Difference between revisions of "2006 AMC 10A Problems/Problem 16"
m (→Solution) |
Dairyqueenxd (talk | contribs) (→Solution 2) |
||
(3 intermediate revisions by the same user not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
<!-- [[Image:2006_AMC10A-16.png]] --> | <!-- [[Image:2006_AMC10A-16.png]] --> | ||
− | A | + | 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> | ||
Line 13: | 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>\textbf{(A) | + | <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 1 == | ||
Line 33: | 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 | + | 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 43: | 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 <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{\ | + | 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 == | == Solution 2 == | ||
Line 72: | Line 72: | ||
Let <math>FC = a</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> | + | 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> | 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> | ||
Line 84: | Line 84: | ||
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> | 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{\ | + | Thus our answer is <math>\boxed{\textbf{(D) }16 \sqrt{2}}</math>. |
~mathboy282 | ~mathboy282 |
Latest revision as of 12:06, 17 December 2021
Contents
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.