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

Latest revision as of 21:02, 26 July 2021

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 $A$ intersecting $BC$ 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 the Pythagorean Theorem (with radius $r$ and height $OD = h$ ) on $\triangle OBD$ and $\triangle ABD$ $\begin{align*} h^2 + 1 & = r^2 \\ (h + r)^2 + 1 & = 9 \end{align*}$

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 intersect 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 values 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 the origin and the base of triangle = $a$ . $1 + a^2 = r^2$ $r + a = 2\sqrt{2}$ $a = 2\sqrt{2} - r$ $9 - 4r\sqrt{2} = 0$ $r = \frac{9\sqrt{2}}{8}$ $\pi r^2 = \boxed{(\text{C})\frac{81}{32}\pi}$

Solution 6

We can also use LOC to solve this problem. If O is the center of the circle, angle $OBC$ is double angle $BAC$ . From law of cosines, we have that $\cos(\angle BAC) = \frac{7}{9}$ . We now apply law of cosines on triangle $OBC$ to get that $4 = r^2+r^2-2r^2\cos(2 \angle BAC)$ but we know that $2 \cos (\angle BAC) = 2 \cdot (\frac{7}{9})^2 - 1 = \frac{17}{81}$ . Plugging this in and simplifying, we get $r = \frac{9}{8} \cdot \sqrt{2}$ and thus the area of the circle is equal to $\frac{81}{32} \pi$ .

@@ Line 119: / Line 119: @@
 <cmath> a = 2\sqrt{2} - r </cmath>
 <cmath>9 - 4r\sqrt{2} = 0 </cmath>
-<cmath>r = (9\sqrt{2})/8</cmath>
+<cmath>r = \frac{9\sqrt{2}}{8}</cmath>
-<cmath>\pi r^2 = 81\pi/32</cmath>
+<cmath>\pi r^2 = \boxed{(\text{C})\frac{81}{32}\pi}</cmath>
+===Solution 6===
+We can also use LOC to solve this problem. If O is the center of the circle, angle <math>OBC</math> is double angle <math>BAC</math>. From law of cosines, we have that <math>\cos(\angle BAC) = \frac{7}{9}</math>. We now apply law of cosines on triangle <math>OBC</math> to get that <math>4 = r^2+r^2-2r^2\cos(2 \angle BAC)</math> but we know that <math>2 \cos (\angle BAC) = 2 \cdot (\frac{7}{9})^2 - 1 = \frac{17}{81}</math>. Plugging this in and simplifying, we get <math>r = \frac{9}{8} \cdot \sqrt{2}</math> and thus the area of the circle is equal to <math>\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
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