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: clarified wording)
m (→‎Solution)
 
(5 intermediate revisions by 4 users not shown)
Line 9: Line 9:
 
\textbf{(E)}\ \frac{4}{3} </math>
 
\textbf{(E)}\ \frac{4}{3} </math>
  
−
== Solution ==
+
== Solution 1 ==
  
 
<asy>
 
<asy>
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 areas of the three triangles determined by the center and the two points of tangency of each circle 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, using Triangle Area by Sine,
  
 
<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 45: Line 45:
 
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>.
 
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>.
  
 +
==Solution 2 (Analytical)==
 +
 +
Let <math>O_1,O_2,O_3</math> be the centers of the circles with radii <math>1,2,3</math>. Notice that the points of tangency of the <math>3</math> circles are also the points of tangency of the incircle of <math>\triangle O_1O_2O_3</math>. Using the radius of an Incircle formula, <math>r = \frac{A}{S}</math> where <math>S</math> is the semi-perimeter, and noting that <math>\triangle O_1O_2O_3</math> is a 3-4-5 right triangle, we see that, <cmath>r = \frac{3 \cdot 4/2}{\frac{3+4+5}{2}} = 1.</cmath>
 +
Now we set <math>\triangle O_1O_2O_3</math> on the coordinate plane with <math>O_1=(0,0),O_2=(3,0), O_3 = (0,4)</math>. So the incenter lies on <math>(1,1)</math>. Let the points of tangency of <math>\triangle O_1O_2O_3</math> with it's incenter are <math>A,B,C</math> with <math>A</math> on <math>O_1O_2</math>, <math>B</math> on <math>O_1O_3</math>, and <math>C</math> on <math>O_2O_3</math>. We have that <math>A = (1,0), B=(0,1)</math>. Since the line defined by <math>C</math> and the incenter is perpendicular to <math>O_2O_3</math> who has equation <math>y = \frac{-4}{3}x + 4</math>, we have it's equation as <math>(y-1) = \frac{3}{4}(x-1) \rightarrow y = \frac{3}{4}x + \frac{1}{4}</math>. We have the intersection of the <math>2</math> lines at, <cmath>\begin{align*}\frac{-4}{3}x + 4 &= \frac{3}{4}x + \frac{1}{4} \\ \frac{15}{4} &= \frac{25}{12}x \\ x &= \frac{9}{5} \\ y &= \frac{-4}{3} \frac{9}{5} + 4 = \frac{8}{5}.\end{align*}</cmath>
 +
Here we can use Shoelace Theorem on the points <math>(\frac{9}{5}, \frac{8}{5}),(0,1),(1,0)</math> we get our areas as <math>\frac{6}{5} \rightarrow \boxed{(D)}.</math>
 +
 +
~Aaryabhatta1
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2011|num-b=16|num-a=18|ab=A}}
 
{{AMC12 box|year=2011|num-b=16|num-a=18|ab=A}}

Latest revision as of 16:49, 3 February 2024

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 1

[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, using Triangle Area by Sine,

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

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

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

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 $6 - 4.8 = 1.2 = \frac{6}{5} \rightarrow \boxed{(D)}$.

Solution 2 (Analytical)

Let $O_1,O_2,O_3$ be the centers of the circles with radii $1,2,3$. Notice that the points of tangency of the $3$ circles are also the points of tangency of the incircle of $\triangle O_1O_2O_3$. Using the radius of an Incircle formula, $r = \frac{A}{S}$ where $S$ is the semi-perimeter, and noting that $\triangle O_1O_2O_3$ is a 3-4-5 right triangle, we see that, \[r = \frac{3 \cdot 4/2}{\frac{3+4+5}{2}} = 1.\] Now we set $\triangle O_1O_2O_3$ on the coordinate plane with $O_1=(0,0),O_2=(3,0), O_3 = (0,4)$. So the incenter lies on $(1,1)$. Let the points of tangency of $\triangle O_1O_2O_3$ with it's incenter are $A,B,C$ with $A$ on $O_1O_2$, $B$ on $O_1O_3$, and $C$ on $O_2O_3$. We have that $A = (1,0), B=(0,1)$. Since the line defined by $C$ and the incenter is perpendicular to $O_2O_3$ who has equation $y = \frac{-4}{3}x + 4$, we have it's equation as $(y-1) = \frac{3}{4}(x-1) \rightarrow y = \frac{3}{4}x + \frac{1}{4}$. We have the intersection of the $2$ lines at, \begin{align*}\frac{-4}{3}x + 4 &= \frac{3}{4}x + \frac{1}{4} \\ \frac{15}{4} &= \frac{25}{12}x \\ x &= \frac{9}{5} \\ y &= \frac{-4}{3} \frac{9}{5} + 4 = \frac{8}{5}.\end{align*} Here we can use Shoelace Theorem on the points $(\frac{9}{5}, \frac{8}{5}),(0,1),(1,0)$ we get our areas as $\frac{6}{5} \rightarrow \boxed{(D)}.$

~Aaryabhatta1

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=214390"