We started with a discussion of the first day handout, then did introductions, before continuing with our first activity.
Our view tube activity involved the concept of similarity: two triangles are similar if they have the same shape. We also discussed what we mean by "equilateral triangle" (a triangle with three congruent sides, equal laterals)) and what we mean by "congruent" (we decided two figures are congruent if they are the same size and shape; i.e., if one can slide one exactly on top of the other, possibly after turning it over or rotating it).
Here is Euclid's proof of his Proposition 1:
Proposition 1: Given a line segment
Strategy: Draw two circles, one centered at A with radius BA (see figure below), and one centered at B with radius AB. The circles meet at a point C above
At lest some of you did the construction in GeoGebra. Here is a GeoGebra file for the construction.
Square Problem: Given a line segment
Strategy: Different groups used different strategies to do this. One strategy went like this. (Given any line L and any point P, let's assume we know how to construct a line L' containing P and perpendicular to L.) Construct a circle centered at A through B, and a circle centered at B through A, then construct perpendiculars to
We can see that ABCD looks like a square and we discussed how to prove it.
To do this, we talked about what we should use as our definition
for a square. One definition we considered was:
A square is a quadrilateral with 4 congruent sides and a right angle.
What we settled on was: A square is a quadrilateral with 4 congruent sides and 4 right angles.
In summary, this construction
starts by constructing a line segment
Here are four other strategies that can be used to construct the square.
(2) construct a line segment
(3) construct a line
(4) construct a line segment
(5) construct a point C on the perpendicular bisector of
Problem 1: Varignon's Theorem: Prove that the midpoints of the sides of any quadrilateral are the vertices of a parallelogram. Hint 1: Consider the diagonals of the quadrilateral and use the a triangle similarity property (for example, two triangles are similar if the ratios of the lengths of two sides are the same, as long as the included angle is also the same).
Sketch of key ideas: People used at least two different approaches. Given a quadrilateral ABCD, one approach was to assign coordinates to the vertices, say A = (x1,y1), B=(x2,y2), C=(x3,y3) and D=(x4,y4). Then the midpoint of side AB is E=((x1+x2)/2,(y1+y2)/2), the midpoint of side BC is F=((x2+x3)/2,(y2+y3)/2), the midpoint of side CD is G=((x3+x4)/2,(y3+y4)/2), and the midpoint of side DA is H=((x4+x1)/2,(y4+y1)/2). The slope of EF is (((y1+y2)/2)-((y2+y3)/2))/(((x1+x2)/2)-((x2+x3)/2)) as long as ((x1+x2)/2)-((x2+x3)/2) is not 0. The slope of GH is (((y3+y4)/2)-((y4+y1)/2))/(((x3+x4)/2)-((x4+x1)/2)) as long as ((x3+x4)/2)-((x4+x1)/2) is not 0. But ((x1+x2)/2)-((x2+x3)/2) simplifies to ((x1-x3)/2) while ((x3+x4)/2)-((x4+x1)/2) simplifies to ((x3-x1)/2). Thus if either is 0, then both are, so if either EF or GH is vertical so is the other so these two sides are parallel in that case. If, however, EF is not vertical (i.e., if ((x1+x2)/2)-((x2+x3)/2) is not 0), then GH is not vertical (since ((x3+x4)/2)-((x4+x1)/2) is not 0), and the slope of EF, (((y1+y2)/2)-((y2+y3)/2))/(((x1+x2)/2)-((x2+x3)/2)), simplifies to ((y1-y3)/2)/((x1-x3)/2), while the slope of GH, (((y3+y4)/2)-((y4+y1)/2))/(((x3+x4)/2)-((x4+x1)/2)), simplifies to the same value, ((y3-y1)/2)/((x3-x1)/2), so EF and GH have the same slope and hence are parallel. The argument to show that FG and HE are parallel works the same way.
Here is the other main argument people used.
In the figure below, ABCD is our given quadrilateral (here is a GeoGebra file for it),
and the midpoints of the sides are E, F, G and H.
Cut the quadrilateral by diagonal
If we double the size of triangle EBF, we get a triangle congruent to
ABC by SAS, thus EBF is similar to ABC. Thus angles BEF and BAC are congruent, so
Problem 2: Look at all triangles two of whose vertices are four inches apart and whose perimeter is 12 inches. You can do this in the following way. Put pins in a piece of cardboard 4 inches apart and make a foot long loop of string around the pins. Use a pencil to pull the loop taut, and keeping it taut, trace out a curve. This curve shows the possible positions for the third vertex of the triangle.
What curve did you get? Can you find a reference on the web to justify your guess? Based on this, what is the largest possible area of a triangle with one side of length 4 and perimeter 12? What is the shape of this triangle? Explain your answer.
Sketch of key ideas: Here's a GeoGebra file for this problem. Many people already knew or suspected the curve was an ellipse. The base of the triangles is always 4, so the largest area is for the triangles of largest height, which occurs when the third vertex is at the top or bottom of the ellipse. By symmetry of the figure when the third vertex is at the very top or bottom of the ellipse you see that the triangle has bilateral symmetry. But having two equal sides with a perimeter of 12 and a base of 4 means that all three sides have length 4, so the triangle is equilateral. So the maximumn height is the altitude of an equilateral triangle of side length 4. Now we just have to figure out what this altitude is to determine the maximum possible area.
Problem 3: Consider a circle X with center C:
Let D, E and F be points on X. Let A be the angle through D and F with vertex at E (i.e., angle DEF). Let m(A) be its angle measure. Assume that m(A) is at most 180 degrees. Let B be the angle DCF. Prove that 2m(A)=m(B). (What this shows is, given a point on a circle and a circular arc, then angle with vertex at the point and subtended by the arc is half the angle of the arc.)
Conclude that m(A) = 90 degrees if the line segment from E to F is a diameter of X. Explain how this is related to the 3 squares problem.
Sketch of key ideas: Triangle CED is isosceles (two of the sides
are radii of the same circle), so angles CED and CDE are congruent.
Likewise for angles CEF and CFE. Also, angles FCD, ECF and ECD together comprise a full 360 degrees around the point C. Now:
m(CEF) + m(ECF) + m(EFC) = 180 and
m(CED) + m(ECD) + m(CDE) = 180, so
2m(DEF) + m(ECF) + m(ECD) = 360 = m(ECF) + m(ECD) + m(DCF), hence
2m(DEF) = m(DCF). (If F and D are on the same side of the diagonal through E and C, this same argument works if you interpret some of the angles as being negative. It is probably clearer just to handle that case with a separate argument where angles all have non-negative measure, in which case some of the angle sums in the original argument become differences.)
We did not get to the 3-squares problem yesterday, so our next activity was to do that.
In the 3 squares problem, D in the figure above
corresponds to the right upper corner of the left
square, F corresponds to the left upper corner of the right square,
Suppose there are n lines in the plane.
(a) What's the maximum number of different points of intersections there can be? What's the minimum number of different points of intersections there can be? First try answering this for small values of n and see if you can get a formula for these numbers in terms of n.
Answer: The minimum number is 0, which you get if all n lines are parallel to each other. The maximum number occurs if every pair of lines intersect, and no three or more lines meet at the same point. Then there is a 1 to 1 correspondence between points of intersection and pairs of lines. But the number of pairs of lines is n choose 2, or n(n-1)/2. So the maximum number of points is n(n-1)/2.
(b) What has to be true about the lines to get the maximum number? What has to be true about the lines to get the minimum number?
Answer: To get the maximum number you can't have any parallel pairs of lines (since then you don't get a point of intersection from those two lines), and you can't have 3 or more lines meet at a single point (since then you get only one point of intersection, instead of at least 3 or more points if the lines didn't all meet at a single point). To get the minimum number none of the lines can meet, so they must all be parallel.
(c) Suppose we start with lines defined by the sides of a regular n-gon (so each side of the n-gon is a line segment; extending the segment to a full line gives us our n lines). How many points of intersection do you get in terms of n? Explain why you get a different formula when n is even versus when n is odd.
Answer: When n is odd, some experimentation shows that none of the lines are parallel and at most two lines intersect at any point, so the number of points of intersection is the maximum, n(n-1)/2. If n is even, there are n/2 pairs of opposite sides, and these are parallel, so the number of points is n(n-1)/2 - n/2.
(d) We'd like to define a new kind of geometry so that we get the same formula
whether n is odd or even. How might this be done?
Look at these jpgs:
More railroad tracks
How are these related to Raphael's famous Vatican fresco, The school of Athens? (Which figure represents Euclid?) How do these jpg's suggest an approach? Here are some more:
Answer: The idea here is that parallel lines to an observer seem to meet in the far distance, at a point which artists call the vanishing point. Two sets of parallel lines which run in different directions have different vanishing points. So in fact there is an entire line of points at infinity. (Euclid is represented in Raphael's fresco by the figure on the right, bending over to make a circle using a compas.)
(e) By adding "vanishing points" to the plane in a way as suggested by these jpgs, we get a geometry called "projective geometry" and instead of the Euclidean plane we get the projective plane, which contains the ordinary Euclidean plane. What can one say about parallel lines in this new larger geometry? What do we mean when we say that two lines in the plane are parallel? I.e., what is the definition of lines being parallel? Does this definition work for lines in space? What happens in the projective plane
Answer: The definition of two lines being parallel is just that they do not intersect. In the projective plane there are no parallel lines; every pair of lines in the euclidean plane either meet in the usual way, or they are parallel and meet at infinity. In ordinary 3 dimensional space, two lines can either intersect (in this case they are coplanar), or they are coplanar but do not intersect (i.e., they are parallel), or they are not coplanar (and hence do not intersect). In this last case they are not regarded as being parallel but instead are called skew. As another note, there is only one vanishing point at infinity for each line, even though a line runs off to infinity in two directions (forward and backward). If you want to have each line have two pints at infinity you get a different geometry, basically the geometry of a hemisphere, hence a subset of spherical geometry, which we started to look at by the end of today's class.
(f) Not every triple of lines in the ordinary Euclidean plane determine sides of a triangle; why not? What about in the projective plane; what kinds of triangles can we get in this new geometry that we couldn't get before? What kind we say about the sum of the degree measures of the angles of these new triangles?
Answer: Three lines in the ordinary plane can either meet at a single point, or they can form a triangle, or two of them can be parallel with the third crossing them both, or all three can be parallel. In the projective plane, three lines can meet in a single point, or they can form a triangle in the usual way (as before), or two of them can be parallel and the third can cross them both (in this case we get a triangle in the projective plane with the third vertex being the point at infinity). Note that if all three lines are parallel, then they meet at the same point at infinity, so this is just a case of three lines through a single point. When two sides of a triangle are parallel in the projective plane, the angle between those sides can be regarded as being 0, so the sum of the angles of any triangle in the projective plane is 180 degrees, just as it is for triangles in the ordinary Euclidean plane.
Marta started us off by presenting Problem 1. She started with a quadrilateral with three right angles and three congruent sides, and the goal was to prove it was a square. After Marta drew in the diagonal between the two opposite right angles. This is isosceles so a 45-90-45 right triangle, and then Allison D. pointed out that we can then apply SAS to the two triangles that the diagonal cuts the quadrilateral into to conclude that the fourth angle is right and the fourth side is also congruent, so we have a square. Emily presented a different case. Her construction gave a rhombus with a right angle. Now put in the diagonal between two of the unknown angles and apply SSS to show that the angle opposite the right angle is also right. giving two 45-90-45 right triangles sharing the diagonal as their hypotenuse. This makes the two unknown angles sums of two 45 degree angles, so they both are also right, so we have a square. Rachel presented Problem 2. The basic idea is that the formula for the circumference C of a circle of radius r is C=2*pi*r, so the ratio C/r has a constant value of 2pi, regardless of the values of C and r. This fact leads to the fact if you increase r by x inches, then C will increase by 2*x*pi inches, regardless of how big r is to start with. Tiffany presented Problem 3. The point here turns out to be that for circles on spheres of radius R the ratio of circumference C to radius r is not constant; in fact the relationship is C = 2*pi*R*sin(r/R). So here is an aspect of geometry that is different on spheres than it is on the Euclidean plane.
Jessica P. presented part (a) of Problem 4. By unfolding the box into a net you can convert paths of the spider on, the 3-dimensional box into paths on the plane, and we know the shortest path on the plane is a straight line. The issue is that there is more than one way to flatten the box; each way has a shortest straight line path, but some unfoldings give shorter paths than others. (One question is how do we know for sure that one of these straight line paths coming from an unfolding is the shortest possible path. The idea is that the shortest possible path must stay on at most two sides; you can check what happens if the two sides are the right end and the front, and for the right end and the top. These are the only two possibilities, and the second one is the shorter. The reason that one of these two is the best possible, is that in order to get to the fly, the spider must at some point leave the right end of the box. Thus the spider must go to a point P on the edge of the right end of the box, and from there it must go to the fly. The shortest way to get to P is a straight line from the starting position to P, and the shortest way to get to the fly from P is another straight line. If P is on the top edge of the right end of the box, then the shortest path to the fly by way of P is for the spider to stay on the right end of the box and the top face, and we know the shortest path in that case. If P is on the vertical edge of the right end of the box, then the shortest path to the fly by way of P is for the spider to stay on on the right end of the box and the front face, and again we know the shortest path in that case. So those two cases are the only ones we need to check to find the absolute shortest path.) Lee then presented part (b). The idea, in essence, is if we replace the rectangular box by a box whose right end is square, there are two shortest paths from one corner to the opposite corner. Thus if we reposition the spider so effectively it's at a position that would correspond to the opposite corner from the fly for a rectangular box with square ends, there will be two shortest paths. Stephen at this point commented that whereas the shortest distance between two points in the plane is always the straight line between the two points, and so is unique, for paths on the box uniqueness can fail. The box is a right model for a sphere, so we might expect that there's not always a unique shortest path between two points on a sphere, and we saw examples of this; there are infinitely many shortest possible ways to get from the North Pole to the South Pole on the Earth. Anessa then presented part (c). After some discussion we saw that there are positions of the spider and the fly such that the shortest path between them traverses all three visible sides of the box. Here is the idea. Whenever both fly and spider are on the same side of the box, the shortest path is a straight line. But say they are both on the same side, but on different edges. If we push them slightly past the edge, the shortest path will be close to what was the shortest before we pushed them, with a little bit extending over the edges to get to the new positions of the arthropods, hence mostly being on one side, but having to have a bit on two other sides.
We then had a brief discussion about grading. A score of check is a fine grade, showing good understanding and clear exposition. A check plus is reserved for solutions that go above and beyond that; this is what we would expect for what is turned in for Sample Six problem solutions.
We next took some time to allow everyone to choose groups for the presentations at the end of week 2, following a protocol suggested by Courtney B. Since several groups suggested interest in presenting on Pythagoras' Theorem, and give that the course was going to cover Pythagoras' Theorem briefly, we moved that up, so that whichever group ended up presenting on Pythagoras' Theorem would know early on what aspects of it would be covered in class. (The point of discussing Pythagoras' Theorem is to provide context for a later discussion of a version of Pythagoras' Theorem on the sphere.) Brian presented a proof, which went like this. Start with an a, b, c right triangle. Arrange four copies of it into a square, as shown:
We then have a big square of side length a+b containing the four triangles and a smaller square of side length c. Computing areas gives (a + b)2 = 4ab/2 + c2, which simplifies to a2 + b2 = c2.
Courtney presented her Venn diagram for plane triangles (the version below gives examples of the kind of triangles each region represents):
In preparation for doing the same thing for spherical triangles, the class worked on a worksheet whose point was to think about how one might define a triangle on a sphere.
The group discussion on this issue gave rise to the following. The sides of a triangle on the sphere should be 3 great circle arcs terminating at 3 vertices, which should be "noncollinear" (hence not all 3 on the same great circle). But between any two points on the sphere there are at least two great circle arcs, since you can go straight from one point to the other, or you can go around the back side of the sphere (i.e., there is a major and minor arc, unless the points are antipodal, like the north and south poles on the Earth, in which case there are infinitely many great circles through the 2 points, like the longitude lines for the two poles on the Earth). There's also the issue that if you choose major arcs for two of the sides, these arcs cross on the far side of the sphere; i.e., the "sides" of our triangle meet at more than just the vertices. We'd like whatever we come up with for triangles on spheres to divide the sphere up into exactly two regions, which we'd like to think of as the interior and the exterior of the triangle, but notice that if two (or more) of the sides are major arcs then we get more than just 2 regions, whereas if all at most one arc we get only two regions, as we see below from the drawings below (produced in Spherical Easel):
Here we have only major arcs, which we see five regions:
Here we have the same two points and only two major arcs, which we see three regions:
Here we have the same three points and only one major arc, which we see two regions:
Here we have the same three points but only minor arcs, which again we see only two regions:
Thus only the last two cases agree with our intuition of how a triangle should work, so Definition 1 on the worksheet (a triangle is the set of three non-collinear points (vertices) which are joined by three straight line segments [great circle segments] (sides), where the endpoints of the line segments are the vertices and it separates the surface into exactly two regions) is equivalent to saying that at most one of the sides is the major arc. Definition 2 (which states that a proper triangle consists of three non-collinear points (vertices), and three line segments[great circle segments] (sides), where the endpoints of the segments are the vertices and the length of each segment is less than one half the circumference of the sphere (minor arc)) is just the case where all 3 sides are minor arcs. As we can see from the drawing above, this case most closely aligns with our usual notion of what a triangle is. So a triangle on the sphere is any 3 vertices not all on a single great circle (hence no two of the points are antipodal) where at most one of the sides is a major arc, while a proper triangle is the same but none of the sides can be a major arc.
We then thought about item 3 on the worksheet, regarding what kinds of spherical triangles there can be.
Emily showed an example of a spherical triangle with 3 right angles and two right angles.
We wondered but did not settle whether having three right angles made the triangle
equilateral. We also noticed that it was possible to have 3 obtuse angles; just take the vertices spread evenly near the equator:
We did not finish discussing the Venn diagram of properties for spherical triangles.
Courtney W. presented her solution to Problem 2, using similar triangles. In the figure below, by similar triangles we know the angle at lower left (at the foot of the line segment of length y) and the angle at the foot of the line segment of length z, are congruent (to avoid using the fraction x/2, we can introduce another variable, t = x/2). Thus they have the same tangent, so (2t+34)/1775 = 20/t, or 2t2 + 34t = 20*1775, so t2 + 17t - 17750 = 0. Solving this gives t=125 and t=-142. We want the positive solution, so x = 250. The town is a 250 by 250 square.
It may be of interest to know that this problem can be solved using the Pythagorean theorem,
but this approach uses some algebra.
We have three equations:
z2 = t2 + 202;
y2 = (2t + 14)2 + (1775 - t)2; and
(y + z)2 = (2t + 34)2 + 17752.
The third equation gives
y2 + 2yz + z2 = (2t + 34)2 + 17752
or 4y2z2 = ((2t + 34)2 + 17752 - y2 - z2)2, so
((2t + 34)2 + 17752 - y2 - z2)2 - 4y2z2 = 0.
We can now substitute in for y2 and z2 using the first two equations to get:
((2t + 34)2 + 17752 - ((2t + 14)2 + (1775 - t)2) - (t2 + 202))2 - 4((2t + 14)2 + (1775 - t)2)(t2 + 202) = 0.
Using algebra (or mathematical software, like Wolfram alpha) we can simplify this to get
-16t4 - 544t3 + 563376t2 + 9656000t - 5041000000 = 0
and using the mathematical software again we can factor this to get
(-16)(t + 142)2(t - 125)2 = 0.
Thus t is either -142 or 125, and since we want a positive solution, we get t = 125 so x = 250.
Leo presented Problem 4. The key idea was that by rotating the shaded region we can cover the circle, whose area we know, but when we do this we overcount the area of the pentagon. Taking into account the area of the circle and the overcount, gives a formula 5S-4P=Pi, which we can solve for the area P of the pentagon in terms of the shaded area S: P=(5S-Pi)/4. The point of this problem, is that it gives a model for finding the area of a proper spherical triangle, but for the area of a proper spherical triangle we'll need to know the surface area of a sphere. Some of the people in the class also found a numerical value for the area of the pentagon. The values found were not always exactly the same but they were all about 0.3. Here is a graphic made from a Geogebra construction:
Geogebra gives the area of the shaded triangle as 0.06938, hence the pentagon has area 5*0.06938 = 0.3469. This can be done by hand since we can write down coordinates for the points of the star; each point is on the unit circle, at a multiple of 72 degrees (the first point of the star is at 0 degrees, the second at 72 degrees, the third at 144 degrees, etc.). Knowing the coordinates of the points, we can get the equations of the lines forming the star. Point B is intersection of two of these lines, so we can solve the equations of these two lines simultaneously to get the coordinates of B. We can see that A is the origin, so knowing A and B we can get the distance d from A to B. The angle at A from C to B is also 72 degrees, so if M is the midpoint of line segment BC, the triangle AMB is a right triangle of hypotenuse d and the angle from M to B with vertex A is 36 degrees (half of 72). We can now get the length from M to B as d*sin(36) and the other leg of triangle AMB as d*cos(36). Thus triangle ABC has area d*sin(36)*d*cos(36) = 0.06938.
After going over the homework, the class experimented using Spherical Easel. We then constructed the Venn diagram for spherical triangles, in analogy with what we did yesterday for planar triangles. Here is what we found:
We then worked on a worksheet the point of which was to develop a formula for the area of a spherical triangle. What we found is that the area of a lune (the region between two great semi-circles which meet at two antipodal points) is twice the angle (in radians) between the two great semi-circles. We used that fact and reasoning modeled on homework Problem 4 (following the steps on the worksheet), to determine the area of a spherical triangle. The formula is that the area is A+B+C-Pi, where A, B and C are the angles (in radians) of the triangle. This quantity is also referred to as the spherical excess, since it is the amount that the angle sum of a spherical triangle exceeds what you'd expect for the angle sum based on what happens in the plane. (Also note that a lune can be regarded as a degenerate spherical triangle in which one angle has measure Pi. Thus the other two angles are congruent, so the spherical excess for a lune is twice the angle, and this is indeed the area.) The spherical excess formula also helps us understand that the angle sum for a spherical triangle is always more than 180 degrees, since the area of a spherical triangle is always positive. It's not too hard to see that the angle sum can be as close to 180 degrees as we like, since this will be the case if he area is small enough (i.e., a very small spherical triangle will have angle sum just slightly larger than 180 degrees).
The rest of the time was spent on having the groups meet to work on their presentations.
Alex had a different way to do the case that the centroid and the orthocenter were the same.
Consider some triangle ABC. If the centroid and the orthocenter are the same point,
call it P, then line AP is both a median and an altitude, so it is orthogonal
One can attempt to do any pair of cases chosen from among the incenter, centroid and orthocenter
in a way similar to what Alex did, since the corresponding lines (angle bisectors, medians and altitudes
all go through a vertex in these cases). So for example, consider
the case that the incenter and the orthocenter are the same point,
call it P. Then mimicking Alex's argument we would say:
line AP is both an angle bisector and an altitude, so it is orthogonal
The case that the centroid and the incenter are the same point P is a little trickier, however.
Mimicking Alex's argument we would say:
line AP is both a median and an angle bisector, so it is bisects
The three cases involving the circumcenter also differ somewhat
from Alex's argument since a priori the perpendicular bisectors do not need to go through a vertex.
For example, consider the case that the circumcenter is also the centroid.
In this case, given a triangle ABC where P is both the centroid and the circumcenter,
let D be the midpoint of side
Consider the case that the incenter is also the circumcenter; i.e.,
consider a triangle ABC and let P be the point which is
both the incenter and the circumcenter. Since P is the incenter, the perpendicular
line segments from P to each side have the same length (they are radii of the inscribed circle).
Since P is the circumcenter, these line segments are perpendicular bisectors of each side.
If D is the midpoint of side AB and E the midpoint of side AC, then angle PAE and angle PAD
are congruent (since
Finally, consider the case that the orthocenter is also the circumcenter; i.e., consider a triangle ABC and let P be the point which is both the orthocenter and the circumcenter. Since P is the orthocenter, the lines perpendicular to each side through P go through the opposite vertex, and since P is the circumcenter, these lines also bisect each side. If D is the midpoint of side BC, then triangle ADB is congruent to triangle ADC by SAS, so AB=AC, so by similar reasoning any two sides of ABC are congruent, so ABC is equilateral.
For Problem 2 we found two ways to cut up the cake.
Alyssa showed how using the incenter and cutting from the incenter
to each edge along a perpendicular we got three kites. The bilateral symmetry
of the kites means each one is the same as when turned over.
The figure below shows the procedure: The cake is given at 1; turn the cake over and cut as specified to get 2, then slide and rotate the pieces to reassemble (as shown, without turning them over again) to get 3.
Lee gave an alternate solution. He noticed that a right triangle needs only one cut, using the circumcenter.
The circumcenter center is at the midpoint of the hypotenuse. Cut from there to the vertex at the right angle.
You get two isosceles triangles which you can reassemble to get the original triangle turned over.
Now consider any triangle. Drop a perpendicular from a vertex to the longest side
(but don't make any cuts yet). That divides the original
triangle into two right triangles; one cut each on those two right triangles gives two isosceles triangles.
But since we made only two cuts (we didn't cut along the perpendicular, shown below as a dashed line), the inner two isosceles triangles give a kite, so we get three bilaterally symmetric pieces (1, 2 and the kite 3) as shown:
The last homework problem was part 5 of the worksheet from yesterday. Jennifer presented it. The first part was to find the area of a triangle with angle measures of 65, 66, and 72 degrees on the Earth, assuming it to be a sphere of radius 6371 km. Using the spherical excess formula for area, we get (65*Pi/180)+(66*Pi/180)+72*Pi/180)-Pi = 23*Pi/180 for the area on a unit sphere. For the sphere of radius 6371 km we have to multiply by 6371^2, so we get (23*Pi/180)6371^2 = 16293726.2 square kilometers. Part 2 was to find the angle measures of an equiangular triangle of area Pi on the unit sphere. If the angle is x, then 3x-Pi=Pi so x=2Pi/3 in radians, or 120 degrees.
We then worked on a handout
(here's a filled out version of the worksheet),
whose goal was to derive a version of Pythagoras' Theorem and a law of cosines
for spherical triangles. For a proper spherical triangle of sides a, b and c
with opposite angles A, B and C as above, the result is:
cos(c) = cos(a)*cos(b) + sin(a)*sin(b)*cos(C), referred to as the law of cosines for sides for a spherical triangle. If C is a right angle, then cos(C) = 0 and we get a version of Pythagoras' Theorem for spherical triangles: cos(c) = cos(a)*cos(b). This can be made to look more like the usual planar version of Pythagoras' Theorem. To do so we square cos(c) = cos(a)*cos(b) to get cos2(c) = cos2(a)*cos2(b) and use 1 - sin2(x) = cos2(x) to get 1 - sin2(c) = (1 - sin2(a))(1 - sin2(b)). This simplifies to sin2(c) = sin2(a) + sin2(b) + sin2(a)*sin2(b). When a, b and c are small, i.e., for a very small triangle on a sphere, the term sin2(a)*sin2(b) will be negligible and from calculus we know that sin2(a) will be close to a2, sin2(b) will be close to b2 and sin2(c) will be close to c2, so what we'll get will be a very close approximation to a2 + b2 = c2.
Using the law of cosines, we worked out Problem 3 from yesterday's homework handout (which wasn't assigned since we hadn't gotten to the law of cosines by the end of class yesterday). The problem was to use the law of cosines for sides to find the distance from New York to Los Angeles, given that the Earth is a sphere of radius approximately 4000 miles and that the latitudes for these cities are respectively about 41 and 34, and the longitudes for these cities are respectively about 74 and 118, to explain your how you obtain your answer, and to compare your result to the actual distance. The idea is to make a triangle on the Earth, two of whose vertices are LA and NY, and the North Pole is the third. The angle at the pole is the difference in longitude (so 118-74=44 degrees), and the sides, given as degrees of lattitude are 90-41=49 for NY and 90-34=56 for LA. Now plug into the law of cosines to get the cosine of the central angle x (i.e., the angle at the center of the Earth) from LA to NY: cos(x) = cos(49)*cos(56) + sin(49)*sin(56)*cos(44) = 0.8169 so x = 0.6147 radians, hence the distance is 4000*0.6147 = 2458.8 miles. Wolfram alpha gives the distance from NY to LA as 2464 miles. Google maps gives the driving distance as 2794 miles.
We also recalled the law of sines. For a planar triangle with angles A, B and C where, a is the length of the side opposite A, b is the side length opposite B and c is the side length opposite C, the law of sines says
sin(A)/a = sin(B)/b = sin(C)/c. The proof is to relate all three ratios to the area of the triangle.
There is also a law of sines for spherical triangles.
Given a proper spherical triangle
with angles A, B and C and sides a, b and c we have:
sin(A)/sin(a) = sin(B)/sin(b) = sin(C)/sin(c).
The proof is to relate all three ratios to the volume of the tetrahedron whose vertices are the vertices of the triangle, together with the center of the sphere. We gave out a handout explaining with details.
The rest of class was spent with groups working on their presentations for next week. There was no homework assignment for Monday.
When you have the data for a congruence property, like if you have two sides and the included angle of some triangle, since it determines a unique triangle, in principle you can figure out (i.e., solve for) the other parts of the triangle. But how can you do this? With SAS you can use the law of cosines to get the third side. For example, say we have a triangle with angles A, B and C where a is the side opposite A, b is the side opposite B and c is the side opposite C. The law of cosines says cos(c)=cos(a)cos(b)+sin(a)sin(b)cos(C). So if you know a, b and C you can get c. And you can do this in reverse: given three sides (which we now would have), you can solve for any of the angles.
But what if you have ASA? It turns out there is a second law of cosines which reverses the roles of the sides and angles. The second law says: cos(C)=-cos(A)cos(B)+sin(A)sin(B)cos(c). So given ASA you can use the second law to solve for the third angles, and then get the other two sides.
Note that you have to be careful using the law of sines, since knowing the sine of an angle does not always tell you the angle. For example, if sin(a)=1/2, a could be either 30 degrees (i.e., 90-60) or 150 degrees (i.e., 90+60). This is one way to see why ASS is not a congruence property: given an angle and the opposite side we have sin(A)/a, so if we know b, then we know sin(B) since sin(B)/b=sin(A)/a, but knowing sin(B) doesn't mean we know B.
We then worked on handout 2. The question here is if we mimic the various constructions we did of squares in the plane, but on the sphere, what happens? What other constructions are possible? Can rectangles be constructed on the sphere? What about rhombi? What about rhombi with two right angles?
For example, suppose we construct a quadrilateral with a right angle and four congruent sides on the sphere: is it a square? We first have to decide what we mean by a square. If we mean a quadrilateral with congruent sides and four right angles, then the answer is no. Four right angles means an angle sum of 360, but we can divide a quadrilateral into two triangles by cutting along a diagonal, and the angle sums for each is more than 180, so a quadrilateral on the sphere can never be a square (or a rectangle). But let's say that a square is a regular quadrilateral; i.e., four congruent sides and four congruent angles (but maybe not right angles). Then a quadrilateral with a right angle and four congruent sides on the sphere will still not be a square but, as Emily pointed out, it will be a rhombus with two opposite right angles (answering a later question), and, as Alex mentioned, when it is very small, it will be close to a square, but it's never quite a square.
If we construct a quadrilateral with two right angles and three congruent sides on the sphere, then it is again not a square, since, as Tyler explained, if we regard the included side as being on the equator, the fourth side will be a latitude line above the equator, and thus will be shorter.
If we construct a quadrilateral with three right angles and the two included sides are congruent, Staci showed why it would not be regular. One way to see this is if it were regular, then it would have two right angles and three congruent sides, which we already saw in the previous case could not be regular.
Finally, Tiffany pointed out that all four sides can be congruent and so can all four angles (they just can't all be right angles). Alyssa gave an example: take points near the cardinal directions on the equator; the sides will be congruent (by symmetry) and the angles will be too (each about equal to 180 degrees).
We then began discussing the parallel postulate. It's helpful to recall what it means for lines to be parallel: it just means they don't intersect. Given any line L and a point P not on L, Euclid asserted that there was a unique line through P parallel to L (actually he asserted something somewhat different which was equivalent to this). On the sphere we know this is false since tow lines always intersect, but for thousands of years people believed that the parallel postulate was so obvious that it must be a consequence of Euclid's simpler assumptions, but no one was ever able to show this. We now know it's impossible since examples can be given, like spherical geometry, where the parallel postulate is false. (On the sphere it's false because no lines are parallel but in hyperbolic geometry we will see it's false because even though there are always lines through a point parallel to any line not containing the point, the lines are not unique.)
As a way of exploring this issue, we started with the proof of why the angles of a Euclidean triangle sum to 180. As Leo mentioned, the proof starts by taking a triangle with vertices A, B and C, and drawing a line through C parallel to AB. Emily and Kyle then gave us the rest of the proof: using alternate interior angles we can see that the three angles we now have at C and which form a straight angle (and so add up to 180) are congruent to the angles of the triangle.
We then went to a worksheet showing that if we don't assume Euclid's parallel postulate but still assume Euclid's other postulates (like two points determine a unique line, etc., which isn't true on a sphere) we can still prove that the angles of a triangle add up to at most 180 degrees. The idea is that given any triangle, one can (using SAS and the construction on the handout, as Julia explained) show that any two angles add up to less than 180. Based on the construction which shows this, given any triangle, we can ( as Emily explained) also show that we can find a new triangle with the same angle sum but which has an angle less than half as big as the smallest angle of the original triangle. Thus for any sum which is an angle sum, we can find a triangle with that sum but which has one angle as small as we like. So, for example, as Tyler and lynn explained, if there really were a triangle with an angle sum of 180.2 degrees, we could find a triangle with the same angle sum, but where the smallest angle was at most .1 degrees. Since the other two angles add up to less than 180, the angle sum is at most 180.1, which contradicts it having the same angle sum (which would be 180.2). Thus it's impossible to have an angle sum bigger than 180.
The rest of the time was spent working in groups on presentations.
Allison presented the second part, in which the sides were as before but now on a sphere of radius 2. Since the sphere is twice as big, the angles subtended by the sides are half as big as before, and what we need for the law of cosines is not the actual length of the sides but the central angle subtended by the side; i.e., we need the radian measure of the sides. The formula is rx=s, where r is the radius of the sphere, x is the radian measure of the side (i.e., the angle subtended by the side measured at the center of the sphere) and s is the arclength of the side (i.e., the actual length, measured in the same units as the radius, inches say, or centimeters, etc.). Thus x = s/r, and so if x = 1.5, then x = 1.5/2 = .75. Likewise, if s = 1, then x = .5 and if s = 2 then x = 1. So we now just repeat what we did for the first part but using .5, .75 and 1 instead of 1, 1.5 and 2. The angles we get are 32.8 degrees, 50 degrees, and 108 degrees (approximately).
Anessa started our discussion of Problem 2. The proof of step 1 of the Saceccheri-Legendre Theorem is false for
spherical triangles in general, as she showed by pointing to an example where the sum of two angles of a spherical triangle
is not less than 180 degrees (her triangle had tow right angles). Emily showed the step which goes awry. Given an angle
CAB with vertex A and a point D in the interior of the angle, all points on the ray starting at A going through D stay
in the interior of the angle for angles in the plane. But for spherical angles this isn't true. Consider angle CAB in the following figure:
Note that point D is in the interior of angle CAB but if we extend it far enough we get to a point E which is not in the interior. In more detail, given triangle ABC, the proof in the plane that the sum of angles ACB and ABC is less than 180 goes by constructing the midpoint D of side BC. Then constructing a point E so that A, D and E lie on a line but AE is twice as long as AD. Triangle ACD is now congruent to triangle DBE by SAS, but in the plane case, line segment AE does not cross line AB, so angle CBE is less than the supplement to the interior angle of triangle ABC at B. But as the figure shows, on a sphere line segment AE can cross line AB if AD is long enough, so angle CBE can be more than the supplement to the interior angle of triangle ABC at B.
Karen presented Problem 3. In the following figure,
the goal is to find d in terms of t and theta. So let Q be the intersection of C_1 and C_2 closest to A, and let r be the radius of C_2. Note that triangle OQP is a right triangle, so 1^2+r^2 = d^2. Consider triangle OAP, and apply the law of cosines to obtain r^2=t^2+d^2 - 2dt cos(theta). Solving the first equation for r^2 and substituting into the second equation, we obtain d^2-1 = t^2 + d^2 -2dt cos(theta), and solving for d we obtain d=(t^2-1)(2t cos(theta)). [Note: this is related to constructions of "lines" in the disk model of the hyperbolic plane.]
Stephen then gave a discussion of the history of the development of the hyperbolic plane. The idea is to ask if it's possible to have a geometry satisfying all of the axioms of the Euclidean plane but one. Instead of the Euclidean parallel postulate (which says that given a line L and a point P not on L, there is a unique line parallel to L through P), assume the hyperbolic parallel postulate (which says that given a line L and a point P not on L, there are at least two lines parallel to L through P).
It was discovered in the 1800s by Gauss, Bolyai and Lobachevksi that such geometries did in fact exist, even though philosophers and their followers claimed that as a matter of pure reason one could "see" that no such geometry was possible. But in fact it's much harder to see that something is not possible than to see that it is possible; for the latter you just have to give an example. So we will look at examples where the hyperbolic parallel postulate is true.
To start it's helpful to have a physical model, like we did sing the Lenart spheres when studying spherical geometry.
For hyperbolic geometry we will make a hyperbolic soccer ball as an approximate physical model for the hyperbolic plane.
An actual soccer ball is made out of pentagons surrounded by hexagons:
The only way for the hexagons and pentagons to fit together is if they curve around, making a ball (i.e., a positively curved surface). If you replace the pentagons by hexagons, they all can fit together while lying flat, and you get a tiling of the plane by hexagons (i.e., a surface of 0 curvature). If you now replace the pentagons by heptagons, it turns out you get a frilly surface that can't lie flat but doesn't close in on itself (i.e., it is negatively curved). This is an approximation to the hyperbolic plane (it's only an approximation since it's made out of flat polygons; all of the curvature come at the joints where the pieces connect to each other).
Here is the handout giving directions for making the hyperbolic soccer ball. Here is the handout with the outlines of the hexagons and heptagons.
We then used the hyperbolic soccer ball to study the area of triangles in the hyperbolic plane; see this handout for guidance for this activity. Here is a file for making protractors to measure angles with. What we found was that the angle sum was always less than 180 degrees, and the area of triangles is about proportional to the hyperbolic deficiency (i.e., the difference Pi-(a+b+c), where a, b and c are the angles measured in radians), just like on a sphere the area is proportional to the spherical excess (specifically the area is (a+b+c-Pi)R^2, where R is the radius of the sphere). In fact, on the hyperbolic plane the area is (Pi-(a+b+c))r^2 for some quantity r that plays the role of the radius in spherical geometry.
We then took a break, after which Brian gave a discussion of two other models for hyperbolic geometry, the Poincare disk model and the upper half plane model. (In fact, you ca think of the two models as being different views of the same model. This is emphasized in Non-Euclid, which allows you to choose the disk view or the upper half plane view of the same construction.) For the disk model, the points of the hyperbolic plane are the points in the interior of the unit circle. The points on the unit circle itself are not points in the hyperbolic plane, but you can think of the as points at infinity with respect to the hyperbolic plane. In the disk model, hyperbolic lines come in two forms, diameters of the unit circle and circular arcs in the interior that run from the unit circle back to the unit circle, and meet it at right angles. So, in the figure above for Karen's solution to Problem 3, the interior of C_1 is the hyperbolic plane and the circular arc running from Q to R on C_2 is a hyperbolic line. For the upper half plane, the points are the points in the usual plane, but having positive y-coordinate; the points on the x-axis again can be thought of as points at infinity. Hyperbolic lines are rays running vertically from the x-axis or semicircles centered on the x-axis. Diameters in the disk view can be regarded as circular arcs of infinite radius. Likewise, vertical rays in the upper half plane view can be regarded as semicircles of infinite radius.
Here are some hyperbolic lines in the disk view and the same lines in the upper half plane view (these figures were made using Non-Euclid):
Note that two lines can either intersect or not, but they intersect at most at a single point. When they don't intersect, they are regarded as being parallel. It's also not too hard to see that there is always exactly one hyperbolic line through any given pair of points.
Here (in both views) is a line and a point A, and various lines through A parallel to the given line. Note that the parallels can be nowhere near the given line, or they can intersect at infinity:
Thus we see that these models do satisfy the Hyperbolic Parallel Postulate (of there being more than one parallel through a point not on a line).
In early paintings figures of about the same size (like two people) were drawn the same size, whether they were in the foreground or the background (but note how the bishop is bigger than the peasants!):
In more modern paintings artists had figured out perspective so that figures of the same size are drawn at different sizes depending on how far away they are:
The two models for the hyperbolic plane are similar in that distances are represented essentially with perspective.
Angles in these models are as they appear, but distances are not as they appear. As a figure in the hyperbolic
plane (inside the unit circle for the disk model, or above the x-axis in the upper half plane model)
gets closer to the boundary (i.e., the unit circle or the x-axis, depending on which model is being used),
it appears to get smaller, because it's getting closer to infinity, and thus is "farther away" so looks smaller.
We can see that in these two Escher drawings, one for the disk model and the other for the upper half plane model; all of the figures are the same size, but don't appear to be:
Here's a handout giving an explicit formula for measuring distances between points in the upper half plane model.
We finished by using Geogebra to model the upper half plane model by constructing hyperbolic lines through pairs of points. (One approach was to draw any circle centered on the x-axis through one of the points, and move the center along the x-axis until the circle went through the other point too. Another method was to draw the perpendicular bisector of the Euclidean line segment between the two points; it intersects the x-axis at the center of the circle giving the hyperbolic line through the two points. A third method found the distance from each point to an arbitrary point whose x-coordinate was x, and then solved for the value of x equidistant to each point.)
Earlier, using the hyperbolic soccer ball, measurements seemed to show that the angle sum for hyperbolic triangles
always were less than 180 degrees. The disk model suggests why this is true.
Not all triangles in the disk model have concave edges; look at triangle BCD for example:
But if we reflect this triangle across line BC (reflections are an operation you can do in Non-Euclid) we get a congruent triangle (so it has the same angle sum) and now the sides are all concave, so it's contained in a Euclidean triangle (shown with dotted lines) but has smaller angles, which thus add up to less than 180 degrees. It turns out that any triangle can, by a sequence of reflections, be made to contain the center of the disk, in which case it's sides will be concave, so in fact every triangle in the hyperbolic plane has angle sum less than 180 degrees. Thus the proof o the Saccheri-Legendre Theorem, that without the Euclidean Parallel postulate we get that triangles have angle sums of at most 180 degrees is the best you can do. Sometimes the angle sum really can be less than 180!
We spent the rest of the time working in groups on the presentations and a=handed out the homework.
Lydia presented Problem 2. From the following Geogebra figure we see angle ABC is hte angle between the tangent
lines, and this angle is a vertical angle for the 45-90-45 right triangle DBE. Thus the angle is 90 degrees.
Courtney B. presented Problem 3, based on the following diagram:
The big circle has radius 2, the small one radius 1, and the angle at point D is a right angle, so triangle BDE is a 30-60-90 right triangle. Using this fact and the various vertical angles at point B we find that the angle between the tangents to the circles at B is 120 degrees; i.e., hyperbolic angle ABC is 120 degrees.
Tiffany did this a different way. She used Geogebra to make a 120 degree angle and then created circles tangent to the angle and put points on the circles.
Lynn did it a third way, similar to Courtney's method, but using different triangles.
Her diagram looked like this:
Her triangle, ABC, is an equilateral triangle, so the angle BCA at point C is 60 degrees, just like the angle at point B in Courtney's diagram was a 60 degree angle, and just like with Courtney, the angle here between the tangents is 120 degrees.
We then worked through the angle sum worksheet to show the existence of hyperbolic triangles with angle sum less than 180 degrees. We saw this yesterday using the disk model of the hyperbolic plane, but this worksheet shows this without having to assume any particular model of the hyperbolic plane. The basic idea of the worksheet is that two lines parallel to a third line don't have to be parallel to each other. The angle between them ives a gap which allows a triangle to be constructed with angle sum less than 180 degrees. (Tyler, Marta, Tiffany and Emily presented different parts of the worksheet.)
We then worked through a worksheet for deriving the hyperbolic deficiency formula for the area of hyperbolic triangles, similar to what we did for spherical triangles.
Finally, we worked through a worksheet on hyperbolic trigonometry, showing how sinh(x) and cosh(x) are to the unit hyperbola exactly as sin(x) and cos(x) are to the unit circle, and giving hyperbolic laws of cosines and sines, in analogy with the ones for the sphere. We then discussed congruence properties for hyperbolic triangles. We decided that the congruences were the same as those for the sphere, and for the same basic reasons, except AAS; AAS is not a congruence property for spherical triangles but it is a congruence property for hyperbolic triangles. In fact, a proof that AAS is a congruence can be given just using the Saccheri-Legendre Theorem; i.e., a proof can be given that it's a congruence for hyperbolic triangles that also works for Euclidean triangles! Here's the proof (refer to the diagram below). Say we have two triangles with AAS. We can lay the one triangle on top of the other so that they share a side and an angle. In the diagram below, the side is PQ and the angle is QPX = QPY. But the other angle in AAS must be the one that does not include the side PQ, so it is QXP for the one triangle and QYP for the other, but assuming AAS means that we're assuming QXP is congruent to QYP. If triangles QPX and QPY are congruent, then PX and PY must have the same length, so X=Y and triangles QPX and QPY are the same triangle. But if triangles QPX and QPY are not congruent, then X and Y must be different points; in this case, we may assume that X is the point closer to P (which is how the diagram below is drawn).
By the Saccheri-Legendre Theorem, the angle sum of triangle XYQ is at most 180 degrees. Angles PXQ and YXQ are supplementary, so angle PXQ is greater than or equal to the sum of angles XYQ and YQX. But since YQX is a positive angle, that means that angle PXQ is strictly greater than angle XYQ. This contradicts our assumption that those angles are the same; i.e., we didn't have AAS in the first place!