Problem Solving. Kurt Bryan. Here s an amusing little problem I came across one day last summer.

Similar documents
C if U can. Algebra. Name

Take the Anxiety Out of Word Problems

Pre-calculus is the stepping stone for Calculus. It s the final hurdle after all those years of

Main topics for the First Midterm Exam

NIT #7 CORE ALGE COMMON IALS

Direct Proofs. the product of two consecutive integers plus the larger of the two integers

Math101, Sections 2 and 3, Spring 2008 Review Sheet for Exam #2:

ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER / Lines and Their Equations

Distributive property and its connection to areas

Sums of Squares (FNS 195-S) Fall 2014

Big-oh stuff. You should know this definition by heart and be able to give it,

Math 300: Final Exam Practice Solutions

Solution to Proof Questions from September 1st

2.1 Definition. Let n be a positive integer. An n-dimensional vector is an ordered list of n real numbers.

A π day celebration! Everyone s favorite geometric constant!

Proofs. 29th January 2014

Chapter 4 Picture proofs

Math 90 Lecture Notes Chapter 1

Algebra & Trig Review

Solving with Absolute Value

STEP Support Programme. Hints and Partial Solutions for Assignment 4

M155 Exam 2 Concept Review

MATH 104, HOMEWORK #3 SOLUTIONS Due Thursday, February 4

Major Ideas in Calc 3 / Exam Review Topics

base 2 4 The EXPONENT tells you how many times to write the base as a factor. Evaluate the following expressions in standard notation.

A Series Transformations

Math 425 Fall All About Zero

1 Review of the dot product

2017 SUMMER REVIEW FOR STUDENTS ENTERING GEOMETRY

, p 1 < p 2 < < p l primes.

Calculus II. Calculus II tends to be a very difficult course for many students. There are many reasons for this.

Prealgebra. Edition 5

Study skills for mathematicians

Lecture 11: Extrema. Nathan Pflueger. 2 October 2013

A Primer on Statistical Inference using Maximum Likelihood

Algebra Exam. Solutions and Grading Guide

The following are generally referred to as the laws or rules of exponents. x a x b = x a+b (5.1) 1 x b a (5.2) (x a ) b = x ab (5.

Student Activity: Finding Factors and Prime Factors

Mon 3 Nov Tuesday 4 Nov: Quiz 8 ( ) Friday 7 Nov: Exam 2!!! Today: 4.5 Wednesday: REVIEW. In class Covers

Horizontal asymptotes

Quadratic Equations Part I

( ) 0. Section 3.3 Graphs of Polynomial Functions. Chapter 3

Basics of Proofs. 1 The Basics. 2 Proof Strategies. 2.1 Understand What s Going On

M5-2 Exact Trig Values

We are going to discuss what it means for a sequence to converge in three stages: First, we define what it means for a sequence to converge to zero

Introduction. So, why did I even bother to write this?

2.4 The Extreme Value Theorem and Some of its Consequences

In this unit we will study exponents, mathematical operations on polynomials, and factoring.

DIFFERENTIAL EQUATIONS

( )( b + c) = ab + ac, but it can also be ( )( a) = ba + ca. Let s use the distributive property on a couple of

Lecture 8: A Crash Course in Linear Algebra

QUADRATICS 3.2 Breaking Symmetry: Factoring

Algebra. Here are a couple of warnings to my students who may be here to get a copy of what happened on a day that you missed.

Physics 6A Lab Experiment 6

II. Unit Speed Curves

MI 4 Mathematical Induction Name. Mathematical Induction

CALCULUS I. Review. Paul Dawkins

What is proof? Lesson 1

Solving Quadratic & Higher Degree Equations

Solutions for practice questions: Chapter 7, Triangle Trigonometry If you find any errors, please let me know at

MAT137 - Term 2, Week 2

Vectors for Beginners

OPTIMATIZATION - MAXIMUM/MINIMUM PROBLEMS BC CALCULUS

DEVELOPING MATH INTUITION

Induction 1 = 1(1+1) = 2(2+1) = 3(3+1) 2

Solving Equations by Adding and Subtracting

Chapter 5 Simplifying Formulas and Solving Equations

Answers for Calculus Review (Extrema and Concavity)

Chapter 1 Review of Equations and Inequalities

Understanding Exponents Eric Rasmusen September 18, 2018

Completeness, Eigenvalues, and the Calculus of Variations

Ch. 3 Equations and Inequalities

MATH 308 COURSE SUMMARY

Physics 2A Chapter 3 - Motion in Two Dimensions Fall 2017

To factor an expression means to write it as a product of factors instead of a sum of terms. The expression 3x

Section 5.4. Ken Ueda

CS1800: Mathematical Induction. Professor Kevin Gold

3 Fields, Elementary Matrices and Calculating Inverses

PHYSICS LAB: CONSTANT MOTION

Linear Algebra, Summer 2011, pt. 2

Solving Quadratic & Higher Degree Equations

4.4: Optimization. Problem 2 Find the radius of a cylindrical container with a volume of 2π m 3 that minimizes the surface area.

POISSON PROCESSES 1. THE LAW OF SMALL NUMBERS

Roberto s Notes on Linear Algebra Chapter 11: Vector spaces Section 1. Vector space axioms

#29: Logarithm review May 16, 2009

For math conventions used on the GRE, refer to this link:

Tips and Tricks in Real Analysis

MIT BLOSSOMS INITIATIVE

Calculus I. Here are a couple of warnings to my students who may be here to get a copy of what happened on a day that you missed.

5.4 Continuity: Preliminary Notions

Preface. Here are a couple of warnings to my students who may be here to get a copy of what happened on a day that you missed.

Math 5a Reading Assignments for Sections

The First Derivative Test

Abstract & Applied Linear Algebra (Chapters 1-2) James A. Bernhard University of Puget Sound

Getting Started with Communications Engineering

Vector Basics, with Exercises

Spring 2018 Math Week Week 1 Task List

Lesson 8: Why Stay with Whole Numbers?

CLASS NOTES: INTERMEDIATE ALGEBRA AND COORDINATE GEOMETRY

1 The Real Number Line

Transcription:

Introduction Problem Solving Kurt Bryan Here s an amusing little problem I came across one day last summer. Problem: Find three distinct positive integers whose reciprocals add up to one. Prove that the solution is unique. Solution: Let s use m for the smallest of the three (they re distinct, so there is a smallest), n for the next largest, and p for the largest, so 0 < m < n < p. By basic algebra then < <. The goal is to find m, n, and p so that p n m m + n + p =. () First, I claim that m = 2 is the only possibility: If you add up three numbers and the sum equals one then at least one of the numbers is greater than or equal to. In this case is the biggest of the three reciprocals, so 3 m we need, or m 3. In fact, we can t even take m = 3, for then we m 3 have to take n 4 and p 5 and then the reciprocals sum to less than one. So m < 3. Since m has to be a positive integer that leaves only m = or m = 2. But m = doesn t work that would require and to be zero! So n p we have no choice: m = 2. This leaves us with finding just two positive integers n and p with 2 < n < p and n + p = 2. The same kind of reasoning works again! If you add two numbers and get then one of the numbers is greater than or equal to. In this case that 2 4 means, or n 4. But in fact we can t take n = 4, for that forces n 4 p = 4 and then n and p aren t distinct. So 2 < n < 4, which again leaves no choice: n = 3. From equation () we now have no choice for p: we need + + = which gives p = 6. The unique solution is m = 2, n = 3, and 2 3 p p = 6.

It took only a moment to figure this solution out. How d I do so fast? Was it a brilliant insight? Hardly. In fact, my approach was quite methodical. Looking back over the problem, you should realize that although it looked hard, the individual steps consisted of nothing more than VERY easy algebra and common sense. You re going to encounter a lot of problems like this in the next year (and beyond), on tests and homework. Perhaps you won t be able to solve a given problem, but after seeing the solution key you ll slap yourself on the forehead and wonder why didn t I think of that? In many cases an inability to solve the problem isn t due to not understanding the material, but rather to poor problem solving skills. The goal of this handout is to boost your problem solving skills, so that you won t have to slap yourself in the head so much! Most of what follows is from the classic book How to Solve It by George Polya. 2 The Four Basic Steps in Problem Solving To illustrate the basic steps in problem solving, let s apply them to the following simple problem: Problem: Consider an equilateral triangle with sides of length one. What are the dimensions of the largest (area) rectangle which can be inscribed in the triangle, where one side of the rectangle lies on one side of the triangle? Step : Understand the Problem. In particular, ask yourself: (a) What is the unknown? Specifically, what kind of object will the answer be? Will it be a number, a set of numbers, a function, a geometric figure, a logical proof? (b) What data is given? (c) What additional conditions are specified? Of the utmost importance are the following two steps. Draw a picture. If the problem is in any way geometric or physical, a picture is the place to start. 2

Introduce suitable notation. If there are any unknown quantities in the problem they need to represented with letters; otherwise, you can t write down equations and inequalities. Example for Step : For the problem above, the unknown is a rectangle, or more specifically, the dimensions of the rectangle. We should draw a picture of an equilateral triangle with sides of length one and an inscribed rectangle (with one side lying on a side of the triangle), as illustrated in Figure : a b Figure : Rectangle inscribed in triangle. We know the rectangle if we know the length of its sides. This is where we need to introduce notation; we ll label the sides of the rectangle a and b. If you don t do this, you can t write down any equations! Of course the area A of the rectangle is just A = ab. It s worth noting at this point that a and b are not independent choosing a dictates the value of b (and vice-versa). Step 2: Devise a Plan. This is the hard part: Figure out how to get from the given data and conditions to the unknown. In an easy problem you might see how to do this instantly. In a hard problem it takes ingenuity, experimentation, and persistence. Here are some general hints. Look at the unknown. Do you know a related problem and solution that you can use? 3

Experiment and guess. The solution to the problem may be hard to find from scratch, but it might be easy to check whether a guess is correct. If the first guess isn t right, refine it and try again. Doing this repeatedly builds your insight into the problem. It might lead to a plan. If you can t solve the given problem, can you solve a related but easier problem? Try dropping one or more of the conditions and solving it may show you how to do the full problem. Or try adding a condition and see if that makes it easier. If the unknown consists of a set of numbers, then write down every equation you can think of which relates the numbers. Remember the general fact that n numbers can be determined from n equations. Don t forget to use meta-information, that is, information you can glean about the problem that is not explicitly stated. For example, if the problem appears at the end of a section on related rates, you can bet that it s a related rate problem! Example for Step 2: The keyword maximize suggests that calculus might be useful: The problem requires maximizing a certain quantity (area A), so if we can write A as a function of some other variable, say a, we can use derivatives to maximize A as a function of a. Specifically, if we can find an equation relating between a and b, we can solve for b in terms of a and then obtain the area A entirely as a function of a, then maximize. If you carefully label more Figure above as I do below in Figure 2 then you see that by a similar triangles argument, b /2 a/2 = H /2 or b = H( a), where H is the height of the triangle. It s easy to work out that H = 3/2 (since (/2) 2 + H 2 = 2 ). So here s the plan: Write A in terms of a and then maximize A for 0 a. Step 3: Carry out the Plan. Since A = ab and b = 3( a), we have 2 3 A = 2 (a a2 ). 4

H a b /2-a/2 a/2 Figure 2: A few labels added. Take the derivative da/da and set it to zero. This is purely mechanical computation you could feed it to a computer algebra system. The only critical point is at a = /2. The maximum value for A must occur at a = /2 or an endpoint, a = 0 or a =. The latter two yield A = 0, so a = /2 yields the max, with A = 3 8. Step 4: Look Back. Make sure your answer is correct. Some tips: Don t throw common sense out the window make sure your answer is reasonable. For example, check that it doesn t violate well known mathematical or physical facts. Can you do a quick and dirty estimate of the answer and compare to your solution in step 3? Did you use all of the data and conditions that were given? Most standard textbook problems don t have any extraneous or irrelevant data. Warning: real problems usually do. Can you derive the answer using a second, independent approach? Example for Step 4: One thing jibes with common sense: the proposed max is between extremes a = 0 and a = which clearly yield zero area, so that s reassuring. The maximum area is 3/8 0.22, which is half the area of the triangle ( 3/4). This conforms to a visual, common sense estimate. 5

3 Summary/Quick Reference The basics steps in problem solving:. Understand the Problem. What is the unknown? What is the data given? What additional conditions are specified? Draw a picture. Introduce suitable notation. 2. Devise a Plan. Think of a related problem. Experiment. Guess and play with the problem. Solve a related but easier problem, by dropping or adding conditions. Remember n unknowns can be determined by n equations. 3. Carry out the Plan. Check each step carefully. Use a computer to do really complicated computations. 4. Look Back. The solution should check with common sense (if applicable). Compare the answer to a quick and dirty estimate. Did you use all of the given data and conditions? Use a second, independent approach. 4 Your Turn Use the four basic steps understand, plan, carry out, look back to solve the following problems. Consider an equilateral triangle with sides of length one. What are the dimensions of the rectangle of largest perimeter which can be inscribed in the triangle, where one side of the rectangle lies on one side of the triangle? 6

Find all triples of positive integers whose reciprocals sum to one. The integers need not be distinct. A single crease is made in an 8.5 by inch sheet of paper, so that the upper left corner lies on the right-hand edge; see Figure 3, with the crease as the dotted line. What is the shortest possible crease that can be created? Figure 3: Creased sheet of paper. 7