Isosceles and Hyperbola

A right isosceles triangle embedded in a hyberbola, solved using three mathematical disciplines

February 20, 2026 · John Peach

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I came across this problem recently:

Find the minimal area of a right isosceles triangle whose vertices lie on the hyperbola xy=1xy=1. The hyperbola xy=1:xy=1: This is a rectangular hyperbola with both coordinate axes as asymptotes. It has two branches: one in the first quadrant (where both xx and yy are positive) and one in the third quadrant (where both are negative). It is also the graph of y=1/x.y=1/x.

The hyperbola xy=1xy=1 is one of the most familiar curves in mathematics — the graph of y=1/xy=1/x, sweeping through two opposite quadrants, never touching either axis, stretching toward infinity in four directions at once. It would seem to be an unlikely home for a triangle. Yet you can place three points on this curve and connect them to form a right isosceles triangle. The question is: how small can that triangle get?

Figure 1.

Isosceles Right Triangle on a Hyperbola

The solution turns out to be a bit intricate, involving some linear algebra, geometry and calculus. We’ll start with linear algebra to set the orthogonality Orthogonality refers to a relationship of perpendicularity or independence between vectors, functions, or system components, typically defined by an inner product (dot product) of zero. In mathematics, it signifies 90-degree angles between vectors. (right angle) conditions, solve for the isosceles An isosceles triangle is a polygon with at least two sides of equal length, known as legs, and two congruent base angles opposite those sides. Key features include a third, unequal side (base) and a vertex angle opposite that base.constraint, and use calculus to find the minimum spacing between the points.

The Right Isosceles Conditions

The problem gives two conditions that must be met. The first is that the triangle is isosceles, meaning that two of the sides have the same length. The distance between two points is derived from the Pythagorean formula, d=(x1−x2)2+(y1−y2)2,d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}, but since we only care that the two sides have the same length, we can use the square of the distance and omit the square root.

The second condition is that the triangle is a right triangle, meaning that the angle ∠BAC=90°\angle BAC = 90\degree. One way to force this condition is to calculate the dot product Dot product: The dot product of two vectors u=[u1​,u2​]u=[u_1​,u_2​] and v=[v1​,v2​]v=[v_1​,v_2​] is u⋅v=u1​v1​+u2​v2u⋅v=u_1​v_1​+u_2​v_2​. When the dot product equals zero, the vectors are perpendicular. This is a compact algebraic way to enforce a right angle. between the vectors AB⇀\overrightharpoon{AB} and AC⇀\overrightharpoon{AC} and set the result AB⇀⋅AC⇀=0.\overrightharpoon{AB} \cdot \overrightharpoon{AC} = 0.

Let the coordinates be A(a,1/a),B(b,1/b),C(c,1/c). We have two equations that must be satisfied:

(b−a)2+(1/b−1/a)2=(c−a)2+(1/c−1/a)2(1)(b-a)^2+(1/b-1/a)^2 = (c-a)^2+(1/c-1/a)^2 \tag{1} [(b−a)(1/b−1/a)][c−a 1/c−1/a]=0(2)\begin{bmatrix}(b-a) & (1/b-1/a) \end{bmatrix} \begin{bmatrix}c-a \\\ 1/c-1/a \end{bmatrix} = 0 \tag{2}

Note that

1b−1a=−b−aab.(3)\frac{1}{b} - \frac{1}{a} = - \frac{b-a}{ab}. \tag{3}

Let β=b−a\beta = b-a and γ=c−a\gamma = c-a so the two equations become

β2+β2(ab)2=γ2+γ2(ac)2(4)\beta^2 + \frac{\beta^2}{(ab)^2} = \gamma^2 + \frac{\gamma^2}{(ac)^2} \tag{4}

and

[β−βab][γ−γac]=0.(5)\begin{bmatrix} \beta & -\frac{\beta}{ab} \end{bmatrix} \begin{bmatrix} \gamma \\ -\frac{\gamma}{ac} \end{bmatrix} = 0. \tag{5}

From the previous equation,

βγ+βγ(ab)(ac)=βγ(1+1(ab)(ac))=0.(6)\beta \gamma + \frac{\beta \gamma}{(ab)(ac)} = \beta \gamma \left( 1 + \frac{1}{(ab)(ac)} \right) = 0. \tag{6}

Since β≠0\beta \neq 0 and γ≠0\gamma \neq 0 then (ab)(ac)=−1(ab)(ac) = -1 because 1(ab)(ac)=−1\frac{1}{(ab)(ac)} = -1 in (6). Thus (ab)(ac)=a2bc=−1(ab)(ac) = a^2bc = -1 so (4) becomes

β2(1+1(ab)2)=γ2(1+(ab)2).(7)\beta^2 \left( 1 + \frac{1}{(ab)^2} \right) = \gamma^2 (1 + (ab)^2). \tag{7}

Rearranging,

β2γ2=1+(ab)21+(ab)2(ab)2=(ab)2(8)\frac{\beta^2}{\gamma^2} = \frac{1 + (ab)^2}{\frac{1+(ab)^2}{(ab)^2}} = (ab)^2 \tag{8}

so

β=±(ab)γ.(9)\beta = \pm (ab)\gamma. \tag{9}

Now, let’s substitute the coordinates back in for β\beta and γ\gamma, and use c=−1a2bc = -\frac{1}{a^2b}

b−a=±(ab)(c−a)=±(ab)(−1a2b−a)=±(−1a−a2b)(10)\begin{aligned} b-a &= \pm(ab)(c-a) \\ &= \pm(ab) \left( -\frac{1}{a^2b} - a \right) \\ &= \pm\left( -\frac{1}{a} - a^2b \right) \\ \end{aligned} \tag{10}

Separating this into two cases and solving for bb gives first

b−a=(−1a−a2b)b+a2b=−1a+ab(1+a2)=−1a+ab=−1a+a1+a2=−1a+a2a1+a2=−1+a2a1+a2=a2−1a(a2+1).(11)\begin{aligned} b-a &= \left( -\frac{1}{a} - a^2 b \right) \\ b+a^2b &= -\frac{1}{a} + a\\ b(1+a^2) &= -\frac{1}{a} + a \\ b &= \frac{-\frac{1}{a} + a}{1+a^2} \\ &= \frac{-\frac{1}{a} + \frac{a^2}{a}}{1+a^2} = \frac{\frac{-1+a^2}{a}}{1+a^2} = \frac{a^2-1}{a(a^2+1)}. \end{aligned} \tag{11}

In the second case,

b−a=−(−1a−a2b)=1a+a2bb−a2b=1a+ab=1a+a1−a2=1a+a2a1−a2=a2+1a1−a2=−a2+1a(a2−1).(12)\begin{aligned} b - a &= -\left( -\frac{1}{a} - a^2 b \right) = \frac{1}{a} + a^2 b \\ b - a^2b &= \frac{1}{a} + a\\ b &= \frac{\frac{1}{a} + a}{1-a^2} \\ &= \frac{\frac{1}{a} + \frac{a^2}{a}}{1-a^2} = \frac{\frac{a^2+1}{a}}{1-a^2} = -\frac{a^2 + 1}{a(a^2 - 1)}. \end{aligned} \tag{12}

Area Minimization

The area of a triangle is half the product of the base times the height. For a right isosceles triangle, this simplifies to half the square of the length of one side. The square of the length of the side AB‾,∥β⇀∥2\overline{AB}, \|\overrightharpoon{\beta}\|^2 is

∥β⇀∥2=(b−a)2+(1/b−1/a)2=(a2−1a(a2+1)−a)2+(a(a2+1)a2−1−1a)2=(a2−1a(a2+1)−a2(a2+1)a(a2+1))2+(a2(a2+1)a(a2−1)−a2−1a(a2−1))2=(a2−1a(a2+1)−a4+a2a(a2+1))2+(a4+a2a(a2−1)−a2−1a(a2−1))2=(−a4+1a(a2+1))2+(a4+1a(a2−1))2(13)\begin{aligned} \Vert \overrightharpoon{\beta} \Vert^2 & = (b-a)^2+(1/b-1/a)^2 \\ &= \left( \frac{a^2-1}{a(a^2+1)} - a \right)^2 + \left( \frac{a(a^2+1)}{a^2-1} - \frac{1}{a} \right)^2 \\ &= \left( \frac{a^2-1}{a(a^2+1)} - \frac{a^2(a^2+1)}{a(a^2+1)} \right)^2 + \left( \frac{a^2(a^2+1)}{a(a^2-1)} - \frac{a^2-1}{a(a^2-1)} \right)^2 \\ &= \left( \frac{a^2-1}{a(a^2+1)} - \frac{a^4 + a^2}{a(a^2+1)} \right)^2 + \left( \frac{a^4 + a^2}{a(a^2-1)} - \frac{a^2-1}{a(a^2-1)} \right)^2 \\ &= \left( -\frac{a^4 + 1}{a(a^2+1)} \right)^2 + \left( \frac{a^4 + 1}{a(a^2-1)} \right)^2 \\ \end{aligned} \tag{13}

for the first case, and

∥β⇀∥2=(b−a)2+(1/b−1/a)2=(−a2+1a(a2−1)−a)2+(−a(a2−1)a2+1−1a)2=(−a2+1a(a2−1)−a2(a2−1)a(a2−1))2+(−a2(a2−1)a(a2+1)−a2+1a(a2+1))2=(−a2+1a(a2−1)−a4−a2a(a2−1))2+(−a4+a2a(a2+1)−a2+1a(a2+1))2=(−a4+1a(a2−1))2+(−a4+1a(a2+1))2(14)\begin{aligned} \Vert \overrightharpoon{\beta} \Vert^2 & = (b-a)^2+(1/b-1/a)^2 \\ &= \left( -\frac{a^2+1}{a(a^2 - 1)} - a \right)^2 + \left( -\frac{a(a^2 - 1)}{a^2+1} - \frac{1}{a} \right)^2 \\ &= \left( -\frac{a^2+1}{a(a^2 - 1)} - \frac{a^2(a^2 - 1)}{a(a^2 - 1)} \right)^2 + \left( -\frac{a^2(a^2 - 1)}{a(a^2+1)} - \frac{a^2 + 1}{a(a^2 + 1)} \right)^2 \\ &= \left( -\frac{a^2 + 1}{a(a^2 - 1)} - \frac{a^4 - a^2}{a(a^2 - 1)} \right)^2 + \left( \frac{-a^4 + a^2}{a(a^2 + 1)} - \frac{a^2 + 1}{a(a^2 + 1)} \right)^2 \\ &= \left( -\frac{a^4 + 1}{a(a^2 - 1)} \right)^2 + \left( -\frac{a^4 + 1}{a(a^2+1)} \right)^2 \\ \end{aligned} \tag{14}

for the second case. After squaring, both sides are equivalent.

Let’s call S(a)S(a) the area of the right isosceles triangle as a function of aa,

S(a)=12(a4+1a(a2+1))2+(a4+1a(a2−1))2.(15)S(a) = \frac{1}{2} \left( \frac{a^4 + 1}{a(a^2+1)} \right)^2 + \left( \frac{a^4 + 1}{a(a^2-1)} \right)^2. \tag{15}

Minimizing the area of the triangle is equivalent to minimizing the area of the square with side length b−a=∥β∥b - a = \Vert \beta \Vert since the triangle is half the square. To simplify the calculation, we can drop the 12\frac{1}{2} multiplier.

Figure 2.

Square area is ∥β∥2\Vert \beta \Vert ^2

Taking the derivative with respect to aa,

S′(a)=2(a4+1)(a6+3a4−3a2−1)a3(a2+1)3+2(a4+1)(a6−3a4−3a2+1)a3(a2−1)3=−4(a4+1)2a(a2−1)3−4(a4+1)2a(a2+1)3+8a(a4+1)(a2−1)2+8a(a4+1)(a2+1)2−2(a4+1)2a3(a2−1)2−2(a4+1)2a3(a2+1)2=4(a4+1)2(a8−10a4+1)a3(a4−1)3.(16)\begin{aligned} S'(a) &= \frac{2(a^4+1)(a^6 + 3a^4 - 3a^2 - 1)}{a^3(a^2 + 1)^3} + \frac{2(a^4+1)(a^6 - 3a^4 - 3a^2 + 1)}{a^3(a^2 - 1)^3} \\ &= -\frac{4 \left(a^4+1\right)^2}{a \left(a^2-1\right)^3}-\frac{4 \left(a^4+1\right)^2}{a \left(a^2+1\right)^3}+\frac{8 a \left(a^4+1\right)}{\left(a^2-1\right)^2}+\frac{8 a \left(a^4+1\right)}{\left(a^2+1\right)^2}-\frac{2 \left(a^4+1\right)^2}{a^3 \left(a^2-1\right)^2}-\frac{2 \left(a^4+1\right)^2}{a^3 \left(a^2+1\right)^2} \\ &= \frac{4 \left(a^4+1\right)^2 \left(a^8-10 a^4+1\right)}{a^3 \left(a^4-1\right)^3}. \end{aligned} \tag{16}

A critical value for SS occurs when S′(a)=0S'(a) = 0 and since a4+1>0a^4 + 1 > 0 for all aa, then the critical point must be when a8−10a4+1=0.a^8 - 10a^4 + 1 = 0. Substituting u=a4u = a^4 gives the quadratic

u2−10u+1=0⇒u=12(10±100−4)=5±26,\begin{aligned} u^2 - 10u + 1 &= 0 \\ \Rightarrow u &= \frac{1}{2} \left(10 \pm \sqrt{100 - 4} \right) \\ &= 5 \pm 2 \sqrt{6}, \end{aligned}

which means a=(5±26)1/4.a = \left(5 \pm 2 \sqrt{6} \right)^{1/4}.

Using Julia, set the value for A=(a,1/a)A = (a,1/a) and define functions for bb and cc:

julia> a = (5+2*sqrt(6))^(1/4)
1.7737712281864233

julia> 1/a
0.563770560774312

julia> b(a) = (a^2-1)/(a*(a^2+1))
b (generic function with 1 method)

julia> b(a)
0.29182911639304016

julia> 1/b(a)
3.426662878467493

julia> c(a,b) = -1/(a^2*b)
c (generic function with 1 method)

julia> c(a,b(a))
-1.0891210895067573

julia> 1/c(a,b(a))
-0.9181715510190711

Using these, you can plot the triangle in Geogebra,

Figure 3.

Isosceles triangle in Geogebra

With the three vertices A,B,CA,B,C defined, draw line segments AB‾\overline{AB} and AC‾\overline{AC}, then use the angle option to show that ∠BAC=90°.\angle{BAC} = 90\degree. You should also notice that the length of the segments are the same, demonstrating the △ABC\triangle{ABC} is, in fact, isosceles. Add a polygon element by clicking on each vertex in sequence (and back to the start AA) to generate a triangle. This will calculate the area, t1=5.19615.t_1 = 5.19615.

We’ve now shown that a minimal area right isosceles triangle exists on a hyperbola and it has two symmetric solutions on the opposite branches of the hyperbola.

Conclusion

Finding the minimal area right isosceles triangle on the hyperbola xy=1xy=1 is an interesting mathematical problem, but there are applications of hyperbolas in physics as well. The hyperbola is the basis of Boyle’s Law, relating pressure and volume in an ideal gas at constant temperature, and they appear in the geometry of gravitational and electrostatic fields, in the shock waves produced by supersonic aircraft, and in the navigation system LORAN, which located ships by measuring differences in arrival times of radio signals.

Orthogonality — the idea that two directions are completely independent of one another — is foundational to signal processing, quantum mechanics, and machine learning. When engineers decompose a complex signal into frequencies using a Fourier transform, they are exploiting the orthogonality of sine waves. When a quantum system is measured, the possible outcomes correspond to orthogonal states.

Experiments

Some experiments you might want to try are:

Code for this article

Working through the equations can be tricky at times, so it might be helpful to use Wolfram Language in a Jupyter notebook. See JupyterLab Desktop for details on installation and use. You can also use Julia in JupyterLab, and to draw the triangle, start Geogebra either online or download the desktop version.

Software

References

Image credits