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

(Solution 3)
(Solution)
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The area of the circle is therefore <math>\pi r^2 = \left(\frac {9\sqrt{2}}{8}\right)^2 \pi = \boxed{\mathrm{(C) \ } \frac {81}{32} \pi}</math>
 
The area of the circle is therefore <math>\pi r^2 = \left(\frac {9\sqrt{2}}{8}\right)^2 \pi = \boxed{\mathrm{(C) \ } \frac {81}{32} \pi}</math>
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===Solution 4===
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First, we extend <math>AD</math> to hit the circle at <math>E.</math>
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<center><asy>
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import olympiad;
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pair B=(0,0), C=(2,0), A=(1,3), D=(1,0), E=(1,-(8^0.5)/8);
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pair O=circumcenter(A,B,C);
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draw(A--B--C--A--E);
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draw(circumcircle(A,B,C));
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dot(O);
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dot(D);
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dot(B);
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dot(C);
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dot(A);
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dot(E);
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label("$A$",A,N);
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label("$B$",B,S); label("C",C,S);
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label("$D$",D,NE);
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label("$O$",O,W);
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label("$E$",E,S);
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label("$3$",(A+B)/2,NW);
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label("$1$",(B+D)/2,N);
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</asy></center>
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By the Pythagorean Theorem, we know that
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<cmath>1^2+AD^2=3^2\implies AD=2\sqrt{2}.</cmath>
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We also know that, from the Power of a Point theorem, <cmath>AD\cdot DE=BD\cdot DC.</cmath>
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We can substitute the ones we know to get
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<cmath>2\sqrt{2}\cdot DE=1</cmath>
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We can simplify this to get that
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<cmath>DE=\dfrac{2\sqrt{2}}{8}.</cmath>
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We add <math>AD</math> and <math>DE</math> together to get the length of the diameter, and then we can find the area.
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<cmath>AE=AD+DE=2\sqrt{2}+\dfrac{2\sqrt{2}}{8}=\dfrac{9\sqrt{2}}{4}.</cmath>
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Therefore, the radius is <math>\dfrac{9\sqrt{2}}{8}</math>, so the area is
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<cmath>\left(\dfrac{9\sqrt{2}}{8}\right)^2\pi=\boxed{(\text{C})\dfrac{81}{32}\pi.}</cmath>
  
 
== See also ==
 
== See also ==

Revision as of 14:53, 30 July 2017

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 \[AD^2 = 3^2 - 1^2\] \[AD = 2\sqrt{2}\]

$\triangle ADC$ 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 \[\left(\dfrac{9\sqrt{2}}{8}\right)^2\pi=\boxed{(\text{C})\dfrac{81}{32}\pi.}\]

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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