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Difference between revisions of "1992 AHSME Problems/Problem 11"

(Created page with "== Problem == <asy> draw(circle((0,0),18),black+linewidth(.75)); draw(circle((0,0),6),black+linewidth(.75)); draw((-18,0)--(18,0)--(-14,8*sqrt(2))--cycle,black+linewidth(.75)); d...")
 
(→‎Solution (Similarity))
 
(2 intermediate revisions by the same user not shown)
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\text{(E) } 26</math>
 
\text{(E) } 26</math>
  
−
== Solution ==
+
== Solution (Similarity) ==
 +
 
 +
We are given that <math>BC</math> is tangent to the smaller circle. Using that, we know where the circle intersects <math>BC</math>, it creates a right triangle. We can also point out that since <math>AC</math> is the diameter of the bigger circle and triangle <math>ABC</math> is inscribed the semi-circle, that angle <math>B</math> is a right angle. Therefore, we have <math>2</math> similar triangles. Let's label the center of the smaller circle (which is also the center of the larger circle) as <math>D</math>. Let's also label the point where the smaller circle intersects <math>BC</math> as <math>E</math>. So <math>ABC</math> is similar to <math>DEC</math>. Since <math>DE</math> is the radius of the smaller circle, call the length <math>x</math> and since <math>DC</math> is the radius of the bigger circle, call that length <math>3x</math>. The diameter, <math>AC</math> is <math>6x</math>. So,
 +
 
 +
<math>\frac{AB}{AC} = \frac {DE}{DC} \Rightarrow \frac{12}{6x} = \frac{x}{3x} \Rightarrow 36x = 6x^2 \Rightarrow 6 = x</math>
 +
 
 +
But they are asking for the larger circle radius, so
 +
 
 +
<math>3x= \fbox{18}</math>
 +
 
 +
 
 
<math>\fbox{B}</math>
 
<math>\fbox{B}</math>
  

Latest revision as of 11:02, 24 November 2016

Problem

[asy] draw(circle((0,0),18),black+linewidth(.75)); draw(circle((0,0),6),black+linewidth(.75)); draw((-18,0)--(18,0)--(-14,8*sqrt(2))--cycle,black+linewidth(.75)); dot((-18,0));dot((18,0));dot((-14,8*sqrt(2))); MP("A",(-18,0),W);MP("C",(18,0),E);MP("B",(-14,8*sqrt(2)),W); [/asy]

The ratio of the radii of two concentric circles is $1:3$. If $\overline{AC}$ is a diameter of the larger circle, $\overline{BC}$ is a chord of the larger circle that is tangent to the smaller circle, and $AB=12$, then the radius of the larger circle is

$\text{(A) } 13\quad \text{(B) } 18\quad \text{(C) } 21\quad \text{(D) } 24\quad \text{(E) } 26$

Solution (Similarity)

We are given that $BC$ is tangent to the smaller circle. Using that, we know where the circle intersects $BC$, it creates a right triangle. We can also point out that since $AC$ is the diameter of the bigger circle and triangle $ABC$ is inscribed the semi-circle, that angle $B$ is a right angle. Therefore, we have $2$ similar triangles. Let's label the center of the smaller circle (which is also the center of the larger circle) as $D$. Let's also label the point where the smaller circle intersects $BC$ as $E$. So $ABC$ is similar to $DEC$. Since $DE$ is the radius of the smaller circle, call the length $x$ and since $DC$ is the radius of the bigger circle, call that length $3x$. The diameter, $AC$ is $6x$. So,

$\frac{AB}{AC} = \frac {DE}{DC} \Rightarrow \frac{12}{6x} = \frac{x}{3x} \Rightarrow 36x = 6x^2 \Rightarrow 6 = x$

But they are asking for the larger circle radius, so

$3x= \fbox{18}$


$\fbox{B}$

See also

1992 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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