Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m
m (→‎Solution 1)
Line 7: Line 7:
 
== Solutions ==
 
== Solutions ==
 
=== Solution 1 ===
 
=== Solution 1 ===
−
Let <math>\triangle ABC</math> have vertex <math>A</math> and center <math>O</math>, with foot of altitude from <math>A</math> at <math>D</math>.  
+
Let <math>\triangle ABC</math> have vertex <math>A</math> and center <math>O</math>, with foot of altitude from <math>A</math> intersecting <math>BC</math> at <math>D</math>.  
 
<center><asy>
 
<center><asy>
 
import olympiad;
 
import olympiad;

Revision as of 12:09, 4 June 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 Pythagorean Theorem (with radius $r$ and 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 \[\left(\dfrac{9\sqrt{2}}{8}\right)^2\pi=\boxed{(\text{C})\dfrac{81}{32}\pi.}\]

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 origin and base of triangle = $a$. \[1 + a^2 = r^2\] \[r + a = 2\sqrt{2}\] \[a = 2\sqrt{2} - r\] \[9 - 4r\sqrt{2} = 0\] \[r = (9\sqrt{2})/8\] \[\pi r^2 = 81\pi/32\]

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_10B_Problems/Problem_11&oldid=155112"