Difference between revisions of "2006 AMC 12A Problems/Problem 17"
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Therefore, <math>\frac{\frac{5}{3}}{3}=\frac{5}{9}=\boxed{B}</math> | Therefore, <math>\frac{\frac{5}{3}}{3}=\frac{5}{9}=\boxed{B}</math> | ||
+ | |||
+ | ===Solution 4 - Alcumus=== | ||
+ | |||
+ | Let <math>B=(0,0)</math>, <math>C=(s,0)</math>, <math>A=(0,s)</math>, <math>D=(s,s)</math>, and <math>E=\left(s+\frac{r}{\sqrt{2}},s+\frac{r}{\sqrt{2}} \right)</math>. Apply the Pythagorean Theorem to <math>\triangle AFE</math> to obtain\[ | ||
+ | r^2+\left(9+5\sqrt{2}\right)=\left(s+\frac{r}{\sqrt{2}}\right)^2+\left(\frac{r}{\sqrt{2}}\right)^2, | ||
+ | \]from which <math>9+5\sqrt{2}=s^2+rs\sqrt{2}</math>. Because <math>r</math> and <math>s</math> are rational, it follows that <math>s^2=9</math> and <math>rs=5</math>, so <math>r/s = \boxed{5/9}</math>. | ||
+ | |||
+ | OR | ||
+ | |||
+ | Extend <math>\overline{AD}</math> past <math>D</math> to meet the circle at <math>G \ne D</math>. Because <math>E</math> is collinear with <math>B</math> and <math>D</math>, <math>\angle EDG = 45^\circ.</math> Also, <math>ED = EG,</math> which implies <math>\angle EGD = 45^\circ</math>, so <math>\triangle EDG</math> is an isosceles right triangle. Thus <math>DG = r\sqrt{2}</math>. By the Power of a Point Theorem,\begin{align*} | ||
+ | 9+5\sqrt{2} &= AF^2 \ | ||
+ | &= AD\cdot AG\ | ||
+ | & = AD\cdot \left(AD+DG\right) \ | ||
+ | &= | ||
+ | s\left(s+r\sqrt{2}\right) \ | ||
+ | &= s^2+rs\sqrt{2}.\end{align*}As in the first solution, we conclude that <math>r/s=\boxed{5/9}</math>. | ||
== See Also == | == See Also == |
Revision as of 15:08, 19 July 2022
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 ?
Solutions
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,
Solution 4 - Alcumus
Let , , , , and . Apply the Pythagorean Theorem to to obtain\[ r^2+\left(9+5\sqrt{2}\right)=\left(s+\frac{r}{\sqrt{2}}\right)^2+\left(\frac{r}{\sqrt{2}}\right)^2, \]from which . Because and are rational, it follows that and , so .
OR
Extend past to meet the circle at . Because is collinear with and , Also, which implies , so is an isosceles right triangle. Thus . By the Power of a Point Theorem,
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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