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

(Solution)
Line 46: Line 46:
 
2R = \frac {AC}{\sin \angle ABC} = \frac {3}{\frac {2\sqrt {2}}{3}} \Longrightarrow R = \frac {9}{4\sqrt {2}}
 
2R = \frac {AC}{\sin \angle ABC} = \frac {3}{\frac {2\sqrt {2}}{3}} \Longrightarrow R = \frac {9}{4\sqrt {2}}
 
</cmath>
 
</cmath>
Answer follows as above.  
+
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 ==
 
== See also ==

Revision as of 19:53, 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 $\pi r^2 = \left(\frac {9\sqrt{2}}{8}\right)^2 \pi = \boxed{\mathrm{(C) \ } \frac {81}{32} \pi}$

See also

2007 AMC 10B (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS