2012 AIME I Problems/Problem 13
Contents
Problem 13
Three concentric circles have radii
and
An equilateral triangle with one vertex on each circle has side length
The largest possible area of the triangle can be written as
where
and
are positive integers,
and
are relatively prime, and
is not divisible by the square of any prime. Find
Solution
Solution 1
Reinterpret the problem in the following manner. Equilateral triangle has a point
on the interior such that
and
A
counter-clockwise rotation about vertex
maps
to
and
to
Note that angle
is
and
which tells us that triangle
is equilateral and that
We now notice that
and
which tells us that angle
is
because there is a
-
-
Pythagorean triple. Now note that
and
so
and
Applying the law of cosines on triangle
yields
and thus the area of equals
so our final answer is
Remark: The new figure (the rotations and the triangle) must be twice the original triangle's area. So it is simply
Solution 2
Here is a proof that shows that there are 2 distinct equilateral triangles (up to congruence) that have the given properties:
We claim that there are 2 distinct equilateral triangles (up to congruence) that have the given properties; one of which has largest area. We have 2 cases to consider; either the center of the circles lies in the interior of triangle
or it does not (and we shall show that both can happen). To see that the first case can occur, refer to Solution 1 above, or for a less creative and more direct approach proceed as follows. Using the notation from Solution 1, let
be the measure of angle
so that angle
has measure
. Let
. The law of cosines on triangles
and
yields
and
. Solving this system will yield the value of
. Since
we have that
. Substituting these into the equation
we obtain
. After clearing denominators, combining like terms, isolating the square root, squaring, and expanding, we obtain
so that by the quadratic formula
. Under the hypothesis that
lies in the interior of triangle
,
must be
. To see this, note that the other value for
is roughly
so that
, but since
and
we have a contradiction. We then obtain the area as in Solution 1.
Now, suppose does not lie in the interior of triangle
. We then obtain convex quadrilateral
with diagonals
and
intersecting at
. Here
. We may let
denote the measure of angle
so that angle
measures
. Note that the law of cosines on triangles
and
yield the same equations as in the first case with
replaced with
. Thus we obtain again
. If
then
, but this is impossible since
but the shortest possible distance from
to
is the height of equilateral triangle
which is
; a contradiction. Hence in this case
. But, the area of this triangle is clearly less than that in the first case, so we are done. Hence the phrasing of the question (the triangle with maximal area) is absolutely necessary since there are 2 possible triangles (up to congruence).
Solution 3
The problem basically asks for the area of such that it is equilateral and there is a point
inside of the triangle which satisfies
,
, and
.
Let . We want the area of the triangle, which is just
. Thus, we want to know
, and then finding the area will be a matter of simple calculation.
By law of cosines, , where
.
Now, if we plug this value in for in the area formula, we get
.
Notice that, in order for this expression to be in the required answer form ,
must involve
. Now, even minimal experience with simple trigonometric functions will instantly make you think of
or
. The former doesn't work, since
is obtuse, so
.
Thus, our area must be , and so our answer is
.
When we were using law of cosines, we instantly went to the sub-triangle . Why?
Notice that is bigger than both
and
because the sides which meet to form the angle are shorter for
than the other two triangles.
Also notice that the three angles add up to radians. One of our angles is
by our reasoning/guessing above. This angle then must be the greatest angle out of the three, since the three angles all must be obtuse, and if one angle was
, and another was greater than
, then the third angle would be less than
, leading to a contradiction, since we know that all of the three angles are obtuse.
Thus, we know that our greatest angle . So, we instantly went to the sub-triangle
when using law of cosines since the other sub-triangles will get us nowhere.
Solution 4
![[asy] import olympiad; import cse5; import graph; dotfactor = 2; unitsize(0.3inch); pair B = (0,0), C= (5,0), A = (sqrt(9-2.4*2.4),2.4); pair D = rotate(60,B)*A, E=rotate(60,A)*C, F=rotate(60,C)*B; pair X = extension(A,F,D,C); pair L = (-1.5,2), M = (6.2,3), N = rotate(-60,L)*M; dot("$C$", C, dir(0)); dot("$A$", A, dir(90));dot("$B$", B, dir(180)); dot("$D$", D, NE); dot("$E$", E, dir(90));dot("$F$", F, dir(270)); dot("$M$", M, NE); dot("$N$", N, dir(270));dot("$L$", L, NW); dot("$X$", X, dir(250)); draw(L--X); draw(M--X); draw(N--X); draw(A--B--C--cycle); draw(A--D--B); draw(B--F--C); draw(A--E--C); draw(A--F,dashed); draw(D--C,dashed); draw(B--E,dashed); draw(L--M--N--cycle); [/asy]](http://latex.artofproblemsolving.com/0/d/a/0da7943d44acdd6a610940b08e776b5b24eb4e48.png)
Let's call the circle center . It has a distance of 3, 4, 5 to an equilateral triangle
. Consider
’s pedal triangle
. Since
’s antipedal triangle is equilateral,
must be the one of the isogonic centers of
. We’ll take the one inside
, i.e., the Fermat point, because it leads to larger
. Now we construct the three equilateral triangles
,
, and
, the same way the Fermat point is constructed. Then we have
. Since
is concyclic with
=4 as diameter, we have
. Similarly,
, and
. So
is a 3-4-5 right triangle with
. With some more angle chasing we get
By Law of Cosines, we have
And the area follows.
By Mathdummy
Solution 5
Let be the equilateral triangle with
Assume the coordinates of the vertices are
and
Let
be such that
and
Then
Subtraction and addition of the first two equations yield
The third equation gives
Then
We can then solve for
in terms of
and have a substitution. We have
Simplify it we have a quadratic equation for
So
The larger one leads to the solution. The smaller one relates to another equilateral triangle, as indicated in Solution 2.
-JZ
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2012aimei/353
~ dolphin7
See also
2012 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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