Some Geometry You Never Met

There is no royal road to geometry.—Euclid, to Ptolemy I, ruler of Egypt

Out of nothing I have created a strange new universe.—Janos Bolyai, referring to the creation

of a non-euclidean geometry

B

A

C

Figure 1. Outline of a Farmer’s Property

I begin this chapter with a challenge exercise. A farmer, who owns a triangular propertythat has the shape of Figure 1, has the plot surveyed. The surveyor reports the anglesat the lot corners. Their measures are ∠A = 78○ 20′ 23.52”, ∠B = 42○ 51′ 3.13” and ∠C =58○ 48′ 33.37”. From these measurements the farmer is able to determine the area of thetriangle. What is that area and how does he calculate it?

1 Triangle area formulas

Students who studied geometry in high school may recall several ways to calculate the areaof a triangle. If it is a right triangle or if you know the triangle’s base and height, you canuse the simple area formula

A = 12bh (1.1)

If you know two sides of a triangle and the angle between them, you can use the formula

A = 12ab sinC (1.2)

A less well known formula for area is the one for a triangle in which you know two angles,A and C and the side between them, b. It is

A = 12b2

tanAtanC

tanA + tanC (1.3)

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And finally you can use Heron’s Formula if you know all three sides. First calculate thesemiperimeter s = (a + b + c)/2 and then use that value in the formula

A =√s(s − a)(s − b)(s − c) (1.4)

Of course, none of those methods addresses the farmer’s problem, because they all requirethe length of at least one side of the triangle. In fact, you were probably cautioned whenyou studied geometry in high school that the so-called AAA case, when you know onlythe triangle’s angles, only allows for similarity — triangles with the same shape but notnecessarily the same size

Figure 1.5 The AAA Case: Similar triangles with different areas

so their area cannot be calculated.

There is, however, something different about the farmer’s problem and we will addressthat difference later in this chapter.

2 Euclid’s Elements

The Greek mathematician, Euclid, lived about 2300 years ago in Alexandria, where hecompiled and organized the most successful and influential mathematics text of all times,probably providing many of the proofs it contained himself. The 13 books1 of his Elementsserved as the basis for understanding mathematics until the early 20th century.

Two Elements propositions remain famous, but for quite different reasons. The first isElements I, 5 (that is, Book I, Proposition 5) which we usually state today as “Base anglesof an isosceles triangle are equal.”2 Because Euclid included in his proof the fact that thesupplements of those angles would also be equal, his diagram took on several additionalline segments, making it appear as in Figure 2.

1The books are more like the chapters of modern texts, but one modern Elements reissue is still in threevolumes.

2In this book I am not adopting today’s formal usage that would require me to restate that theoremeither replacing “equal” with “congruent” or adding “The measure of” at the beginning. I can imagine noreader who does not understand what I am talking about when I say that angles of a triangle are equal.

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!

!

!

!

A

CB

Figure 2. The Diagram for Elements I, 5

That structure gave the diagram the appearance of a bridge with trusses and the propositionbecame known to teachers and students alike as Pons asinorum, Latin for Bridge of Asses(a.k.a. fools). In this sense, Elements I, 5 was said to separate the students who were ableto continue from those who would not make it past this point in the course.

!

!

!

!

A

CB

Figure 3. The Diagram for Elements I, 47

The other proposition is Elements I, 47, the Pythagorean Theorem, easily the mostfamous theorem of all time. Euclid knew it in terms of areas: “The sum of the areas of thesquares on the legs of a right triangle is equal to the area of the square on the triangle’shypotenuse.” Today we usually think of this theorem in algebraic terms: for legs a and band hypotenuse c, a2 + b2 = c2.

Although algebra and even our Hindu-Arabic numeration system would not be inventedfor centuries after Euclid’s time, many other algebraic ideas are presented in his books ingeometric form. For example, using arguments involving rectangular areas, Elements II,1 proves a form of the distributive law a(b + c + d) = ab + ac + ad and the theorems thatfollow prove such complex identities as (2a + b)2 + b2 = 2(a2 + (a + b)2). Later books provetheorems normally found in other areas of math. For example, another famous theoremyou will meet again, Elements IX, 20 proves that the number of primes is infinite.

Today, although Euclid is scarcely mentioned in school geometry texts, the secondaryschool course follows much of the development of the early books in the series.

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3 Problems with Euclid

Despite the power and influence of the Elements, it is easy today to identify problems withsome of the foundations from which Euclid developed his proofs. The whole idea of hisdevelopment was to list a series of definitions and axioms that are at once simple and easyto understand and accept. He would then use these as the basis for proving what was tofollow.

Some of the definitions are strange and of little value: “A point is that which has nopart” and “A line is breadthless length.” Others are oddly stated: “A circle is a planefigure contained by one line such that all the straight lines falling upon it from one pointamong those lying within the figure are equal to one another,” but that one is at leastpartly straightened out by the following definition: “And the point is called the center ofthe circle.”

A more significant problem arises in Euclid’s approach to parallel lines. He defines themin what seems a reasonable way, as “straight lines which, being in the same plane and beingproduced indefinitely in both directions, do not meet one another in either direction.” Butthen he lists his five basic postulates:3

1. To draw a straight line from any point to any point.2. To produce a finite straight line continuously in a straight line.3. To describe a circle with any center and distance.4. That all right angles are equal to one another.5. That, if a straight line falling on two straight lines make the interior angles on thesame side less than two right angles, the two straight lines, if produced indefinitely,meet on that side on which are the angles less than the two right angles.

Now think about those postulates. The first three merely describe the basic constructionsdone with compasses and straightedge. The fourth is evident. But the fifth is quite unlikeall the others. At 45 words, it is longer than all the others combined. And it is quitecomplicated. Figure 4 shows what it is saying.

3Postulates and axioms, like definitions, are accepted statements on which subsequent proofs are based.They are supposed to be self-evident as the Postulate 5 clearly is not.

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!

Figure 4. A diagram illustrating Elements Postulate 5

According to the postulate, when those two angles marked in red sum to less than 180○, thelines will meet on that side of the transverse line. The smaller figure illustrates a situationwhen the two angles do add to 180○.

Because of the strange characteristics of Postulate 5, mathematicians tried first to proveit based on the other postulates and definitions. Failing this, they identified postulatesequivalent to it, that is, statements that can be proved on the basis of Postulate 5 andthat, at the same time, can be used to prove Postulate 5. The list of equivalent theoremsis quite remarkable. It includes:

The sum of the angles in every triangle is 180.There exists a pair of similar, but not congruent, triangles.Two lines that are parallel to the same line are also parallel to each other.Even the Pythagorean Theorem

Still another equivalent to Proposition 5 is called the Playfair Postulate. It states thata line can be drawn parallel to a given line through a given point outside that line. Thuswe have the situation of Figure 5.

!

A B

C

Figure 5. The Playfair Postulate equivalent of Euclid’s Postulate 5

The Playfair Postulate or equivalently Elements Postulate 5 allows exactly one line to bedrawn through C parallel to AB. This should be quite familiar to you as you almostcertainly learned in school mathematics to construct that line.

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A number of mathematicians, however, saw in that new formulation the possibility ofdifferent answers. Among them were the 12th century Persian, Omar Khayyam;4 the 17thcentury Italian, Giovanni Girolamo Saccheri; the 18th century German, Carl FriedrichGauss; and in the 19th century the Russian Nikolai Ivanovich Lobachevsky and the Hun-garian, Janos Bolyai.

What these famous mathematicians finally found was that the Playfair Postulate’s “one”line was arbitrary and that there are, in fact, three alternatives to this postulate:

(1) Playfair: Through a given point one line can be drawn parallel to a given line.(2) Through a given point less than one line (that is, no lines) can be drawn parallelto a given line.(3) Through a given point more than one line can be drawn parallel to a given line.

Your first response to (2) and (3) may be that they are simply nonsense. Surely (2) iswrong because we learned to construct that line through C on Figure 5. And (3) wouldrequire us to have additional lines through C. What you have to understand, however, isthat (2) and (3) define entirely different geometries, in a real sense different worlds. And,believe it or not, one of those geometries is all around us. We will explore that geometrynow and in doing so find the answer to the challenge with which this chapter began.

4 Geometry on the Surface of a Sphere

The plane geometry of our high school studies makes perfect sense because locally —within a few miles, that is — we can consider ourselves living on a plane. But our Earthis a sphere and geometry on a sphere is different from geometry on a plane. A majordifference becomes immediately apparent. There are no straight lines like those of planegeometry on a sphere. A line of sight looking toward the horizon, for example, would shootaway from the Earth on a tangent.

We need to modify the concept of a straight line in the plane to give us a reasonablesubstitute on a sphere. We still want to be able to travel “in a straight line” in any directionalong the earth’s surface and have our path along that line shorter than any alternate path.5

In order to do this, we substitute for a straight line on a sphere a line following what iscalled a great circle. Indeed, great circles do represent the shortest distance between twopoints on them.6

4Omar Khayyam is better known as the poet who wrote the Rubiayat, a series of quatrains translatedinto English by Edward FitzGerald in 1859. Often quoted are the lines, “A Flask of Wine, a Book of Verse— and Thou Beside me singing in the Wilderness — Oh, Wilderness were Paradise enow!”

5Shortest paths are called geodesics.6There is an exception to this. If the points on the sphere are at opposite ends of a diameter there are

many equal paths.

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You are familiar with one great circle, the equator. Lines of longitude (those north-southlines measured east and west from Greenwich, England) also run along great circles. It isimportant to note, however that lines of latitude (east-west lines), except for the equatoritself, do not follow great circles. They do not represent shortest paths. It is easy to showthis with a globe. You need only connect two points at some distance apart with a string.When pulled taught, in seeking the minimal path, the string will lie along a great circleand not along a line of latitude other than the equator.

Most maps and charts display geographic features with the lines of latitude horizontal.For larger areas, especially those involving continents and oceans, this significantly distortsboth shapes and measurements. One consequence on those maps is that Greenland andAntarctica appear much larger than they really are. Another is that minimum distances arenot along straight lines. Consider, for example, the shortest (great circle) route comparedwith the route that follows a line of latitude between New York’s Kennedy Airport and theCiampino Airport of Rome, Italy as shown in Figure 6. Although that figure shows howthe great circle route takes you from JFK north over Labrador and Newfoundland, thatroute is shorter by more than 1400 miles than a route that simply takes the flight along aline of latitude.

Figure 6. The Great Circle Route from New York to Rome.

Once we have identified great circles with straight lines, we can examine what wouldreplace on spheres those five Elements postulates:

1’. Construct a segment between two points (not its opposite pole).2’. Extend a line.3’. Construct a circle given its center and radius.4’. All right angles are equal.5’. That any two lines intersect.

It takes little thought to see that the first four postulates remain essentially as in theElements and that they are reasonable on a sphere. A segment is, of course, an arc ofa great circle just as in plane geometry a segment is a piece of a straight line. The finalpostulate is, however, very different. It tells us that we have no parallel lines on a sphere.

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In fact, any two distinct great circles intersect in two points that are poles (that is, 180○)apart. This statement may seem strange, but you need only do some experimenting witha globe to see that it is indeed the case as in Figure 7.

a

b

A B

C

Figure 7. Postulate 5’ on a Sphere7

In order to continue, we need some vocabulary. Where two great circles meet on a sphere,they form an angle between those curves. This angle corresponds to and is measured bythe plane angle between the tangents to the great circles at that point, as in Figure 8.

Figure 8. Spherical Angle APB

A lune is a region bounded by two great circles as in Figure 9.8 The name lune isetymologically related to lunar, pertaining to the moon, and the lune’s shape clearly justifiesthis name.

Figure 9. Lune PAP ′B

7Figure 7 was drawn with the Cinderella software as are some other figures that follow. In these viewspoints and lines on the back of the sphere also appear. These include points at the opposite ends ofdiameters through points on the front of the sphere.

8Some authors refer to a lune as the solid slice taken from the sphere — like a section of an orange.Here we follow other authors in considering it only the surface region.

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Note that the angle AOB suggests a way of identifying the area of a lune. That angle,which is the same as the angle at P , is a fraction of the distance around O. Thus we needonly find the angle of a lune, say x○, on a sphere with area S to calculate the lune’s area.It is x

360 ⋅S. To calculate the area of a lune on a sphere with radius r, you need the formulafor S. It is S = 4πr2, a formula derived by Archimedes 2200 years ago.9

Finally, a spherical triangle is a 3-sided figure bounded by arcs of great circles. Figure9 shows a spherical triangle together with the image created on the opposite side of thesphere by diameters from each corner of the original △ABC.

a

b

c

A

B

C

Figure 10. Spherical triangle ABC.

There is an interesting feature of that shadow triangle on the opposite side of the sphere.Although its angles are equal to those of the original triangle and it is equal in area to△ABC, it is not congruent to it. It is a mirror image with orientation reversed and like aglove, it cannot be made to fit its partner.10

In Figure 9, △PAB is also a spherical triangle and I will use that triangle to suggest theformula for the area of any spherical triangle. Clearly, the area of that triangle is half thearea of the lune, whose area we determined is x

360S. Thus the area of △PAB is x720S.

Now consider what kind of angles are ∠PAB and ∠PBA. They are like meridian linesmeeting the equator, where I hope you will accept that they are both right angles. Thistells us something very interesting about △ABC on Figure 9. The sum of its angles is morethan 180○. In fact, if ∠P is x○, the angle sum is 180+ x degrees. That x, the amount over180○ for a spherical triangle is called the spherical excess. It turns out that every sphericaltriangle has such a spherical excess, whether or not they have right angles in them, andthat spherical excess is related to the area of the triangle.

9By fortunate chance the Danish philologist Johan Ludvig Heiberg discovered parchments in Con-stantinople (now Istanbul) that included the only known copy of The Method of Mechanical Theoremsof Archimedes. In it Archimedes describes how he found the formulas for the area, S = 4πr2, and thevolume of a sphere, V = 4

3πr3, as well. His method used calculus-like techniques long before the calculus

was invented.10For vocabulary enthusiasts, a mirror image is an enatiomorph and enatiomorphisms play an important

role in crystallography.

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a

b

cA

B

C

Figure 10.5 Spherical triangle ABC with sides extended.

On Figure 10.5, the great circle sides of spherical ∆ABC have been extended. Noticethat each of those angles determines a lune. If you add all of those lunes together you willhave the surface of half of the sphere, but the triangle itself will have been counted twoextra times.11Thus you have the geometric situation regarding areas:

the hemisphere + 2 ∆ABC = Lune A + Lune B + Lune C.

This translates into algebraic terms with S the surface of the sphere:

12S + 2∆ABC = A

360S + B

360S + C

360S

and we have the following algebraic simplification:

2∆ABC = A

360S + B

360S + C

360S − 1

2S (4.1)

∆ABC = A

720S + B

720S + C

720S − 180

720S (4.2)

∆ABC = A +B +C − 180720

S (4.3)

Recall that we named A +B + C − 180 the triangle’s spherical excess. If you designatethat spherical excess in degrees, x, and the area of any spherical triangle T , you have thefollowing:

T = x

720S and T = x

7204πr2 = x

180πr2

the latter formula the result of the fact that the area of a sphere is S = 4πr2.

This finally provides you with the tools necessary to solve the farmer’s problem withwhich you began this chapter. You need only find the spherical excess for those surveyor’sangle measures and use the formula to calculate the area of the property. The actualcalculation is left for the exercises.

11In making up the area of the hemisphere, it is necessary to take the shadow area of one of the lunes.

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5 The Other Alternative to Postulate 5

Recall that a third alternative to Euclid’s Postulate 5 was offered. It was:

(3) Through a given point more than one line can be drawn parallel to a given line.

The geometry that this leads to is called hyperbolic geometry and it is more difficult todemonstrate. Figure 11 gives one way of doing so.

a

b

c

d

A B

C

Figure 11. Postulate 5 in the Hyperbolic Plane

As on the sphere, hyperbolic lines are curved but, if you examine them closely, you willsee that each of them is perpendicular to the edge of the region containing them. Thinkof that region made infinitely large so that the circle represents the “edge” of space. Thenyou can think of those lines as being perpendicular to that “edge”, as they are on thisdiagram. In order for that to be the case, the lines through most given points will appearas curved, as they are on the figure. Only lines through the center of the figure wouldappear straight.

Under these conditions we do indeed have more than one line through a point outside agiven line that does not intersect it. In particular, two of those lines would meet the linethrough A and B at the edge of space. as in Figure 12. Those lines we would designate asparallel to line AB.

a

c

bA

B

C

Figure 12. Two lines parallel to AB in the Hyperbolic Plane

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This is like those parallel railroad tracks that appear to meet at the horizon.

6 Distance on a Sphere

As a kind of bonus, I offer, this time without derivation, a formula for calculating distanceon a sphere. With it you can calculate the mileage between locations for which you havelongitude and latitude. The formula as I will write it will give the distance in nauticalmiles. Nautical miles are used — as you might expect — by sailors. Ships’ speeds aregiven in knots and a knot is one nautical mile per hour.

A nautical mile is the length of one minute of arc along the equator. You will be askedin the exercises to calculate how many feet there are in a nautical mile.

One other thing is necessary to translate longitude and latitude into values acceptableto mathematical computation. Longitude values are given in east or west coordinatesmeasured from Greenwich and latitude values north and south of the equator. In orderto make these measures acceptable to calculation (and GPS devices) west and south arenegative, east and north positive.

Here then is the formula. Given points with longitude and latitude coordinates (a, b)and (c, d), the distance in nautical miles along the shorter great circle arc joining them is:

60 cos−1[sin b ⋅ sind + cos b ⋅ cos d ⋅ cos(a − c)]It is this kind of information that is calculated very quickly and very accurately by GPS

devices.

Exercises

(1.1) There are many proofs of the Pythagorean Theorem. In fact, Elisha Scott Loomis offered367 of them — enough, that is, for a year of daily proofs — in his book The PythagoreanProposition, first published in 1927. That edition is now worth a great deal of money tohistorians of science. Some of these proofs are very simple and everyone should know at leastone of them. Perhaps the simplest was proved by United States President James Garfield.Look up this proof and write it out.

(1.2) Look up and follow the proof of the Elements. Wikipedia is a good source for this proof andprovides a thorough explanation. Reduce this explanation to its key steps.

(1.3) Look up the proof of the Pythagorean Theorem in the Heath translation of Euclid’s Elements.You can access a version of this text through Gutenberg.

(2.1) The words axiom and postulate are synonyms. See if you can determine a technical differencebetween the words.

(2.2) Access the 1723 translation of the Elements by John Keil through Google. Where does Keillist what we have called Proposition 5 of Book I? Why do you suppose this difference arises?

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(3.1) Of course, the Earth is only approximately a sphere. There are two ways that it differs frombeing exactly spherical. Consider the first of those ways. The highest mountain on earthis Mount Everest at about 29,000 feet and the farthest down you can get, in the MarianaTrench, is about 36,200 feet deep. What percent of the earth’s radius, about 3960 miles, isthis difference?

(3.2) You can buy raised relief globes that have mountain ranges raised above the surface of theoceans. If the water were removed, we would have deep trenches in the ocean and these couldbe represented as indentations. Toy rubber balloons have a skin thickness of less than .5 mm≈ .02 inch. Find the percent this is for a 5-inch diameter balloon and compare this with youranswer in 3.1. What does this answer say about the accuracy of raised relief globes?

(3.3) How far can you see on the Earth? The answer is not far when landforms and vegetationget in the way, but it is a reasonable question at sea or from the seashore. Suppose your eyesare about 50 feet above sea level. This might be on land along the seashore or from a shipcrow’s nest. Find approximately how far away is your horizon.

(3.4) Devise a general formula to use to estimate the distance you can see in miles, given yourheight in feet above sea level.

(3.5) The earth is coordinated so that, as Global Positioning System (GPS) devices make clear,any point on it may be located with extreme accuracy. Find the coordinates of where youare currently residing. Give those coordinates in three forms: (a) degrees, (b) degrees andminutes, and (c) degrees, minutes and seconds.

(3.6) Most people know that the so-called prime meridian of longitude passes through Greenwich,England, a London suburb. Where is the longitude-latitude origin, the point that correspondsto (0,0) on graphs? Where is the point (180○,0○)?

(3.7) In one of the later Oz books by Frank Baum, Dorothy climbs down through the center of theEarth and comes out in Oz, which is in the middle of the Great Sandy Desert of Australia.She started, of course, from Kansas. Determine if these two locations are indeed poles apart.

(3.8 Calculate the area of the farmer’s land with which this chapter began.a b

c

A

BC

Figure 13. A spherical triangle with three right angles

(3.9) I did not prove the formula for area of a spherical triangle on a sphere with area S. To furtherjustify that formula, however, consider the spherical triangle of Figure 13. Each of its anglesis a right angle.(a) Find its area in terms of S using the area formula.(b) How many of these triangles could you fit on the sphere without overlapping?(c) Show that your answer in (b) supports your answer in (a).

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(4.1) We considered in this chapter the only three possibilities for the number of lines througha given point parallel to a given line: none, 1, and more than one. There are no otherpossibilities. This approach is similar to a method of proof called proof by cases or proof byexhaustion. You list all the possibilities and eliminate all but one, then that one must provideyour solution. The problem is that we often forget possibilities. (Politicians and lawyers oftenencourage you to forget them.) In the following mathematical examples, supply the missingalternative. Then think up examples of your own:(a) x > 4 or x < 4.(b) y is either an integer or a fraction.(c) Positive integer z is either prime or composite.

(6.1) The length of the equator is 24,901.5 statute miles. (Statute miles are the miles with whichwe make land measurements.)(a) Using this value, what is the radius of the earth at the equator?(b) We noted that a minute of arc along the equator is equal to a nautical mile. How manyfeet are there then in a nautical mile?(c) How does the nautical mile compare with the statute mile? Is it the same, shorter orlonger?(d) If there is a difference in (c), what factor can be used to convert nautical miles to statutemiles?

(6.2) Write a calculator program to accept longitude and latitude for two different locations and useyour program to calculate the length of flights between the following cities whose approximatevalues are given. Have your calculator provide both nautical miles and statute miles betweenfive of these locations. (Be especially careful to get longitude and latitude in the correct orderand with the correct sign.)

City Longitude LatitudeNew York 74.0○ W 40.7○ NLos Angeles 118.3○ W 34.2○ NLondon 7.6○ W 51.5○ NMoscow 37.6○ E 55.8○ NTokyo 139.7○ E 35.7○ NRio de Janeiro 43.2○ W 22.9○ SSydney 151.2○ E 33.9○ SJohannesburg 28.1○ W 26.2○ S

(6.3) Use Google Earth or some other program to find the exact location in degrees, minutes andseconds of where you live. Then find another place of special interest to you and record it inthe same way. Use your program to find the distance between these locations. (You must bevery careful in recording these values this, because coordinates are usually given in the orderlatitude, longitude, which does not conform to our usual sense of (x,y) graph coordinates.)

(6.4) Earlier in this chapter I claimed that the great circle flight from New York to Rome savedover 1400 miles against a flight following closely a line of latitude. Determine the length ofthe great circle flight.

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(6.5) Find a way to determine the length of a New York-Rome trip that approximately followsthe 41○ latitude line and compare your answer with the great circle distance you calculatedin 5.4.

(6.6) We have taken great circle routes as our “straight line” routes between places on the earthbecause those are the shortest routes. More generally shortest routes are called geodesics. Inthe plane, of course, the geodesic is a straight line. Here is a famous geodesic problem posedoriginally by the famous puzzle constructor, Henry Ernest Dudeney, in 1903: In a rectangularroom with dimensions, 30’ x 12’ x 12’, a spider is located in the middle of one 12’?12’ wallone foot away from the ceiling. A fly is in the middle of the opposite wall one foot away fromthe floor. If the fly remains stationary, what is the shortest total distance (i.e., the geodesic)the spider must crawl in order to capture the fly? He can crawl along walls, floor or ceiling.This problem is best solved by drawing and cutting out the six walls with correct proportionsand rearranging them in various ways that would allow you to calculate the spider’s route.

ceiling

wall

wallwall

wall

floor

fly spider

The spider and the fly

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