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

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== Problem ==
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==Problem==
A [[circle]] passes through the three [[vertex|vertices]] of an [[isosceles triangle]] that has sides of length <math>3</math> and a base of length <math>2</math>. What is the area of this circle?
 
  
<math>\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</math>
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A circle passes through the three vertices of an isosceles triangle that has two sides of length <math>3</math> and a base of length <math>2.</math> What is the area of this circle?
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<math>\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</math>
  
__TOC__
 
 
== Solution ==
 
== Solution ==
 
=== Solution 1 ===
 
=== Solution 1 ===
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</cmath>
 
</cmath>
  
Substituting and solving gives <math>r = \frac {9}{4\sqrt {2}}</math>. Then the area of the circle is <math>r^2 \pi = \left(\frac {9}{4\sqrt {2}}\right)^2 \pi = \frac {81}{32} \pi \Rightarrow \mathrm{(C)}</math>.
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Substituting and solving gives <math>r = \frac {9}{4\sqrt {2}}</math>. Then the area of the circle is <math>r^2 \pi = \left(\frac {9}{4\sqrt {2}}\right)^2 \pi = \boxed{\mathrm{(C) \ } \frac {81}{32} \pi}</math>.
  
 
=== Solution 2 ===
 
=== Solution 2 ===
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R = \frac {abc}{2Bh} = \frac {3 \cdot 3 \cdot 2}{2(2)(2\sqrt {2})} = \frac {9}{4\sqrt {2}}
 
R = \frac {abc}{2Bh} = \frac {3 \cdot 3 \cdot 2}{2(2)(2\sqrt {2})} = \frac {9}{4\sqrt {2}}
 
</cmath>
 
</cmath>
and the answer is <math>R^2 \pi = \mathrm{(C)}</math>
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and the answer is <math>R^2 \pi = \boxed{\mathrm{(C) \ } \frac {81}{32} \pi}</math>
  
 
Alternatively, by the Extended [[Law of Sines]],
 
Alternatively, by the Extended [[Law of Sines]],

Revision as of 15:30, 5 June 2011

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.

See also

2007 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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All AMC 10 Problems and Solutions
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