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>? | 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>? | ||
+ | |||
+ | [asy] | ||
+ | pair A,B,C,D,I,F; | ||
+ | A=(0,10); B=(0,0); C=(10,0); D=(10,10); | ||
+ | |||
+ | I=(14,13); F=(11,17); | ||
+ | draw(A--B--C--D--cycle,linewidth(0.7)); | ||
+ | draw(Circle(I,5),linewidth(0.7)); | ||
+ | draw(A--F,linewidth(0.7)); | ||
+ | label("<math>A</math>",A,NW); | ||
+ | label("<math>B</math>",B,SW); | ||
+ | label("<math>C</math>",C,SE); | ||
+ | label("<math>D</math>",D,SW); | ||
+ | label("<math>F</math>",F,N); | ||
+ | label("<math>E</math>",I,E); | ||
+ | dot(I); | ||
+ | [/asy] | ||
+ | |||
<center>[[Image:AMC12_2006A_17.png]]</center> | <center>[[Image:AMC12_2006A_17.png]]</center> |
Revision as of 16:07, 19 July 2022
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
?
[asy] pair A,B,C,D,I,F; A=(0,10); B=(0,0); C=(10,0); D=(10,10);
I=(14,13); F=(11,17);
draw(A--B--C--D--cycle,linewidth(0.7));
draw(Circle(I,5),linewidth(0.7));
draw(A--F,linewidth(0.7));
label("",A,NW);
label("
",B,SW);
label("
",C,SE);
label("
",D,SW);
label("
",F,N);
label("
",I,E);
dot(I);
[/asy]
![AMC12 2006A 17.png](https://wiki-images.artofproblemsolving.com//e/e2/AMC12_2006A_17.png)
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,
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.