The joy of problem solving

THE JOY OF PROBLEM SOLVING IN MATHEMATICS: FROM FUN PUZZLES TO CHALLENGING PROBLEMS. BOYAN KOSTADINOV 1. Introduction I strongly believe that advanced math can be taught at an early age. I know from experience that high school students are capable of understanding college math if it is presented from first principles in an intuitive way, motivated by practical examples that stimulate the imagination. My lesson attempts to demonstrate that this belief is attainable as long as we make a good effort to explain hard math in an easy way. I would discuss the following strategies, in terms of examples: • • • • • • • • • • Account for all possibilities or outcomes. Make an intelligent guess and test it. Use logical reasoning to connect unknowns in equations. Look for a pattern and make a deduction. Use mathematical induction to prove algebraic formulas. Exploit Symmetry. Make a table (or make an organized list). Draw a diagram or picture to facilitate the logical thought process. Adapt another point of view. Use numerical approximations. I would provide step by step solutions to the problems marked as Example while the problems marked as Homework would be left for independent work. None of the problems requires Calculus but only algebra and geometry, basic probability, linear functions and some basic knowledge of rate of change and the chain rule. 2. Fun Puzzles with little Math Example 1. $ Matters A man called his daughter to ask her to buy a few things he needed for a trip. He told her she would find enough dollar bills for the purchases in an envelope on his desk. She found the envelope with 98 written on it. She then bought $90 worth of things, but when it was time to pay, not only did she not have $8 left over but she was even short. The question is how much she was short and why? Date: April 4 2009. 1 2 BOYAN KOSTADINOV Solution 1. There is no math here and the point is to question the assumption that we make about how much money we think there is in the envelope. The symmetry of 98 suggests that she simply saw the envelope from the opposite side, so the actual number was 86 and that’s why she was short $4. Example 2. Where is the boat? Two Coast Guard radar stations receive distress radio signals from a boat. The first radar station estimates that the boat is 75 miles away from the station and the second radar station estimates that the boat is 90 miles away from them. How can you use this information to find the precise location of the boat? Figure 1. Where is the boat? Solution 2. We only know the distance and all possible locations at the same distance from the radar station form a circle, whose radius is that distance. So, we can find the precise location by drawing two circles centered at the two radar stations and where the two circles intersect, we find the precise location of the boat. The two circles will intersect at two points, but only the point at sea is of interest to us, unless there is a lake inland where the boat might also be located. Example 3. How old am I? When my father was 42, I was 6. Now, he is twice as old as I am. How old am I now? Solution 3. The difference in ages is preserved over time. On the one hand, the difference in ages in the past was 42 − 6 = 36. On the other hand, if x is my age now, then this same difference now is 2x − x = x. Thus, x = 36. Homework 1. Burning ropes You have two ropes, each of which takes 1 hour to burn. But ether rope has different densities at different points, so there is no guarantee of consistency in the time it takes different sections within the rope to burn. How do you use these two ropes to measure 45 min? Hint: Think first how you would measure 30 min. THE JOY OF PROBLEM SOLVING IN MATHEMATICS:FROM FUN PUZZLES TO CHALLENGING PROBLEMS. 3 Homework 2. Counterfeit coins There are 10 bags with 100 identical coins in each bag. In all bags but one, each coin weighs 10 grams. However, the coins in the counterfeit bag, all of which have the same weight, weigh either 9 or 11 grams. Can you find the counterfeit bag in only one weighing, using a digital scale that tells the exact weight? Hint: Think of a way to be able to identify different bags. 3. Account for all outcomes Example 4. Girls or Boys (1) I have 2 children. They are not both boys. What is the probability that both children are girls? (2) A math teacher has 2 children. The older is a boy. What is the probability that both children are boys? Solution 4. We have 4 possible outcomes in both cases: (1) We can list these possibilities as ordered pairs. For example, the pair (g,b) means that the younger child is a girl and the older is a boy. So, we have (g,g),(g,b),(b,g),(b,b). We know that not both of them are boys. Thus, we can remove the last possibility (b,b) and we have only 3 possibly outcomes. Since there is only one outcome when both kids are girls, the probability that both children are girls is one out of 3 cases, that is 31 . (2) The new information now is that the older kid is a boy. So, we have now only two possible outcomes: (g,b), (b,b). Thus the probability that both kids are boys is 21 . Homework 3. Probability Practice (1) You flip a fair coin twice and you know that at least one of the flips is a head. Knowing that, what is the probability that both flips turn out to be heads? (2) You flip a fair coin twice. The first time you see that it comes up heads. Knowing that, what is the probability that both flips turn out to be heads? 4. Intelligent Guessing and Testing Example 5. Exam Time Rahul took a 20-question multiple-choice test. The test was scored +5 for each correct answer, -2 for each wrong answer, and 0 if the question was skipped. Rahul scored a 44, even though he skipped some questions. How many questions did Rahul skip? 4 BOYAN KOSTADINOV Solution 5. Let us attempt to solve this problem algebraically. Let x = the number of questions answered correctly. y = the number of questions answered incorrectly. z = the number of questions skipped. Then, based on the given information, we have x + y + z = 20, 5x − 2x + 0z = 44. Effectively, we have only two equations containing three variables. Students would expect to find another equation if they are to solve the system and arrive at a unique set of answers but this is not really the case, which may puzzle some students. The important thing to realize here is that we are looking for a positive integer solution not any real solution and the integrality of the solution is this missing extra constraint that would lead to a unique solution. Also, we know that there must be some nonzero value for z, because the problem stated that Rahul skipped some questions. If we now express y = 20 − x − z, we obtain the equation 7x + 2z = 84, getting z= 84 − 7x . 2 We must have z > 0 integer, so 84 − 7x must be even or x must be even. This gives that x < 12 and it must be even but even if we don’t solve the inequality z > 0 for x, we can simply make some guesses and check what we get x = 12, z = 0 x = 11, z = 3.5 x = 10, z = 7, y = 3 x = 9, z = 10.5 x = 8, z = 14, y = −2 The conditions of the problem now lead us to the unique answer that Rahul skipped 7 questions, because we know he did not skip 0 nor could he skip a negative or noninteger number of questions. Homework 4. Find the smallest integer n such that the sum of the first n positive integers exceeds 1000. THE JOY OF PROBLEM SOLVING IN MATHEMATICS:FROM FUN PUZZLES TO CHALLENGING PROBLEMS. 5 5. Linear Relations Example 6. On a foot-long ruler, a rubber band stretches from the 1-inch mark to the 10-inch mark. Suppose the ends of the rubber band are further stretched so that the rubber band lies exactly over the entire 12-inch ruler. During the stretching, points near the left end of the rubber band move left, and those near the right end move right. Somewhere in between, there has to be some point that ends up where it started. Where on the ruler is that point? Figure 2. Rubber Band Stretching → Linear Map Solution 6. Think of the Ruler as the x-axis. Stretching the rubber band is equivalent to having a map (function) f that sends the points 1 and 10 to 0 and 12, respectively. f f 1 → 0, 10 → 12 In general, any linear function can be written as f (x) = a + bx, so we have 2 parameters to be determined, a and b. Since we have 2 conditions, we can get 2 equations, from which we can find the 2 unknown parameters. So, we look for a linear function f that will give us the solution, namely: 0 = f (1) = a + b, 12 = f (10) = a + 10b The first equation gives us b = −a and substituting into the second we get a = − 43 , so the linear map that models the stretching is given by 4 f (x) = (x − 1) 3 We could have also guessed this form since we know that x = 1 must go to 0, so we must have the factor (x − 1). For x = 10 this factor gives 9, so we must divide by 9 and multiply by 12 so that we can also send 10 to 12. The point x0 that will not change its position under stretching is mapped by f to itself, that is 4 f (x0 ) = x0 ⇒ (x0 − 1) = x0 ⇒ x0 = 4 3 Such a point is called a fixed point and in our case we get x0 = 4. 6 BOYAN KOSTADINOV Homework 5. A candle is burning. It originally had a length of 15 cm. After 20 minutes it is only 13 cm long. How long will the candle be burning altogether? How long will the candle remain burning (after these 20 minutes)? Hint: Draw a picture and exploit the linear geometry. 6. Look for Patterns, Make Deductions and Prove by Induction Example 7. Number Patterns (1) Around 200 years ago, a math teacher asked his students to add the first 100 integers, hoping that this would take them so much time so that he could relax for a bit and read his book. Imagine his surprise, when one of his students, named Gauss (who later became a world famous mathematician), came up with the answer in less than a minute. Instead of adding the numbers one by one, Gauss derived a closed formula for the sum of the first n integers: n(n + 1) 2 We shall discuss how Gauss solved the problem in less than a minute. (2) We will look at the usual multiplication table from an unusual point of view and we will discover patterns that will lead us to the famous ancient formula 2  n(n + 1) 3 3 3 3 2 1 + 2 + 3 + · · · + n = (1 + 2 + 3 + · · · + n) = 2 1 + 2 + 3 + ··· + n = Solution 7. Number Patterns (1) The story says that Gauss had the following idea. Let the sum be S and let’s write it in two different ways, then add the two rows: 1 + 2 + 3 + ··· + n = S n + (n − 1) + (n − 2) + · · · + 1 = S −−−−−−−−−−−−−−−−−− (n + 1) + (n + 1) + (n + 1) + · · · + (n + 1) = 2S {z } | n n(n + 1) = 2S ⇒ S = n(n + 1) 2 (2) Consider the multiplication table for the first n integers: THE JOY OF PROBLEM SOLVING IN MATHEMATICS:FROM FUN PUZZLES TO CHALLENGING PROBLEMS. 7              1 2 3 4 2 4 6 8 5 ... n  10 . . . 2 n    3 6 9 12 15 . . . 3 n    4 8 12 16 20 . . . 4 n   5 10 15 20 25 . . . 5 n    ... ... ... ... ... ... ...  n 2 n 3 n 4 n 5 n . . . n2 In this table, the product, say 15, is at the intersection of the row and the column of its factors, 3 and 5, or 5 and 3. One can observe the following patterns: (a) The sum of the numbers in a square array, with 1 at the upper-left corner, is itself an exact square: 1 = 12 , 1st square 1 + 2 + 2 + 4 = 9 = 32 = (1 + 2)2 , 2nd square 1| +{z 2 + 3} + |2 +{z 4 + 6} + |3 +{z 6 + 9} = 36 = 62 = (1 + 2 + 3)2 , 3rd square 1st column 2nd column 3rd column (b) Imagine that we have made corridors bending around at a right angle at the diagonal of the table.The sum of the numbers in a corridor surprisingly turns out to be a cube: 1 = 13 , 1st corridor 2 + 4 + 2 = 8 = 23 , 2nd corridor 3 + 6 + 9 + 6 + 3 = 27 = 33 , 3rd corridor (c) The n × n square array consists of n corridors. On the one hand, based on observation (a), the sum of all numbers in the n × n array is (1 + 2 + 3 + · · · + n)2 On the other hand, using observation (b), the sum of all n corridors is 13 + 23 + 33 + · · · + n3 By equating both, we can make the conjecture that 13 + 23 + 33 + · · · + n3 = (1 + 2 + 3 + · · · + n)2 Now, one can prove this conjecture by induction on n. 8 BOYAN KOSTADINOV This is how proving by induction works: • The result holds for n = 1 : 13 = 12 . • (induction hypothesis) We assume the result holds for n > 1 13 + 23 + 33 + · · · + n3 = (1 + 2 + 3 + · · · + n)2 • Finally, using the induction hypothesis, we want to show that the result holds for n + 1: ? 13 + 23 + · · · + n3 + (n + 1)3 = (1 + 2 + · · · + n + (n + 1))2 Let’s expand the RHS above and check if it equals the LHS: (the right-hand side) = (1 + 2 + · · · + n + (n + 1))2 = = (1 + 2 + · · · + n)2 + 2 (1 + 2 + · · · + n)(n + 1) + (n + 1)2 = | {z } n(n+1) 2 = (1 + 2 + · · · + n)2 + n(n + 1)2 + (n + 1)2 = (by induction hypothesis) = (1 + 2 + · · · + n)2 +(n + 1)3 = | {z } 3 3 3 13 +23 +33 +···+n3 3 = 1 + 2 + · · · + n + (n + 1) = (the left-hand side) This proves by induction the famous ancient formula 3 3 3 3 2 1 + 2 + 3 + · · · + n = (1 + 2 + 3 + · · · + n) =  n(n + 1) 2 2 In his historic book on arithmetic ‘Liber abaci’, which appeared in 1202, the Italian mathematician Leonardo of Pisa, known later by his nickname Fibonacci, which means ‘son of Bonacci’, gives the following rabbit problem: Homework 6. Rabbits...and lots of them! How many pairs of rabbits can be produced from a single pair in 6 months if every month each pair produces a new pair, which from the second month on becomes productive? Hint: Denote by F (n) the number of pairs of rabbits in n months and try to relate it to the number of pairs, one month earlier F (n − 1) and two months earlier F (n − 2), using the given information. In this work, Fibonacci introduced to Europe the Arabic numerals that we use today, which he had learned by studying with Arabs while living in North Africa with his father. Liber Abaci was among the first Western books to describe Arabic numerals, the first being by Pope Silvester II in 999. THE JOY OF PROBLEM SOLVING IN MATHEMATICS:FROM FUN PUZZLES TO CHALLENGING PROBLEMS. 9 7. Look for a Different Point of View Example 8. Algebra versus Geometry We will give a geometric proof (using √ only squares and the area of a square) of the famous algebraic result that 2 is an irrational number, that √ is, it cannot be written as the ratio of two positive and 2 6= m , where m, n are positive and relatively prime relatively prime integers: n integers; integers are relatively prime if they have no common factor other than 1 or, equivalently, if their greatest common divisor is 1 and it is written as gcd(m, n) = 1. Figure 3. Algebra vs. Geometry √ Solution 8. If 2 = m , then m2 = 2n2 . Let’s look at the problem in a geometric n way. Could there be two squares with side equal to an integer number n whose total area 2n2 is identical to that of a single square with side equal to another integer number m? Let’s assume that it can be done. Then there must be a smallest integer number m for which it can be done. Let’s draw a picture using that smallest possible m. Let’s stick the two small grey squares in the top right and bottom left corners of the big square. Now, part of the big square is covered twice, and part of the big square isn’t covered at all, by the smaller squares. The part that’s covered twice is shown in dark grey, and the bits that are not covered are shown in white. Since we assumed the area of the original big white square is exactly equal to the total area of the light grey squares, the area of the bit that’s covered twice must be exactly equal to the area of the bits that are not covered. Now, what are the sizes of these three areas? The dark grey bit is a square, and the size of that square is a whole number, equal to 2n − m > 0 and the two white areas are also squares, with a side equal to m − n. So, starting from the alleged smallest possible whole number m, such that m2 is twice the square of a whole number, we’ve found that there is an even smaller whole number having this property, namely: (2n − m)2 = 2(m − n)2 10 BOYAN KOSTADINOV So there can be no smallest solution. Remember, if there are any solutions, one of them must be the smallest since we are dealing with positive integers. So we conclude that there are no solutions. This result shows that not all real numbers are the ratio of whole numbers. Example 9. No Calculus, just Algebra The profit P of a company depends on two variables x and y according to the formula P = 20x − 3y. The variables x and y are restricted to the ranges 0 ≤ x ≤ 3, 0 ≤ y ≤ 7, 2x ≤ 5y. What choice of x and y achieves the maximum profit? Figure 4. No Calculus, just Algebra Solution 9. The given inequalities restrict the variables (x, y) to the region on Figure4. The 3D graph of P (x, y) is a plane (linear in both x and y) lying over this region. Intuitively, it is clear that the maximum value of P over this region must occur for some value on the boundary of this region. So, we need to examine the function P (x, y) = 20x − 3y on the boundary. On side I, x = 0 and y varies from 0 to 7, so the maximum value of P on this boundary is 0. On side II, y = 7 and x varies from 0 to 3, so the max value of P is 60 − 21 = 39. On side III, x = 3 and y varies from 56 to 7. The max value here is 6 (20)(3) − (3) = 56.4 5 THE JOY OF PROBLEM SOLVING IN MATHEMATICS:FROM FUN PUZZLES TO CHALLENGING PROBLEMS. 11 attained at (3, 65 ). Finally, on side IV, y = 25 x, so 94 6 P = 20x − x = x. 5 5 Since x runs from 0 to 3, the max value is again 56.4, achieved at the corner (3, 65 ). Homework 7. Solve for x: (x − 3)3 + (x − 7)3 = (2x − 10)3 , by expanding only the right-hand side. Hint: Use the formula (a + b)3 = a3 + b3 + 3a2 b + 3ab2 , and the new variables a = x − 3 and b = x − 7. 8. Use Symmetry Example 10. Continued Fraction Find the value of the infinite fraction: 2 y =2+ 2 + 2+2 2 ··· Solution 10. The thing to spot here is that what appears under the first division line is merely another copy of y, so that y satisfies the equation: 2 y =2+ , y which gives a quadratic equation y 2 − 2y − 2 = 0. The two roots are √ √ −( 3 − 1), 3 + 1 √ and our answer is the positive solution y = 3 + 1. Example 11. Nature likes to take the shortest path Two points A and B lie on the same side of a line L. Show how to construct the shortest path from A to B which touches the line L. Hint: If the point A were a source of light, what do you think would happen then? Solution 11. One can solve this problem using Calculus but then one needs to know how to minimize a function plus it is hard work this way. There is much simpler way based on every-day intuition. Think of the line L as a mirror and ask yourself: What would a ray of light do to get from A to B? The answer, as we all know, is that the light reflects from the mirror at the point C, where the angle of incidence is equal to the angle of reflection. The question now is why AC + CB represents the shortest 12 BOYAN KOSTADINOV Figure 5. Reflection in L path? Let’s consider the mirror image B ′ of B w.r.t. the mirror L, that BD = B ′ D and the angle at D is a right angle. Draw the line segment AB ′ , which intersects L at C. Then, it’s easy to see that C is the point of reflection if you think about the angles at C. Indeed, ∠ACC ′ = ∠DCB ′ and ∠BCD = ∠DCB ′ , so the angle of incidence ∠ACC ′ is equal to the angle of reflection ∠BCD. If C ′ is any point different from C on the line L, then because of the triangle inequality for △AC ′ B ′ AC ′ + C ′ B ′ > AB ′ = AC + CB ′ , but since C ′ B ′ = C ′ B and CB ′ = CB, since B ′ is the mirror image of B, we get AC ′ + C ′ B > AC + CB, for any C ′ 6= C This proves that the point of reflection C is the best choice minimizing the distance. Homework 8. Golden Ratio Find the value of the following special fraction: 1 α=1+ 1 + 1+1 1 ··· The number α is the famous Golden Ratio. Homework 9. Infinite sequence Find x if xx x··· =2 9. Numerical Methods: The Bisection Method Example 12. The Bisection method Show that the equation x3 −2x2 +5x−7 = 0 has a positive root, and find its value to the nearest integer. THE JOY OF PROBLEM SOLVING IN MATHEMATICS:FROM FUN PUZZLES TO CHALLENGING PROBLEMS. 13 Solution 12. Consider the graph of the function f (x) = x3 − 2x2 + 5x − 7. When x = 0, f (0) = −7, and when x is big, say 100, f (x) is positive. Since the graph of f (x) is below the x-axis at x = 0, and above the x-axis at x = 100, and since the graph of f (x) is continuous, it must cross the x-axis between x = 0 and x = 100, and so it has a positive root. Locating the root is done by bisecting the initial interval and iterating this process until a given accuracy is reached. We basically hunt for a change in the sign of f (x) : f (1) = −3, f (2) = 3. These values show that there is a = 1.5 : root between 1 and 2. We then evaluate at the mid-point 1+2 2 5 f (1.5) = − < 0 ⇒ the root x0 ∈ (1.5, 2), f (x0 ) = 0 8 Thus, the approximate value of the root, to the nearest integer, is 2. Note that we can use the bisection method to find the value of the root to any given accuracy. Homework 10. Use the Bisection method to find an approximate value of 3 digit accuracy, without using a calculator. √ 2 to a 10. Rate of Change and Applications Example 13. Spider’s skills A hollow cone of semi-vertical angle arctan( 21 ) is fixed vertex-downwards with its axis vertical and is being filled with water at a constant rate k. A spider which can run at speed u is asleep at the end of a web, hanging at a height h0 vertically above the vertex. Find, in terms of k and h0 , the minimum value of u that will enable the spider to escape a soaking, assuming that he starts to run upwards as soon as the water touches his feet. Figure 6. Spider’s skills 14 BOYAN KOSTADINOV Solution 13. This problem requires more solid knowledge in derivatives. Let the height of the water in the cone after time t be h(t). The volume V (t) of the water cone at time t is then 1 V (t) = πr(t)2 h(t), 3 where r(t) is the radius at time t of the cross-sectional circle, representing the surface of the water. Let the semi-vertical angle be α = arctan(1/2). From the figure h(t) r(t) = h(t) tan(α) = 2 Thus the volume becomes a function of h(t) only π V (t) = h(t)3 . 12 We are given that the rate of change of the volume w.r.t. time is k, that is dV (t) = constant k= dt The key observation is that if the spider is to escape the coming water he should run upwards at least at the speed of the rising water as soon as the water reaches level h0 . The speed of the rising water is the rate of change of the water height h(t) w.r.t. time, that is dh(t) , so we must have dt dh(t) dt t0 ≤ u, | is the derivative evaluated at time t0 when the water reaches levelh0 . where dh(t) dt t0 We assume the students are familiar with derivatives and the chain rule, so that they can relate the two derivatives using the formula for the volume V (t), namely
d π π d dh(t) h(t)3 = 3h(t)2 V (t) = dt 12 dt 12 dt We evaluate the expression above at t = t0 , using that h(t0 ) = h0 k= π π dh(t) ≤ h20 u (3h20 ) 12 dt t0 4 and so to avoid soaking, the speed of the spider must satisfy the inequality k= u≥ 4k πh20 Homework 11. A spherical balloon is inflated so that its volume V increases at a constant rate of 3 cm3 per second. Find the rate at which the radius r is changing 3 when r = 2.5 cm. Answer: r is increasing at a rate of 25π cm/s when r = 2.5cm.