Find the volume of this right triangle pyramid with height 13' and AB = 2sqrt(6).
(AB is the hypotenuse of the base, a triangle with a 45-degree angle.)
So the base is clearly another special triangle -- 45-45-90. We determine that each of its legs must have length 2sqrt(6)/sqrt(2) = 2sqrt(3). The area of the right triangle is half the product of its legs, and so it's 2sqrt(3)2sqrt(3)/2 = 6 square feet.
The volume of the pyramid is one-third the product of its base and height, and so it's (6)(13)/3, which works out to be 26. Therefore the volume is 26 cubic feet -- and of course, today's date is the 26th.
Today I subbed in a high school economics class. It's the first of a five-day assignment, since the teacher is on an unexpected trip out of the country. In fact, one reason I couldn't have stayed in the middle school math class after Tuesday is that I was already assigned this class.
The original plan is no "Day in the Life" for non-math classes (unless it's middle school), but I do wish to have one for the first day of a multi-day assignment. The focus resolution is the first one:
1. Implement classroom management based on how students actually think.
And once again, it's because if I don't manage well on the first day, I'll have lost control of the class by the fifth day. So let's begin:
7:05 -- Yes, it's yet another zero period class -- officially called "first period" in high school. Here in California, Econ is generally a one-semester class for seniors (with Government usually as the other semester class).
Today, the students are watching a DVD episode of Frontline from about twenty years ago. The episode is on the history of the stock market. Students are to take notes and answer fifteen questions about the video. The assignment is worth one point per question.
It takes a while for me to figure out where the DVD player even is, and with the episode being just under an hour in length, I need to have started the video as soon as the class begins in order to reach Question #15 by the end of class. The video is in the middle of Question #13 when I realize that I must collect a homework assignment that's due today. So in my notes for the teacher, I inform him that students should only be responsible for Questions #1-12, since he's grading it a point a question.
As far as classroom management is concerned, the key rules to enforce are "no cell phones" and "no gum," since these rules are explicitly written on the side board.
8:00 -- First period leaves and second period arrives. In this Econ class, I know to start playing the video right away.
8:55 -- This school has a tutorial period. Only four students, all girls, attend it. As often happens in tutorial, I see one girl taking out her math assignment for an IB Math Studies class. The question she's on happens to be a stats question -- but unfortunately, stats is my weakest math class. I let her know this, and she replies that she'll just ask her teacher for help. I'm hoping that she'll move on to the next question, which is more geared towards algebra. But she succumbs to temptation -- she starts talking to the other three girls, who simply put their own (non-math) assignments away. In the end, the quartet accomplishes nothing academic during the tutorial period.
Here's what I should have done -- I should have insisted on helping the first student with her math assignment, even if the question is stats. I know enough about stats that I could assist her with parts of that question, and we could have struggled through the rest together. And by sitting next to the first girl, the others would be more encouraged to work on their own assignments.
9:30 -- Third period is slightly longer than the other classes, because this is like homeroom. The video announcements are shown at the beginning, and at the end is a senior survey where they vote for their favorite teacher. So there's just as much time for the Econ video as there is in second period.
10:30 -- Third period leaves for snack.
10:50 -- Fourth period arrives. This is the only non-Econ class -- instead, it's World History, a sophomore class in California. The students are labeling maps of Europe -- latitude and longitude, bodies of water, countries and their capitals, and so on. I tell the students that they must at least fill in all the countries today -- the rest is a homework assignment due tomorrow.
This is the period where I'm strict about students who go to the restroom right after a snack break. As it happens, two girls ask to leave -- one to go to her locker to get her history textbook (even though the teacher yesterday told the class to bring their texts the rest of the week), and the other to correct an issue with yesterday's attendance (but the regular teacher isn't here to verify that attendance).
11:45 -- Fifth period is the last Econ class. Each class runs more smoothly than the previous Econ class, since by now I know what to expect. I start the video as soon as the tardy bell rings. Question #1 isn't answered until eight minutes into the DVD -- so I have these eight minutes talk over the video, in order to take attendance and collect last night's homework. Question #15, meanwhile, is answered at the 48-minute mark -- just two minutes before the end of class. I have the students leave their worksheets on their desks, since there's no time to collect it before the dismissal bell rings.
Meanwhile, as far as enforcing the two main rules is concerned, phone use and gum chewing seem to increase as the day goes on. I point to the side board where the rules are written -- and most of all, I try to avoid arguments and yelling. Also, in one of the classes, one guy lays his head on the desk during the DVD. I remind him that many of the students are just four months away from their first college class, and napping in class won't set them up for success in college. (I don't remember which class this is, but I know it's not first period -- the class you'd expect student to enter half-asleep.)
12:35 -- Fifth period leaves for lunch, and so I collect the DVD worksheets. As it turns out, sixth period is the teacher's conference period, and most teachers who have a first period don't have a seventh (unless they're an athletic coach). Thus my day actually ends at lunch!
As it turns out, most of what Econ's doing this week is on video -- including Monday's video which will be on YouTube. On the other hand, the sophomores get only one video tomorrow -- next week, they'll start a multi-day project.
I hope I will have a successful week in the classroom, even though the class isn't math. But don't expect any more "Day in the Life" posts until I'm done with this assignment.
Chapter 9 of Wayne Wickelgren's How to Solve Problems is called "Relations Between Problems":
"When mankind has a satisfactory theory of problems, it will be possible to state many deep and detailed relations between different types of problems. But even without such a theory, we can still state certain basic types of relations between different problems."
Here are the five basic relations Wicklegren defines in this chapter:
- Problem a is unrelated to problem b.
- Problem a is equivalent to problem b.
- Problem a is similar to problem b.
- Problem a is a special case of problem b.
- Problem a is a generalization of problem b.
He continues:
"In determining whether any of these five relations holds between two problems, it is important to note that the critical problem elements concern the types of operations and the relations that can obtain between different expressions or things, not the specific expressions or things themselves."
For example, we can compare Wickelgren's five-disk Tower of Hanoi problem to Brian Harvey's tower of six disks:
"Any two special cases of the general Tower of Hanoi problem are similar problems, though I would hesitate to call them equivalent problems. Similarly, in the nim problem...."
...well, I already linked to a similar, yet not equivalent, problem involving flags.
The author's first problem in this chapter is the fox, goose, corn problem:
A man went on a trip with a fox, a goose and a sack of corn. He came upon a stream which he had to cross and found a tiny boat to use to cross the stream. He could only take himself and one other - the fox, the goose, or the corn - at a time. He could not leave the fox alone with the goose or the goose alone with the corn. How does he get all safely over the stream?
"Stop reading and try to solve the problem by recalling the methods used to solve a similar (partly analogous) problem."
The similar problem Wickelgren has in mind is the missionaries-and-cannibals problem from all the way back in Chapter 5. In both problems, a detour is necessary to obtain the solution:
The answer is: Take the goose over first and come back. Then take the fox over and bring the goose back. Now take the corn over and come back alone to get the goose. Take the goose over and the job is done!
In this case, "bring the goose back" to the original side of the river is the detour.
Yesterday's second Square One TV song tells us to solve "simpler but similar problems," and yes, Wickelgren would agree, but he warns us that a "simpler" problem might turn out to be harder:
(Yes, there's more on the Cut the Knot site besides 100+ Pythagorean proofs!)
There are 10 stacks of 10 identical-looking coins. All of the coins in one of these stacks are counterfeit, and all the coins in the other stacks are genuine. Every genuine coin weighs 10 grams, and every fake weighs 11 grams. You have an analytical scale that can determine the exact weight of any number of coins. What is the minimum number of weighings needed to identify the stack with the fake coins?
Stop reading and try to solve the problem.
Now the author tells us that we might try to simplify the problem by having only one coin in each stack instead of ten. But in the end, that would increase the number of weighings. On the other hand, if we reduce the number of stacks, say to three, then we can reach a solution in just one weighing:
Number the coin stacks from 1 to 10. Take 1 coin from the first stack, 2 coins from the second, and so on, until all 10 coins are taken from the last stack. Weigh all these coins together. The difference between this weight and 550, the weight of (1 + 2 + ... + 10) = 55 genuine coins, indicates the number of the fake coins weighted, which is equal to the number of the stack with the fake coins. For example, if the selected coins weigh 553 grams, 3 coins are fake and hence it is the third stack that contains the fake coins.
The next problem is a Geometry example. It's not mentioned in the U of Chicago text, but it ought to be (and it does appear in other texts -- possibly the Glencoe text):
A ray of light travels from point A to B in (the figure) by bouncing off a mirror represented by the line CD. Determine the point X on the mirror such that the distance traveled from point A to point B is a minimum. What is the relationship between the angles alpha and beta?
Wicklegren tells us that the easier problem to solve is to replace B with E, where E is the reflection image of B over line CD. Then it's trivial to find the shortest distance from A to E through CD -- the straight-line distance AE. Since reflections preserve distance, X, the intersection of AE and CD, must be the desired point in the original problem. And angle alpha must equal angle gamma (the reflection image of angle beta) as they are vertical angles. Since reflections preserve angle, alpha = beta -- that is, the angle of incidence equals the angle of reflection.
This idea isn't directly stated in the U of Chicago text -- it's only hinted at in Lesson 6-4, the lesson on miniature golf and billiards.
There's one more Geometry problem in this chapter -- the walking-fly problem:
In a rectangular room (a cuboid) with dimensions 30' * 12' * 12', a spider is located in the middle of one 12' * 12' wall one foot away from the ceiling. A fly is in the middle of the opposite wall one foot away from the floor. If the fly remains stationary, what is the shortest total distance (i.e., the geodesic) the spider must crawl along the walls, ceiling, and floor in order to capture the fly?
Stop reading and try to solve the problem. (In Wickelgren's example, the fly is the hunter, not the target -- instead, a lollipop is the target.)
As it turns out, the solution is to make a room into net, using the ideas of Lesson 9-7. Click on the link above, since I won't reproduce the net here.
Let's skip to the next Geometry example:
You are given the following: (a) A straight line equals an angle of 180 degrees, (b) A right angle equals 90 degrees. (c) If two parallel lines are cut by a transversal, the alternate interior angles are equal. Prove that the sum of the angles of any triangle equals 180 degrees. Stop reading and try to prove this theorem, making use of the method of special case.
This is, of course, the Triangle-Sum Theorem of Lesson 5-7. We already know how to prove this theorem from the U of Chicago text. But that proof doesn't involve a special case. Since one of the givens, namely (b), mentions a right angle, perhaps the special case is a right triangle.
And this is exactly what the author does. There's not much difference between the right angle case and the usual proof of the general case. But perhaps for those who don't already know the proof, the right angle case might be easier to prove first.
Here's another Geometry example given later in this chapter:
A cylindrical hole 10 inches long is drilled through the center of a solid sphere, as shown in [the figure]. What volume remains in the sphere? Stop reading and try to solve the problem, using the method of special case.
Wickelgren tells us that this problem makes an unstated assumption -- that the volume remaining in the sphere is independent of the diameter of the hole (or of the sphere). And so it's possible to consider a special degenerate case -- when the diameter of the hole is zero.
Then the length of the hole equals the diameter of the sphere -- 10 inches. A hole of zero diameter has no volume, and so the volume remaining in the sphere must its entire volume. Using the sphere volume formula, (4/3)pi r^3 = (4/3)pi 5^3 = 500pi/3. Therefore, provided that a well-defined answer exists, it must necessarily be equal to 500pi/3.
Triangle ABC is formed by three tangents to a circle, as shown in [the figure]. Angle DAE = 26 degrees. Solve for angle COB.
Let me describe the diagram, which contains important given information. The entire circle, except for the points of tangency, lies in the interior only of angle A, not the other two angles of the original given circle O. Thus Triangle ABC lies completely in the exterior of circle O, except for a single point of tangency. The points of tangency on segmentsAB, AC, and BC are D, E, and F respectively. Stop reading and try to solve the problem.
The statement of the problem implies that the measure of angle COB depends only on the measure of angle DAE and not on the relative location of the points. A special case to consider is to let the three points O, F, and A be collinear.
The first step the author uses is a tricky one. He writes:
"Since angle AED and angle ADE intersect the same arc of the circle, these two angles are equal."
This is tricky, since this is implied by a theorem that appears later in Chapter 15. But the subgoal here is to prove that Triangles AOE and AOD are congruent. We can do this by using HL instead -- angles AEO and ADO are both right angles (Radius-Tangent Theorem, Lesson 13-5), they each have a radius for one leg and a common hypotenuseOA. Let's pick up the rest of Wickelgren's solution from here:
Let's skip to the next Geometry example:
You are given the following: (a) A straight line equals an angle of 180 degrees, (b) A right angle equals 90 degrees. (c) If two parallel lines are cut by a transversal, the alternate interior angles are equal. Prove that the sum of the angles of any triangle equals 180 degrees. Stop reading and try to prove this theorem, making use of the method of special case.
This is, of course, the Triangle-Sum Theorem of Lesson 5-7. We already know how to prove this theorem from the U of Chicago text. But that proof doesn't involve a special case. Since one of the givens, namely (b), mentions a right angle, perhaps the special case is a right triangle.
And this is exactly what the author does. There's not much difference between the right angle case and the usual proof of the general case. But perhaps for those who don't already know the proof, the right angle case might be easier to prove first.
Here's another Geometry example given later in this chapter:
A cylindrical hole 10 inches long is drilled through the center of a solid sphere, as shown in [the figure]. What volume remains in the sphere? Stop reading and try to solve the problem, using the method of special case.
Wickelgren tells us that this problem makes an unstated assumption -- that the volume remaining in the sphere is independent of the diameter of the hole (or of the sphere). And so it's possible to consider a special degenerate case -- when the diameter of the hole is zero.
Then the length of the hole equals the diameter of the sphere -- 10 inches. A hole of zero diameter has no volume, and so the volume remaining in the sphere must its entire volume. Using the sphere volume formula, (4/3)pi r^3 = (4/3)pi 5^3 = 500pi/3. Therefore, provided that a well-defined answer exists, it must necessarily be equal to 500pi/3.
Triangle ABC is formed by three tangents to a circle, as shown in [the figure]. Angle DAE = 26 degrees. Solve for angle COB.
Let me describe the diagram, which contains important given information. The entire circle, except for the points of tangency, lies in the interior only of angle A, not the other two angles of the original given circle O. Thus Triangle ABC lies completely in the exterior of circle O, except for a single point of tangency. The points of tangency on segments
The statement of the problem implies that the measure of angle COB depends only on the measure of angle DAE and not on the relative location of the points. A special case to consider is to let the three points O, F, and A be collinear.
The first step the author uses is a tricky one. He writes:
"Since angle AED and angle ADE intersect the same arc of the circle, these two angles are equal."
This is tricky, since this is implied by a theorem that appears later in Chapter 15. But the subgoal here is to prove that Triangles AOE and AOD are congruent. We can do this by using HL instead -- angles AEO and ADO are both right angles (Radius-Tangent Theorem, Lesson 13-5), they each have a radius for one leg and a common hypotenuse
- Angle EAO = DAO (CPCTC) = 13 (since angle DAE = 26)
- Angle AEO = ADO = 90 (Radius-Tangent Theorem)
- Angle AOE = AOD = 90 - 13 = 77 (Triangle-Sum Theorem)
- Triangles OEC and OFC are congruent (They are right triangles by Radius-Tangent, each with a radius for a leg and sharing a hypotenuse. It's interesting that Wickelgren uses HL here, but not to prove AOE and AOD congruent earlier.)
- Triangles ODB and OFB are congruent (same reason)
- Angle COF = 1/2 EOF, angle FOB = 1/2 FOD (implied by CPCTC)
- Angle COB = 1/2 EOD = 1/2 (77 + 77) = 77 degrees
and the problem is solved.
The final idea mentioned in this chapter is generalization. An example from Geometry is given by -- don't tell SteveH -- Polya:
The problem is to find a plane that passes through a given line and bisects the volume of a given octahedron.
The solution is to consider the center of symmetry of the octahedron. Then since a line and a point not on the line determine a plane (asserted by Euclid but only implied in the U of Chicago), this plane is the solution. (If the center lies on the given line, then any plane containing that line works.) This solution generalizes to any solid with a center of symmetry -- which we'll define here loosely as the intersection of all planes of symmetry, if it exists.
Wickelgren concludes the chapter:
"Thus, once again we see that the role of generalization and the role of representation of information (as discussed in Chapters 3 and 10) are very closely linked and perhaps identical."
In [the figure], angle BAD = 20, AB = AC, and AD = AE. Solve for angle CDE.
In the diagram, we have Triangle ABC with D between B and C, and E between A and C. Stop reading and try to solve the problem, making use of the method of a special case.
Wickelgren points out that not only are we not given any side lengths, but it's possible to draw many non-similar triangles satisfying these conditions --in other words, angle DAE can take on a wide range of values. In the special case, he lets DAE = 20 (that is, equal to BAD). Then he obtains:
- Angle BAC = 40 (Angle Addition)
- Angle ABD = ACD (Isosceles Triangle Theorem) = 70 (Triangle-Sum Theorem)
- Angle ADE = AED = 80 (same reason)
- Angle DEC = 100 (Linear Pair Theorem)
- Angle CDE = 180 - 100 - 70 (Triangle-Sum) = 10
The author tells us that we can choose DAE to be many other values, or even DAE = y. But it's easier just to plug in one specific value and find a solution
This is what I wrote last year about today's lesson:
Lesson 15-3 of the U of Chicago text is on the Inscribed Angle Theorem. I admit that I often have trouble remembering all of the circle theorems myself, but this one is the most important:
Inscribed Angle Theorem:
In a circle, the measure of an inscribed angle is one-half the measure of its intercepted arc.
Inscribed Angle Theorem:
In a circle, the measure of an inscribed angle is one-half the measure of its intercepted arc.
2018 Update: By a fortuitous coincidence, this just happens to be one of the problems given in the current Wickelgren chapter! So if you thought we were done with Wickelgren today, guess again!
"Stop reading and try to solve the problem by first considering a special case."
Well, the special case is when one of the given sides of the angle contains a diameter. This is given as Case I in the U of Chicago text:
Given: Angle ABC inscribed in Circle O
Prove: Angle ABC = 1/2 * Arc AC
Proof:
Case I: The auxiliary segmentOA is required. Since Triangle AOB is isosceles [both OA and OB are radii of the circle -- dw], Angle B = Angle A. Call this measure x. By the Exterior Angle Theorem, Angle AOC = 2x. Because the measure of an arc equals the measure of its central angle, Arc AC = 2x = 2 * Angle B. Solving for Angle B, Angle B = 1/2 * Arc AC. QED Case I.
Notice that the trick here was that between the central angle (whose measure equals that of the arc) and the inscribed angle is an isosceles triangle. We saw the same thing happen in yesterday's proof of the Angle Bisector Theorem -- the angle bisector of a triangle is a side-splitter of a larger triangle, and cutting out the smaller triangle from the larger leaves an isosceles triangle behind.
Given: Angle ABC inscribed in Circle O
Prove: Angle ABC = 1/2 * Arc AC
Proof:
Case I: The auxiliary segment
Notice that the trick here was that between the central angle (whose measure equals that of the arc) and the inscribed angle is an isosceles triangle. We saw the same thing happen in yesterday's proof of the Angle Bisector Theorem -- the angle bisector of a triangle is a side-splitter of a larger triangle, and cutting out the smaller triangle from the larger leaves an isosceles triangle behind.
"Now stop reading and extend your solution to the general case."
Wickelgren admits that the general case divides into two subcases -- where the diameter is in the interior of the angle and whether is not in the interior. The U of Chicago calls these Cases II and III:
Let's move onto Case II. Well, the U of Chicago almost gives us a two-column proof here, so why don't we complete it into a full two-column proof. For Case II, O is in the interior of Angle ABC.
Statements Reasons
1. O interior ABC 1. Given
2. Draw ray BO inside ABC 2. Definition of interior of angle
3. Angle ABC = Angle ABD + Angle DBC 3. Angle Addition Postulate
4. Angle ABC = 1/2 * Arc AD + 1/2 * Arc DC 4. Case I and Substitution
5. Angle ABC = 1/2(Arc AD + Arc DC) 5. Distributive Property
6. Angle ABC = 1/2 * Arc AC 6. Arc Addition Property and Substitution
The proof of Case III isn't fully given, but it's hinted that we use subtraction rather than addition as we did in Case II. Once again, I bring up the Triangle Area Proof -- the case of the obtuse triangle involved subtracting the areas of two right triangles, whereas in the case where that same angle were acute, we'd be adding the areas of two right triangles.
Wickelgren admits that the general case divides into two subcases -- where the diameter is in the interior of the angle and whether is not in the interior. The U of Chicago calls these Cases II and III:
Let's move onto Case II. Well, the U of Chicago almost gives us a two-column proof here, so why don't we complete it into a full two-column proof. For Case II, O is in the interior of Angle ABC.
Statements Reasons
1. O interior ABC 1. Given
2. Draw ray BO inside ABC 2. Definition of interior of angle
3. Angle ABC = Angle ABD + Angle DBC 3. Angle Addition Postulate
4. Angle ABC = 1/2 * Arc AD + 1/2 * Arc DC 4. Case I and Substitution
5. Angle ABC = 1/2(Arc AD + Arc DC) 5. Distributive Property
6. Angle ABC = 1/2 * Arc AC 6. Arc Addition Property and Substitution
The proof of Case III isn't fully given, but it's hinted that we use subtraction rather than addition as we did in Case II. Once again, I bring up the Triangle Area Proof -- the case of the obtuse triangle involved subtracting the areas of two right triangles, whereas in the case where that same angle were acute, we'd be adding the areas of two right triangles.
Wickelgren writes:
"In proving the theorem for each of these two more general special cases, the truth of the theorem for the special case was used in the proof."
Here's the rest of what I wrote last year:
The text mentions a simple corollary of the Inscribed Angle Theorem:
Theorem:
An angle inscribed in a semicircle is a right angle.
The text motivates the study of inscribed angles by considering camera angles and lenses. According to the text, a normal camera lens has a picture angle of 46 degrees, a wide-camera lens has an angle of 118 degrees, and a telephoto lens has an angle of 18. I briefly mention this on my worksheet. But a full consideration of camera angles doesn't occur until the next section of the text, Lesson 15-4 -- but we're only really doing Lesson 15-3 today.
"In proving the theorem for each of these two more general special cases, the truth of the theorem for the special case was used in the proof."
Here's the rest of what I wrote last year:
The text mentions a simple corollary of the Inscribed Angle Theorem:
Theorem:
An angle inscribed in a semicircle is a right angle.
The text motivates the study of inscribed angles by considering camera angles and lenses. According to the text, a normal camera lens has a picture angle of 46 degrees, a wide-camera lens has an angle of 118 degrees, and a telephoto lens has an angle of 18. I briefly mention this on my worksheet. But a full consideration of camera angles doesn't occur until the next section of the text, Lesson 15-4 -- but we're only really doing Lesson 15-3 today.
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