Difference between revisions of "2006 AMC 12A Problems/Problem 17"
(→See also) |
|||
Line 7: | Line 7: | ||
== Solution == | == Solution == | ||
+ | ===Solution 1=== | ||
One possibility is to use the [[coordinate plane]], setting <math>B</math> at the origin. Point <math>A</math> will be <math>(0,s)</math> and <math>E</math> will be <math>\left(s + \frac{r}{\sqrt{2}},\ s + \frac{r}{\sqrt{2}}\right)</math> since <math>B, D</math>, and <math>E</math> are [[collinear]] and contain a diagonal of <math>ABCD</math>. The [[Pythagorean theorem]] results in | One possibility is to use the [[coordinate plane]], setting <math>B</math> at the origin. Point <math>A</math> will be <math>(0,s)</math> and <math>E</math> will be <math>\left(s + \frac{r}{\sqrt{2}},\ s + \frac{r}{\sqrt{2}}\right)</math> since <math>B, D</math>, and <math>E</math> are [[collinear]] and contain a diagonal of <math>ABCD</math>. The [[Pythagorean theorem]] results in | ||
Line 18: | Line 19: | ||
This implies that <math>rs = 5</math> and <math>s^2 = 9</math>; dividing gives us <math>\frac{r}{s} = \frac{5}{9} \Rightarrow B</math>. | This implies that <math>rs = 5</math> and <math>s^2 = 9</math>; dividing gives us <math>\frac{r}{s} = \frac{5}{9} \Rightarrow B</math>. | ||
+ | |||
+ | ===Solution 2=== | ||
+ | First draw the line <math>\overline{AE}</math>. Since <math>\overline{AF}</math> is tangent to the circle, <math>\angle AFE</math> is a right angle. Using the Pythagorean Theorem on <math>\triangle AFE</math>, then, we see | ||
+ | <cmath>AE^2 = 9 + 5\sqrt{2} + r^2.</cmath> | ||
+ | |||
+ | But it can also be seen that <math>\angle ADE = 45^\circ</math> since <math>E</math> lies on <math>BD</math>. Therefore, <math>\angle ADE = 135^\circ</math>. Using the Law of Cosines on <math>\triangle ADE</math>, we see | ||
+ | |||
+ | <cmath>AE^2 &= s^2 + r^2 - 2sr\cos(135^\circ)</cmath> | ||
+ | <cmath>AE^2 = s^2 + r^2 - 2sr\left(-\frac{1}{\sqrt{2}}\right)</cmath> | ||
+ | <cmath>AE^2 = s^2 + r^2 + \sqrt{2}sr</cmath> | ||
+ | <cmath>9 + 5\sqrt{2} + r^2 = s^2 + r^2 + \sqrt{2}sr</cmath> | ||
+ | <cmath>9 + 5\sqrt{2} = s^2 + \sqrt{2}sr.</cmath> | ||
+ | |||
+ | Thus, since <math>r</math> and <math>s</math> are rational, <math>s^2 = 9</math> and <math>sr = 5</math>. So <math>s = 3</math> and <math>r = \frac{5}{3}</math> and <math>\frac{r}{s} = \frac{5}{9}</math>. | ||
== See also == | == See also == |
Revision as of 18:09, 8 December 2013
Contents
[hide]Problem
Square has side length , a circle centered at has radius , and and are both rational. The circle passes through , and lies on . Point lies on the circle, on the same side of as . Segment is tangent to the circle, and . What is ?
Solution
Solution 1
One possibility is to use the coordinate plane, setting at the origin. Point will be and will be since , and are collinear and contain a diagonal of . The Pythagorean theorem results in
This implies that and ; dividing gives us .
Solution 2
First draw the line . Since is tangent to the circle, is a right angle. Using the Pythagorean Theorem on , then, we see
But it can also be seen that since lies on . Therefore, . Using the Law of Cosines on , we see
\[AE^2 &= s^2 + r^2 - 2sr\cos(135^\circ)\] (Error compiling LaTeX. Unknown error_msg)
Thus, since and are rational, and . So and and .
See also
2006 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.