Difference between revisions of "2017 AIME I Problems/Problem 15"
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− | Let the vertices of the right triangle be <math>(0,0),(5,0),(0,2\sqrt{3}),</math> and let <math>(a,0),(0,b)</math> be two of the | + | Let the vertices of the right triangle be <math>(0,0),(5,0),(0,2\sqrt{3}),</math> and let <math>(a,0),(0,b)</math> be the two vertices of the equilateral triangle on the legs of the right triangle. Then, the third vertex of the equilateral triangle is <math>\left(\frac{a+\sqrt{3}b}{2},\frac{\sqrt{3}a+b}{2}\right)</math>. This point must lie on the hypotenuse <math>\frac{x}{5} + \frac{y}{2\sqrt{3}} = 1</math>, i.e. <math>a,b</math> must satisfy |
<cmath> \frac{a+\sqrt{3}b}{10}+\frac{\sqrt{3}a+b}{4\sqrt{3}} = 1,</cmath> | <cmath> \frac{a+\sqrt{3}b}{10}+\frac{\sqrt{3}a+b}{4\sqrt{3}} = 1,</cmath> | ||
which can be simplified to | which can be simplified to |
Revision as of 22:48, 2 August 2020
Contents
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
Let's start by proving a 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 .
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Let the vertices of the right triangle be and let be the two vertices of the equilateral triangle on the legs of the right 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
Let be the right triangle with sides , , and and right angle at .
Let an equilateral triangle touch , , and at , , and respectively, having side lengths of .
Now, call as and as . Thus, and .
By Law of Sines on triangles and ,
and .
Summing,
.
Now substituting , , and and solving, .
We seek to minimize .
This is equivalent to minimizing .
Using the lemma from solution 1, we conclude that
Thus, and our final answer is
- Awsomness2000
Solution 4
We will use complex numbers. Set the vertex at the right angle to be the origin, and set the axes so the other two vertices are and , respectively. Now let the vertex of the equilateral triangle on the real axis be and let the vertex of the equilateral triangle on the imaginary axis be . Then, the third vertex of the equilateral triangle is given by: .
For this to be on the hypotenuse of the right triangle, we also have the following:
Note that the area of the equilateral triangle is given by , so we seek to minimize . This can be done by using the Cauchy Schwarz Inequality on the relation we derived above:
Thus, the minimum we seek is simply , so the desired answer is .
Solution 5 (Alcumus)
In the complex plane, let the vertices of the triangle be and Let be one of the vertices, where is real. A point on the line passing through and can be expressed in the form We want the third vertex to lie on the line through and which is the imaginary axis, so its real part is 0. Since the small triangle is equilateral, or Then the real part of is Solving for in terms of we find Then so so This quadratic is minimized when and the minimum is so the smallest area of the equilateral triangle is
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.