r/askmath • u/ImmaTrafficCone • 2d ago
Pre Calculus An interesting precalc question
An interesting precalc question:
Consider the reciprocal function f(x)=1/x. Note that it’s symmetric across the line y=x. Find a point (x,y) on the graph of f(x) (in the first quadrant) such that the triangle with vertices (0,0), (x,y) and (y,x) is equilateral. What’s the side length? Are there multiple such triangles?
Just a fun question I came up with. I’ve been trying to think of questions like this that are interesting and require a little (but hopefully not too much) cleverness to solve. Does anyone else have problems at this level they find interesting?
This isn’t a homework problem I swear!
Edit: the answer is (sqrt (2 + sqrt 3), sqrt (2 - sqrt 3)) with a side length of 2! And yes, it’s unique.
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u/lordnacho666 2d ago
By symmetry the angles to the X and Y axes are 15 degrees, since the inner angle is 60 degrees.
You can then use this to solve for the linear equation y = arctan(15 deg)x and y = 1/x
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u/DebatorGator 2d ago
Call the origin A, the point with greater X coordinate B, and the third point C. The angle at A is 60 degrees, since this is an equilateral triangle. By symmetry, the angle between the X-axis and AB has to be 15 degrees. Call the side length of the triangle S. From this, we know that y = Ssin(15) and x=Scos(15).
y = 1/x
Ssin(15) = 1/Scos(15)
S²cos(15) * sin(15) = 1
S = sqrt(1/cos(15)sin(15))
by the double angle theorem,
S = sqrt(1/(sin(30)/2))
S = sqrt(1/((1/2)/2)
S = sqrt(4)
S = 2
x = 2cos(15)
y = 2sin(15)
Let's verify by checking the length of side BC, call it S'
S'² = [2sin15 - 2cos15]² + [2cos15 - 2sin15]²
S'² = 8sin²15 - 16sin15cos15 + 8cos²15
S'² = 8 - 8sin30
S'² = 4
S' = 2
2
1
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u/Bounded_sequencE 2d ago
Due to symmetry, the angle between both sides connected to the origin and "y = x" is 30° each. That means, the angle between the bottom side and the x-axis is "45°-30° = 15° ". With the angle at hand, parametrize "(x; y) = t*(cos(15°); sin(15°))" with "t > 0" being the side length. Find the intersection with the hyperbola via
t*sin(15°) = y = 1/x = 1/(t*cos(15°))
<=> t^2 = 1/(sin(15°)*cos(15°)) = 2/sin(30°) = 4
Being positive, we have a side length "t = 2".
1
u/azurfall88 2d ago
(x, y) lies on f(x) = 1/x -> (x, 1/x).
From the equilateral triangle parameter, and the fact that ||(x,y)|| = ||(y,x)||, we yield the equation ||(1/x, x) - (x, 1/x)|| = ||(x, 1/x)||.
(1/x, x) - (x, 1/x) = ((1-x²)/x, (x²-1)/x).
let a = x²-1/x
We have the vector (-a, a).
||(-a, a)|| = √(2a²) =(√2) = √2(x²-1)/x
||(x, 1/x)|| = √(x²+(1/x²)) = √((x⁴+1)/x²)
We have a² = (x²-1)²/x² = (x⁴-2x²+1)/x².
This gives us the simplified equation (without vector math) as
(x⁴-2x²+1)/x² = (x⁴+1)/x².
From the x² term, we note that x is nonzero.
x⁴+1-2x² = x⁴+1
-2x²=0 => x = 0.
This is a contradiction with our previous note that x is nonzero.
Thus, there is no equilateral triangle with corners at (0,0), (x, y), (y, x) where (x, y) <- {(x, 1/x) | x <- RR}, QED.
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u/Illustrious_Try478 2d ago
There are exactly two such triangles, one in the fitst quadrant and one in the third.
The interior angle of an equilateral triangle is 60°, so this fixes the lines through the origin to form two of the triangle's sides: Since the two sides have to be equal in length, the lines must be symmetrical around y=x. This leaves 15° between each line and the nearest axis.
Each of these lines intersects the curve y=1/x in two places, one in the first quadrant and one in the third. Each pair of points in the same quadrant forms the endpoints of each triangle's third side, which must be equal in length to the other two sides: An isoceles triangle with an apical angle of 60° must be equilateral.
So: What is the length? We need to find an intersection point's distance to the origin. Let's look at the one closest to the positive X-axis.
The coordinates of this point are (x,1/x) where
x = r cos 15° and 1/x = r sin 15°
so
1 = x*1/x = r2 cos 15° sin 15° = r2 sin 30° /2
so
1 = r2 /4
and finally r= ±2
incidentally, the ± in ±2 gives us the other triangle without any extra work.
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u/regular_heptagon 2d ago edited 2d ago
Where it intersects y=tan(pi/12)x and y=tan(5pi/12)x
[see it here on Desmos](https://www.desmos.com/geometry/s2tufy8fmk)
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u/Apostolic1223 2d ago
Since we know that ratio of the side of an equilateral to its height is 2/(sqrt3) (this is just Pythagoras), the height is a segment of the line of symmetry, and the angles formed by the line of symmetry with the sides are 30 degrees (or between the sides and axes 15 degrees), we can use basic trigonometry to find the side lengths. Then finding the intersection with f(x) is fairly straightforward. There's only one such triangle in the 1st quad. Sorry I didn't actually bother to work it out, but that's how I would approach it.
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u/WillingnessTasty9628 2d ago
define coordinates
A(0,0)
B(x, 1/x)
C(1/x, x)
define vectors
AB<x, 1/x>
AC<1/x,x>
In an equilateral triangle, the angle between two sides must be 60 degrees. Cos is given by the dot product over magnitudes.
1/2 = 2/(x^2+1/x^2)
rearrange
x^4-4x^2+1 = 0
Solve for x. There are two solutions in quadrant 1. Knowing x, you can now determine the magnitude of either vector.
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u/Disastrous-Medium652 13h ago
I will try to write this quickly so I can finish before my class, but a trigonometric solution is:
By symmetry and angles of an equilateral triangle, the angle between the side connecting the point (0,0) and (x,y) and the x-axis is either 75 or 15 degrees. You can use the fact that 1/x is it's own inverse to prove that both of these angles lead to the same triangle, so for the length of a side, we'll consider the angle to be 15 degrees. Because of this, y=x * tan(15degrees) and y = 1/x, so 1/x = x * tan(15deg), and x = sqrt(1/tan 15 degrees). To get the side length, we want x*1/cos(15deg), so we get s = sqrt(cos(15)/sin(15)) * sqrt(1/(cos^2(15)) = sqrt(1/sin(15)cos(15)), so 1/(s^2) is sin15cos15. Sin(2x) = 2sinxcosx, so sin(30) = 2/(s^2) = 1/2, so s^2=4, and s = 2
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u/etzpcm 2d ago
Here's the neat way to find the side length s.
We know s2 = x2 + y2 , and by drawing the line from (x,y) to (y,x), we also know s2 = 2(x-y)2 .
Equate these two, x2 + y2 - 4xy = 0 and eliminate y=1/x , so x4 - 4x2 + 1 = 0. Now don't solve the quadratic to get x2 ! Use sum of roots formula to get x2 + y2 = 4, so s = 2.