2017 AIME I Problems/Problem 15
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
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
Let , lies on , lies on and lies on
Set as the origin, , can be expressed as in argand plane, the distance of is
We know that . We know that the slope of is , we have that , after computation, we have
Now the rest is easy with C-S inequality, so the smallest area is , and the answer is
~bluesoul
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 (Trigonometry)
Let
Then
By Law of Sines on triangle we get The smallest area is vladimir.shelomovskii@gmail.com, vvsss
Note
follows from Cauchy-Schwarz. ~mathboy282
Solution 5 (Complex numbers)
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 6
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
Solution 7
We can use complex numbers. Set the origin at the right angle. Let the point on the real axis be and the point on the imaginary axis be . Then, we see that Now we switch back to Cartesian coordinates. The equation of the hypotenuse is This means that the point is on the line. Plugging the numbers in, we have Now, we note that the side length of the equilateral triangle is so it suffices to minimize that. By Cauchy-Schwarz, we have Thus, the area of the smallest triangle is so our desired answer is .
(Solution by Pleaseletmetwin, but not added to the Wiki by Pleaseletmetwin)
Solution 8
Employ the same complex bash as in Solution 5, but instead note that minimizing is the same as minimizing the distance from 0,0 to x,y, since they are the same quantity. We use point to plane instead, which gives you the required distance.
Solution 9 (Non Analytic)
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 10
Let the right triangle's lower-left point be at . Notice the 2 other points will determine a unique equilateral triangle. Let 2 points be on the -axis () and the -axis () and label them and respectively. The third point () will then be located on the hypotenuse. We proceed to find the third point's coordinates in terms of and .
1. Find the slope of and take the negative reciprocal of it to find the slope of the line containing . Notice the line contains the midpoint of so we can then have an equation of the line.
2. Let For to be an equilateral triangle, the altitude from to must be
We then have two equations and two variables, so we can solve for 's coordinates.
We can find Also, note that must be on the hypotenuse of the triangle We can plug in and as the coordinates of , which simplifies to
We aim to minimize the side length of the triangle, which is Applying the Cauchy inequality gives us
From which we obtain Thus, the area of the triangle = which leads to the answer
-hi_im_bob
Solution 11
The general solution to the minimal area is as following:
where and are the two legs of the right triangle. In this particular case and . When we plug in these two values, we recover the correct answer of .
The contour of the minimal area is plotted as a function of leg lengths and , as shown on the right hand side.
Note
The proof of the formula can be done in a similar fashion as Solution 4, where instead of using specific values, we define variables, and in this case, as per the formula.
Solution 12 (Geometry)
Let Then
midpoint midpoint circumcenter Points and lies on bisectors of and so
The smallest area is vladimir.shelomovskii@gmail.com, vvsss
Solution 13 (Kinematics+Geometry)
Let Then
Let the required triangle with minimal sides be constructed.
Let us vary its position in an acceptable way, that is, we will perform a movement in which the vertices of the remain on the sides of the Any motion of a solid plane figure can be considered as rotation around some point, the center of rotation. The speed of movement of any point is perpendicular to the segment from the center of rotation to this point. With an acceptable variation, the velocities of the vertices of the are directed along the sides Consequently, there is a point located at the intersection of perpendiculars to the sides of around which rotates. The bases of the perpendiculars dropped from to the sides of form a regular
Therefore is the first isodynamic point of
It is known that
The points and are concyclic, is the diameter, so Similarly, The smallest area is vladimir.shelomovskii@gmail.com, vvsss
See Also
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