Today on her Mathematics Calendar 2018, Theoni Pappas writes:
Segments m and n are parallel.
Well, if we think in terms of Wayne Wickelgren's book, this isn't a fully stated problem yet because I listed neither the givens nor the goal. The givens and the goal are provided only in the diagram, which I don't post on the blog. Thus I must describe them to you. None of the vertices in this problem are labeled, so I must create point names in order to describe the givens and goal to you:
The givens:
Angle A measures 87 degrees. Lines m and n intersect the sides of angle A in two points each -- we'll call these B and C, D and E respectively (that is, line BC is line m, while line DE is line n). Point F is chosen such that both B and D lie between A and F. Point G is chosen on line n such that D lies between G and E. Angle FDG measures 75 degrees.
The goal:
Angle AED measures x degrees. Find x.
To solve this problem, we must use allowable operations -- previously proved theorems. Two such operations are needed to solve this problem.
Operation #1: Vertical Angle Theorem
Angles FDG and ADE are vertical angles. Angle FDG is 75 degrees. Thus angle ADE is 75 degrees.
Operation #2: Triangle-Sum Theorem
In Triangle ADE, angle A (that is, DAE) is 87 degrees, angle ADE is 75 degrees, and angle AED, our goal angle, is x degrees. Therefore we write:
x + 87 + 75 = 180
x + 162 = 180
x = 18
Here we used more operations -- but these are operations of arithmetic and algebra. We have reached our solution -- x = 18, and of course, today's date is the eighteenth.
We know that of the two geometric operations used here, Vertical Angle appears in Lesson 3-2, and Triangle-Sum is in Lesson 5-7. So this is essentially a Chapter 5 problem. We also notice how the given information that m | | n has nothing to do with the solution of problem. Lines m and n could have intersected -- indeed, line m could have been left out of the problem entirely -- and this would have made no difference to our solution of 18. Perhaps Pappas should have asked for angle ACB, not AED, but our job is to solve the problem at hand rather than create one.
Chapter 3 of Wayne Wickelgren's How to Solve Problems is called "Inference." Here's how it begins:
"Virtually all problems present some of the relevant information in implicit, rather than explicit, form. That is, some of the information concerning givens, operations, or occasionally even goals is presented in a subtle manner that may not strongly attract your attention, unless you know what to look for."
Well, in today's Pappas problem, all of the givens and the goal are explicitly given in the diagram. I suppose we could say that the operations weren't explicitly given -- but we know where to find the allowable operations in our textbook.
He continues:
"Problems often evolve from (a) vaguely formulated to (b) semiprecisely formulated to (c) precisely to partly implicitly formulated to (d) precisely and explicitly formulated stages."
So I suppose that today's Pappas problem counts as stage (c). Of course, without the diagram, it would be a stage (a) problem.
This chapter is all about making inferences, or drawing conclusions from the given information. As Wickelgren tells us:
"Thus, the general problem-solving method described in this chapter may be stated as follows: Draw inferences from explicitly and implicitly presented information that satisfy one or both of the following two criteria:"
(a) the inferences have frequently been made in the past from the same type of information;
(b) the inferences are concerned with properties (variables, terms, expressions, and so on) that appear in the goal, the givens, or inferences from the goal and the givens.
In today's Pappas problem, the inferences made from Vertical Angle and Triangle-Sum satisfy criterion (a), while the others satisfy criterion (b).
The author now describes insight problems:
"Difficult insight problems are often difficult precisely because they require you to draw an inference that is not too close to the top of your hierarchy of inferences from this type of information [criterion (a)]."
Today's Pappas problem would not qualify as an insight problem. But that "sixth grade" problem that I mentioned in my Palm Sunday post (goal: find the area of S) probably does. A clever insight was needed to show that S has the same area as a certain circle sector.
Wickelgren points out that even the goal might be only implicitly stated:
"The goal of the problem is occasionally not completely clear, and the solver must get a precise and correct definition of the goal."
But naturally, most implicit information concerns the other two inputs:
"This other kind of implicit information concerns the properties possessed by each of the givens or operations in a problem."
OK, so let's finally get to some real examples of problems. Wickelgren's first example involves numerical properties -- Karl Duncker's famous 13 problem:
- Prove that all six-place numbers of the form abcabc (for example, 416416 or 258258) are divisible (evenly) by 13.
"Stop reading and try to solve this problem, then read on."
The author tells us that the key to this problem is that abcabc = (abc)(1001), and since 1001 is a multiple of 13, so is abcabc. In fact, since 1001 = (7)(11)(13), we learn that abcabc must be divisible by 7 and 11 as well as 13. It's possible to make divisibility rules for 7, 11, and 13 based on this fact -- in my July 31st post, I referred to it as "cube alpha" and provided the following link:
Many problems involve topological properties. I've written about topology in many previous posts, most recently in my side-along reading book on the famous topologist Poincare last fall. Wickelgren defines topology as the study of the properties of geometric figures that remain unaltered when the figures are stretched, shrunk, and twisted in any regular or irregular way.
The first example of a topological problem is the notched checkerboard problem (often called the mutilated chessboard problem by some sources). This problem is so famous that I'd rather cut and paste from a link than retype Wickelgren's words:
Suppose a standard 8×8 chessboard has two diagonally opposite corners removed, leaving 62 squares. Is it possible to place 31 dominoes of size 2×1 so as to cover all of these squares?
"Stop reading and try to solve this problem," repeats Wicklegren after each problem. I really must emphasize it here -- you should try to solve it before clicking on the solution at the above link.
Wickelgren tells us that, though the colors on the board make it obvious that it's impossible, there remains no solution even if the board is colorless. For example, we might try labeling the squares by ordered pairs from (0, 0) to (7, 7) instead:
"Now one might eventually look for some property common to all pairs of squares that a single domino could cover."
And this property is that one square must have an even sum while the other has an odd sum. This restores the idea that there are two types (or "colors") of squares without actually coloring them.
Wicklegren's next example is the cube-cutting problem:
Can a cube be cut in 27 smaller cubes in less than 6 cuts?
Prove or disprove that it is possible.
Prove or disprove that it is possible.
"Stop reading and try to solve the problem." And this is especially important now, since the solution is in the second post in the thread at the link above. According to the author, that same trick can be used to solve other cutting problems:
"For example, a cube cut into four subcubes requires three cuts, since each of the four subcubes has three unexposed faces that must be cut."
Wicklegren now describes the hardest part of problem solving -- performing the right operations:
"Many practical problems require you to think of a type of operation that will solve the problem. The operation is usually one with which you might be familiar, but thinking of that operation may be far from trivial."
Wicklegren's next example is the one-heavy-coin problem:
https://www.algebra.com/algebra/homework/word/numbers/Numbers_Word_Problems.faq.question.324693.html
"Stop reading and try to solve the problem."
He mentions a common incorrect answer followed by a correct solution, and both of these are listed at the link above (with the wrong answer first). So I'll keep this post spoiler-free -- just go to the link above for the solution.
The next problem is a liar/truth teller problem, also known as knights and knaves. There are so many knights and knave problems out there that it's hard to find an exact link for this one. So let me just give Wicklegren's problem as written (except I changed the authors "liars" and "truars" to "knaves" and "knights" respectively for familiarity:
- Joan says, "Shawn and Peter are both knaves."
- Shawn says, "I am not a knave."
- Peter says, "Shawn is indeed a knave."
- How many of the three are knaves and how many are knights?
"Stop reading and try to solve this problem, then read on."
Wickelgren points out that trying to exactly determine which ones are knaves is impossible, but it's possible to determine exactly how many knaves there are -- two. That is, the goal of determining who the knight is an impossible goal, but the goal of counting the knaves is possible.
At this point Wickelgren quotes Polya -- the same Polya whom traditionalist SteveH holds in disdain:
"As Polya (1962, p. 7) has emphasized, it may be useful to imagine for a moment that you have already solved the problem and ask yourself, 'What would I have?'" Polya variously calls this exercise wishful thinking or taking the problem as solved."
And a natural place for seeing what the goal looks like first before starting is -- Geometry:
"As Polya (1962, p. 7) has emphasized, it may be useful to imagine for a moment that you have already solved the problem and ask yourself, 'What would I have?'" Polya variously calls this exercise wishful thinking or taking the problem as solved."
And a natural place for seeing what the goal looks like first before starting is -- Geometry:
- Given an acute angle UVW and a point P within the angle, use a compass and straightedge to construct a segment
QRpassing through P, such that QP and PR stand in the ratio 2:1, Q and R lying onUVandVWrespectively.
It's easy to make a diagram with an angle V and possible locations for Q and R drawn in, before we attempt the construction. The author tells us that he won't give the solution until Chapter 4, so you'll have to wait until tomorrow.
But here's Wickelgren's second Geometry example:
- Can two triangles have five of their six parts (three sides and three angles) be equal and yet the triangles not be congruent?
"Stop reading and try to solve this problem, using the problem-solving method of explicit representation of the goal and deriving properties of the goal."
To solve this problem, Wickelgren makes several inferences:
- There are two possibilities -- either we have three sides (SSS) and two angles (AA), or else we have two sides (SS) and three angles (AAA). Since SSS is already enough to guarantee congruence, we must be in the second case, SS and AAA. "If you did not solve the problem previously, stop reading and try again."
- The second inference is that since we have AAA, the triangles must be similar -- in fact, AA is already enough to guarantee similarity. "If you did not solve the problem so far, stop reading and try again."
- The next thing to do is draw two similar triangles and label corresponding parts. If we label one triangle as ABC, then the other can be labeled A'B'C'. We can imagine A'B'C' as the image of ABC under a similarity transformation such as a dilation. The sides of ABC can have lengths a, b, c, while the sides of A'B'C' have length a', b', c'. "If you have not solved the problem already, stop reading and try again."
- At this point you might wonder why we can't use ASA (or SAS) to conclude that the triangles are in fact congruent. The reason is that the equal sides are not corresponding sides. "If you still have not solved the problem, stop reading and try again."
- Since the triangles are similar, the sides are in proportion, Thus, a'/a = b'/b = c'/c. "If you have not solved the problem, stop reading and try again."
- Without loss of generality, we let c be the longest side -- c > b > a, c' > b' > a'. "If you have not solved the problem thus far, stop reading and try again."
- The use of the non-strict inequality > implies that the triangles might be isosceles, or possibly even equilateral. But it's easy to show that if the triangles were equilateral, corresponding sides would also be equal, and the same thing happens even if they're merely isosceles. Thus the triangles must be scalene. "If you have not already done so, stop reading and prove this and attempt to solve the rest of the problem."
- Without loss of generality, we assume that A'B'C' is the larger triangle -- that is, the dilation mapping ABC to A'B'C' has scale factor greater than 1. In order to make non-corresponding sides equal, we set a' = b and b' = c. Mercifully, I'm going to skip several of Wickelgren's "if you have not solved...." lines here. (Yes, he really does keep repeating that line!)
- In fact, if we let k > 1 be the scale factor of the dilation, then k = a'/a = b'/b = c'/c. In other words, b = a' = ak and c = b' = bk = ak^2.
This tells us that the three sides of the original triangle are a, ak, ak^2. Scaling this up by k gives us a triangle with sides ak, ak^2, ak^3. Since the sides are in proportion, the triangles are similar and so we don't even need to check the angles to be sure that they are equal. The three sides are said to follow a geometric progression -- the middle side is the geometric mean (as defined in the current chapter, Lesson 14-2) of the other two sides.
Wicklegren warns us that there's just one more thing to check -- the Triangle Inequality. If k is too large, the Triangle Inequality will fail. We must make sure that ak^2 doesn't exceed ak + a -- in other words, that k^2 doesn't exceed k + 1. Notice that the positive solution of k^2 = k + 1 happens to be Phi, the golden ratio. So k must be strictly more than 1 and strictly less than Phi. Thus, the original problem is solved.
We can find specific solutions for various values of k between 1 and Phi. If we try k = 3/2, then we might look at the geometric sequence 8, 12, 18, 27. So the smaller triangle has sides 8-12-18 and the larger triangle has sides 12-18-27, and these are similar triangles.
The most interesting case is k = sqrt(Phi). Our smaller triangle will have sides 1-sqrt(Phi)-Phi -- and notice that this is a Pythagorean triple since 1 + sqrt(Phi)^2 = 1 + Phi = Phi^2. Therefore, this triangle is a right triangle. The larger triangle has sides sqrt(Phi)-Phi-Phi sqrt(Phi). If we follow Lesson 14-2 and draw the altitude to the larger triangle, then the altitude is 1 and the larger of the two resulting triangles is the 1-sqrt(Phi)-Phi triangle. The smaller triangle is phi sqrt(Phi)-1-sqrt(Phi). (Recall that lowercase phi equals Phi - 1 = 1/Phi.)
OK, this post has gone on long enough. Wickelgren provides examples of chain -cutting problems before ending the chapter with the same chess example from yesterday. I'll just provide a link -- no pun intended:
and then conclude with what he writes about the chess problem:
"The essential strategy for solving the chess problem comes by conjecturing that you wish to have the white rook at h1, without the possibility of black blocking the attack by his bishop, and then working forward to determine what white must do at each move in order to achieve that terminal checkmate position."
Lesson 14-7 of the U of Chicago text is on adding vectors using trigonometry -- and we can't skip it because it appears in the following Common Core standard:
CCSS.MATH.CONTENT.HSN.VM.B.4.B
Given two vectors in magnitude and direction form, determine the magnitude and direction of their sum.
Notice that we are essentially converting the vectors from polar to rectangular form, adding their components, and then converting the sum back to polar form. And all of this is done without the students' even knowing what polar coordinates are.
In an actual trig course (as part of precalculus), we find out that we can avoid thinking about angles larger than 90 degrees by considering the reference angle -- the angle formed by the x-axis and the terminal side of the original angle. We see that even though the U of Chicago's Geometry text doesn't teach reference angles, the angles shown in the text are always formed using the x-axis -- that is, the west-east axis -- and never using the y-axis. So the U of Chicago, while not explicitly teaching reference angles, clearly has these reference angles in mind when writing this chapter.
The text states that one of the two components is found using the sine of the angle, while the other is using the cosine. But because of the way the angles are drawn, the horizontal component will always use the cosine and the vertical component will always use the sine -- just as they would be for polar coordinates in a trig class.
To convert from rectangular back to polar coordinates, we use the distance formula and the inverse tangent function. This is the only time that an inverse trigonometric function appears in the U of Chicago text -- although many other geometry texts discuss inverse trig in more detail.
CCSS.MATH.CONTENT.HSN.VM.B.4.B
Given two vectors in magnitude and direction form, determine the magnitude and direction of their sum.
Notice that we are essentially converting the vectors from polar to rectangular form, adding their components, and then converting the sum back to polar form. And all of this is done without the students' even knowing what polar coordinates are.
In an actual trig course (as part of precalculus), we find out that we can avoid thinking about angles larger than 90 degrees by considering the reference angle -- the angle formed by the x-axis and the terminal side of the original angle. We see that even though the U of Chicago's Geometry text doesn't teach reference angles, the angles shown in the text are always formed using the x-axis -- that is, the west-east axis -- and never using the y-axis. So the U of Chicago, while not explicitly teaching reference angles, clearly has these reference angles in mind when writing this chapter.
The text states that one of the two components is found using the sine of the angle, while the other is using the cosine. But because of the way the angles are drawn, the horizontal component will always use the cosine and the vertical component will always use the sine -- just as they would be for polar coordinates in a trig class.
To convert from rectangular back to polar coordinates, we use the distance formula and the inverse tangent function. This is the only time that an inverse trigonometric function appears in the U of Chicago text -- although many other geometry texts discuss inverse trig in more detail.
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