MathCounts Preparation · MathCounts Preparation How to Excel at Middle School Math Competitions By Huasong Yin

MathCounts Preparation How to Excel at Middle School Math Competitions

By

Huasong Yin

www.jacksonareamath.com

2014, Huasong Yin

ALL RIGHTS RESERVED

This book contains material protected under International and Federal Copyright Laws and Treaties. Any unauthorized reprint or use of this material is prohibited. No part of this book may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or by any information storage and retrieval system without express written permission from the author.

ISBN-13: 978-1-62890-725-4

ISBN-10: 1-62890-725-8

Printed in the United States

Print Date: 12/26/2013

Email [email protected] for any errors or mistakes

Table of Contents Chapter 1 Number Sense and Speed Calculation .................................................................................. 2

Addition ........................................................................................................................................... 2

Subtraction ...................................................................................................................................... 3

Multiplication ................................................................................................................................... 4

Division ............................................................................................................................................ 7

Mental Skills for Problem Solving ..................................................................................................... 8

Translating from Verbal to Algebra ................................................................................................. 14

Chapter 2 Exponents and Fractions .................................................................................................... 17

Exponential Notation and Properties of Exponents ........................................................................ 17

Place Values and Number Systems ................................................................................................. 21

Powers of Ten and Scientific Notation ............................................................................................ 23

Divisibility, Prime Numbers and Prime Factorization ...................................................................... 28

Fractions ........................................................................................................................................ 42

Ratio, Proportion and Percent ........................................................................................................ 51

Speed - Distance Time ................................................................................................................. 60

Chapter 3 Probability, Counting and Basic Statistics ........................................................................... 64

Sets, Subsets, Set Operations and Logic .......................................................................................... 64

Counting Techniques, Permutations and Combinations ................................................................. 72

Probability ...................................................................................................................................... 91

Mean, Median, Mode, Range and Other Means ........................................................................... 101

Chapter 4 Elementary Algebra ......................................................................................................... 110

Variables and Algebraic Expressions ............................................................................................. 110

Using (a + b)2 = a2 + 2ab + b2 ......................................................................................................... 112

Difference of Two Squares: Using (a + b)(a b) = a2 b2 .............................................................. 116

Square Roots and Simplifying Square Root Expressions ................................................................ 119

Chapter 5 Geometry ........................................................................................................................ 122

Perimeter and Area ...................................................................................................................... 122

Angles, Degree Measurement and Polygons ................................................................................ 127

Triangles ...................................................................................................................................... 134

Pythagorean Theorem and Special Right Triangles ....................................................................... 139

Chapter 1: Number Sense and Speed Calcula on --- Addi on

1

Triangles and Trapezoids .............................................................................................................. 149

Circles .......................................................................................................................................... 155

Solid Objects, Polyhedrons, Volumes and Euler Theorem ............................................................. 159

Chapter 6 Sequences........................................................................................................................ 163

Sequences and Pattern Recognition ............................................................................................. 163

The Story of Little Gauss and Arithmetic Sequences ..................................................................... 164

The Smart Pizza Eater and Geometric Sequences ......................................................................... 168

Other Sequences .......................................................................................................................... 171

Chapter 7 Equations, Inequalities and Functions .............................................................................. 180

Linear Equations........................................................................................................................... 180

Quadratic Equations ..................................................................................................................... 188

Inequalities .................................................................................................................................. 195

Functions ..................................................................................................................................... 199

Chapter 8 A Little Bit of Analytic Geometry ...................................................................................... 202

Number Lines and the Coordinate System .................................................................................... 202

Lines and Circles in a Coordinate System ...................................................................................... 213

Chapter 9 A Little Bit of Number theory ........................................................................................... 226

Repeating Decimals ...................................................................................................................... 226

Modular Arithmetic ...................................................................................................................... 230

Chapter 10 Practice Tests ................................................................................................................. 234

Some Test-Taking Tips .................................................................................................................. 234

Practice Test #1 ............................................................................................................................ 235

Practice Test #2 ............................................................................................................................ 240

Practice Test #3 ............................................................................................................................ 245

Practice Test #4 ............................................................................................................................ 250

Practice Test #5 ............................................................................................................................ 255

Solutions to Practice Test 1 .......................................................................................................... 260





Chapter 1: Number Sense and Speed Calcula on --- Mul plica on

4

Solu on: If we take $1.99 off the Quarterly, he will only need to pay a total of 17.89 - 1.99 = 17.89 2 + 0.01 = 15.89 + 0.01 = 15.90 dollars. Due to this price decrease, the two magazines cost the same. So

95.705.082

1.016290.15

dollars is the price of the MathTown Monthly. Answer: 7.95 dollars

Multiplication Multiplication is commutative and associative, i.e. abba and cbacba for any numbers a , b and c . So when we multiply a bunch of numbers, we can rearrange the orders. For example,

100000100010012582541252584 . Relating multiplication to addition and

subtraction, we have the distributive property, i.e. cabacba . For example,

461064106 and this is how we multiply 6 with 14 . Most of the interesting results in algebra are derived from the distributive property.

Using the distributive property we will see many mental skills involving multiplications.

Example 1: ?31258542

Solu on: 248000248100103182545231258542

Mental Skill: Mul plying by 9, 99, 999 etc. 1, 99 1 and use the distributive property. This method sometimes can be applied to multiplying by 98, 97, 998, 997, 9998, etc.

783 = 782217

527 × 998 = 527 × (1000 2) = 527 × 1000 527 × 2 = 527000 1054 = 525946

Mental Skill: Mul plying by 11, 101, 1001 etc. Think 11 as 10 + 1, 101 as 100 + 1, 1001 as 1000 + 1 and use the distributive property. This method sometimes can be applied to multiplying by 12, 102, 1002, 1003, etc.

34 × 11 = 34 × (10 + 1) = 34 × 10 + 34 × 1 = 340 + 34 = 374. Note the middle digit is 3 + 4 = 7 and the result is obtained by inserting 7 in the middle of 34.

12 × 11 = 132. Just insert 1 + 2 = 3 in the middle of 12.

62 × 11 = 682. Just insert 6 + 2 = 8 in the middle of 62.


5

78 × 11 = 78 × (10 + 1) = 78 × 10 + 78 × 1 = 780 + 78 = 858. Note 7 + 8 = 15. Insert 5 in the middle and add 1 to the digit 7.

67 × 101 = 67 × (100 + 1) = 67 × 100 + 67 × 1 = 6700 + 67 = 6767

673 × 101 = 673 × (100 + 1) = 673 × 100 + 673 × 1 = 67300 + 673 = 67973

726 × 1001 = 726 × (1000 + 1) = 726000 + 726 = 726726

Mental Skill: Mul plying by Using the Difference of Two Squares.

This technique is derived from the formula 22 bababa .

It works when a is a good number whose square is easy to see and b is small. a is the average of the two factors (the middle number between the two factors).

2 12

2 22

497 × 503 = 2 32

Example 2: If 2998×3002 4 is simplified what is the sum of the digits?

Solu on: 2998× 4 = (3000 - 2) × (3000 +2) 4 = 30002 22 4 8 = 8999992

The sum of the digits = 8 + 9 + 9 + 9 + 9 + 9 + 2 = 9×5 +10 = 55

Example 3: Two numbers have a total of 80 and a difference of 4. What is the product of the two numbers?

Solu on: The average of the two numbers is half of the total, which is 40. So the two numbers are 38 and 42. 38 × 42 = 402 22 = 1600 4 = 1596 Answer: 1596

Example 4: In a rectangle whose perimeter is 400, the difference between the width and the length is 6. Find the area of the rectangle.

Solu on: The sum of the width and the length is half of the perimeter, which is 200. The middle number between the width and the length is 100. Because the difference between the width and the length is 6, they are 103 and 97. 103 × 97 = 1002 32 = 10000 9 = 9991 Answer: 9991

Mental Skill: Squaring a 2-digit number ending with 5

This technique is derived from the formula (a + b)2 = a2 + 2ab + b2. This formula works perfectly if all of the three terms on the right side are easy to see. To demonstrate this 2.

452 = (40 + 5)2 = 402 + 2×40×5 + 52 = 1600 + 400 + 25 = 2025.

If the last digit is 5, we will have 251001005510210510 22222 aaaaa


6

251001251002 aaaa , where a is obtained from the number to be squared by deleting the last digit 5. The square of an integer whose last digit is 5 will end with 25 and start with

1aa or aa2 with no digits between. (Note: if an integer has units digit 5 and tens digit not 2, it is not a perfect square!)

Example 5: Find 852

Solu on: Answer is 7225, because 8×(8 + 1) = 8×9 = 72 and 52 = 25. 72 followed by 25 is 7225.

Example 6: Find 1152

Solu on: Answer is 13225, because 11 × 12 = 132 and 52 is 25. 132 followed by 25 is 13225.


Solu on: Answer is 4020025, because 200 × 201 = 40200 and 52 = 25. 40200 followed by 25 is the answer.


Solu on: 15×16 = 15×4×4 = 60×4 = 240. 240 followed by 25 is the answer: 1552 = 24025.

Example 9: How many four-inch by four-inch square tiles are needed to cover a four-foot by three-foot rectangular section of floor?

Solu on: Each one-foot by one-foot square can be divided into 3x3 = 9 four-inch by four-inch squares. We have 4x3 = 12 one-foot by one-foot squares in a four-foot by three-foot rectangular section of floor. We need 12x9 = 108 four-inch by four-inch square tiles to cover the floor. Answer: 108

Example 10: Suppose the 12×12 multiplication grid, shown here, were filled in completely. What would be the sum of the 144 products?

Solu on: The sum of the 144 products is:

1x(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12)

+ 2×(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12)

+ 3×(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12)

+ 12×(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12)

= (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12) × (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12)

= (13x6) × (13x6) = 78x78 = 6084 Answer: 6084

Chapter 5: Geometry --- Triangles

136

Solu on As shown in the figure on the right side, BD bisects angle B into two equal angles measuring in x degrees. Because the triangle ABC is an isosceles triangle angle C measures 2x degrees. BD = BC implies that triangle CBD is also an isosceles triangle. So BDC also measures 2x degrees. ADB measures 3x because the exterior angle of a triangle is the sum of the other two interior angles: ADB = DBC + BCD = 3x. 3x + 2x = 180 because ADC is a line segment. 5x = 180, x = 36. A = 180 4x = 5x - 4x = x = 36 degrees. Answer: 36

Centroid Theorem: The three medians (the lines drawn from the vertices to the middle points of the opposite sides) meet in the centroid. The centroid divides each median in a ratio of 2:1. The three medians divide the triangle into six small triangles of equal area.

A shown on the left ABC is any triangle. M is the middle point of BC; N is the middle point of AC; R is the middle point of AB. AM, BN and CR meet at one point O. This point is called the centroid of ABC. AO:OM = 2:1, BO:ON = 2:1 and CO:OR = 2:1. If the triangle has a mass and the mass is evenly distributed on the triangle (same material with same thickness) then the center of the mass is the centroid.

Proof: Let AM and BN be two medians of the triangle ABC and they meet at point O as shown on left. Triangle CNM is similar to CAB with a side proportion 1:2, so NM is parallel to AB and NM has half the length of AB. Let P and Q be the middle points of AO and BO, respectively. Triangle OPQ is similar to triangle OAB with side ratio 1:2, so PQ is parallel to AB also with QP has half the length of AB as well. This shows that MNPQ is a parallelogram. Since the two diagonals of a parallelogram bisect each other

we have MO = OP = AP or AO is twice of MO. Since this is true for any two medians, draw a median from C to AB and it will intersect AM at the 2:1 distance mark, i.e. the point O.

To see that the three small triangles have the same area, I shaded four of the triangles, as shown on the left side. The two lightly shaded triangles have equal area because M is the middle point of BC. Same reason shows that the two heavily shaded triangles have equal area and the two un-shaded triangles have equal area. Since triangle ARC and BRC have equal area also we conclude that the un-shaded part and the lightly shaded part have equal area. This implies that triangle CNO and CMO have equal area. Similarly ANO and ARO have equal area.

Chapter 5: Geometry --- Circles

155

Circles A circle consists of all points that are of the same distance to a fixed point. The fixed point is called the center of the circle and the distance from any point on the circle to the center is called the radius. We also use radius to refer the line segment that connects the center to any point on the circle. The diameter is the line segment that passes though the center and connects two opposite points on the circle. Diameter is also used to refer the length of the diameter, which is twice the radius.

If a line touches a circle at exactly one point we call the line a tangent line to the circle.

Theorem 1 on Tangent Lines: Through any given point on a circle there is exactly one tangent line and the tangent line is perpendicular to the radius passing though the touching point.

Proof: As shown, draw the line l passing through A that is perpendicular to OA. If B is any point on the line other than A, the triangle OAB is a right triangle. The hypotenuse OB is longer than OA. This shows that B is outside the circle. So the line through A and B only touches the circle at A and therefore it is the tangent line.

Mental image: Think of a bicycle with tires touching a level ground. The center of each tire is exactly above the touching point.

Theorem 2 on Tangent Lines: From any point C outside a circle two tangent lines can be drawn to the circle, as shown. The two line segments CA and CB have equal length. Proof: The two right triangles COB and COA share the same hypotenuse x and OA = OB = r as the radius of the circle. By the

Pythagorean Theorem, both CA and CB have length 22 rx .

Theorem 3: An angle inscribed in a semicircle is always a right angle. As shown, if AB is the diameter of the circle O and C is any point on the circle other than A and B, then the angle ACB is a right angle. Proof: OA, OB and OC are all radius of the circle. AOC and BOC are both isosceles triangles. So the two angles labeled with 1 are equal and the two angles labeled with 2 are equal. The sum of the three angles in ABC is 180 . 1802211 . 180212 ,

9021 , 90ACB .

Chapter 5: Geometry --- Circles

156

A secant line of a circle is a line that intersects the circle in two points. An inscribed angle is formed when two secant lines intersect at a point on the circle. A central angle is an angle with vertex at the center of the circle.

Here the angle ACB is an inscribed angle. It is an inscribed angle subtended by the arc AB.

Three distinct points on a circle form a triangle. The triangle is called an inscribed triangle. All the three angles of the triangle are inscribed angles.

Here the angle AOB is the central angle subtended by the arc AB. O is the center of the circle.

Inscribed Angle Theorem: An inscribed angle is half of the central angle subtended by the same arc. This implies that all inscribed angles subtended by the same arc are equal.

As shown, if C is a point on the circle but outside of the arc AB, then AOBACB21

.

Case 1: When the center O is on one side of the inscribed angle. The two angles labeled are equal because OB = OC. The inscribed angle is and the central angle is 2 because

180BOC .

Case 2: When the center O is inside the inscribed angle. Draw the line CO and it divides the inscribed angle into two angles and The inscribed angle is and the central angle is

222 .

Case 3: When the center O is outside the inscribed angle ACB . Extend BO to E and AO to D.

OCB and OBC are labeled . xCOD 2 and 2COE . Then

xDOE 22 . The inscribed angle is x and the central angle is xDOEAOB 2 .

Chapter 6: Sequences --- The Story of Li le Gauss and Arithme c Sequences

164

The Story of Little Gauss and Arithmetic Sequences Carl Friedrich Gauss (1777 - 1855) was one of those remarkable infant prodigies whose natural aptitude for mathematics soon becomes apparent that he is one of the world's most famous mathematicians ever. An anecdote of his goes like this: Gauss was about 9 years old, already a super genius. The new substitute teacher walked into the classroom and gave them a "busy work" arithmetic problem. They were to work on it and not bother her because she wanted to have a nap. The problem was to find the sum of all the numbers from 1 to 100. She thought that they would have 99 additions to do and this would keep them busy for a while. No sooner had she finished writing the question on the chalkboard, Gauss got the correct answer. The teacher was astonished and could not understand how Gauss could figure it out in seconds.

The young Gauss later confessed to having recognized the pattern: 101=100+1 , 101=99+2 , 101=98+3 , 101=97+4 , 101=52+49 , 101=51+50 . Since there are 50 pairs of numbers, each

of which adds up to 101, the sum of all the numbers must be 5050 = 10150 .

This technique will work for any sequence of numbers provided that they are evenly spaced. For example: 2, 4, 6, 8, , 100 or 3, 6, 9, 12, , 300 or, 5, 10, 15, 20, , 500 etc. In every case, to find the sum you need only add the first and the last and then multiply by the number of pairs in the sequence.

This method provides a way of deriving the formula 2

1nn =n3 2 1 for the sum of the first

n positive integers. One need only display the consecutive integers 1 through n in two rows as follows:

1 2 3 n-2 n-1 n n n-1 n-2 3 2 1

Addition of the vertical columns produces n terms, each of which is equal to n+1; when these terms are added, we get the value n(n+1). Because the same sum is obtained on adding the two rows horizontally, what occurs is the formula 2(1 + 2 + 3 + ... + n) = n(n + 1)

An even spaced sequence is called an arithmetic sequence. Equivalently an arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant

Examples of arithme c sequences

This is the sequence of positive odd integers. The common difference is 2.

Chapter 6: Sequences --- Other Sequences

171

Other Sequences Sum of Special Frac ons: The fractions in the following sequence can all be written as the difference of two fractions in a nice way.

21

, 61

, 121

, 201

, 301

, , 1

1nn

,

21

11

21

, 31

21

61

, 41

31

121

, 51

41

201

, 61

51

301

, , 1

111

1nnnn

,

The sum of the first n terms is 1

1111

141

31

31

21

21

11

nnnn

111

1111

31

31

21

211

nnnn. So we have the following formula:

111

11

121

61

21

nnn

Example 13 What is the value of

561

421

301

201

121

61

21

? Express your answer as a decimal

to the nearest thousandth. Solu on

875.0125.01811

81

71

41

31

31

21

211

561

121

61

21

Sum of Squares: This formula gives the sum of the first n perfect squares.

61211321 22222 nnnnn

Proof: Take the difference of the two identities: 1331 233 xxxx and

1331 233 xxxx , and we get: 2611 233 xxx . We will let x to be 1, 2, 3, etc. until n and get a series of n identities, then add both sides of these identities:

21602 233 22613 233 23624 233

2162 233 nnn

2611 233 nnn

Add the left side of these n equations we see that 32 , 3331n are all cancelled out.

3333 011 nn nnn 213216 22222

nnnnn 21113216 3322222

1323221133 223323 nnnnnnnnnnn

121 nnn Divide both side by 6 we get the right formula.

Chapter 10: Prac ce Tests --- Prac ce Test #3

249

25. _________ What is the smallest possible value of x for the equation

2032

28432

284 2

xx

xx

?

26. _________ Find the number of unordered pairs (a, b) of two-digit positive integers with their product a multiple of 9.

27. _________ In the figure shown, ABCD is a square. Inside ABCD

another square is formed with the middle points of the four sides of ABCD as vertices. The diagonals of the two squares are drawn. How many triangles are in the figure?

28. _________ From each side of an equilateral triangle ABC a square

is constructed outwards. Then a hexagon is created using the vertices of the squares, as shown. If the area of the equilateral triangle ABC is 1, what is the area of the hexagon DEFGHI? Express your answer in simplest radical form.

29. _________ Four circles are inscribed in a bigger circle. The two

median- sized circles touch at the center of the outer circle and the two smallest circles are tangent to the neighboring circles. If a point is randomly chosen in the biggest circle, what is the probability that the chosen point falls into the shaded region? Express your answer as a common fraction.

30. _________ 30. A structure consists of 4 unit cubes, as shown.

An ant decides to travel from vertex A to vertex B and it can only move right, up or forward one unit at a time. How many possible routes can the ant take from A to B if it can only use the top and the four side faces of the structure?

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MathCounts Preparation · MathCounts Preparation How to Excel at Middle School Math Competitions By Huasong Yin