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 "2011 AMC 12A Problems/Problem 17"

(→‎Solution)
(→‎Solution: clarified wording)
Line 35: Line 35:
 
The centers of these circles form a 3-4-5 triangle, which has an area equal to 6.
 
The centers of these circles form a 3-4-5 triangle, which has an area equal to 6.
  
−
The 3 triangles determined by one center and the two points of tangency that particular circle has with the other two are, by Law of Sines,
+
The areas of the three triangles determined by the center and the two points of tangency of each circle are, by Law of Sines,
  
 
<math>\frac{1}{2} \cdot 1 \cdot 1 \cdot 1 = \frac{1}{2}</math>
 
<math>\frac{1}{2} \cdot 1 \cdot 1 \cdot 1 = \frac{1}{2}</math>
Line 43: Line 43:
 
<math>\frac{1}{2} \cdot 3 \cdot 3 \cdot \frac{3}{5} = \frac{27}{10}</math>
 
<math>\frac{1}{2} \cdot 3 \cdot 3 \cdot \frac{3}{5} = \frac{27}{10}</math>
  
−
which add up to <math>4.8</math>. Thus the area we're looking for is <math>6 - 4.8 = 1.2 = \frac{6}{5} \rightarrow \boxed{(D)}</math>.
+
which add up to <math>4.8</math>. The area we're looking for is the large 3-4-5 triangle minus the three smaller triangles, or <math>6 - 4.8 = 1.2 = \frac{6}{5} \rightarrow \boxed{(D)}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 16:09, 20 January 2014

Problem

Circles with radii $1$, $2$, and $3$ are mutually externally tangent. What is the area of the triangle determined by the points of tangency?

$\textbf{(A)}\ \frac{3}{5} \qquad \textbf{(B)}\ \frac{4}{5} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ \frac{6}{5} \qquad \textbf{(E)}\ \frac{4}{3}$

Solution

[asy] unitsize(.5cm); defaultpen(linewidth(.8pt)); dotfactor=4; pair A=(0,0), B=(3,0), C=(0,4); dot (A); dot (B); dot (C); draw(A--B); draw(A--C); draw(B--C); draw(Circle(A,1)); draw(Circle(B,2)); draw(Circle(C,3)); [/asy]

The centers of these circles form a 3-4-5 triangle, which has an area equal to 6.

The areas of the three triangles determined by the center and the two points of tangency of each circle are, by Law of Sines,

$\frac{1}{2} \cdot 1 \cdot 1 \cdot 1 = \frac{1}{2}$

$\frac{1}{2} \cdot 2 \cdot 2 \cdot \frac{4}{5} = \frac{8}{5}$

$\frac{1}{2} \cdot 3 \cdot 3 \cdot \frac{3}{5} = \frac{27}{10}$

which add up to $4.8$. The area we're looking for is the large 3-4-5 triangle minus the three smaller triangles, or $6 - 4.8 = 1.2 = \frac{6}{5} \rightarrow \boxed{(D)}$.

See also

2011 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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 12 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=2011_AMC_12A_Problems/Problem_17&oldid=58758"