Difference between revisions of "2006 AMC 12A Problems/Problem 17"
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== Problem == | == Problem == | ||
− | {{ | + | Square <math>ABCD</math> has side length <math>s</math>, a circle centered at <math>E</math> has radius <math>r</math>, and <math>r</math> and <math>s</math> are both rational. The circle passes through <math>D</math>, and <math>D</math> lies on <math>\overline{BE}</math>. Point <math>F</math> lies on the circle, on the same side of <math>\overline{BE}</math> as <math>A</math>. Segment <math>AF</math> is tangent to the circle, and <math>AF=\sqrt{9+5\sqrt{2}}</math>. What is <math>r/s</math>? |
− | + | <center>[[Image:AMC12_2006A_17.png]]</center> | |
<math> \mathrm{(A) \ } \frac{1}{2}\qquad \mathrm{(B) \ } \frac{5}{9}\qquad \mathrm{(C) \ } \frac{3}{5}\qquad \mathrm{(D) \ } \frac{5}{3}\qquad \mathrm{(E) \ } \frac{9}{5}</math> | <math> \mathrm{(A) \ } \frac{1}{2}\qquad \mathrm{(B) \ } \frac{5}{9}\qquad \mathrm{(C) \ } \frac{3}{5}\qquad \mathrm{(D) \ } \frac{5}{3}\qquad \mathrm{(E) \ } \frac{9}{5}</math> | ||
== Solution == | == Solution == | ||
− | One possibility is to use the [[coordinate plane]], setting B at the origin. Point A will be | + | === 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 | ||
+ | |||
+ | <cmath>AF^2 + EF^2 = AE^2</cmath> | ||
+ | |||
+ | <cmath>r^2 + \left(\sqrt{9 + 5\sqrt{2}}\right)^2 = \left(\left(s + \frac{r}{\sqrt{2}}\right) - 0\right)^2 + \left(\left(s + \frac{r}{\sqrt{2}}\right) - s\right)^2</cmath> | ||
+ | |||
+ | <cmath>r^2 + 9 + 5\sqrt{2} = s^2 + rs\sqrt{2} + \frac{r^2}{2} + \frac{r^2}{2}</cmath> | ||
+ | |||
+ | <cmath>9 + 5\sqrt{2} = s^2 + rs\sqrt{2}</cmath> | ||
+ | |||
+ | 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 note that angle <math>\angle AFE</math> is right since <math>\overline{AF}</math> is tangent to the circle. Using the Pythagorean Theorem on <math>\triangle AFE</math>, then, we see | ||
+ | <cmath>AE^2 = 9 + 5\sqrt{2} + r^2.</cmath> | ||
− | <math> | + | But it can also be seen that <math>\angle BDA = 45^\circ</math>. Therefore, since <math>D</math> lies on <math>\overline{BE}</math>, <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> | ||
− | <math>r^2 | + | 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>, <math>r = \frac{5}{3}</math>, and <math>\frac{r}{s} = \frac{5}{9}</math>. |
− | <math> | + | === Solution 3 === |
+ | (Similar to Solution 1) | ||
+ | First, draw line AE and mark a point Z that is equidistant from E and D so that <math>\angle{DZE} = 90^\circ</math> and that line <math>\overline{AZ}</math> includes point D. Since DE is equal to the radius <math>r</math>, <math>DZ=EZ=\frac{r}{\sqrt2}=\frac{r\sqrt2}{2}.</math> | ||
− | + | Note that triangles <math>\triangle AFE</math> and <math>\triangle AZE</math> share the same hypotenuse <math>(AE)</math>, meaning that | |
− | < | + | <cmath>AZ^2+EZ^2=AF^2+EF^2</cmath> |
+ | Plugging in our values we have: | ||
+ | <cmath>\left(s+\frac{r\sqrt{2}}{2}\right)^2+\left(\frac{r\sqrt{2}}{2}\right)^2=\left(\sqrt{9+5\sqrt{2}}\right)^2+r^2</cmath> | ||
+ | <cmath>s^2+sr\sqrt{2}+\frac{r^2}{2}+\frac{r^2}{2}=9+5\sqrt{2}+r^2</cmath> | ||
+ | <cmath>s^2+sr\sqrt{2}=9+5\sqrt{2}</cmath> | ||
+ | By logic <math>s=3</math> and <math>sr=5 \implies r=5/3.</math> | ||
− | == | + | Therefore, <math>\frac{\frac{5}{3}}{3}=\frac{5}{9}=\boxed{B}</math> |
− | |||
− | {{ | + | == See Also == |
+ | {{AMC12 box|year=2006|ab=A|num-b=16|num-a=18}} | ||
+ | {{MAA Notice}} | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] |
Revision as of 01:00, 19 October 2020
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 note that angle is right since is tangent to the circle. Using the Pythagorean Theorem on , then, we see
But it can also be seen that . Therefore, since lies on , . Using the Law of Cosines on , we see
Thus, since and are rational, and . So , , and .
Solution 3
(Similar to Solution 1) First, draw line AE and mark a point Z that is equidistant from E and D so that and that line includes point D. Since DE is equal to the radius ,
Note that triangles and share the same hypotenuse , meaning that Plugging in our values we have: By logic and
Therefore,
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 |
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