# Rethinking Pythagoras

Author(s):
Daniel J. Heath (Pacific Lutheran University)

Abstract: The Pythagorean Theorem, perhaps the most widely known result in mathematics, has been proven in countless ways, and remains a basic building block of Euclidean geometry. Pythagorean Theorem analogs in non-Euclidean geometries can provide a gateway into understanding those geometries.

This article includes activities which run in Java WebStart -- you will need the Java Runtime Environment, version 1.5 or later, in order to run these activities.

# Rethinking Pythagoras

Author(s):
Daniel J. Heath (Pacific Lutheran University)

Abstract: The Pythagorean Theorem, perhaps the most widely known result in mathematics, has been proven in countless ways, and remains a basic building block of Euclidean geometry. Pythagorean Theorem analogs in non-Euclidean geometries can provide a gateway into understanding those geometries.

This article includes activities which run in Java WebStart -- you will need the Java Runtime Environment, version 1.5 or later, in order to run these activities.

# Rethinking Pythagoras - 1. Introduction

Author(s):
Daniel J. Heath (Pacific Lutheran University)

How can we use calculus to enhance geometric understanding across multiple geometries? Here we discuss the Pythagorean Theorem and its "evil twins" in alternate geometries, ending, in the appendix, with discovery activities appropriate either for college geometry students or for accelerated calculus students wanting to use their knowledge to explore alternate geometries.

Many of the proofs of the Pythagorean Theorem begin with Figure 1a, showing a right triangle with squares constructed on each side. These proofs usually proceed by matching pieces from the smaller squares to sub-regions of the larger square. However, since squares do not exist in non-Euclidean geometries, we use circles to relate lengths of sides to areas, as in Figure 1b. In this case, if we call the lengths of sides of the triangle $$a$$, $$b$$, and $$c$$ in ascending order of length, then it is easy to see that the circles have area $$\pi a^2$$, $$\pi b^2$$, and $$\pi c^2$$, at least in Euclidean geometry. We note that the switch from squares to circles comes at a cost: figure 1b is both harder to interpret than 1a, and does not constitute a rigorous proof in the Euclidean case as does 1a. However, the universality of circles will make up for the messiness of the picture, and will allow us to conduct numerical explorations that lead us to conjectures in Euclidean, spherical, and hyperbolic geometries. These conjectures can be followed up with rigorous proof, and much of this article describes the "rigorization" of some of those conjectures. Figure 1a. Figure 1b.

In Section 2, we discuss the Pythagorean Theorem in the Euclidean plane. We move to its twin in spherical geometry in Section 3, rigorously treating circumference and area of circles on the unit sphere, then use dynamic software, together with our formula for area, to conjecture the Spherical Pythagorean Theorem. We repeat the process in the Poincaré Disc Model of hyperbolic geometry in Section 4a, and in the hyperboloid model (also known as the Minkowski-Weierstrass model) in Section 4b (both of which may be considered advanced). In all cases, we make use of the tools of calculus and offer evidence in support of our conjectures (in the way of other known calculus formulas that are implied by our answers), though we do not prove all of the conjectures. We summarize in Section 5.

In the Appendix, we give discovery activities that could be used in the classroom or as an assignment for students; in fact the author has used them at the end of a non-Euclidean Geometry course. Although they are targeted to post-calculus geometry students, the activities are gradual, and could be made accessible to students at lower levels by cutting some of the last activities from each non-Euclidean geometry, and/or cutting one or more of the non-Euclidean geometries from the activity. Indeed, the weaker students in the author's classes usually do not complete the hyperbolic elements of the discovery activity. The activities also seem appropriate as extra-credit problems for calculus students wanting to peek at more advanced material.

We use Geometry Playground , as the basis for our discovery activities. Although there are other packages that allow discovery in non-Euclidean geometries (Cinderella , Spherical Easel , NonEuclid , Geometry Explorer ), Geometry Playground combines all of the "standard" geometries (and a few unusual ones) in one package, and is available freely to anyone with an internet connection, allowing students to use it in a take-home activity.

We would like to thank the referees for many useful suggestions.

# Rethinking Pythagoras - 2. The Plane Truth

Author(s):
Daniel J. Heath (Pacific Lutheran University)

We begin our explorations in Euclidean geometry, the most familiar geometry to most of us. Euclidean geometry might best be explored via the wonderful Euclid's Elements , a reproduction of Euclid's own work, written in English, and published with an applet accompanying each theorem to aid in visualization. Wikipedia's Euclidean Geometry page  and Wolfram MathWorld's Euclidean Geometry page  are also good references.

Most students have seen the Pythagorean Theorem so many times that they have lost count. Many have seen at least two or three proofs of it as well, though most would be hard pressed to recall one, much less elucidate what, exactly, constitutes a proof. While dynamical software allows discovery of concepts for making conjectures, many students are more convinced by their "discoveries" via dynamic software than by mathematically rigorous proofs. We hope to use that surety students have with their own discoveries to convince them of the need for actual proof.

We first consider Figure 1b in the Euclidean plane.  Although it is easy to evaluate the area of a circle using the well-known formula $$A=\pi r^2$$ (where $$A$$ represents the area and $$r$$ the length of the radius of the circle, both as measured in Euclidean geometry), finding the formula in the first place requires some knowledge of calculus.  (It is true that the formula was discovered long before calculus, but the methods involved used limits, a concept that is now taught as part of calculus.)

One such method is to cut the circle into evenly many pie sections, and rearrange the sections as shown in Figure 2.  In Figure 2a, the circle has been cut into 4 pie sections, and in Figure 2b into 10.  The idea is clear when viewed in the linked animation .  It can be seen that the resulting figure approaches a parallelogram (indeed, a rectangle) as the number of slices becomes large. The limiting rectangle has height $$r$$ and width equal to half the circumference of the circle. Since the number $$\pi$$ is often defined as the ratio of the circumference of a circle to its diameter, half the circumference is $$\pi r$$. The area of the limiting rectangle, and thus the circle itself, must therefore be $$\pi r^2$$.

Here we have assumed that $$C=2\pi r$$. However, this is often used as the definition of the number $$\pi$$. We refer to Playground Tutorial Video 3 for an exploration of the constancy of the circumference to radius ratio in Euclidean geometry. Figure 2a. Figure 2b.

### 2.1 Using Differential Calculus

Another method might be to recognize that as the radius of the circle changes by a small amount, say $$\Delta r$$, the area grows at approximately the circumference times $$\Delta r$$, or $$\Delta A=2\pi r\Delta r$$. (Note again the assumption that the circumference of a circle is $$2\pi r$$.) The relationship $$\Delta A=2\pi r\Delta r$$ is shown in Figure 3, where in Figure 3a, circles are shown with radius $$r$$ and $$r+\Delta r$$, in Figure 3b the strip between them is beginning to be "unrolled," and in Figure 3c it has been completely flattened, showing the approximate relationship. Moving to the limit, we obtain $$\frac{dA}{dr}=2\pi r$$. Taking the antiderivative, we then obtain $$A=\pi r^2+K$$, for some constant $$K$$. Looking at the $$r=0$$ case, we quickly see that $$K=0$$, in other words, $$A=\pi r^2$$. Figure 3a. Figure 3c. Figure 3b.

### 2.2 Using Integral Calculus

A third method is to set up and evaluate a definite integral. In this case, we assume that the circle of radius $$r$$ is centered at the origin, so that we are considering the curve $$x^2+y^2=r^2$$.  Solving for $$y$$, we obtain $$y=\pm(r^2-x^2)^{1/2}$$. We want to find the area under this curve, or twice the area under the top half, between $$x=-r$$ and $$x=r$$, in other words, $$A=2\int_{-r}^{r} (r^2-x^2)^{1/2} dx$$. After making the trigonometric substitution $$x=r\sin \theta$$, and a little simplifying, we obtain $$A=2r^2\int_{-\pi/2}^{\pi/2}\cos^2 \theta d \theta$$. This can be evaluated quite simply, giving us the standard area formula $$A=\pi r^2$$. We could also use an arclength integral to show that $$C=2\pi r$$, or simply differentiate the area formula.

Alternately, we can use polar coordinates. In this case, we need to integrate $$A=\int_0^{2\pi} \frac12 r^2 d \theta$$, which again gives us the standard area formula $$A=\pi r^2$$.

### 2.3 The Pythagorean Theorem

Once we have agreed on an algebraic formula for the area of a circle, we can move to the Pythagorean Theorem. In Geometry Playground , we construct a right triangle with circles along each edge, as appears in Figure 1b.

To construct a right triangle, construct a segment, then a perpendicular through one of the endpoints. Construct a point on that perpendicular line, and then hide the line. Now it is possible to construct the two segments joining the point on the perpendicular to the endpoints of the original segment. It is straightforward to construct circles along each edge such as appear in Figure 1b.

For those who prefer to have a jump-start, you can open Geometry Playground with a pre-constructed Euclidean right triangle. Again, it is straightforward to construct circles along each edge.

Measuring the areas of the circles and exploring the relationships between those areas, hopefully the relationship $$\pi a^2+\pi b^2=\pi c^2$$ will eventually stand out. Dividing by $$\pi$$ yields the Pythagorean Theorem. Hurray! We have proven the Pythagorean Theorem! Or have we? Really, what we have done is discover that Geometry Playground seems to imply the Pythagorean Theorem, at least in the cases we have looked at so far. This is far from a proof! It is time to go exploring in a world closer to our own.

# Rethinking Pythagoras - 3. Rounding Things Out

Author(s):
Daniel J. Heath (Pacific Lutheran University)

We assume that spherical geometry takes place on the unit sphere, that is, the sphere of radius 1 unit centered at the origin. If we later want to assume that the sphere is our planet, we just set 1 unit=6378.1 km. We will thus assume that the units of measurement (of lengths and areas) are chosen so that the radius of the sphere itself is 1 unit. For more background on spherical geometry, see Wikipedia's Spherical Geometry page , Wolfram MathWorld's Spherical Geometry page , or John Polking's Spherical Geometry Page .

When we measure the distance between points $$A$$ and $$B$$ in spherical geometry, we measure the shortest distance between the points as measured on the surface of the sphere. It is important to notice that, although the distance from New York to Melbourne through the interior of the earth is shorter than the distance along the surface, beings living in spherical geometry, like humans living on earth, cannot travel through the interior. When we speak of a circle of radius $$r$$ in spherical geometry, we refer to the set of points on the sphere that are of distance $$r$$ from a given point (also on the sphere), as measured on the surface of the sphere.

In spherical geometry, we do not calculate the area of a circle using the method of Subsection 2.1, since we cannot lay our slices out flat. We can, however, use variants of the other two methods.

### 3.1 Using Differential Calculus

The circumference of a circle begins at 0 and grows until the radius is $$\pi/2$$, when the circumference is $$2\pi$$. After this it shrinks until the radius is $$\pi$$ (as large as possible), when the circumference is 0. Area of a circle also begins at 0, increases to $$2\pi$$ (half the sphere) when $$r=\pi/2$$, and further increases to $$4\pi$$ (the whole sphere) when $$r=\pi$$. It is difficult to find a "nice" formula for area that fits these criteria without thinking a bit further. Figure 4.

Rotate the sphere so that the center of the circle in question is at the point $$(1,0,0)$$ . The circle then must lie inside the Euclidean plane $$x=x_0$$  for some appropriate $$x_0$$, so the projection of the circle onto the $$xz$$-plane looks like the vertical line in Figure 4 (see also Figure 5 for the three dimensional picture). The vertical line also represents a diameter for the Euclidean circle corresponding to the spherical circle. Note that since we are working with a sphere of radius 1, the radius $$r$$ of the spherical circle is equal to the angle measure $$r$$ (measured in radians) shown in Figure 4, so that the radius of the Euclidean circle in the plane $$x=x_0$$ ($$=\cos r$$) is $$\sin r$$. Then it is straightforward to calculate the circumference of the circle: $$C=2\pi\sin r$$.

Note that for the same reason as in Subsection 2.1, we can say that $$dA/dr$$ is the length of the circumference. However, the circumference is now $$2\pi\sin r$$. Hence we obtain $$dA/dr=2\pi\sin r$$. Antidifferentiation gives us $$A=-2\pi\cos r+K$$. Inspection of the known values of $$A$$ gives us $$K=2\pi$$, so that $$A=2\pi-2\pi\cos r$$ (this is equivalent to $$4\pi\sin^2(r/2)$$, given by some texts).

From our own experience, we know that when the radius of a circle in spherical geometry is small, the actual area is very nearly the Euclidean area $$\pi r^2$$ and the actual circumference is very nearly the Euclidean circumference $$2\pi r$$. Why? Because we live on a sphere, and a small radius on the unit sphere corresponds to a radius here on earth that is small compared to the radius of the earth. When we measure circles, we usually don't notice the curvature of the earth, obtaining (incorrectly) the Euclidean formulas. However, the Euclidean formulas and the (correct) spherical formulas are so close that we cannot tell the difference. In other words, we expect that the spherical circumference and area formulas are asymptotic to the Euclidean formulas as $$r \rightarrow 0$$. We can check that this is true of our formulas by examining their MacLaurin series:

$C=2\pi\sin r=\frac{2\pi}{1!}r-\frac{2\pi}{3!}r^3+\frac{2\pi}{5!}r^5-\dots\approx2\pi r$

$A=2\pi-2\pi\cos r=\frac{2\pi}{2!}r^2-\frac{2\pi}{4!}r^4+\frac{2\pi}{6!}r^6-\dots\approx\pi r^2$

the latter approximations true when $$r$$ is small. This verifies that our calculated formulas are reasonable. Note that the former implies

$\lim_{r \rightarrow 0}\left(\frac{\sin r}{r} \right)=1$

while the latter implies

$\lim_{r \rightarrow 0}\left(\frac{1-\cos r}{r^2} \right)=\frac12$

both familiar formulas from differential calculus.

### 3.2 Using Integral Calculus

We can calculate area of a spherical circle by using integral calculus as well. We assume that our circle is centered at the point $$(1,0,0)$$, and that the circle lies in the plane $$x=x_0$$, parallel to the $$yz$$-plane, as in Figure 5. Then the interior of the circle is a surface of rotation, its area given by the integral

$A=2\pi\int_{x_0}^1 y\left(1+\left[\frac{dy}{dx} \right]^2 \right)^{1/2} dx$

Substituting $$y=(1-x^2)^{1/2}$$ into this equation gives us a particularly simple integral, evaluating to

(1)$A=2\pi(1-x_0)$ Figure 5.

Since we are interested in the relationship between area and radius, we now calculate the radius of the circle in terms of $$x_0$$. Recall, radius in spherical geometry is as measured on the surface of the sphere. In other words, the radius can be seen as the arclength of the curve $$y=(1-x^2)^{1/2}$$ from $$x=x_0$$ to 1. Then we obtain the integral

$r=\int_{x_0}^1\left(1+\left[\frac{dy}{dx} \right]^2 \right)^{1/2}dx$

Making the appropriate substitutions yields $$r=\pi/2-\arcsin x_0$$, or in reverse, $$x_0=\sin(\pi/2-r)=\cos r$$. Note that substitution of this value into Equation (1) gives us

(2)$A=2\pi-2\pi\cos r$

just as in Subsection 3.1. From here we can differentiate to verify our formula for circumference.

### 3.3 The Pythagorean Theorem

For a spherical right triangle, we again denote the lengths of the sides $$a$$ and $$b$$, and the hypotenuse, that across from "the" right angle, we denote $$c$$. If our spherical triangle has more than one right angle, we will have to agree on which right angle we are considering "the" right angle when we examine the "Spherical Pythagorean Theorem." Note that $$0\leq a\leq \pi$$, and similarly for $$b$$ and $$c$$.

It is unlikely that students will discover the Spherical Pythagorean Theorem by trial and error as they might by simply examining the areas of circles constructed on each side as in Subsection 2.3. In Geometry Playground , we choose spherical geometry and construct a right triangle and its edge-circles as in Figure 1b, or simply open Geometry Playground with a pre-constructed spherical right triangle, and then construct the edge-circles on that right triangle as in Figure 1b.

Measuring the area $$A$$ of each circle, and calculating $$\cos r=(2\pi-A)/(2\pi)$$ from Equation (2), it shouldn't take long for the relationship $$\cos a\cdot \cos b = \cos c$$ to stand out. In fact, this is the "Spherical Pythagorean Theorem."

However, we live on the surface of a sphere, and usually do not calculate the length of the hypotenuse of a right triangle using this formula; we use the Pythagorean Theorem (we rarely need the Spherical version, as the sides of our right triangles are usually quite small in comparison to unit length, the radius of the sphere!). This tells us that as $$a$$, $$b$$, and $$c$$ decrease, we should expect the Spherical Pythagorean Theorem to approach the Euclidean one. This can again be verified via MacLaurin expansion. Substituting MacLauren series $$\cos r=1-r^2/2!+r^4/4!-\dots$$ for $$\cos a$$, $$\cos b$$, and $$\cos c$$, we obtain

$\left(1-\frac1{2!}a^2+\frac1{4!}a^4-\dots \right)\left(1-\frac1{2!}b^2+\frac1{4!}b^4-\dots \right)=1-\frac1{2!}c^2+\frac1{4!}c^4-\dots$

Distributing the left, we arrive at $$1-\frac1{2!}a^2-\frac1{2!}b^2+\dots=1-\frac1{2!}c^2+\dots$$, with all remaining terms of degree 4 or higher. Since we are assuming $$a$$, $$b$$, and $$c$$ are all small, we ignore the higher order terms. Canceling the constant terms and multiplying by $$-2$$, we obtain the Euclidean Pythagorean Theorem.

# Rethinking Pythagoras - 4a. Stretching the Truth (Poincaré Disc Model)

Author(s):
Daniel J. Heath (Pacific Lutheran University)

We define distances and circles in hyperbolic space analogously to how we did on the sphere, and assume that the reader is familiar with the Poincaré Disc Model of hyperbolic geometry, in particular, its (distance-inducing) differential $$ds=2(dx^2+dy^2)^{1/2}/(1-x^2-y^2)$$. For those who need a refresher, Wikipedia has a reference page for the Poincaré Disc Model , Wolfram Math World  has another, and the Geometry Center  has an archive containing descriptions, pictures, equations, and an applet concerning it.

### 4a.1 Using Differential Calculus

Let $$A$$ and $$C$$ represent the area and circumference, respectively, of a circle in the hyperbolic plane. Repeating an idea from Subsections 2.1 and 3.1, we note that $$dA/dr=C$$. It is harder to conjecture a formula for circumference in hyperbolic geometry. Using Geometry Playground, we note that for small $$r$$, $$A\approx \pi r^2$$ and $$C\approx \pi r^2$$, but for large $$r$$, both $$A$$ and $$C$$ seem to approach $$\pi e^r$$. Those familiar with hyperbolic trigonometric functions are probably already beginning to see the "parallel," and are conjecturing that $$C=2\pi\sinh r$$, where $$\sinh r=(e^r-e^{-r})/2$$ (and similarly $$\cosh r=(e^r+e^{-r})/2$$). If this is true, then a quick repeat of the steps of Subsection 3.1 shows that $$A=2\pi\cosh r-2\pi$$, a remarkable resemblance to the spherical case.

It is elementary to verify that $$\frac{d}{dx}\sinh x=\cosh x$$ and $$\frac{d}{dx}\cosh x=\sinh x$$, so that calculating MacLaurin series for each is a straightforward matter. Then we can see that

$A=2\pi\cosh r-2\pi=2\pi\left(1+\frac1{2!}r^2+\frac1{4!}r^4+\dots\right)-2\pi\approx\pi r^2$

$C=2\pi\sinh r=2\pi\left(\frac1{1!}r^1+\frac1{3!}r^3+\frac1{5!}r^5+\dots\right)\approx 2\pi r$

the latter approximations true for small $$r$$ . Again it is clear that both expressions are asymptotic to their Euclidean counterparts when distances are small. In addition, they imply hyperbolic trigonometric analogs for the familiar limit formulas from calculus:

$\lim_{r\to 0}\left(\frac{\cosh r-1}{r^2}\right)=\frac12$

implied by the former, and

$\lim_{r\to 0}\left(\frac{\sinh r}{r}\right)=1$

implied by the latter. It would be difficult to determine whether your universe were Euclidean, spherical, or hyperbolic if you were trapped in only one small corner of it, since in any of the three geometries, measurements on a small scale are either Euclidean or almost Euclidean. Indeed, physicists have not yet agreed on the overall shape of our universe , , .

### 4a.2 Using Integral Calculus

We take a circle in the Poincaré Disc model, and for convenience center it at the origin. In the model, we assume that it passes through the point $$(x_0,0)$$. We can find its area by integrating the square of the line element, $$ds^2=4(dx^2+dy^2)/(1-x^2-y^2)^2$$ over the region inside the circle. This is easier in polar coordinates, where

$A=\int_0^{2\pi}\int_0^{x_0} \frac4{(1-r^2)^2} rdrd\theta$

This is relatively simple as double integrals go. After substitution $$u=1-r^2$$, $$du=-2rdr$$, we can move quickly to the solution

(3)$A=(4\pi x_0^2)/(1-x_0^2)$

However, we want to know the area in terms of the radius. To find the radius, we need to calculate the distance between the origin and the point $$(x_0,0)$$. This requires a hyperbolic arclength integral. Thus:

$r=\int_0^{x_0} ds$

$=\int_0^{x_0} \frac{2(dx^2+dy^2)}{(1-x^2-y^2)}$

$=\int_0^{x_0}\frac{2dx}{1-x^2}$

the latter because $$y=0$$  on the geodesic joining the two points. This integral can be evaluated using the method of partial fractions, but we prefer to use the hyperbolic trigonometric substitution $$x=\tanh\theta$$ , $$dx=\mathrm{sech}^2\theta d\theta$$, giving us, after simplification:

$r=\int_0^{\tanh^{-1}(x_0)}\frac{2\mathrm{sech}^2\theta d\theta}{\mathrm{sech}^2\theta}$

$=2\tanh^{-1}(x_0)$

Then we have $$x_0=\tanh(r/2)$$ . Substitution into Equation (3) yields:

$A=4\pi\tanh^2(r/2)/(1-\tanh^2(r/2))$

$=4\pi\tanh^2(r/2)/(\mathrm{sech}^2(r/2))$

$=4\pi\sinh^2(r/2)$

(4)$A=2\pi\cosh r -2\pi$

the latter equality via hyperbolic trigonometric identity $$\sinh(r/2)=\frac12(\cosh r -1)$$. This is precisely what we conjectured in Subsection 4a.1. From here we can differentiate to verify our circumference conjecture of the same subsection.

### 4a.3 The Pythagorean Theorem

After seeing the similarities between spherical and hyperbolic area, we begin with a headstart; our initial conjecture at a Hyperbolic Pythagorean Theorem is $$\cosh a\cdot\cosh b=\cosh c$$. In Geometry Playground , we choose hyperbolic geometry and construct a right triangle and its edge-circles as in Figure 1b, or simply open Geometry Playground with a pre-constructed hyperbolic right triangle in the Poincaré Disc Model, and then construct the edge-circles on that right triangle as in Figure 1b.

Measuring the area $$A$$ of each circle, and calculating $$\cosh r=(A+2\pi)/(2\pi)$$ from Equation (4), it shouldn't take long for the relationship $$\cosh a\cdot\cosh b=\cosh c$$ to stand out, as expected. In fact, though we leave it without proof, this is the "Hyperbolic Pythagorean Theorem."

We again note that as $$a$$, $$b$$, and $$c$$ decrease, the Hyperbolic Pythagorean Theorem approaches the Euclidean one. Substituting MacLauren series $$\cosh r=1+r^2/2!+r^4/4!+\dots$$ for $$\cosh a$$, $$\cosh b$$, and $$\cosh c$$, we obtain

$\left(1+\frac1{2!}a^2+\frac1{4!}a^4+\dots\right)\left(1+\frac1{2!}b^2+\frac1{4!}b^4+\dots\right)=1+\frac1{2!}c^2+\frac1{4!}c^4+\dots$

Distributing the left, we arrive at

$1+\frac1{2!}a^2+\frac1{2!}b^2+\dots=1+\frac1{2!}c^2+\dots$

with all remaining terms of degree 4 or higher. Since we are assuming $$a$$, $$b$$, and $$c$$ are all small, we ignore the higher order terms. Canceling the constant terms and multiplying by 2, we obtain the Euclidean Pythagorean Theorem.

# Rethinking Pythagoras - 4b. Stretching the Truth (Hyperboloid Model)

Author(s):
Daniel J. Heath (Pacific Lutheran University)

We define distances and circles in hyperbolic space analogously to how we did on the sphere, and assume that the reader is familiar with the Hyperboloid Model of hyperbolic geometry, in particular, that we envision it as the points on the upper half of the hyperboloid of two sheets $$-x^2-y^2+z^2=1$$ , with (distance-inducing) differential $$ds=({dx}^2+{dy}^2-{dz}^2)^{-1/2}$$ , and the fact that the upper half of the hyperbola $$y^2-x^2=1$$  can be parametrized by hyperbolic trigonometric functions $$( \sinh t, \cosh t)$$ in much the same way that the unit circle $$x^2+y^2=1$$ can be parametrized by standard trigonometric functions $$(\cos t, \sin t)$$. For those who need a refresher, Wikipedia has a reference page for the Hyperboloid Model , and the Geometry Center has an archive containing descriptions and images concerning the Hyperboloid (Minkowski) Model .

### 4b.1 Using Differential Calculus

Let $$A$$ and $$C$$ represent the area and circumference, respectively, of a circle in the hyperbolic plane. Repeating an idea from Subsections 2.1 and 3.1, we note that $$dA/dr=C$$. However, it takes a bit more thinking to find a formula for circumference of a hyperbolic circle. Figure 6.

For convenience, we consider the circle with center $$(0,0,1)$$. The circle then must lie inside the Euclidean plane $$z=z_0$$ for some appropriate $$z_0$$, so the projection of the circle onto the $$xz$$-plane looks like the horizontal segment in Figure 6 (see also Figure 7 for the three dimensional picture). This segment also represents a diameter for the Euclidean circle coinciding with the hyperbolic circle.

The radius of the (hyperbolic) circle in question is the arclength, along the hyperbola in the $$xz$$-plane, from $$(0,1)$$ to $$(\sinh r,\cosh r)$$, where we use the hyperbolic differential to measure length. In this case, since $$y=0$$, the differential is $$ds=({dx}^2+{dz}^2)^{1/2}$$, so we obtain:

hyperbolic radius $$=\int_0^r (\cosh^2(t)-{\sinh^2(t)})^{1/2} dt$$

It is easy to check that $$\cosh^2 t-\sinh^2 t=1$$, so the integral in question evaluates to just $$r$$. Then the radius of the hyperbolic circle is $$r$$, and the same circle, considered as a Euclidean circle in the plane $$z=z_0$$ ($$=\cosh r)$$ has radius $$\sinh r$$. Hence the circumference of either circle (hyperbolic or Euclidean) is $$C=2 \pi \sinh r$$ (since in the plane $$z=z_0$$, $$dz=0$$, and the Euclidean and hyperbolic distances are equal). Note the resemblance to the spherical case, both in process and final answer.

Now since $$dA/dr=C$$ and $$C=2 \pi \sinh r$$, we obtain $$dA/dr=2 \pi \sinh r$$. Antidifferentiation gives us $$A=2 \pi \cosh r+K$$. Inspection of the known value of $$A$$ (i.e., when $$r=0$$) gives us $$K=-2 \pi$$, so that $$A=2 \pi \cosh r-2 \pi$$, a remarkable resemblance to the spherical case!

It is elementary to verify that $$\frac{d}{dx} \sinh x=\cosh x$$ and $$\frac{d}{dx} \cosh x=\sinh x$$, so that calculating MacLaurin series for each is a straightforward matter. Then we can see that:

$$A=2 \pi \cosh r-2 \pi=2 \pi (1+\frac{1}{2!}r^2+\frac{1}{4!}r^4+\cdots )-2 \pi \approx \pi r^2$$

$$C=2 \pi \sinh r=2 \pi (\frac1{1!}r^1+\frac1{3!}r^3+\frac1{5!}r^5+\cdots )-2 \pi \approx 2\pi r$$

the latter approximations true for small $$r$$. Again it is clear that both expressions are asymptotic to their Euclidean counterparts when distances are small. In addition, they imply hyperbolic trigonometric analogs for the familiar limit formulas from calculus:

$$\lim_{r \rightarrow 0}(\frac{\cosh r-1}{r^2} )=\frac{1}{2}$$

implied by the former, and

$$\lim_{r \rightarrow 0}(\frac{\sinh r}{r} )=1$$

implied by the latter. It would be difficult to determine whether your universe were Euclidean, spherical, or hyperbolic if you were trapped in only one small corner of it, since in any of the three geometries, measurements on a small scale are either Euclidean or almost Euclidean. Indeed, physicists have not yet agreed on the overall shape of our universe , , .

### 4b.2 Using Integral Calculus

We can calculate area of a hyperbolic circle by using integral calculus as well. We assume that our circle is centered at the point $$(0,0,1)$$, and that the circle lies in the plane $$z=z_0$$, parallel to the $$xy$$-plane, as in Figure 7. Then the interior of the circle is a surface of rotation, its area given by the integral:

$$A=2 \pi \int_1^{z_0} x ([\frac{dx}{dz} ]-1)^{1/2}$$

Substituting $$x=(z^2-1)^{1/2}$$ into this equation gives us a particularly simple integral, evaluating to:

(5)$$A=2 \pi(z_0-1)$$ Figure 7.

Since we are interested in the relationship between area and radius, we now calculate the radius of the circle in terms of $$z_0$$. The radius can be seen as the arclength of the curve $$x=(z^2-1)^{1/2}$$ from $$z=1$$ to $$z_0$$, using the hyperbolic differential $$ds=(dx^2-dz^2)^{1/2}$$ (since $$y=0$$) to measure distance. Then we obtain the integral:

$$r=\int_1^{z_0}([\frac{dx}{dz} ]-1)^{1/2} dz$$

Substitution gives us:

$$r=\int_1^{z_0} \frac{1}{(z^2-1)^{1/2}} dz$$

From here, we make the hyperbolic trigonometric substitution $$z=\cosh t$$, $$dz=\sinh t, dt$$ and simplify, obtaining:

$$r=\int_0^{\cosh^{-1}(z_0)} 1dt$$

Hence we have $$r=\cosh^{-1}(z_0)$$, or in reverse, $$z_0=\cosh(r)$$. Note that substitution of this value into Equation (5) gives us:

(6)$$A=2 \pi \cosh r-2 \pi$$

just as in Subsection 4b.1. From here we can differentiate to verify our formula for circumference.

### 4b.3 The Pythagorean Theorem

After seeing the similarities between spherical and hyperbolic area, we begin with a headstart; our initial conjecture at a Hyperbolic Pythagorean Theorem is $$\cosh a \cdot \cosh b=\cosh c$$. In Geometry Playground , we choose hyperbolic geometry and construct a right triangle and its edge-circles as in Figure 1b, or simply open Geometry Playground with a pre-constructed hyperbolic right triangle in the Minkowski-Weierstrass Model, and then construct the edge-circles on that right triangle as in Figure 1b.

Measuring the area $$A$$ of each circle, and calculating $$\cosh r=(A+2 \pi)/(2 \pi)$$ from Equation (6), it shouldn't take long for the relationship $$\cosh a \cdot \cosh b=\cosh c$$ to stand out, as expected. In fact, though we leave it without proof, this is the "Hyperbolic Pythagorean Theorem."

We again note that as $$a$$, $$b$$, and $$c$$ decrease, the Hyperbolic Pythagorean Theorem approaches the Euclidean one. Substituting MacLauren series $$\cosh r=1+r^2/{2!}+r^4/{4!}+\cdots$$ for $$\cosh a$$, $$\cosh b$$, and $$\cosh c$$, we obtain:

$$(1+\frac{1}{2!}a^2+\frac{1}{4!}a^4+\cdots )(1+\frac{1}{2!}b^2+\frac{1}{4!}b^4+\cdots)=1+\frac{1}{2!}c^2+\frac{1}{4!}c^4+\cdots$$

Distributing the left, we arrive at

$$1+\frac{1}{2!}a^2+\frac{1}{2!}b^2+\cdots=1+\frac{1}{2!}c^2+\cdots$$

with all remaining terms of degree 4 or higher. Since we are assuming $$a$$, $$b$$, and $$c$$ are all small, we ignore the higher order terms. Canceling the constant terms and multiplying by $$2$$, we obtain the Euclidean Pythagorean Theorem.

# Rethinking Pythagoras - 5. Summary

Author(s):
Daniel J. Heath (Pacific Lutheran University)

The ideas in the previous sections weave together several themes. Differential and integral calculus, Taylor series, Euclidean, spherical, and hyperbolic geometries, and measurement of length and area are all explicitly discussed, but straightedge and compass construction, discovery via dynamic software, and the difference between discovery and proof are also implicit in the discussion. Accordingly, the author has used the discovery activity given in the appendix during three of the last four days of the College Geometry class he teaches (the last day is alloted for review). That activity leads students to discover and prove some relationships between radius and circumference and area of circles in three geometries, and to discover (without proof) Pythagorean Theorem analogs in the non-Euclidean geometries.

There are several reasons why this activity is useful. First, all students can at least begin the activity, constructing and exploring figures in the dynamic software. While stronger students usually finish the entire activity in the alloted three class periods, most weaker students only complete the Euclidean and spherical parts, and need quite a bit of prompting from their instructor on some of the deeper questions. All of the author's students have needed assistance with step 2h, which utilizes the identity: $$A(S)=2\pi\cos(r)-2\pi$$, implying that $$\cos(r)=[2\pi-A(S)]/(2\pi)$$.

In fact weaker students have been unusually engaged in the activity, and often related "aha!" moments having to do with a new and improved understanding of calculus (rather than geometry). For example, when integrating $$2\pi\int\sin(r)dr$$ in order to find the area of a spherical circle, one student at first complained that "the" solution, $$-2\pi\cos(r)$$, didn't work. However, when he discovered that addition of the appropriate constant gave a workable solution, he was ecstatic: "That's why we needed that constant back in Calc 2! I never understood that before!"

Stronger students, on the other hand, have usually expressed "aha!" moments having to do with their understanding of geometry. One student, after discovering the spherical analog to the Pythagorean Theorem, wondered why we use the Pythagorean Theorem at all, since we live on a sphere and should be using the spherical version. After working through step 2j in the appendix, she realized why we use the simplified Euclidean Pythagorean Theorem: "Oh, so it really doesn't matter which one we use unless we either need a lot of accuracy or are traveling a long distance." A few (juniors) have been inspired to undertake senior capstone projects related to non-Euclidean geometry in their subsequent year.

Secondly, the activity gives students an opportunity to use skills from prerequisite classes in a new context. We already mentioned how weaker students have often expressed increased understanding of calculus through this activity. Even several stronger students struggled to remember calculus concepts that were becoming rusty with disuse.

A common coffee-shop topic of discussion at math conferences is students' difficulty in applying knowledge in new contexts. It makes sense to create opportunities for students to practice applications, such as by using MacLaurin series to explore the spherical Pythagorean Theorem analog in their geometry class.

Thirdly, the activities are designed to force students to the conclusion that numeric explorations can be misleading. For example, the relationship between radius and circumference of a circle in spherical geometry is explored numerically beginning on a very tiny scale. Students conclude, incorrectly, that $$C=2\pi r$$, and then explore larger circles, discovering that their original conclusion was incorrect. This is a point that the instructor can use quite effectively to point out the weakness of numeric explorations and the importance of proof; if the students had stopped their explorations after looking at only small-scale circles, they would probably remain convinced that circles in spherical geometry have the same area and circumference formulas as their Euclidean counterparts.

In fact, the author has noticed that his students often are more convinced by numerical explorations than by rigorous proofs. After trying many methods to convince students of the need for proof, he finally saw the obvious: "If they are more convinced by software than by proof, then I can convince them of the need to go beyond the software by making use of their trust in the software." An exploration that leads students to create a conjecture that later turns out to be demonstrably false is one way of doing this.

Lastly, students find these numerical explorations fun. Many students are clearly more engaged in the class material while doing this kind of exploration than with any other method of instruction. While this may be because of the superficial resemblance of exploration using dynamic software to playing video games, it can nonetheless be used to encourage students to immerse themselves in geometric concepts. In particular, the author has found that the weakest students usually have the most to gain from this type of exploration.

Perhaps the most amusing evidence of the effectiveness of this activity came from this year's final exam. On it, I included the following problem, which was intended to be a simple "warm up" type problem, and at the same time to address the fact that most of my students plan to teach high school math.

• The following problem comes directly from a sample problem found at yourteacher.com , and was meant for a high school algebra class. "Raul is 6 feet tall, and notices that his shadow is 5 feet long. The shadow of his school building is 30 feet long. How tall is his school building?"

In fact, the problem assumes knowledge of geometric ideas. Solve this problem, explaining at which steps you are using algebra and at which steps you are using geometry.

One student gave the following response, with a sketch of two appropriate triangles (parenthetical clarification added).

• We know angles with the ground and [angles with the] sun are the same. Since AA [two angles are congruent], the triangles are similar. The similar triangles theorem says 6/5 = x/30, so x=36. The building is 36 feet. Except we live on a sphere, so we'd have to use a spherical similar triangles theorem, which is probably different. But since we don't care about hundredths of an inch, I guess Euclidean rules will give a good approximation. 36 feet is about right.

### Bibliography

1. Geometry Playground v. 1.2, by Daniel J. Heath and Joshua Jacobs, 2010.
2. Cinderella, Springer, 2006.
3. Spherical Easel, by David Austin and Will Dickinson, 2009.
4. NonEuclid, by Joel Castellanos, Joe Dan Austin, and Ervan Darnell, 2009.
5. Geometry Explorer, by Michael Hvidsten
6. Euclid's Elements, by Euclid, adapted by David E. Joyce, 1996, 1997, 1998.
7. Wikipedia: Euclidean Geometry
8. Euclidean Geometry, by Weisstein, Eric W. From MathWorld--A Wolfram Web Resource.
9. National Curve Bank: Animations of Two Classics: Derivation of the Formula for the Area of a Circle and the Pythagorean Theorem, by Tom and Bettina Richmond.
10. Wikipedia: Spherical Geometry
11. Spherical Geometry, by Weisstein, Eric W., From MathWorld--A Wolfram Web Resource.
12. The Geometry of the Sphere, by John C. Polking, Park City Mathematics Institute, 1996.
13. Wikipedia: The Poincaré Disc Model
14. Poincaré Hyperbolic Disk, by Weisstein, Eric W. From MathWorld--A Wolfram Web Resource.
15. The Geometry Center: The Poincaré Disc Model, 1996.
16. Wikipedia: The Hyperboloid Model
17. The Geometry Center: The Minkowski Model, by John Hartman, 1996.
18. Principles of Physical Cosmology, by P. J. E. Peebles, Princeton University Press, 1993.
19. WMAP website at NASA.
20. The Shape of Space lecture by Jeffrey Weeks, Institute for Mathematics and Its Applications, 2009.
21. yourteacher.com

# Rethinking Pythagoras - Appendix: Discovering Pythagoras

Author(s):
Daniel J. Heath (Pacific Lutheran University)

Here we give a discovery activity that we have used, in its entirety, with undergraduate students studying non-Euclidean geometry. We use freeware Geometry Playground  for these investigations; simply select the main link on the Geometry Playground page to run it. Geometry Playground allows straightedge and compass construction in Euclidean and non-Euclidean geometries such as spherical and hyperbolic. We let $$r$$ represent the radius of a circle as measured in each geometry, while $$C$$ and $$A$$ represent the circumference and area of that circle, again as measured within that geometry.

1. This link will open Geometry Playground in Euclidean geometry with a pre-constructed right triangle having vertices $$A$$, $$B$$, and $$C$$, the last corresponding to the right angle, and a constant term of $$\pi$$ that can be used in forming sums, differences, products, and ratios. Construct three circles, one with center $$A$$ and containing point $$B$$, another with center $$C$$ containing $$A$$, and the third with center $$B$$ containing point $$C$$. Measure the areas of these circles.
1. Denote the length of the side with vertices $$B$$ and $$C$$ by $$a$$, the side with vertices $$A$$ and $$C$$ by $$b$$, and the third, the hypotenuse, by $$c$$. Write the areas of the circles in terms of $$a$$, $$b$$, and $$c$$. How do you know that the expressions you wrote down are correct?
2. Find a relationship between the areas of the circles. Make sure that the relationship you find remains after moving points in the triangle about. Write the relationship in terms of $$a$$, $$b$$, and $$c$$.
Hint: Sum the areas of the two smaller circles (you can do this by selecting Measure → Sum) and compare the sum to the area of the larger circle, perhaps by finding the ratio or the difference.
2. Select Spherical geometry and choose the plane model by selecting Display → Model → Plane Model. Scale up as far as possible, so that you are looking at a very, very tiny part of the unit sphere. Construct a circle, and measure its radius, circumference, and area. Do $$C=2\pi r$$ and $$A=\pi r^2$$? (Note that you can study the ratios of measures dynamically by selecting Measure → Ratio.) What happens as you scale down and look at larger and larger circles?
1. Note that when $$r$$ is very small, we can estimate circumference and area using familiar formulas. What are they?
2. Switch back to the standard (sphere) model of spherical geometry. When $$r=\pi/2$$, the circle is a great circle; what are circumference and area? How about when $$r=\pi$$?
3. Use these known facts to conjecture an expression for $$C$$ with respect to $$r$$. Test your conjecture for several values of $$r$$ using Geometry Playground.
Hint: Since circumference grows and then shrinks, a trigonometric function might be useful here.
4. Note that in Euclidean geometry, $$dA/dr=C$$. Explain why. A picture may be helpful.
5. Explain why your answer to part (d) remains true in spherical geometry (even though the expression for $$C$$ is different).
6. Use the results of (c)-(e) to find an expression for $$A$$ with respect to $$r$$. Test your answer for several values of $$r$$ using Geometry Playground.
7. This link will open Geometry Playground in spherical geometry with a pre-constructed right triangle having vertices $$A$$. $$B$$, and $$C$$, the last corresponding to the right angle, and a constant term of $$2\pi$$ that can be used in forming sums, differences, products, and ratios. Again construct three circles with radii the sides of the triangle, and measure the areas of those circles. For each circle $$S_i$$, calculate $$[2\pi-A(S_i)]/(2\pi)$$, hiding all intermediate calculations.
8. If we denote by $$S_0$$ and $$S_1$$ the circles constructed with radii the legs of the right triangle, and $$S_2$$ the circle whose radius is the hypotenuse, note that $$[2\pi-A(S_0)]/(2\pi)\cdot [2\pi-A(S_1)]/(2\pi)=[2\pi-A(S_2)]/(2\pi)$$, that is, try it.
(Note: based on your results for part (f), you should be able to find a simple expression for $$[2\pi-A(S_i)]/(2\pi)$$ in terms of the length of the radius $$r_i$$ of the circle $$S_i$$.) Find an equivalent expression in terms of the lengths of the sides ($$a$$, $$b$$) and hypotenuse ($$c$$) of the triangle. This is the "Spherical Pythagorean Theorem."
9. Find the Taylor series for the trigonometric expressions in (h). Verify that by ignoring the terms of degree larger than 2 (for example, when $$a$$, $$b$$, and $$c$$ are small), we obtain the standard Pythagorean Theorem.
10. In Geometry Playground, spherical length measurements are all with respect to the radius of the sphere itself. That is, the sphere's radius is assumed to be 1 unit. Hence, if the radius of the sphere is 6371 kilometers (such as for the sphere we live on), then all measurements in kilometers should be divided by 6371 to obtain their equivalent in terms of radii length, and measurements in terms of sphere radius length should be multiplied by 6371 to obtain their kilometer equivalents. Find the length of the hypotenuse of a triangle on the surface of the earth that has sides of length 5000 km and 8000 km. How about 5 km and 8 km? How about 5 m and 8 m? At what point can we ignore the fact that we live on a sphere when making calculations?
3. Select Hyperbolic geometry and choose the model by selecting Display → Model → Minkowski-Weierstrass Model (also known as the hyperboloid model). Scale up as far as possible, so that you are looking at a very, very tiny part of the hyperbolic plane. Construct a circle, and measure its radius, circumference, and area. Do $$C=2\pi r$$ and $$a=\pi r^2$$? What happens as you scale down and look at larger and larger circles?
1. Explain why the relationship in 2(d) still holds in hyperbolic geometry.
2. We note that the formula for circumference in hyperbolic geometry is $$C(r)=2\pi\sinh r$$. Use this and your answer to part (a) to find an expression for the area of a circle in hyperbolic geometry in terms of the length of the radius. Note the similarity to your answer for 2(f). Test your answer for several values of $$r$$ using Geometry Playground.
3. This link will open Geometry Playground in the Minkowski-Weierstrass model of hyperbolic geometry with a pre-constructed right triangle having vertices $$A$$, $$B$$, and $$C$$, the last corresponding to the right angle, and a constant term of $$2\pi$$ that can be used in forming sums, differences, products, and ratios. Again construct three circles with radii the sides of the triangle, and measure the areas of those circles. For each circle $$S_i$$, calculate $$[A(S_i)+2\pi]/(2\pi)$$, hiding all intermediate calculations. Note: based on your results for part (b), you should be able to find a simple expression for $$[A(S_i)+2\pi]/(2\pi)$$ in terms of the length of the radius $$r_i$$ of the circle $$S_i$$.
4. If we denote by $$S_0$$ and $$S_1$$ the circles constructed with radii the legs of the right triangle, and $$S_2$$ the circle whose radius is the hypotenuse, note that $$[A(S_0)+2\pi]/(2\pi)\cdot [A(S_1)+2\pi]/(2\pi)=[A(S_2)+2\pi]/(2\pi)$$, that is, try it. Find an equivalent expression in terms of the lengths of the sides ($$a$$, $$b$$) and hypotenuse ($$c$$) of the triangle. This is the "Hyperbolic Pythagorean Theorem."
5. Find the Taylor series for the hyperbolic trigonometric expressions in (d). Verify that by ignoring the terms of degree larger than 2 (for example, when $$a$$, $$b$$, and $$c$$ are small), we obtain the standard Pythagorean Theorem.