Difference between revisions of "2017 AIME I Problems/Problem 15"
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+ | == Solution 3 == | ||
==See Also== | ==See Also== | ||
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Revision as of 13:39, 12 March 2017
Problem 15
The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths and
as shown, is
where
and
are positive integers,
and
are relatively prime, and
is not divisible by the square of any prime. Find
Solution 1
Lemma. If satisfy
, then the minimal value of
is
.
Proof. Recall that the distance between the point and the line
is given by
. In particular, the distance between the origin and any point
on the line
is at least
.
---
Let the vertices of the right triangle be and let
be two of the vertices of the equilateral triangle. Then, the third vertex of the equilateral triangle is
. This point must lie on the hypotenuse
, i.e.
must satisfy
which can be simplified to
By the lemma, the minimal value of is
so the minimal area of the equilateral triangle is
and hence the answer is
.
Solution 2 (No Coordinates)
Let be the triangle with side lengths
and
.
We will think about this problem backwards, by constructing a triangle as large as possible (We will call it , for convenience) which is similar to
with vertices outside of a unit equilateral triangle
, such that each vertex of the equilateral triangle lies on a side of
. After we find the side lengths of
, we will use ratios to trace back towards the original problem.
First of all, let ,
, and
(These three angles are simply the angles of triangle
; out of these three angles,
is the smallest angle, and
is the largest angle). Then let us consider a point
inside
such that
,
, and
. Construct the circumcircles
and
of triangles
and
respectively.
From here, we will prove the lemma that if we choose points ,
, and
on circumcircles
and
respectively such that
,
, and
are collinear and
,
, and
are collinear, then
,
, and
must be collinear. First of all, if we let
, then
(by the properties of cyclic quadrilaterals),
(by adjacent angles),
(by cyclic quadrilaterals),
(adjacent angles), and
(cyclic quadrilaterals). Since
and
are supplementary,
,
, and
are collinear as desired. Hence,
has an inscribed equilateral triangle
.
In addition, now we know that all triangles (as described above) must be similar to triangle
, as
and
, so we have developed
similarity between the two triangles. Thus,
is the triangle similar to
which we were desiring. Our goal now is to maximize the length of
, in order to maximize the area of
, to achieve our original goal.
Note that, all triangles are similar to each other if
,
, and
are collinear. This is because
is constant, and
is also a constant value. Then we have
similarity between this set of triangles. To maximize
, we can instead maximize
, which is simply the diameter of
. From there, we can determine that
, and with similar logic,
,
, and
are perpendicular to
,
, and
respectively We have found our desired largest possible triangle
.
All we have to do now is to calculate , and use ratios from similar triangles to determine the side length of the equilateral triangle inscribed within
. First of all, we will prove that
. By the properties of cyclic quadrilaterals,
, which means that
. Now we will show that
. Note that, by cyclic quadrilaterals,
and
. Hence,
(since
), proving the aforementioned claim. Then, since
and
,
.
Now we calculate and
, which are simply the diameters of circumcircles
and
, respectively. By the extended law of sines,
and
.
We can now solve for with the law of cosines:
Now we will apply this discovery towards our original triangle . Since the ratio between
and the hypotenuse of
is
, the side length of the equilateral triangle inscribed within
must be
(as
is simply as scaled version of
, and thus their corresponding inscribed equilateral triangles must be scaled by the same factor). Then the area of the equilateral triangle inscribed within
is
, implying that the answer is
.
-Solution by TheBoomBox77
Solution 3
See Also
2017 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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