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

Revision as of 18:53, 23 December 2013

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?

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

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

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 $AD = 2\sqrt{2}$

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

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

@@ Line 46: / Line 46: @@
 R = \frac {AC}{\sin \angle ABC} = \frac {3}{\frac {2\sqrt {2}}{3}} \Longrightarrow R = \frac {9}{4\sqrt {2}}
 </cmath>
 Answer follows as above.
+=== Solution 3 ===
+Extend segment <math>AD</math> to <math>R</math> on Circle <math>O</math>.
+<center><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></center>
+By the Pythagorean Theorem <cmath>AD = 2\sqrt{2}</cmath>
+<math>\triangle ADC</math> is similar to <math>\triangle ACR</math>, so <cmath>\frac {2\sqrt{2}}{3} = \frac {3}{2r}</cmath> which gives us <cmath>2r = \frac {9}{2\sqrt{2}} = \frac {9\sqrt{2}}{4}</cmath> so <cmath>r = \frac {9\sqrt{2}}{8}</cmath>
+The area of the circle is therefor <math>\pi r^2 = \left(\frac {9\sqrt{2}}{8}\right)^2 \pi = \boxed{\mathrm{(C) \ } \frac {81}{32} \pi}</math>
 == See also ==

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

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