Notes — Chapter 15: Geometric Solids

To the Student. We live in a three-dimensional world and can deal with three-dimensional shapes with ease. Here we learn how mathematical ideas from plane geometry allow us to work out the properties of threedimensional shapes and calculate surface areas and volumes. Euclid’s Elements build toward solid geometry (Book IX) and the regular polyhedra (Book XIII), which we cover in this chapter.

▶ Postulates 11 and 12. Postulate 11 says that if two points on the line ℓ lie in the plane II, then @ Cc II. We need this in order to know that lines do not somehow stick out of the planes they should lie in. Postulate 12 says that if two planes intersect, they intersect in a line. This tells us that ours is a three-dimensional space.

In four dimensions, the two-dimensional planes

Th 4.4, 0:0) | u,v real numbers} and

D3 (0, 0,20, 2) | w,z real numbers}

intersect in a single point, (0,0,0,0), not a line. This cannot happen in three dimensions.

These are the only postulates we need to add for the study of solid geometry. Given these, Euclid proved the results in the supplemental notes for this chapter and outlined in Solid Geometry as a Deductive System, pages 626 and 627 of Geometry. _ til

Lesson 1. Lines and Planes in Space. This lesson and the summary on pages 626 and 627 Of Geometry are the basis for our study of solid geometry. Our

approach is based on intuition, with all assumptions clearly stated. Students will need to use the tools of plane geometry freely as they study solid geometry.

We know that lines that are parallel have the same direction. When thought of this way, it 7s immediate that two lines parallel to a third line must be parallel to each other. However, this transitivity of parallelism is not easy to prove from the formal definition. Teaching the spirit of parallelism is more important than getting hung up on the letter of parallelism. We want students to trust their intuitions, but also to be able to check them, as needed.

Likewise, planes are parallel if and only if they are perpendicular to the same line or to lines in the same family of parallel lines.

This lesson introduces lines and planes in space and their relationships. The goal is to link intuition and common understanding to geometric terminology. This lesson sharpens vision and builds vocabulary.

Set I. These are exercises in seeing what is portrayed and interpreting it as solid geometry: visual thinking, three-dimensional relationships, passing from a twodimensional picture to the three-dimensional geometry, and reasoning about shapes in space.

I like all of these exercises, and I was tickled to see Kilroy, of “Kilroy was here,” put to good use in Kilroy in Space (24-29).

Set II. Triangles in Perspective (31-37). This is Desargues’s Theorem, discussed at the beginning of Chapter 13, The Concurrence Theorems. In its projective geometry form this theorem says: Two triangles are perspective from a point if and only if they are perspective from a line.

Matsumassa Anno’s Impossible Slice (38 and 39) is a delight and gives us a chance to think clearly about what is clearly impossible. A calculator is needed in Parallel Bars (40-43). Lopsided Box is excellent as an exercise in visual thinking and deductive skills. You cannot just go with intuition here.

Cube and Hypercube (54-57) shows how we can reason from the net for a cube to get properties of the cube and extends this reasoning to draw conclusions about hypercubes.

Set III. Escher’s Belvedere lets students find and describe geometric impossibilities.

Geometry as a Deductive System. This brief outline of the basis for solid geometry sets out the essentials.

Lesson 2. Rectangular Solids. The subject of this lesson is as solid and common as a brick, and as crucial to building our understanding of geometry in three

dimensions as building stones and bricks are to architecture.

Vocabulary introduced or reviewed here includes: polyhedron, faces, edges, rectangular solid, diagonal, and the common uses of length, width, and height in describing rectangular solids.

Theorem 79, which gives the formula for the diagonal of a rectangular solid, also tells you that the distance between points P = (z,y,z) and P’ = (z/, y/, z/) is given by

2

IP SP ye 2) eee

where we use a set of Cartesian Coordinates in space. Such axes are mutually perpendicular and have equal units. This distance, written ||P — P’||, is just the diagonal of a rectangular solid whose faces are parallel to the coordinate planes and whose opposite vertices are Pand!Ps

Set I. Inside or Outside? (1-8) is an exercise in visual thinking. Deck of Cards (9-13) will need a calculator. Integer Lengths (14-17) acquaints students with an open research problem. Slicing a Cube (20-28) shows how to figure out what plane slices of a solid look like. For more complex and less familiar three-dimensional shapes the problem of determining what various slices can look like is quite difficult. For instance, what is the maximum number of pieces into which a doughnut can be sliced by three planes? This is Macalester College’s Problem of the Week for Week 866. In 2003 it was on the Math Forum website. Search on Stan Wagon and see Which Way Did the Bicycle Go? by Joseph D. E. Konhauser, Dan Velleman, and Stan Wagon, Mathematical Association of America, 1996 for more problems from the Macalester group.

Set II. An Inside Job (29-31) is problem-solving with an Aha! at the end. Diagonals (32-37) involves visual thinking, three-dimensional relationships, a review of angles and trigonometric ideas, and use of a calculator. Hole through a Cube (38-42) has a conclusion that is quite surprising, until you understand what is going on. Bond Angles (43-47) involves visual thinking, spatial sense, and use of a calculator. Soma Puzzles (48-54) involves visual thinking, spatial sense, logical reasoning, and problem-solving. This sequence shows how local analysis of the pieces of a puzzle allows you to figure out what is happening. This sort of reasoning is the key to solving “hidden block puzzles.”

Set III. Flattened Cube. This is an fresh sort of problem based on nets for a cube. It is within the reach of anyone who will draw the nets on paper and work at it.

Lesson 3. Prisms. These are the next step in building our tool kit of solid shapes and their properties.

Set I gets off to an easy start thinking about 3-D Lettering (1-3). Simple, but it forces students to think about what they see and what gives things their 3-D

look. A Hexagonal Surprise (19-28) is surprising. See Avoiding Trouble.

lt” Avoiding Trouble. This Hexagonal Surprise is a fact whose proof depends on tracking relations between _ quantities that are interdependent. There must be a purely geometric approach that will prove that this hexagonal configuration always gives two congruent triangles, associated with the two sets of non-neighboring vertices of the hexagon, and that these two triangles are translates of each other. I gave up looking for such a proof.

The geometry suggests more complexity than is there. It looks as if we need to consider any six points in the plane, but this is not so because our construction and our conclusions can be stated only using parallel lines. Working with parallel lines alone, we cannot pin down congruence, but we can determine whether a transformation is a translation or not, and that is what we need here.

Simplify. Parallelism can be checked by seeing that slopes are the same. Slopes are the same in one coordinate system iff they are the same in all coordinate systems (where we allow m = oo for the moment). Suppose our hexagon is ABCDEF. We simplify the problem by setting up coordinates so that F = (0,1), A = (0,0), B= (1,0), C= (ci, C²), DD (dy, d²), b= (€1, 2). The axes here will not necessarily be at right angles nor will they necessarily have scales that correspond to the distance in the plane, but that is of no matter, because all we are interested in is parallelism, something that is determined by the same algebraic condition in all such sets of coordinates.

Now we find the coordinates of the points A’, B’,…F’ of the Hexagonal Surprise. Please refer to Hexagonal Surpriise for what follows. FABA’ is a parallelogram, and the diagonals of a parallelogram bisect each other. The bisector of FB is the point whose coordinates are determined by $(F + B). Likewise, the bisector of AA’ is (A +A’). Setting these equal, and solving for A’ we get

(F +B) = 5(A +A), so A’=F+B-A.

hole

And the analogous algebra gives us all of the other primed points. Then calculation gives

(A Pe) Ger 1-05),

CL = (1 —-di= c1,d² — ca);

D/ = (ey +c, —d,,e€2 + c² — dz),

E’ = (dy — e1,1 + d² — 2), and F’ = (e1,e² — 1).

Checking coordinates we find

Le RE a EN REM 2g

= (1—c, + dy — e1,1 — cg + dz — €2).

So AD‘F’B’ is a translate of AA’C’E’, and we have a proof of the property observed in this series.

This would be an interesting problem to explore using technology. I worked out an approach using vectors, but I was unable to reconstruct it. It is interesting that there is no generalization to octagons, let alone 2n-gons for n > 3.

This result can be proven by algebra, and a computer algebra system might help get the result without simplifying things by choosing a convenient coordinate system. Transformations could be used to bring the original hexagon into a standard position, but this is equivalent to the simplification of choosing the coordinates as is done above.

If there is a neat and purely geometric way to do this, I would like to see it. For problems of this nature, the simplest approach is often to grind out the algebra —after setting things up so as to simplify the work.

Set IT. Magic Box (42-49). Andrew M. Gleason mentioned that he had built such a box and that it worked well.

Euler’s Discovery (50-53). For connected planar maps, if we count the number of Countries, Edges, and Vertices we find:

(*) Countries — Edges + Vertices = 1.

There are simple inductive proofs of this result. There are so many good sources that I hesitate to single out one. Euler’s formula, (*), for planar graphs can be related to the Euler characteristic for polyhedra by means of the Schlegel diagrams introduced on page 688 of Geometry. In this interpretation, the Faces of P translate into the Countries of our map. See also the sequence More on the Euler Characteristic, page 651 of Geometry.

Set III. This is a simple start at understanding how the observable geometry crystals can tell us something about the arrangement of atoms. When this work was done devices such as the atomic force microscope had not been developed and our understanding of crystal structure came from indirect observation.

Lesson 4. The Volume ofa Prism. This is a concrete and simple introduction to volume. All geometric measurement comes down to counting how many congruent copies of the basic unit it takes to fill out the shape being measured. We must allow fractional parts of the unit to be used: tenths, thousandths, and so on, ad infinitum. This is where the decimal or other notations come in.

This is how we know that the volume of a rectangular solid of length J, width w, and height h is

24. Volume = Length x Width x Height, or V = lwh.

The beauty is that this is obvious when the dimensions are integral multiples of the unit length. Because of the way our notation works, this ensures that the same formula holds when we add in decimal submultiples of the unit length. It is important that we use the same units for the dimensions being multiplied if we want our answer in standard cubic units. There are other sorts of units, such as acre-feet, but they are uncommon.

Postulate 13, Cavalieri’s Principle, is intuitive and gives us the volumes of any solid shape we are likely to deal with. This principle underlies the slicing methods used to find volumes in calculus. See Keng-chih’s Principle, Exercises 26-28. Postulate 13 follows from Cavalieri’s Principle by the argument given in the last paragraph of page 641 of Geometry.

We need Postulate 13 because we rely on intuitive ideas of volume, namely the properties suggested by the top two figures in the margin of page 641 and referred to in the starred paragraph. To make Postulate 13 into a theorem, we would need to develop a theory of volume, following the idea that volumes can be defined by measuring how many unit cubes or thousandths of a unit cube, and so on can be packed into a given solid. Using this idea we could prove the results in (*), and have Postulate 14 as a consequence of Cavalieri’s Principle. In calculus or analysis, a theory of volume is developed, and these results are theorems.

Set I. These are lovely exercises: concrete, interesting, and thought provoking. You learn something from each of them. The Weathering of Rock sequence (16- 25) describes a recursive fragmentation process. This is a fractal idea, and there is considerable interest in fractal geometric processes in geology, especially in subsurface flows through fractal fissures in connection with recovery of oil from oil fields and in connection with the spread of pollutants in ground water.

Set II. Ream of Paper (38-40) gives students an idea of how thick a sheet of paper is. Emilio Segré told me that when he worked with Enrico Fermi’s group in Rome, they made it a practice to think about common things that could be used to get a quantitative grip on things in everyday life and in the laboratory. Segré mentioned the thickness of sheets of paper, the volume of a drop of water, and so on. But note that some drops are larger than others and some sheets are thinner than others. Geometry by Jacobs is almost 800 pages (400 sheets) and it is less than an inch thick, not counting the covers, but the 500 sheet ream of paper in these exercises is two inches thick. See also Set III, below.

Binomial Cube (41-46) brings algebra and geometry together with visual thinking. Advertising Claim (57-60) brings critical thinking into play, part of the numeracy or quantitative literacy theme discussed following Set III in these notes.

Set ITI. Molecule Experiment. The picture here is from the WGBH series The Ring of Truth, discussed below.

The pond is in a park in central Boston and the oil slick is of the finest olive oil. The idea of this experiment has been around for ages, but it is difficult to get consistent experimental results, and the results are hard to interpret. In the book by Frederick P. Greenleaf, mentioned below, this was done as an experiment in a laboratory section. Greenleaf’s lab manual describes some of the difficulties in getting consistent results.

Irving Langmuir and Katherine B. Blodgett did work on thin films at GE. Out of this work came the idea of antireflective coatings for optics, which has wide uses today. Accounts of their work can be found in the library or on the Web.

lv Quantitative Literacy. This is an emerging theme in the curriculum, and Geometry teaches it throughout the book. In this lesson, we are working out volumes of real things: candy bars, suitcases, swimming pools, reams of paper, and the volume of individual molecules.

It is important to understand that quantitative literacy means having a general idea of what numbers make sense for various quantities, not simply in the ability to figure out volumes once someone else has given you numbers to work with. For example, in Innumeracy John Allen Paulos mentions a question on page 12, said to be used by a consultant from MIT to weed out candidates for jobs. (In 2003, Innumeracy was available from Hill & Wang in a 2001 paperbacked edition.) Here is the question: Suppose trucks come every fifteen minutes, twenty four hours a day without interruption, are instantly filled with earth and rock, and carry their loads away without getting in each other’s way. How long would it take to cart away an isolated mountain, say Mount Fuji, to ground level? The only information Paulos gives is that Mount Fuji is 12,000 feet high, so this must be used to get our answer.

A few pages later, and without explanation, Paulos gives his answer: 5000 years. This seemed too short to me, so I decided to work it out myself. There are three things we need to know in order to work this out, starting from the reasonable assumption that Mount Fuji can be considered to be a cone. 1. What is the altitude of this cone? This includes deciding what we consider “ground level” for this problem. 2. What is the area of the base of this cone? Given the answer to 1, this comes down to deciding how steep the sides of Mount Fuji are. The answer we get will depends on our assumptions here. 3. How much rock can a truck carry in one load?

Looking at a contour map of Mount Fuji suggests that clearing it from its present height of 3776 meters down to the 1000 meter level would make the remnant blend in with the surrounding country side. The map also shows that at the 1000 meter contour we can consider Mount Fuji’s base to be a circle with a radius of 12 kilometers. A call to a local importer of heavy Japanese earth-moving trucks got the information that the largest of these could carry 18.5 cubic yards. I rounded up to

20 cubic meters. Do the calculation and you will find that it would take 600,000 years to truck away the top 2776 meters of Mount Fuji.

Paulos did not tell me all of his assumptions, but he did _ say that he assumed Mount Fuji was a cone as wide at its base as it was high. This would make its sides slope upward at a 60° angle. Very impressive. Here is where his quantitative literacy should have made him check it out, always a good thing to do if you think a calculation is really important.

The above analysis is from my article, “Thoughts on Innumeracy: Mathematics Versus the World,” published in The American Mathematical Monthly, October 1993. A reply from John Allen Paulos followed my article.

Paulos did not name his consultant from MIT. But I know of a similar question asked in a job interview. David Root was teaching at Purdue University around 1970, when he decided to leave academia and work on the problem of energy supplies.

A reigning power then was M. King Hubbert, who correctly predicted in 1956 that US oil output would hit a maximum in the early 1970s and then decline (this was “Hubbert’s Peak”). Having studied energy-related matters for a year at Purdue, Root flew to interview with Hubbert. When they met, the first question Hubbert asked Root was: How many population doublings did it take to go from Adam and Eve to the current population? (then about 3.5 billion) Root answered “31,” not missing a beat. Hubbert said something like, “Closer to 32.” This is practical numeracy. David Root made a career with the US Geological Survey.

We should teach numeracy. I believe that Jacobs has done as much as anyone to help students see how they can use mathematical ideas to make sense of the world around them. He began with his students, for whom he wrote Mathematics: A Human Endeavor, now in its third edition. Many other teachers have taught from this book. Jacobs has made realistic and interesting examples a hallmark of his texts. Working such examples shows the value numeracy and teaches it at the same time.

Numeracy, quantitative reasoning, and critical thinking are terms you can search on the Web to see what is available in this area. For a broad appreciation of measurement and calculation in science, I recommend The Ring of Truth by Philip and Phylis Morrison, both a book (Random House, 1987) and a television series (WGBH, available in VHS format in 2003). Also of interest is Quantitative Reasoning: Understanding the Mathematical Patterns of Nature, the text developed for a core course at New York University by Frederick P. Greenleaf and available from McGraw Hill, Primus in 2003.

Numeracy is important in evaluating economic, social, educational, and medical ideas, but data on things

such as student achievement are difficult to gather and subject to varying interpretation. This takes us away from geometry, where things are easier. Here is what Lynn Steen wrote in the Epilogue to Mathematics and Democracy: The Case for Quantitative Literacy, which he edited for the Woodrow Wilson Foundation and the National Council on Education and the Disciplines.

“… [NJumeracy is not so much about understanding abstract concepts as about applying elementary tools in sophisticated settings…. [T]his is no simple feat. Numeracy takes years of study and experience to achieve.”

A Web search on various key words and names mentioned above will turn up current material on numeracy and quantitative literacy. The best approach to these skills is to use real numbers and find interesting examples that can be dealt with using numbers in subjects across the curriculum. Geometry has such examples, so Jacobs is on the right track. t

Lesson 5. Pyramids. Pyramids are a perennial favorite topic and a Web search on the names of any of them (Bent Pyramid, Pyramid of Cheops) will turn up interesting material. Imhotep is the legendary architect of pyramids.

Thinking about the middle line of figures on page 648 of Geometry the right way will let you see why these three pyramids all have the same volume. Using Cavalier’s Principle, all you need to notice is that each pair of these pyramids can be seen as having bases that are congruent and apexes that are identical. The trick is that you must think of the bases as being on different rectangular sides of the prism when you switch to a different pair of pyramids. This insight will reward the industrious.

Set I. These exercises cover the lesson. Leonardo’s Claim (18-20) involves visual thinking, number sense, and the ability to decide whether something looks like a reasonable picture of what is described.

Set II. Divided Cube (27-32) is a geometric discovery that we would think of as requiring visualization. It is notable that it was made by a blind mathematician. The physicist Plateau, who studied the geometry of soap films (Web-search on Plateau’s Problem) was also blind, and there are a number of successful blind mathematicians today, among them Lawrence Wos, who works on computer-aided reasoning at Argonne National Laboratories, and Lawrence Baggett, who served as Chairman of the Mathematics Department at the University of Colorado, Boulder. See Allyn Jackson’s article from the November 2002 Notices of the American Mathematical Society, titled “The World of Blind Mathematicians” for more about blind mathematicians present and past. This was available from the AMS website over in 2003.

Pyramid Numbers (33-37). Although the cited reference does not say where this is to be found in Herodotus,

digging reveals that it is in Book II (Euterpe), paragraph 124, where you will find “Each face is eight plethra and the height is the same.” Looking under “Weights and Measures” in The Oxford Companion to Classical Literature, Oxford University Press, 1989 for “Plethron” sheds light on this matter. A plethron is both a measure of length and of area. As a measure of length, it is 100 Greek feet, a Greek foot being 16 Greek fingers. As a unit of area, it is the Greek acre. That is, a plethron is the nominal area for a field that a man can plow in one day, and this source gives it as 100 feet square, that is, 10,000 Greek square feet. This area seems low compared to the modern acre, which is 1 chain by 10 chains or 66 by 660 feet and is the nominal amount of land that a man can plow in a day. But these “feet” may not be the same, and we had better plows than the ancient Greeks, even before we got tractors. If we think of both plethera in the quotation as being areas, then that second must be gotten somehow from the height, and squaring is the only reasonable way —and the numbers work out reasonably closely, so this might be what was intended, or it might not.

Richard Gillings, in his book Mathematics in the Time of the Pharaohs, available in 2003 from Dover Publications, says that he sees no evidence that the pyramids were constructed on the area principle given above. It is interesting that he also suggests that Egyptians may have used the method of the Divided Cube above to figure the volume of pyramids.

See Gillings also relating to The Frustum of a Pyramid (48-57), which was something the Egyptians seem to have known about.

Set III. How Many Faces? Student versus SAT. Student wins. Inspiring.

Lesson 6. Cylinders and Cones. The general notion of a cone C over a set S with apex P is that C is the union of all segments sP, such that se S. The results in this lesson are true for volumes of cylinders and cones, whether or not their bases are circular, as can be seen by Cavalieri’s Principle.

▶ Considering Cones. Take a pyramid P whose base has the same area, A, and lies in the same plane as the base of the cone C. Let the apex of P be at the same height above the common base plane of P and C as the apex of C is. Taking a slice parallel to the base plane at distance  below the common plane of the apexes of P and C, we find that the areas are ?A for both the cross-section of P and that of C. This is true because of the general scaling principle for areas under similarity (Lesson 6, Chapter 10). Because slices at equal heights have equal areas, the volumes are equal. Both volumes are 3Bh, the formula for pyramids. “

Set I. Primitive Perspective (1 and 2), visual thinking and three-dimensional geometry. Italian into English (3-9) involves general problem-solving. Solids of

Revolution (10-20) involves visual thinking and threedimensional geometry. Sliced Solids (23-26) brings in key ideas relating to contour maps. These sorts of slicings are used to build three-dimensional models by sticking together slabs cut to the contours. There is a lot out there. I found a nice contour map of Mount Fuji on the Web. Sausages in a Can (27-31) is a practical sort of calculation.

Set II. This set is loaded with quantitative comparisons of geometrical quantities arising in everyday life. An understanding of Rolling Cones plays a key role in Michael Gilbert’s classic 1950 murder mystery, Smallbone Deceased.

Set III. Iron Ore, photographic interpretation again. See also What time is it? on page 453 and my notes on it. In 2003 the Smithsonian Institution had some coverage on Photo Interpretation and there were many college-level and commercial websites devoted to this topic. Books such as Below from Above by George Gerster, Abbeyville Press, 1986, cited on page 453 ofGeometry and Earth from Above Yann Arthus- Bertrand, Harry N. Abrams, 2002 (photographs also available in a large-sized postcard book of the same title; see www.yannarthusberthold.com) offer many striking images that could be the basis for photo-interpretation exercises.

Lesson 7. Spheres. The material of this lesson and the arguments supporting it are at just the right level of detail and completeness. Cavalieri’s Principle is the tool to find the volumes that Archimedes determined by his method of exhaustion and that we find today by integration. The volume of a sphere is a third of its surface area times its radius, for reasons given on page 664. Without making any argument, one can jump to this conclusion by the analogy outlined below. This may be easy for your students to grasp and remember because it is outrageous.

ll” Volume of the Sphere. We may view the circle as a triangle whose base is its circumference and having one vertex at its center, with its altitude being r, so

1

Area = 5 [Base x Altitude] = sl (2πr)r] = πr².

In the same spirit, we may think of a spherical ball as a pyramid whose base is its surface and whose apex lies at its center, then its volume is given by

1 Volume = 3 [BaseArea x Altitude] =;[Spheredrea xr}.

li” Area of the Sphere. Consider a sphere like our earth, with a given axis and equator. Archimedes knew that if you project the points of a sphere outward from its axis onto a cylinder tangent to the sphere at its equator, then this mapping preserves areas. In map making this is Lambert’s Equal Area Projection. If r is the radius of the sphere, the image of the sphere lies on a

cylinder of diameter 2r and height 2r. The area of this cylinder is 7 x 2r x 2r = 4nr?, and if this map projection preserves area, then the area of a sphere of radius r must be 4rr?, as given by Theorem 84 of Geometry.

_The argument that Lambert’s Projection is area-preserving involves looking at an narrow symmetric strip with a north-south width of w about a parallel of latitude. To fix the geometry so a diagram can be drawn, assume this parallel is at a north latitude of 8 on our sphere of radius r. This parallel is a circle of radius rcos@ about the axis and lying distance rsin@ north of the equatorial plane. For small w, its area on the sphere can be thought of as that of a rectangle of length é = 2nrcos@ and width w, so we may take its area to be Area = @ x w = 2mrcos6 x w = 2nrwcos 8.

Our strip projects to a strip around the cylinder tangent to the equator. This cylinder has radius r, and so the image is a strip of length @/ = 2zr. The north-south distance w lies in the surface of the sphere and makes an angle of @ with the vertical generators of the cylinder we are projecting to, so width w on the sphere projects to a width w’ = wcos@ on the cylinder. The area of the strip as it is projected onto the cylinder is then Area’ = ’ x w! = 2πr x wcosé = 2πrweos8, which is the same as the area before it was projected, and the Lambert Projection is area-preserving.

This is a calculus argument. Archimedes used his method of exhaustion, which is a geometric form of integration. See his On the Sphere and Cylinder or, better, Sherman Stein’s book Archimedes: What Did He Do Besides Shout Eureka?, cited earlier. When we worked out the area of our strip on the sphere, we approximated the curved area on the sphere, call it SphereStrip, by the area of a flat strip of the same length and width, 2nmrcos@ x w. This works out because

4 SphereStrip =

w−0 2mrcos@ xX w

As w — 0, the approximation becomes better and better and the ratio of the original and Lambert-projected areas is 1.

Carlos A. Furuti had a nice website covering map projections in 2003; search for map projections on the Web or at Eric Weisstein’s MathWorld. “

Set I. Archimedes’ Discoveries (7-10). A good source on this material is Stein’s book on Archimedes, cited earlier. Flatland (11-15). Abbott’s book is available in an inexpensive Dover edition as well as in an elaborate, annotated version by Ian Stewart. These exercises show how three-dimensional geometry can be worked out using what we know about two-dimensional geometry. Sound Wave (16-18). The intensity of the wave at radius r from the source varies as 1/r². The intensity integrated over a spherical shell gives the outward flow of sound power through the sphere of radius r about

the source. This flow will be independent of the radius r under steady-state conditions.

Volume and Density (19-23). This sequence helps students understand the difference between density and mass — commonly a confusing matter. Chocolate Packing (24-27). See Center-Pivot Irrigation, Chapter 14, Lesson 5, page 601 of Geometry and my notes on it.

Hollow Earth (32-37). Points P and P’ are interchanged by inversion with respect to the sphere S with center O and radius OA. Inversion in S is a conformal mapping that interchanges the outside of S with the inside of S. The origin is interchanged with the point at infinity. Conformal mappings preserve angles but may distort lines and distances. The symmetries of the Poincaré model of the Lobachevskian non-Euclidean plane are inversions in circles that meet bounding circle of this model at right angles. See the Lobachevsky Meets Escher sequence on page 716 of Geometry. Search the Web for inversive geometry for more. You will find Java applets generated by Cabri Geometry or Geometer’s Sketchpad™.

Melting Snowball (38-43). This is a related-rate problem. The assumption is that the volume is decreasing at a constant rate, which means the flow of calories into the snowball is constant over time until the snowball melts away. There is the sublimating mothball problem: the rate of sublimation (in grams per second) is assumed to be proportional to the surface area. If it loses half its mass in two months, how long before it evaporates completely?

Earth Under Water (44 and 45). The assumptions are clear and the fact that there is so much surface water on the earth is surprising. A calculator shows us how these two estimates compare. They agree to about 4 parts in 10,000. We treat the earth as if it were not rotating, so it has no equatorial bulge.

Set III. Raiders of the Lost Ark should give your student’s brains some exercise, and it is fun to recall this action movie.

Lesson 8. Similar Solids. Scaling was one the “two new sciences” Galileo wrote about in his dialogues. There are many sources on scaling, but a favorite for me is On Size and Life by Thomas McMahon and John Tyler Bonner, SCIENTIFIC AMERICAN Library, W. H. Freeman and Company, New York, 1985.

The bottom line for this lesson is that if S; ∼ So in three-dimensional space with ratio of similarity r, so that rS,; = So, and if the areas of these shapes are A, and Ag and their volumes are V; and V9, then,

In words, areas scale as the square and volumes scale as the cube, as found in Theorems 83 and 84.

▶ Scaling and Strength. The tensile strength of a cable scales in proportion to its cross-sectional area. The resistance to bending for a beam is proportional to the (second) moment of inertia of its cross-section, and hence, for similar beams, the resistance to bending scales as the fourth power of their linear sizes. See J. E. Gordon’s Structures; Or Why Things Don’t Fall Down, Pelican Books, 1978.

If a cable is made sufficiently long, it will break under its own weight — independent of its width. It is weight per unit area that determines the failure point in tensile stress, and that is the same for thick and thin cables of the same length hanging under their own weight. Likewise, a pile of bricks sufficiently high will crush the bricks at the bottom of the stack, regardless of the width of the pile. The thinner pile is more subject to toppling, however. Scaling arguments let us understand why there are no giant humans, geometrically similar to us but three times our size.

Twenty six of the exercises in this lesson deal with questions of what happens when you change the scale of a creature or device. This sort of thinking is very useful, but some caution is needed. If you take a bridge design and linearly scale it up to cross a wider span, we know that the weight of the bridge will go up as the cube of the linear scale, while the bearing area for this weight, the footing area, will only go up as the square of the linear scale. We conclude that geometrically similar designs will fail at some larger scale.

In the same way, you know that 0.0001r? will eventually become larger than 10, 000r? if r is made large enough, but you have to go a long way to see this effect. If the bridge was designed to carry large loads compared to its weight, then scaling it up will initially result in increased carrying capacity. It is a mistake to think that larger things of a similar shape must be weaker.

See the note on Scaling and Fractals at the end of the notes on this lesson. t

Set I. Similar or Not? (1-5) involves visual thinking and drawing three-dimensional shapes. Scaling Up (6- 11) involves understanding how strength of support and weight scale up from a model to a full-scale structure. Giant Windmill (12-15) continues this exploration of scaling, this time for renewable energy windmills. Gulliver in Lilliput (16-18) is a classic example of scaling. Note that Johnathan Swift knew his maths!

Set II. Doubling the Cube (29-36) covers basic scaling calculations. Plato’s Method (37-46) is somewhat more theoretical, with a review of some ideas from right triangle similarity. Rubber Bands (44-46) is a wonderful practical example, and utterly original. Height and Weight (51-54) is very practical. Heated Ring (55-56) is a classic. Ask your physics colleagues about this. Teddy Bear Trouble (57-60 ) is about how things of different dimensionality scale: Buttons are like points,

0-dimensional, and when you scale by a factor of r you need r° = 1 times as many as you had. For cloth, which is 2-dimensional, you need to multiply by a factor of r². In general the factor is r¢, where the thing you are concerned with is d-dimensional. See Scaling and Strength above and Scaling and Fractals below.

Set III. Biological scaling, and an example within the reach of every student. It would be natural for students to look up material about giant clams to see how reasonable their estimates were.

▶ Scaling and Fractals. Some things do not follow scaling laws with simple or expected exponents. The lungs of mammals of various sizes are similar in function and may look like geometric scale models of each other. But their development allows the surface area for the exchange of gasses to increase by a ratio greater than the r² that geometric scaling would suggest. The reason is that as the lung grows to a scale appropriate to a larger mammal, it develops more branches in its air passages and more tiny breathing chambers (alveoli) to fill out the larger volume of a larger lung. See, for example, Plates 164 and 165, of Benoit Mandelbrot’s The Fractal Geometry of Nature, W. H. Freeman and Company, New York, 1982 for a two-dimensional analogy of the bronchial tree. See also Mandelbrot’s discussion of lungs on pages 157-160.

For lungs of mammals, the area available for exchange of gasses between blood and air scales more as r° than as r². This is contrary to geometric scaling laws. But fractal geometry comes to the rescue. Benoit Mandelbrot showed that such anomalous scaling was a sign that we should think of the exchange surface of the lung as a surface whose dimension is closer to 3 than to 2. Here he used the idea of fractal dimension. Initially, he used Hau8dorf-Besicovitch (HB) dimension, a way of defining the dimension of any set lying in a space that allows you to find distances between any pair of its points (a metric space). The tricky HB dimension is often replaced by the easier to understand and compute selfsimilarity dimension, the most commonly encountered fractal dimension.

The boundaries of fractal shapes typically have peculiar dimensions. The Koch snowflake is a familiar example. Starting with an equilateral triangle as the Oth stage, you erect outward equilateral triangles on the middle third of each of its faces to get the 1st stage, and continue in this manner to get the 2nd stage, and so on, to get the snowflake as the limit. See Mandelbrot or other sources. The boundary curve of this snowflake has (similarity) dimension ees ∼ 1.2618 versus the dimension of 1 we would expect for a smooth curve.

Similarity dimension is defined for shapes that have a strict sort of fractal symmetry, namely they can be broken into a finite number of pieces so that when these pieces are reduced in scale by a factor of r and you take N sets of the reduced pieces, then you can reassemble

these pieces to get the original shape. In short, the thing is the sum of N copies of itself reduced in scale by r. An example is the square which is the sum of N = 4 copies of itself at scale r = t A shape allowing such a decomposition has

Similarity Dimension = Oilog Nig

log(1/r)

A quick check will show that this gives the right dimension for familiar shapes such as an interval, a square, and a cube.

The limit for the Sierpinski polygon found on pages 144 and 145 of Geometry is a continuous image of a circle. Imagine the circumcircle of the first stage being deformed to match this first polygon. At each stage the deformation becomes more complicated. In the limit, you get a map of the circle onto a curve that covers every point of a square. You can see the outline of this square in the fourth stage shown in Geometry. This limiting “curve” has dimension 2, versus 1, which is the dimension of any smooth image of the circle.

Fractals are intriguing and their shapes are beautiful. They are practical as well, because they give us a geometric way of understanding things that were mysterious before the fractal approach was developed.

Kleiber’s Law is an empirical observation that the metabolic capacity of an organism scales as m?/4, where m is the organism’s mass. Many efforts have been made to derive this law from first principles. The latest direction is work by Geoffrey West, James Brown, and Brian Enquist in 1997 that sees this law as arising from the fractal dimension of the capillary beds in the body. See the McMahon-Bonner book cited above for Kleiber’s Law or look on the Web, where you will find current material on the fractal aspects of this law. In April 2003, New Scientist ran an article by Phillip Hunter, “The Power of Power Laws” on allometry, scaling, and their fractal interpretations. See also “Elephants, Mice, and Red Herrings” by Gary P. Burness, Science, 17 May 2002. This pot will be bubbling for years.

Fractals are extensively used in geosciences, in meteorology, and in practical matters such as designing better battery plates. My introduction to these shapes came more than a decade before Mandelbrot showed that they were useful. It was in a real variables class taught by A. S. Besicovitch (the B of HB dimension). Besicovitch gave us the definition of fractional dimension and showed us how it applied to various shapes. He also cautioned us that this was merely a mathematical curiosity and had no physical meaning or practical use. You can know all about something without being able to see what it might be good for. tl

Lesson 9. The Regular Polyhedra. The tetrahedron, the cube, the octahedron, the dodecahedron, and the icosahedron: these are the five convex regular

polyhedra. The geometry of the first three can be understood by starting with the cube, whose properties are easy to grasp. Notice that if you take the two skew diagonals on opposite faces of a cube, these are opposite skew edges of a tetrahedron, all of whose edges are diagonals of the same cube. The octahedron is related to the cube as shown in Octahedron and Cube (37-42) in Set II. as shown in Octahedron and Cube (37-42) in Set IT.

Euclid relates the dodecahedron to the cube by the Dodecahedron Construction (43-46), in Set II. (See Proposition 17, Book XIII, especially the Porism and its figure in the Heath edition.) It is a surprise that the cube can be used in constructing a dodecahedron because their symmetries seem so different. Once we have the dodecahedron, we can get the icosahedron by taking the centers of the faces of the dodecahedron as the vertices of our icosahedron. The dodecahedron and icosahedron are duals. If you take the centers of the faces of one, these points will serve as the vertices of the other. See the exercise sequence Octahedron and Cube at the start of Set IT.

The regular polyhedra can be constructed starting from a cube, which has an insphere and a circumsphere. With some care we can see that all the vertices of the constructed regular polyhedra are equidistant from the center of these spheres. With a bit more care, you can figure out just what symmetries each regular polyhedron has. All of these will be rotations about axes going through the center O of the circumsphere, or reflections in planes passing through that center, or compositions of these (e.g., the mapping that sends every point to its antipodal point with respect to O).

We defined regularity for these convex polyhedra by the local conditions that all the faces should be congruent regular polygons and that all the vertices should be surrounded by an equal number of these polygons. We find that this makes these regular polyhedra so symmetric that there are isometries that send the polyhedron onto itself and interchange any given pair of faces or any given pair of vertices. This is global regularity.

Set I. Numbers and Names (1-5) the essential vocabulary. Two Tetrahedra (10-11). Figuring out that the ratio of similarity for these two tetrahedra is be 5 requires looking at the configuration and seeing how to use what we know about plane geometry in several intersecting planes to get the result. You can start with a plane containing an edge FE of the larger tetrahedron and one e of the smaller tetrahedron, with E ∥ e. This gives us a triangle with e as a segment to which the Side-Splitter Theorem applies. Because the centroids of the faces lie 2rds of the way from vertex to the base, you can conclude that e is ard the length of E. Da Vinci Polyhedron (12-19) involves visual thinking, counting, calculating, and seeing how these numbers give the number of edges and vertices of the dodecahedron. Escher Box (20-27) does the same thing for the icosahedron. Note that the

earlier set has a hint after Exercise 17 that is not found in the latter set. Assign these together?

Euclid’s Claim (30-36) is the argument that shows no more than five regular polyhedra can be made. The problem is that we cannot even get one vertex to work out with an angle sum less than 360° except for the arrangements that we know give us the tetrahedron, cube, octahedron, dodecahedron, and icosahedron. If the angle sum exceeds 360° then the polyhedron cannot be convex at that vertex. This is Proposition 21, Book XI of Euclid. Euclid proves this by cutting off the vertex from the interior of the convex polyhedron. This gives us a pyramid with the vertex as its apex. Projecting the sides of the pyramid perpendicularly onto its base projects each angle at the vertex into an angle larger than it in the plane of the base, and the sum of these projected angles is 360°, so the sum of the angles at the apex must be less than 360°. The reason that the angles at the vertex are smaller than the angles projected perpendicularly into the base is because their corresponding sides are longer, as can be seen by using the Pythagorean Theorem.

Set II. Octahedron and Cube (37-42) another pair of dual polyhedra. The geometry here is easier to work out than in the case of the tetrahedron and its 1/3rd scale dual tetrahedron. Dodecahedron Construction (43-46). Here we work from the diagram and the assumption that this construction works to get the ratio of the diagonal of a regular pentagon to its side. Euclid worked the other way around. He knew that the ratio of the diagonal of a regular pentagon to its side was the Golden Number, ¢. We know this as well. (For us, ¢ is the positive root of 2? —x−1 = 0.) See Set III, Golden Heather, for Lesson 2 of Chapter 14, page 584, and my notes.

Using this he showed that the arrangement of roofed structures pictured here fits together exactly as shown. Here we take the side-lengths AB, and so on, to be 1, and the side-lengths of the cube to be ¢. Our task is to show that the pentagons of the figure are really planar. One way to check this is to compute the slopes of the planes of the two portions of these faces that join on an edge. This computation shows that each of the faces made up of an isosceles trapezoid and an isosceles triangle in this figure lies in one plane. For me this came down to seeing that 1/¢ = @— 1. Algebra proved that this rather surprising construction works. To verify this, you must work algebraically, without approximations.

Round About (47-60). The fact that regular polyhedra have inspheres and circumspheres is essential to our understanding of their symmetry. The results found in this series are counterintuitive, which is an attraction, but it takes many steps to get to the punch line.

Set III. Tetrahedron Puzzle is just baffling enough, and it involves a model, which is good.

▶ Models. To understand geometry, you must put your hands to the subject, not just think about it. Constructions and models are essential. How far to go? There are many good sources on the Web, and in 2003 George W. Hart’s Virtual Polyhedra: The Encyclopedia of Polyhedra was a good place to start. Hart and Henry Picciotto have a 2001 Key Curriculum black-line masters book, Zome Geometry, of projects for people using the Zome geometric model-building system.

Older technology is featured in Polyhedral Models by Magnus J. Wenninger, Cambridge University Press, 1971, and in Mathematical Models, Revised edition, by H. Martyn Cundy and A. P. Rollett, Oxford University Press, 1960. Cundy and Rollett cover all sorts of models, including mechanical ones.

Sometimes looking at pictures and thinking work — if the pictures and words are good enough. They are in Shapes, Space, and Symmetry by Alan Holden (since 1991 a Dover Publications paperback).

Summary and Review. The areas of right circular cones and cylinders were covered in exercises. See Cylinder and Cone Areas, page 660 in Geometry. These might be added to the list of formulas, although these areas are easy to work out when the occasion arises. It is important to distinguish between the lateral and total areas for these shapes. Confusion on this could give trouble on facing standardized tests.

If you have given more time to coordinate methods, then you might mention the equation of a sphere S of radius r about point O = (20, yo, zo) namely

Sphere Equation (x — Zo)” +(y- yo)? +(z- z)° =r.

Set I. Flipping Cubes (1-6) and Seeing Things (7 and 8) are about visual thinking and three-dimensional representations. Box Pattern (11 and 12) is about a polyhedral net. Octagonal Building (13-16) is about volumes. Screen Illumination (17-19) touches on geometry, scaling, and the intuition behind the inverse square law for intensity of light. Stars (20 — 24) is about scaling applied to astronomy. Divided Cube (25-31) is a nice study of the regular tetrahedron as inscribed in a cube.

Set II. According to Archimedes (32-42) is a nice review of volume and area results for the sphere, right circular cone, and right circular cylinder. Twisted Ring (43-45) and SAT Problem (46-48) involve visual thinking and problem-solving. Wine barrel (49-56) involves thinking about why various volume formulas work and how they relate to other formulas. Euler characteristic (57-61) introduces Schlegel diagrams.

▶ Solid Geometry as a Deductive System. In Books XI-XIII of his Elements, Euclid developed solid geometry up to the construction of the regular polyhedra found in Lesson 9 of this chapter. At the start, his reach exceeds his grasp, as Heath notes in his edition of Euclid. Euclid gives some shaky arguments attempting to prove our Postulates 11 and 12. From that point on

he is solid in his arguments. I am struck by how neatly he got all of this down. The following is largely a distillation of his work. We have advantages over Euclid because once we have established a few key facts, we can use coordinates and algebraic insights and tools to. simplify things.

The first step is to understand how planes can be determined. I set these results out as a series of exercises, with the solutions in brief following the statements. These exercises could be used as honors assignments.

Determining Planes.

Three noncollinear points determine a unique plane P. This is Postulate 2. Show that each of the following configurations determines a unique plane P containing it.

1. A line m and a point O not onm. Points A and B on line m are not collinear with O. Why? The plane P determined by these points contains two points of m, and hence, by Postulate 11, all of m, and P contains O. Using Postulate 12, show that any plane 2 that contains O and m must be identical with P. So P is unique.

Proof. O does not lie on m by hypothesis, so O, A, and B are noncollinear. Remark, P contains A and B, and hence it contains line AB, which is identical with m. Reasons: two points determine a line and Postulate se

If plane Q were not identical to P, then P intersects Q in a line, by Postulate 12. This contradicts the fact that both P and Q contain m and the point O, not on m. Hence P² and Q must be identical, and the plane P found by using A, B, and O is the unique plane determined by point O and line m which misses O.

2. Intersecting lines m and n. Take A on m and B on n, different from the intersection, O. Show that P, determined by O, A, and B, is the unique plane containing lines m and n.

Proof. This is Euclid X1.2. Postulate 12 is Euclid XI.3 First, P contains the line OA = m, because the two points O and A on m determine m. Second, because O and A are points in P, the line OA lies in P, by Postulate 11. The same reasoning shows that P contains line OB = n. If Q were another plane containing lines m and n, then P would intersect Q in a line (Postulate 12), which contradicts the fact that both P and Q are assumed to contain distinct lines m and n. Hence, no such second plane, Q, exists, and the plane P we have found is the unique plane determined by intersecting lines m and n.

3. Parallel lines m and n. To be parallel, these lines must lie in a common plane, which we will call P. Show no other plane, 2, can contain m and n.

Proof. This is Euclid XI1.7. If lines m and n also lie in the distinct plane Q, then, by Postulate 12, P intersects Q in a line, contrary to the fact that planes P and Q

both contain lines m and n. So no such distinct plane Q containing m and n can exist, and P is the only plane containing both m and n.

Determining a Perpendicular Plane.

The plane perpendicular to a line ℓ at O is determined by any pair of lines perpendicular to / at O. Proof. Take points A and A’ equidistant from O on ℓ, as in your diagram. Let the blue lines m and n be perpendicular to line ℓ at O. These lines will be perpendicular bisectors of segment AA’ in two different planes. The blue lines determine a plane P passing through O. To show that P is perpendicular to 2, we must show that any line OX, where X is in P, is perpendicular to ℓ. We draw a green line in the plane P, passing through X, and crossing m and n at points we name W and Y.

This ts Euclid XI.4. Your figure should show @ looking vertical and blue lines m and n shown stretching to the right with X on a green line running front to back at an angle but seeming in the plane P. The green line runs through X and crosses n at a point labeled Y and n at a point labeled W.

4. AW = A’W and AY = A’Y. Why?

Proof. Because n and m are perpendicular bisectors of AA’, so points W on n and Y on m are equidistant from A and A’.

5. △AWY & AA‘WY, so ∠AWY = ∠A’WY. Why?

Proof. Since WY = WY, this congruence follows from 4 by SSS. These angles are equal by CPCTE.

6. △AWX & AA’/WX, hence, AX = A’X. Why?

Proof. The triangles are congruent by SAS. The segments are then equal by CPCTE.

7. In the plane determined by points A, A’, and X, OX | AA’. Why? How does this show that the plane P, that is determined by lines m and n, both perpendicular to 2, is itself perpendicular to @?

Proof. We showed that a point X in the plane P is equidistant from A and A’, hence it lies on the perpendicular bisector of AA’ in the plane determined by ¢ and X, so OX | @. But X was any point in the plane Pf determined by m and n, so any line in the plane P passing through O must be perpendicular to ¢. By definition, this means that plane P is perpendicular to line ie

Perpendicular Lines are Coplanar.

If three lines are all perpendicular to a line ℓ at point O, the three lines are coplanar.

Let lines k, m, and n be perpendicular to line ℓ at point O. We show that line k lies in the plane P determined by mand n. Let 2 be the plane determined by intersecting lines k and @. P intersects Q at O. Hence, by Postulate 12, the intersection of P and Q is a line, which we call j-

8. Because 7 lies in P, 7 1 2 at O. Why?

Proof. This is Euclid XI.5. Plane P is determined by two lines m and n perpendicular to @ at O. Hence P 1 ℓ. Because j lies in P and intersects @ at O, 7 1 & at ©}

9. Line 7 is the intersection of planes P and Q, so it lies in plane Q with line k. Because both j and k are perpendicular to @ at O, we can conclude j = k. Why?

Proof. Because in the plane Q there is a unique perpendicular to @ at O.

10. So line k, m, and n lie in P. Why?

Proof. Because k = j, and 7 is the intersection of Q and P, and P is the plane determined by m and n. So k(= 7) lies in the plane P of m and n.

Planes Perpendicular to a Line.

Two planes perpendicular to the same line are parallel to each other.

11. Let plane P be perpendicular to line ℓ at point O, and plane P’ be perpendicular to @ at O’. If P and P’ intersect at some point X, lines XO and XO’ would be distinct perpendiculars from X to line OO’. Is this possible? Why does this show that P and P’ must be parallel?

Proof. This is Euclid XI.14. No. Three points determine a plane, so O, O’, and X would lie in some plane, and in any plane there is just one line perpendicular to a given line (OO’) through a given point (X). Hence no such point of intersection X of planes P and P’ can exist, so these planes must be parallel, by the definition of parallel planes.

Perpendicular Lines.

Lines perpendicular to the same plane are parallel. Let @ and @’ be lines perpendicular to plane P, intersecting P at points O and O’, respectively. We will use the distance OO’ as our unit, and take point Y on line ℓ so that OY = 1 unit. In plane P, take point X so that XO’ 1 OO! and O’X = 1 unit. Then by the Pythagorean Theorem, OX = V² units = YO’, as you will see when you draw the figure. Because ¢ 1 P and OX is in P, OX 1 OY, and by the Pythagorean Theorem, we find that XY = V3 units.

This 1s Euclid XI.6. This needs a figure. Lines @ and t’ are perpendicular to a seemingly horizontal plane P, and both should look vertical. The points O and O/ are horizontal across from each other on the page. The configuration is such that Y might be the upper-backleft corner of a cube, O the lower-back-left corner of the same cube, O’ the lower-back-right corner of the cube, and X the lower-front-right corner of the cube. The lengths mentioned in the description should be on the diagram and the right angles marked. These would include the two angles made by segment YO where it

hits segments OX and OO’, the right angle between @’ and OO’, in the plane of @ and @’, and the right angle in the plane P between OO’ and OX. Angle XO’Y is shown to be a right angle in 12, below, but this is not given in the setup. Color could help make your figure easier to read.

12. Why must ∠YO’X be a right angle?

Proof. Because (YO’)’ + (O’/X)? = 2+1=3 = (YX), so, by the converse of the Pythagorean Theorem AYO’X is a right triangle, and ∠YO’X is a right angle.

13. Lines @’, O’O, and O’Y must in a single plane. Why? We call this plane 2.

Proof. Because these three lines are perpendicular to line O’X at O’. If three lines are all perpendicular to a line 2 at point O, the three lines are coplanar.

14. Why must @ and ¢’ be parallel in Q?

Proof. Because they are perpendicular to line OO’ in Q. Lines in the same plane perpendicular to the same line are parallel.

Perpendicular Planes & Parallel Lines.

A plane perpendicular to one of two parallel lines is perpendicular to both of them.

This is proven using a figure with @ and @’ as the parallel lines, with P the plane perpendicular to @ at point O, and with X, O, O’, and Y arranged as for the previous result. The proof uses the Pythagorean Theorem and its converse.

This is Euclid XI.8. The next result proves the transitivity of parallelism for lines in space, something not entirely obvious from the definition.

Parallel Lines in Space.

Given lines ¢ ∥ m, m ∥ n, n ||… there is a plane P perpendicular to all of these lines. Any two lines in such a collection must be parallel.

Let P be a plane perpendicular to 2. Because P | ℓ, we know P 1 m, and hence then P 1 n, and so on, by the previous result. So P is perpendicular to all these lines. Then any two lines in this family are parallel, because they are perpendicular to the same plane.

Perpendicular Bisection in Space. The plane perpendicular to segment AA’ at its midpoint consists of the points equidistant from the A and A’.

Extensions of Plane Geometry to Space.

Like perpendicular bisection, other notions of plane geometry extend to space. Reflection in a plane can be defined analogously to reflection in a line. Reflections in planes are isometries, as are translations (which move all points an equal distance in the same direction or are given by reflections in two parallel planes), rotations about an axis, and so on. The Cartesian coordinates extend to space, using three axes, each at right angles to the other two.

To use the methods of plane geometry in three dimensions, we look at relations in the various planes determined by the points and lines of our problem in space. Within each of these planes we use results of plane geometry, then we put these results together to solve our three-dimensional problems.

Euclid XI.11 to XI.13 concern the construction and uniqueness of perpendiculars to planes in space. These results can be followed easily in Heath’s edition of Euclid or as outlined under subheads below.

Euclid’s XI.10 and XI.15 tell us that if intersecting pairs of lines m and n, and m’ and n’ are such that the corresponding lines are parallel, then the planes these intersecting pairs determine are parallel and the angles these make are equal. We have different definitions of prism, so these results are not as important for us as they were for Euclid. However, we need to know that the figures traced out in the two base planes of a prism by lines parallel to a fixed line are congruent. This fact follows from arguments based on translations in threedimensional space, and the details can be established by reasoning outlined here.

The basic insight is that if two planes P and P’ are parallel, any plane Q containing line ℓ that intersects P in a line m must intersect P’ in a line m’ that is parallel to m and hence makes the same angles with @ in Q as m does. By Postulate 12 we know that if two planes intersect, their intersection gives a line.

To see why the above basic insight must be true, we begin by assuming that Q is determined by ¢ and some point O’ in P not on @. Q intersects both P and P’, so it intersects these planes in our m and m’. Because m and m/’ lie in parallel planes P and P’, these lines do not intersect, and because both lines also lie in the plane Q, they must be parallel. Since @ is transversal to m, it is also transversal to m’ in plane Q, making equal angles with both lines, as was to be shown.

This argument shows that @ intersects P’. Hence, any plane containing @ intersects P’ in a line, and hence every such plane is determined by ¢ and some point @! in P’, and our argument covers all planes that contain ℓ. This result is Euclid X1.16.

If 2 1 P, then the above shows that ¢ 1 Q, because plane will 2 contains many lines perpendicular to where they intersect, and thus @ 1 Q.

Constructing Perpendiculars.

From O not on plane P, drop a perpendicular to some line m in P, and let point O’ be the foot of this perpendicular. If OO’ is perpendicular to P, we have our perpendicular from O to P. Otherwise, find the perpendicular n to m at O’ in the plane determined by O and m. Now, in the plane of O and n drop the perpendicular to n from O’ to locate O”. I say, m 1 OO’and m 1 O’O”, so m 1 Q, which we use to denote the plane determined by OO/O”. Now m’ ∥ m, so m’/ 1 Q, and hence m’ 1 OO”. But n 1 O’O” as well. So O’O”

▶ P, which is the plane determined by m’ and n. Thus we have constructed the perpendicular from O to P, as required. This is Euclid XI.11.

Constructing a plane parallel to a given plane through a given point. The plane parallel to P through a point O not in P can be found by dropping the perpendicular from O to P (with its foot at O’) and considering the unique plane determined by any two lines perpendicular to OO’ at O’.

Coordinates in Space.

From the above results we see that given any plane P in space and a point O in P, we may establish a Cartesian coordinate system with origin at O in P and any pair of perpendicular axes through O in P as our z- and yaxes. Taking our z- axis to be a line perpendicular to P at O, we know that every point in space lies on a plane parallel to P and hence perpendicular to our z-axis.

Using a ruler with the same units as on our z- and yaxes along our z-axis and with its 0 at the point O, we define the z-coordinate of a point P in space as the signed distance from O at which the plane parallel to plane P through point P cuts the z-axis. The z- and y- coordinates of P are defined to be the coordinates of the foot of the perpendicular (whose foot we will denote by Pz,) from point P to plane P.

By the Pythagorean Theorem applied in the plane P,

(OBE =a +y²,

where zx and y are the z and y coordinates of Pz, in P. By the Pythagorean Theorem applied in the plane of the z-axis and point Pz,, we see that

(OP)? = (OPzy)? + 27,

where z is the z-coordinate. So the distance from O to any P in space can be found by using:

(OP)? = 2? + y² +27.

In a like way, the distance between any two points P = (Tet, 2) and = (2.5.4, 2.) 18 given by:

2

PP! = /(4—-2’)? + (y-y’)? + (z-2’)’,

where these Cartesian coordinates are given in the usual way. The equation of a sphere was given at the discussion of the notes on the Summary and Review for Chapter 15. It is a consequence of the distance formula established above.

Analytic Geometry in Space. This is an extension of analytic geometry in the plane. The coordinates of the points lying in a given plane Q all satisfy:

Az+By+Cz=D

for some constants A, B, C, and D. The vector point of view gives the nice interpretation that the vector from the origin O to the point (A, B, C) is perpendicular to the plane Q. Furthermore, if A² + B² + C² = 1, then D is the signed distance from O to the plane Q.

There are many results of this type. These are usually covered as part of calculus or linear algebra. The links with geometry rest on the introduction of coordinates as outlined above.