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 "2006 AMC 10A Problems/Problem 16"

m (→‎Solution)
(→‎Solution 1)
 
(14 intermediate revisions by 7 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
 
<!-- [[Image:2006_AMC10A-16.png]] -->
 
<!-- [[Image:2006_AMC10A-16.png]] -->
−
A [[circle]] of [[radius]] 1 is tangent to a circle of radius 2. The sides of <math>\triangle ABC</math> are [[tangent]] to the circles as shown, and the sides <math>\overline{AB}</math>  and <math>\overline{AC}</math> are [[congruent]]. What is the area of <math>\triangle ABC</math>?
+
A circle of radius <math>1</math> is [[tangent]] to a circle of radius <math>2</math>. The sides of <math>\triangle ABC</math> are tangent to the circles as shown, and the sides <math>\overline{AB}</math>  and <math>\overline{AC}</math> are congruent. What is the area of <math>\triangle ABC</math>?
 +
 
 
<asy>
 
<asy>
 
size(200); pathpen = linewidth(0.7); pointpen = black;
 
size(200); pathpen = linewidth(0.7); pointpen = black;
Line 12: Line 13:
 
MP('2', (2*t,3), W); MP('1',(2*t, 5.5), W);</asy>
 
MP('2', (2*t,3), W); MP('1',(2*t, 5.5), W);</asy>
  
−
<math>\mathrm{(A) \ } \frac{35}{2}\qquad\mathrm{(B) \ } 15\sqrt{2}\qquad\mathrm{(C) \ } \frac{64}{3}\qquad\mathrm{(D) \ } 16\sqrt{2}\qquad\mathrm{(E) \ } 24\qquad</math>
+
<math>\textbf{(A) } \frac{35}{2}\qquad\textbf{(B) } 15\sqrt{2}\qquad\textbf{(C) } \frac{64}{3}\qquad\textbf{(D) } 16\sqrt{2}\qquad\textbf{(E) } 24\qquad</math>
  
−
== Solution ==
+
== Solution 1 ==
  
 
Let the centers of the smaller and larger circles be <math>O_1</math> and <math>O_2</math> , respectively.
 
Let the centers of the smaller and larger circles be <math>O_1</math> and <math>O_2</math> , respectively.
Line 32: Line 33:
 
MP('2', (2*t,3), W); MP('1',(2*t, 5.5), W); MP("F",(2*t,0)); MP("O_1",(2*t,5),W); MP("O_2",(2*t,2),W); MP("D",B,NE); MP("E",A,NE); </asy>
 
MP('2', (2*t,3), W); MP('1',(2*t, 5.5), W); MP("F",(2*t,0)); MP("O_1",(2*t,5),W); MP("O_2",(2*t,2),W); MP("D",B,NE); MP("E",A,NE); </asy>
  
−
We see that <math>\triangle ADO_1 \sim \triangle AEO_2 \sim \triangle AFC</math>. Using the first pair of [[similar triangles]], we write the [[proportion]]:
+
We see that <math>\triangle ADO_1 \sim \triangle AEO_2 \sim \triangle AFC</math>. Using the first pair of [[similar triangles]], we write the proportion:
  
 
<div style="text-align:center;"><math>\frac{AO_1}{AO_2} = \frac{DO_1}{EO_2} \Longrightarrow \frac{AO_1}{AO_1 + 3} = \frac{1}{2} \Longrightarrow AO_1 = 3</math></div>
 
<div style="text-align:center;"><math>\frac{AO_1}{AO_2} = \frac{DO_1}{EO_2} \Longrightarrow \frac{AO_1}{AO_1 + 3} = \frac{1}{2} \Longrightarrow AO_1 = 3</math></div>
Line 42: Line 43:
 
<div style="text-align:center;"><math>\frac{AD}{AF} = \frac{DO_1}{FC} \Longrightarrow \frac{2\sqrt{2}}{8} = \frac{1}{FC} \Longrightarrow FC = 2\sqrt{2}</math></div>
 
<div style="text-align:center;"><math>\frac{AD}{AF} = \frac{DO_1}{FC} \Longrightarrow \frac{2\sqrt{2}}{8} = \frac{1}{FC} \Longrightarrow FC = 2\sqrt{2}</math></div>
  
−
Hence, the area of the triangle is <math>\frac{1}{2}\cdot AF \cdot BC = \frac{1}{2}\cdot AF \cdot (2\cdot CF) = AF \cdot CF = 8\left(2\sqrt{2}\right) = \boxed{16\sqrt{2}\ \mathrm{(D)}</math>
+
Hence, the area of the triangle is <cmath>\frac{1}{2}\cdot AF \cdot BC = \frac{1}{2}\cdot AF \cdot (2\cdot CF) = AF \cdot CF = 8\left(2\sqrt{2}\right) = \boxed{\textbf{(D) } 16\sqrt{2}}</cmath>
 +
 
 +
== Solution 2 ==
 +
 
 +
<asy>
 +
size(200); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10);
 +
real t=2^0.5;
 +
D((0,0)--(4*t,0)--(2*t,8)--cycle);
 +
D(CR(D((2*t,2)),2));
 +
D(CR(D((2*t,5)),1));
 +
D('B', (0,0),SW); D('C',(4*t,0), SE); D('A', (2*t, 8), N);
 +
pair A = foot((2*t,2),(2*t,8),(4*t,0)), B = foot((2*t,5),(2*t,8),(4*t,0));
 +
D((2*t,0)--(2*t,8)); D((2*t,2)--D(A)); D((2*t,5)--D(B)); D(rightanglemark((2*t,2),A,(4*t,0))); D(rightanglemark((2*t,5),B,(4*t,0)));
 +
MP('2', (2*t,3), W); MP('1',(2*t, 5.5), W); MP("F",(2*t,0)); MP("O_1",(2*t,5),W); MP("O_2",(2*t,2),W); MP("D",B,NE); MP("E",A,NE); </asy>
 +
 
 +
Since <math>\triangle{A O_1 D} \sim \triangle{A O_2 E},</math> we have that <math>\frac{A O_1}{A O_2} = \frac{O_1 D}{O_2 E} = \frac{1}{2}</math>.
 +
 
 +
Since we know that <math>O_1 O_2 = 1 + 2 = 3,</math> the total length of <math>A O_2 = 2 \cdot 3 = 6.</math>
 +
 
 +
We also know that <math>O_2 F = 2</math>, so <math>A F = A O_2 + O_2 F = 6 + 2 = 8.</math>
 +
 
 +
Also, since <math>\triangle{ABF} \sim \triangle{A E O_2},</math> we have that <math>\frac{AC}{A O_2} = \frac{FC}{O_2 E}.</math>
 +
 
 +
Since we know that <math>A O_2 = 6</math> and <math>O_2 E = 2,</math> we have that <math>\frac{AC}{6} = \frac{FC}{2}.</math>
 +
 
 +
This equation simplified gets us <math>AC = 3 \cdot FC.</math>
 +
 
 +
Let <math>FC = a</math>
 +
 
 +
By the [[Pythagorean Theorem]] on <math>\triangle{AFC},</math> we have that <math>AF^2 + FC^2 = AC^2.</math>
 +
 
 +
We know that <math>AF = 8</math>, <math>FC = a</math> and <math>AC = 3a</math> so we have <math>8^2 + a^2 = (3a)^2.</math>
 +
 
 +
Simplifying, we have that <math>64 = 8a^2 \Rightarrow a^2 = 8 \Rightarrow a = \sqrt{8} = 2 \sqrt{2}.</math>
 +
 
 +
Recall that <math>FC=a</math>.
 +
 
 +
Therefore, <math>BC = 2 \cdot FC = 2 \cdot 2 \sqrt{2} = 4 \sqrt{2}.</math>
 +
 
 +
Since the height is <math>AF = 8,</math> we have the area equal to <math>\frac{4 \sqrt{2} \cdot 8}{2}=16\sqrt{2}.</math>
 +
 
 +
Thus our answer is <math>\boxed{\textbf{(D) }16 \sqrt{2}}</math>.
 +
 
 +
~mathboy282
  
 
== See also ==
 
== See also ==

Latest revision as of 22:19, 19 September 2023

Problem

A circle of radius $1$ is tangent to a circle of radius $2$. The sides of $\triangle ABC$ are tangent to the circles as shown, and the sides $\overline{AB}$ and $\overline{AC}$ are congruent. What is the area of $\triangle ABC$?

[asy] size(200); pathpen = linewidth(0.7); pointpen = black; real t=2^0.5; D((0,0)--(4*t,0)--(2*t,8)--cycle); D(CR((2*t,2),2)); D(CR((2*t,5),1)); D('B', (0,0),SW); D('C',(4*t,0), SE); D('A', (2*t, 8), N); D((2*t,2)--(2*t,4)); D((2*t,5)--(2*t,6)); MP('2', (2*t,3), W); MP('1',(2*t, 5.5), W);[/asy]

$\textbf{(A) } \frac{35}{2}\qquad\textbf{(B) } 15\sqrt{2}\qquad\textbf{(C) } \frac{64}{3}\qquad\textbf{(D) } 16\sqrt{2}\qquad\textbf{(E) } 24\qquad$

Solution 1

Let the centers of the smaller and larger circles be $O_1$ and $O_2$ , respectively. Let their tangent points to $\triangle ABC$ be $D$ and $E$, respectively. We can then draw the following diagram:

[asy] size(200); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10); real t=2^0.5; D((0,0)--(4*t,0)--(2*t,8)--cycle); D(CR(D((2*t,2)),2)); D(CR(D((2*t,5)),1)); D('B', (0,0),SW); D('C',(4*t,0), SE); D('A', (2*t, 8), N); pair A = foot((2*t,2),(2*t,8),(4*t,0)), B = foot((2*t,5),(2*t,8),(4*t,0)); D((2*t,0)--(2*t,8)); D((2*t,2)--D(A)); D((2*t,5)--D(B)); D(rightanglemark((2*t,2),A,(4*t,0))); D(rightanglemark((2*t,5),B,(4*t,0))); MP('2', (2*t,3), W); MP('1',(2*t, 5.5), W); MP("F",(2*t,0)); MP("O_1",(2*t,5),W); MP("O_2",(2*t,2),W); MP("D",B,NE); MP("E",A,NE); [/asy]

We see that $\triangle ADO_1 \sim \triangle AEO_2 \sim \triangle AFC$. Using the first pair of similar triangles, we write the proportion:

$\frac{AO_1}{AO_2} = \frac{DO_1}{EO_2} \Longrightarrow \frac{AO_1}{AO_1 + 3} = \frac{1}{2} \Longrightarrow AO_1 = 3$

By the Pythagorean Theorem, we have $AD = \sqrt{3^2-1^2} = \sqrt{8}$.

Now using $\triangle ADO_1 \sim \triangle AFC$,

$\frac{AD}{AF} = \frac{DO_1}{FC} \Longrightarrow \frac{2\sqrt{2}}{8} = \frac{1}{FC} \Longrightarrow FC = 2\sqrt{2}$

Hence, the area of the triangle is \[\frac{1}{2}\cdot AF \cdot BC = \frac{1}{2}\cdot AF \cdot (2\cdot CF) = AF \cdot CF = 8\left(2\sqrt{2}\right) = \boxed{\textbf{(D) } 16\sqrt{2}}\]

Solution 2

[asy] size(200); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10); real t=2^0.5; D((0,0)--(4*t,0)--(2*t,8)--cycle); D(CR(D((2*t,2)),2)); D(CR(D((2*t,5)),1)); D('B', (0,0),SW); D('C',(4*t,0), SE); D('A', (2*t, 8), N); pair A = foot((2*t,2),(2*t,8),(4*t,0)), B = foot((2*t,5),(2*t,8),(4*t,0)); D((2*t,0)--(2*t,8)); D((2*t,2)--D(A)); D((2*t,5)--D(B)); D(rightanglemark((2*t,2),A,(4*t,0))); D(rightanglemark((2*t,5),B,(4*t,0))); MP('2', (2*t,3), W); MP('1',(2*t, 5.5), W); MP("F",(2*t,0)); MP("O_1",(2*t,5),W); MP("O_2",(2*t,2),W); MP("D",B,NE); MP("E",A,NE); [/asy]

Since $\triangle{A O_1 D} \sim \triangle{A O_2 E},$ we have that $\frac{A O_1}{A O_2} = \frac{O_1 D}{O_2 E} = \frac{1}{2}$.

Since we know that $O_1 O_2 = 1 + 2 = 3,$ the total length of $A O_2 = 2 \cdot 3 = 6.$

We also know that $O_2 F = 2$, so $A F = A O_2 + O_2 F = 6 + 2 = 8.$

Also, since $\triangle{ABF} \sim \triangle{A E O_2},$ we have that $\frac{AC}{A O_2} = \frac{FC}{O_2 E}.$

Since we know that $A O_2 = 6$ and $O_2 E = 2,$ we have that $\frac{AC}{6} = \frac{FC}{2}.$

This equation simplified gets us $AC = 3 \cdot FC.$

Let $FC = a$

By the Pythagorean Theorem on $\triangle{AFC},$ we have that $AF^2 + FC^2 = AC^2.$

We know that $AF = 8$, $FC = a$ and $AC = 3a$ so we have $8^2 + a^2 = (3a)^2.$

Simplifying, we have that $64 = 8a^2 \Rightarrow a^2 = 8 \Rightarrow a = \sqrt{8} = 2 \sqrt{2}.$

Recall that $FC=a$.

Therefore, $BC = 2 \cdot FC = 2 \cdot 2 \sqrt{2} = 4 \sqrt{2}.$

Since the height is $AF = 8,$ we have the area equal to $\frac{4 \sqrt{2} \cdot 8}{2}=16\sqrt{2}.$

Thus our answer is $\boxed{\textbf{(D) }16 \sqrt{2}}$.

~mathboy282

See also

2006 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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=2006_AMC_10A_Problems/Problem_16&oldid=198426"