2018 AMC 12B Problems/Problem 25
Contents
Problem
Circles ,
, and
each have radius
and are placed in the plane so that each circle is externally tangent to the other two. Points
,
, and
lie on
,
, and
respectively such that
and line
is tangent to
for each
, where
. See the figure below. The area of
can be written in the form
for positive integers
and
. What is
?
Solution 1
Let and
be the centers of
and
respectively and draw
,
, and
. Note that
and
are both right. Furthermore, since
is equilateral,
and
. Mark
as the base of the altitude from
to
Since
is a 30-60-90 triangle,
and
. Also, since
and
, we can find
. Thus,
. This makes
So, our answer is
.
Note by diyarv:
A way to see that is through similar triangles. Call the intersection between
and
. Since angles
and
are both right angles, and
and
are congruent, triangles
and
are similar by AA similarity. And, since
is
and
is 2, the common ratio is
. Using the fact that
, we see that these lengths are
and
respectively. Using the Pythagorean theorem, we see that
and
are
and
respectively, giving us a sum of
.
Solution 2
Let be the center of circle
for
, and let
be the intersection of lines
and
. Because
, it follows that
is a
triangle. Let
; then
and
. The Law of Cosines in
gives
which simplifies to
. The positive solution is
. Then
, and so the area of
is
The requested sum is
.
Solution 3
Let be the center of circle
for
. Let
be the centroid of
, which also happens to be the centroid of
. Because
and
,
.
is
the height of
, thus
is
.
Applying cosine law on , one finds that
. Multiplying by
to solve for the height of
, one gets
. Simply multiplying by
and then calculating the equilateral triangle's area, one would get the final result of
.
This makes the answer .
~AlbeePach~
Solution 4
First, note that because the , the arcs inside the shaded equilateral triangle are each
. Also, the distances between the centers of any two of the
given circles are each
.
Draw the circle
concentric with
with radius
. Because the arc of
inside said triangle is
,
touches
, say at a point
. Thus,
is a common tangent of
and
, and it can be seen from inspection of the given diagram that the line is an common internal tangent.
The length of the common internal tangent segment
of
and
is then
, and it is easily seen that
.
Because
, the area of the shaded equilateral triangle is
. We get
~crazyeyemoody907
Solution 5 (Pure Coordinate Bash)
This seems like a coordinate bashable problem. To do this, we notice that it is easier to graph the equilateral triangle first, then the circles, rather than the other way around. Let's forget the lengths of the radius for a moment and focus instead on the ratio of the circles' radius to 's side length.
WLOG let . Place triangle
on the coordinate plane with
,
, and
. Let
be the radius of the circles.
Now, we find the coordinates of the centers of circles and
. Since
is tangent to the x-axis at
, the center of
is
. Draw a right triangle with legs parallel to the x and y axes, and with hypotenuse as the segment from the center of
to
. Since the slope of
is
, the slope of the hypotenuse is
, so the right triangle is
. It's easy to see that the center of
is
.
Since and
are tangent, the distance between the centers is
, so we have
By the Quadratic Formula,
.
We take the positive value to get the radius of the circle is
.
Therefore, the ratio of the radius to the side length of the equilateral triangle is .
The side length of the equilateral triangle is , so its area is
.
~rayfish
See Also
2018 AMC 12B (Problems • Answer Key • Resources) | |
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