Difference between revisions of "2006 AMC 12A Problems/Problem 17"
(→Solution 2) |
(→Solution) |
||
Line 33: | Line 33: | ||
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>. | 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>. | ||
+ | |||
+ | ===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{DEZ} = 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>AFE</math> and <math>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>(s+\frac{r\sqrt{2}}{2})^2+(\frac{r\sqrt{2}}{2})^2=(\sqrt{9+5\sqrt{2}})^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 == | == See also == |
Revision as of 19:38, 21 November 2018
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.