxed{\mathrm{(C) \ } \frac {81}{32} \pi}$

### Solution 4

First, we extend $AD$ to hit the circle at $E.$

$[asy] import olympiad; pair B=(0,0), C=(2,0), A=(1,3), D=(1,0), E=(1,-(8^0.5)/8); pair O=circumcenter(A,B,C); draw(A--B--C--A--E); draw(circumcircle(A,B,C)); dot(O); dot(D); dot(B); dot(C); dot(A); dot(E); label("A",A,N); label("B",B,S); label("$$C$$",C,S); label("D",D,NE); label("O",O,W); label("E",E,S); label("3",(A+B)/2,NW); label("1",(B+D)/2,N); [/asy]$

By the Pythagorean Theorem, we know that $$1^2+AD^2=3^2\implies AD=2\sqrt{2}.$$ We also know that, from the Power of a Point Theorem, $$AD\cdot DE=BD\cdot DC.$$ We can substitute the ones we know to get $$2\sqrt{2}\cdot DE=1$$ We can simplify this to get that $$DE=\dfrac{2\sqrt{2}}{8}.$$ We add $AD$ and $DE$ together to get the length of the diameter, and then we can find the area. $$AE=AD+DE=2\sqrt{2}+\dfrac{2\sqrt{2}}{8}=\dfrac{9\sqrt{2}}{4}.$$ Therefore, the radius is $\dfrac{9\sqrt{2}}{8}$, so the area is $$\left(\dfrac{9\sqrt{2}}{8}\right)^2\pi=\boxed{(\text{C})\dfrac{81}{32}\pi.}$$

### Solution 5

Another possible solution is to plot the circle and triangle on a graph with the circle having center (0,0). Let the radius of the circle = $r$. Let the distance between origin and base of triangle = $a$. $$1 + a^2 = r^2 r + a = 2\sqrt(2) a = (2)\sqrt(2) - r 9 - (4r)\sqrt(2) = 0$$ $$r = ((9)\sqrt(2))/8 \pi r^2 = 81\pi/32$$