Difference between revisions of "2022 AIME I Problems/Problem 8"
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filldraw(X--Y--Z--cycle,green,dashed); | filldraw(X--Y--Z--cycle,green,dashed); | ||
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− | draw(Circle(W,18)); | + | draw(Circle(W,18)^^A--B--C--cycle); |
dot("$A$",A,1.5*dir(A),linewidth(4)); | dot("$A$",A,1.5*dir(A),linewidth(4)); | ||
dot("$B$",B,1.5*dir(B),linewidth(4)); | dot("$B$",B,1.5*dir(B),linewidth(4)); |
Revision as of 05:01, 21 February 2022
Contents
Problem
Equilateral triangle is inscribed in circle
with radius
Circle
is tangent to sides
and
and is internally tangent to
Circles
and
are defined analogously. Circles
and
meet in six points---two points for each pair of circles. The three intersection points closest to the vertices of
are the vertices of a large equilateral triangle in the interior of
and the other three intersection points are the vertices of a smaller equilateral triangle in the interior of
The side length of the smaller equilateral triangle can be written as
where
and
are positive integers. Find
Diagram
~MRENTHUSIASM ~ihatemath123
Solution 1
We can extend and
to
and
respectively such that circle
is the incircle of
.
Since the diameter of the circle is the height of this triangle, the height of this triangle is . We can use inradius or equilateral triangle properties to get the inradius of this triangle is
(The incenter is also a centroid in an equilateral triangle, and the distance from a side to the centroid is a third of the height). Therefore, the radius of each of the smaller circles is
.
Let be the center of the largest circle. We will set up a coordinate system with
as the origin. The center of
will be at
because it is directly beneath
and is the length of the larger radius minus the smaller radius, or
. By rotating this point
around
, we get the center of
. This means that the magnitude of vector
is
and is at a
degree angle from the horizontal. Therefore, the coordinates of this point are
and by symmetry the coordinates of the center of
is
.
The upper left and right circles intersect at two points, the lower of which is . The equations of these two circles are:
We solve this system by subtracting to get . Plugging back in to the first equation, we have
. Since we know
is the lower solution, we take the negative value to get
.
We can solve the problem two ways from here. We can find by rotation and use the distance formula to find the length, or we can be somewhat more clever. We notice that it is easier to find
as they lie on the same vertical,
is
degrees so we can make use of
triangles, and
because
is the center of triangle
. We can draw the diagram as such:
.
.
Finally, the answer is
~KingRavi
Solution 2
Let bottom left point as the origin, the radius of each circle is , note that three centers for circles are
It is not hard to find that one intersection point lies on since the intersection must lie on the angle bisector of the bigger triangle, plug it into equation
, getting that
, the length is
, leads to the answer
~bluesoul
See Also
2022 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.