Difference between revisions of "2006 AMC 10A Problems/Problem 16"
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== Problem == | == Problem == | ||
+ | <!-- [[Image:2006_AMC10A-16.png]] --> | ||
+ | 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> | |
+ | 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> | ||
− | + | <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 1 == | |
− | == Solution == | ||
− | + | Let the centers of the smaller and larger circles be <math>O_1</math> and <math>O_2</math> , respectively. | |
+ | Let their tangent points to <math>\triangle ABC</math> be <math>D</math> and <math>E</math>, respectively. | ||
+ | We can then draw the following diagram: | ||
− | == | + | <!-- [[Image:2006_AMC10A-16a.png]] --> |
+ | <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 <math>\triangle ADO_1 \sim \triangle AEO_2 \sim \triangle AFC</math>. Using the first pair of [[similar triangles]], we write the proportion: | |
− | + | ||
− | * [[2006 | + | <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> |
+ | |||
+ | By the [[Pythagorean Theorem]], we have <math>AD = \sqrt{3^2-1^2} = \sqrt{8}</math>. | ||
+ | |||
+ | Now using <math>\triangle ADO_1 \sim \triangle AFC</math>, | ||
+ | |||
+ | <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 <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 == | ||
+ | {{AMC10 box|year=2006|num-b=15|num-a=17|ab=A}} | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
+ | [[Category:Circle Problems]] | ||
+ | |||
+ | {{MAA Notice}} |
Revision as of 22:19, 19 September 2023
Contents
[hide]Problem
A circle of radius is tangent to a circle of radius . The sides of are tangent to the circles as shown, and the sides and are congruent. What is the area of ?
Solution 1
Let the centers of the smaller and larger circles be and , respectively. Let their tangent points to be and , respectively. We can then draw the following diagram:
We see that . Using the first pair of similar triangles, we write the proportion:
By the Pythagorean Theorem, we have .
Now using ,
Hence, the area of the triangle is
Solution 2
Since we have that .
Since we know that the total length of
We also know that , so
Also, since we have that
Since we know that and we have that
This equation simplified gets us
Let
By the Pythagorean Theorem on we have that
We know that , and so we have
Simplifying, we have that
Recall that .
Therefore,
Since the height is we have the area equal to
Thus our answer is .
~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 |
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