Difference between revisions of "2009 AIME II Problems/Problem 5"
(→Solution 2 (NO TRIGONOMETRY!!!)) |
(→Solution 2 (NO TRIGONOMETRY!!!)) |
||
Line 64: | Line 64: | ||
Since <math>CEF</math> is a right triangle, by the Pythagorean theorem <math>EF^2 + CF^2 = EC^2</math>. | Since <math>CEF</math> is a right triangle, by the Pythagorean theorem <math>EF^2 + CF^2 = EC^2</math>. | ||
− | <math>(8 - r)^2 + (4\sqrt{3})^2 = (r + 2)^2</math> | + | <math>(8 - r)^2 + (4\sqrt{3})^2 = (r + 2)^2</math>: <math>r = \frac {27}{5}</math> and the answer is <math>27 + 5 = \boxed{032}</math>. |
-unhappyfarmer | -unhappyfarmer |
Revision as of 18:44, 22 March 2025
Problem 5
Equilateral triangle is inscribed in circle
, which has radius
. Circle
with radius
is internally tangent to circle
at one vertex of
. Circles
and
, both with radius
, are internally tangent to circle
at the other two vertices of
. Circles
,
, and
are all externally tangent to circle
, which has radius
, where
and
are relatively prime positive integers. Find
.
Solution 1
Let be the intersection of the circles with centers
and
, and
be the intersection of the circles with centers
and
. Since the radius of
is
,
. Assume
=
. Then
and
are radii of circle
and have length
.
, and angle
degrees because we are given that triangle
is equilateral. Using the Law of Cosines on triangle
, we obtain
.
The and the
terms cancel out:
. The radius of circle
is
, so the answer is
.
Solution 2 (NO TRIGONOMETRY!!!)
Draw from circle center
to center
. Let its midpoint be point
.
Draw perpendicular to line segment
and intersecting
at point
.
Let be the common external tangent point of circle
and circle
.
Circle centers and
lie on
.
Draw from circle center
to center
.
has length
.
Triangle is a right triangle with
.
(If you take the original equilateral triangle with centroid A and draw the medians, dividing the triangle into 6 30 degree triangles, you find that triangle is similar to one of the triangles).
and
.
Let the radius of circle have length
. Draw
from circle center
to center
.
has length
.
has length
.
Since is a right triangle, by the Pythagorean theorem
.
:
and the answer is
.
-unhappyfarmer
Video Solution
https://youtu.be/fZAChuJDlSw?si=wJUPmgVRlYwazauh
See Also
2009 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.