What is the interior of a circle? Obvious.

What is the interior of a parabola? Not quite as obvious.

What is the interior of a hyperbola? Not at all obvious.

Is it possible to define interior in a way that applies to all conic sections?

Circles

If you remove a circle from the plane, there are two components left. Which one is the interior and which one is the exterior?

Obviously the bounded part is the interior and the unbounded part is the exterior. But using boundedness as our criteria runs into immediate problems.

Parabolas

If you remove a parabola in the plane, which component is the interior and which is the exterior? You might say there is no interior because both components of the plane minus the parabola are unbounded. Still, if you had to label one of the components the interior, you’d probably say the smaller one.

But is the “smaller” component really smaller? Both components have infinite area. You could patch this up by taking a square centered at the origin and letting its size grow to infinity. The interior of the parabola is the component that has smaller area inside the square all along the way.

Hyperbolas

A hyperbola divides the plane into three regions. Which of these is the interior? If we try to look at area inside an expanding square, it’s not clear which component(s) will have more or less area. Seems like it may depend on the location of the center of the square relative to the position of the hyperbola.

Tangents to a circle

Here’s another way to define the interior of a circle. Look at the set of all lines that are tangent to a point on the circle. None of them go through the interior of the circle. We can define the interior of the circle as the set of points that no tangent line passes through.

This clearly works for a circle, and it’s almost as clear that it would work for an ellipse.

How do we define the exterior of a circle? We could just say it’s the part of the plane that isn’t the interior or the circle itself. But there is a more interesting definition. If the interior of the circle consists of points that tangent lines don’t pass through, the exterior of the circle consists of the set of points that tangent lines do pass thorough. Twice in fact: every point outside the circle is at the intersection of two lines tangent to the circle.

To put it another way, consider the set of all tangent lines to a circle. Every point in the plane is part of zero, one, or two of these lines. The interior of the circle is the set of points that belong to zero tangent lines. The circle is the set of points that belong to one tangent line. The exterior of the circle is the set of points that belong to two tangent lines.

Tangents to a parabola

If we apply the analogous definition to a parabola, the interior of the parabola works out to be the part we’d like to call the interior.

It’s not obvious that every point of the plane not on the parabola and not in the interior lies at the intersection of two tangent lines, but it’s true.

Tangents to a hyperbola

If we look at the hyperbola x² − y² = 1 and draw tangent lines, the interior, the portion of the plane with no crossing tangent lines, is the union of two components, one containing (−∞, −1) and one containing (1, ∞). The exterior is then the component containing the line y = x. In the image above, the pink and lavender components are the interior and the green component is the exterior.

It’s unsatisfying that the interior of the hyperbola is disconnected. Also, I believe the exterior is missing the origin. Both of these annoyances go away when we add points at infinity. In the projective plane, the complement of a conic section consists of two connected components, the interior and the exterior. The origin lies on two tangent lines: one connecting (−∞, −∞) to (∞, ∞) and one connecting (−∞, ∞) to (∞, −∞).

Is the shape of a triangle determined by its perimeter and area? In other words, if two triangles have the same area and the same perimeter, are the triangles similar? [1]

It’s plausible. A triangle has three degrees of freedom: the lengths of the three sides. Specifying the area and perimeter removes two degrees of freedom. Allowing the triangles to be similar rather than congruent accounts for a third degree of freedom.

Here’s another plausibility argument. Heron’s formula computes the area of a triangle from the lengths of the sides.

Here s is the semi-perimeter, half of the sum of the lengths of the sides. So if the perimeter and area are known, we have a third order equation for the sides:

If the right-hand side were 0, then we could solve for the lengths of the sides. But the right-hand side is not zero. Is it still possible that the sides are uniquely determined, up to rearranging how we label the sides?

It turns out the answer is no [2], and yet it is not simple to construct counterexamples. If all the sides of a triangle are rational numbers, it is possible to find a non-congruent triangle with the same perimeter and area, but the process of finding this triangle is a bit complicated.

One example is the triangles with sides (20, 21, 29) and (17, 25, 28). Both have perimeter 70 and area 210. But the former is a right triangle and the latter is not.

Where did our algebraic argument go wrong? How can a cubic equation have two sets of solutions? But we don’t have a cubic equation in one variable; we have an equation in three variables that is the product of three linear terms.

What third piece of information would specify a triangle uniquely? If you knew the perimeter, area, and the length of one side, then the triangle is determined. What if you specified the center of the triangle? There are many ways to define a center of a triangle; would some, along with perimeter and area, uniquely determine a triangle while others would not?

Related posts

• Six triangle properties in one equation
• Perimeter, radii, and sides
• How you define center matters a lot

[1] Two triangles are similar if you can transform one into the other by scaling and/or rotation.

[2] Mordechai Ben-Ari. Mathematical Surprises. Springer, 2022. The author sites this blog post as his source.

Last week Heron’s formula came up in the post An Unexpected Triangle. Given the lengths of the sides of a triangle, there is a simple expression for the area of the triangle.

where the sides are a, b, and c and s is the semiperimeter, half the perimeter.

Is there an analogous formula for the area of a quadrilateral? Yes and no. If the quadrilateral is cyclic, meaning there exists a circle going through all four of its vertices, then Brahmagupta’s formula for the area of a quadrilateral is a direct generalization of Heron’s formula for the area of a triangle. If the sides of the cyclic quadrilateral are a, b, c, and d, then the area of the quadrilateral is

where again s is the semiperimeter.

But in general, the area of a quadrilateral is not determined by the length of its sides alone. There is a more general expression, Bretschneider’s formula, that expresses the area of a general quadrilateral in terms of the lengths of its sides and the sum of two opposite angles. (Either pair of opposite angles lead to the same value.)

In a cyclic quadrilateral, the opposite angles α and γ add up to π, and so the cosine term drops out.

The contrast between the triangle and the quadrilateral touches on an area of math called distance geometry. At first this term may sound redundant. Isn’t geometry all about distances? Well, no. It is also about angles. Distance geometry seeks results, like Heron’s theorem, that only depend on distances.

Related posts

• Shoelace formula for polygon area
• Area of a triangle in the complex plane
• Area of a quadrilateral as a determinant

The basic idea of GPS is that if you know the distance to several satellites, you can figure out your position. But you don’t actually know, or need to know, the distance to the satellites: you know the time (according to each satellite’s clock) when the signals were sent, and you know the time (according to your clock) when the signals arrived.

The atomic clocks on satellites are synchronized with each other to within a nanosecond, but they’re not synchronized with your clock. There is some offset t between your clock and the satellites’ clocks. Presumably t is small, but it matters.

If you observe m satellites, you have a system of m equations in 4 unknowns:

|| a_i − x || = t_i − t

where a_i is the known position of the ith satellite in 3 dimensions, x is the observer’s position in three dimensions, and t_i is the difference between the time when the signal left the ith satellite (according to its clock) and the time when the signal arrived (according to the observer’s clock). This assumes we choose units so that the speed of light is c = 1.

So we have a system of m equations in 4 unknowns. It’s plausible there could be a unique solution provided m = 4. However, this is not guaranteed.

Here’s an example to suggest why there may not be a unique solution. Suppose t is known to be 0. Then observing 3 satellites will give us 3 equations in 3 unknowns. Each t_i determines a sphere of radius t_i. Suppose two spheres intersect in a circle, and the third sphere intersects this circle in two points. This means we have two solutions to our system of equations.

In [1] the authors thoroughly study the solution to the GPS system of equations. They allow the satellites and the observer to be anywhere in space and look for conditions under which the system has a unique solution. In practice, GPS satellites are approximately confined to a sphere (more on that here) and the observer is also approximately confined to a sphere, namely the earth’s surface, but the authors do not take advantage of these assumptions.

The authors also assume the problem is formulated in n dimensional space, where n does not necessarily equal 3. It’s complicated to state when the system of equations has a unique solution, but allowing n to vary does not add to the complexity.

I’m curious whether there are practical uses for the GPS problem when n > 3. There are numerous practical problems involving the intersections of spheres in higher dimensions, where the dimensions are not Euclidean spacial dimensions but rather abstract degrees of freedom. But off hand I cannot think of a problem that would involve the time offset that GPS location finding has.

[1] Mireille Boutin, Gregor Kemperc. Global positioning: The uniqueness question and a new solution method. Advances in Applied Mathematics 160 (2024)

Let a, b, and c be the sides of a triangle. Let r be the radius of an inscribed circle and R the radius of a circumscribed circle. Finally, let p be the perimeter. Then the previous post said that

2prR = abc.

We could rewrite this as

2rR = abc / (a + b + c)

The right hand side is maximized when a = b = c. To prove this, maximize abc subject to the constraint a + b + c = p using Lagrange multipliers. This says

[bc, ac, ab] = λ[1, 1, 1]

and so ab = bc = ac, and from there we conclude a = b = c. This means among triangles with any given perimeter, the product of the inner and outer radii is maximized for an equilateral triangle.

The inner radius for an equilateral triangle is (√3 / 6)a and the outer radius is a/√3, so the maximum product is a²/6.

Related posts

• Computing inscribed radius and circumscribed radius
• Johnson circle theorem
• Nine-point circle theorem

Let a, b, and c be the sides of a triangle.

Let p be perimeter of the triangle.

Let r be the radius of the largest circle that can be inscribed in the triangle, and let R be the radius of the circle through the vertices of the triangle.

Then all six numbers can be related in one equation:

2prR = abc.

A golden ellipse is an ellipse whose axes are in golden proportion. That is, the ratio of the major axis length to the minor axis length is the golden ratio φ = (1 + √5)/2.

Draw a golden ellipse and its inscribed and circumscribed circles. In other words draw the largest circle that can fit inside and the smallest circle outside that contains the ellipse.

Then the area of the ellipse equals the area of the annulus bounded by the two circles. That is, the area of the green region

equals the area of the orange region.

The proof is straight forward. Let a be the semimajor axis and b the semiminor axis, with a = φb.

Then the area of the annulus is

π(a² − b²) = πb²(φ² − 1).

The area of the ellipse is

πab = πφb².

The result follows because the golden ratio satisfies

φ² − 1 = φ.

Related posts

• Posts on ellipses
• Gold, Silver, and Bronze ratios

Barycentric coordinates are sometimes called area coordinates or areal coordinates in the context of triangle geometry. This is because the barycentric coordinates of a point P inside a triangle ABC correspond to areas of the three triangles PBC, PCA and PAB.

(This assumes ABC has unit area. Otherwise divide the area of each of the three triangles by the area of ABC. We will assume for the rest of this post that the triangle ABC has unit area.)

Areal coordinates take three numbers two describe a point in two dimensional space. Why would you do that? It’s often useful to use an overdetermined coordinate system. The benefit of adding one more coordinate is that you get a coordinate system matched to the geometry of the triangle. For example, the vertices of the triangle have coordinates (1, 0, 0), (0, 1, 0), and (0, 0, 1), regardless of the shape of the triangle.

Here is an example of a theorem [1] that is convenient to state in terms of areal coordinates but that would be more complicated in Cartesian coordinates.

First we need to define the midpoint triangle, also called the medial triangle. This is the triangle whose vertices are the midpoints of each side of ABC. In terms of areal coordinates, the vertices of this triangle are (0, ½, ½), (½, 0, ½), and (½, ½, 0).

Now let P be any point inside the midpoint triangle of ABC. Then there is a unique ellipse E inscribed in ABC and centered at P.

Let (α, β, γ) be the areal coordinates of P. Then the area of E is

Because P is inside the medial triangle, each of the areal coordinates are less than ½ and so the quantity under the square root is positive.

Finding the equation of the inscribed ellipse is a bit complicated, but that’s not necessary in order to find its area.

Related posts

• Trilinear coordinates
• Converting between areal and trilinear coordinates
• Triangle centers

[1] Ross Honsberger. Mathematical Plumbs. 1979

I keep running into Edward John Routh (1831–1907). He is best known for the Routh-Hurwitz stability criterion but he pops up occasionally elsewhere. The previous post discussed Routh’s mnemonic for moments of inertia and his “stretch” theorem. This post will discuss his triangle theorem.

Before stating Routh’s theorem, we need to say what a cevian is. Giovanni Ceva (1647–1734) was an Italian geometer, best known for Ceva’s theorem, and for a construction in that theorem now known as a cevian.

A cevian is a line from the vertex of a triangle to the opposite side. Draw three cevians by connecting each vertex of a triangle to a point on its opposite side. If the cevians intersect at a point, Ceva’s theorem says something about how the lines divide the sides. If the cevians form a triangle, Routh’s theorem find the area of that triangle.

Routh’s theorem is a generalization of Ceva’s theorem because if the cevians intersect at a common point, the area of the triangle formed is zero, and then Routh’s area equation implies Ceva’s theorem.

Let A, B, and C be the vertices of a triangle and let D, E, and F be the points where their cevians intersect the opposite sides.

Let x, y, and z be the ratios into which each side is divided by the cevians. Specifically let x = FB/AF, y = DC/BD, and z = EA/CE.

Then Routh’s theorem says the relative area of the green triangle formed by the cevians is

If the cevians intersect at a point, the area of the triangle is 0, which implies xyz = 1, which is Ceva’s theorem.

Related posts

• Routh’s mnemonic
• Kepler triangle
• Triangle centers and Page Rank