Some Articles to Read and Ponder after Mathcounts state competition

For those students who really love problem solving, who care and are inquisitive but are disappointed at their chapter's/state's performance, I want to let you know that it's an honor meeting you online and learning from/along with you.

I know no matter how sincere I write here, it won't help much, so I'll just shut up.

However, I'm also disappointed at some students who said they need to take a break for the foreseeable future.

Well, if you love ___ (fill in the blank), you won't stop if you don't get to the chapter/state/nationals.
So there...

Here are some articles that after your taking a break from problem solving (I hope it's not too long), hope to see you come back and read them. Best of luck !!  Keep me posted !!

Pros and Cons of Math Competitions

Dealing with Hard Problems

Life After Mathcounts

Great Mathematicians on Math Competitions and "Genius"

Math Contests Kind of Suck from Mathbabe

TEDxCaltech -Jordan Theriot- The Pleasure of Finding Things Out

Why Physics? Skateboarding Physicist and Educator Dr. Yung Tae Kim

Richard Feyman --The Pleasure of Finding Things Out

Hope it helps !! 

2015 Mathcounts State/National Prep

Harder concepts from Mathcounts Mini :
Also try the follow-up problems with detailed solutions. Don't do every question. Just the ones you think is hard.

Geometry :
#27 : Area and Volume 

#30: Similar Triangles and Proportional Reasoning

#34: Circles and Right Triangles

#35: Using Similarity to Solve Geometry Problems

#41: Analytic Geometry : Center of Rotation

More questions to practice from my blog post. 

From Mathcounts Mini : Video tutorials on counting and probability for Mathcounts state/national prep concepts are in the of difficulty.

Counting the Number of Subsets of a Set

Constructive Counting

More Constructive Counting 

Probability and Counting

Probability with Geometry Representations  : Oh dear, the second half part is hilarious.

Probability with Geometry Representations : solution to the second half problem from previous video

Try this one from 1998 AIME #9. It's not too bad.

From Mathcounts Mink : Center of rotation, equal distance from a point to other two points or two lines, angle bisector (last problem on the follow-up worksheet)

Mathcounts Mini : #41 - Analytic Geometry

2013 AMC 10 A and B : Solutions to the Harder Problems

From AoPS videos :

• 2013 AMC 10 A #21 - This is also #17 on the AMC 12 A.
• 2013 AMC 10 A #22 - This is also #18 on the AMC 12 A.
• 2013 AMC 10 A #23 - This is also #19 on the AMC 12 A.
• 2013 AMC 10 A #24
• 2013 AMC 10 A #25

2013 AMC-10 B #23  It's a good idea to be able to get all those split-side lengths, height to the hypotenuse fast and right. Those are basic skills.

Also, you should be able to get the area, in-radius and others fast as well of 13-14-15 and 10-17-21
triangles.

2013 AMC 10 A Math Jam from AoPS  

2013 AMC 10 B Math Jam from AoPS 

Problem Solving Strategies: Applications of the “Choose 2” method

1. Example: How many diagonals can be drawn for a polygon with “n” sides?

Method I:

The number of diagonals in a polygon = n(n-3)/2, where n is the number of polygon sides.

For a convex n-sided polygon, there are n vertexes, and from each vertex you can draw n-3 diagonals, so the total number of diagonals that can be drawn is n (n-3).

However, this would mean that each diagonal would be drawn twice, (to and from each vertex), so the expression must be divided by 2.

Method II:

nC2 (choose 2) - n sides = n(n-1)/ 2 – n sides


2. Example: There are n people at a party, each person shakes hands with the every other person once. How many handshakes?

Method I:
nC2 in this case (10 x 9) /2 =45
__ __ First slot you have 10 persons to choose from, second slot 9 persons. Since A shakes hands with B is the same as B shakes hands with A, so you divide the number by 2 and get the answer.

Method II:
Sum of the first consecutive Natural numbers: n (n+1) /2
The first person shakes hands with 9 other person; the second person shakes hands with 8 other person, etc…
So 9 + 8 + 7 + …= (9 x 10)/ 2 = 45

3. Example: N dots evenly spaced on a circle. How many chords can you make using those dots?

Methods: This is very similar to hand-shaking questions.

I: nC2

II: Sum of the first consecutive (N-1) Natural numbers

Word problems: Answers below.

#1 : A convex polygon with n sides has 20 diagonals. How many diagonals does an (n+1)-sided convex polygon have?

#2: A polygon has n sides and n diagonals. What is n? 

#3: How many diagonals does a decagon have?

#4: How many diagonals does a dodecagon have? 

#5: How many line segments have both their endpoints located at the vertexes of a given cube?

#6: There are 8 points on a circle, how many lines can you make? How many triangles can you make?

#7: If each of the interior angles of two regular polygons adds up to 255 degrees and their diagonals add up to 29, what is the sum of their sides?

Answers: 

# 1- 27    #2 -5     #3-35      #4-54 [12C2 -12]     #5 28 [There are 8 vertices, so 8C2]



#6  8C2 = 28 for lines and 8C3 = 56 triangles   

#7: 6[hexagon] + 8[octagon] = 14 sides

Unit digit, Tenth digit and Digit Sum

Word problems on unit digit, tenth digit or digit sum.

#1: How many digits are there in the positive integers 1 to 99 inclusive? 

Solution I:  From 1 to 9, there are 9 digits.
From 10 to 99, there are 99 - 10 + 1 or 99 - 9 = 90 two digit numbers. 90 x 2 = 180
Add them up and the answer is 189.

Solution II:  ___ There are 9  one digit numbers (from 1 to 9).
___ ___ There are 9 * 10 = 90 two digit numbers (You can't use "0" on the tenth digit but you
can use "0" on the unit digit.) 90 * 2 + 9 = 189

# 2: A book has 145 pages. How many digits are there if you start counting from page 1?

There are 189 digits from page 1 to 99. (See #1, solution I)
From 100 to 145, there are 145 - 100 + 1 or 145 - 99 = 46 three digit numbers.
189 + 46*3 = 327 digits.

#3: "A book has N pages, number the usual way, from 1 to N. The total number of digits in the page number is 930. How many pages does the book have"?  Similar to one Google interview question.

Read the questions and others here from the Wall Street Journal.

Solution I: 

930 - 189 (digits of the first 99 pages) =741
741 divided by 3 = 247. Careful since you are counting the three digit numbers from 100 if the book has N
pages N - 100 + 1 or N - 99 = 247. N = 346 pages.

Solution II: 

930 - 189 (total digits needed for the first 99 pages) = 741
741/3 = 247 (how far the three digit page numbers go).
247 + 99 = 346 pages

#4: If you write consecutive numbers starting with 1, what is the 50th digit you write? 

Solution I:
50 - 9 = 41, 9 being the first 9 digits you need to use for the first 9 pages.

Now it's 2 digit. 41/2 = 20.1 , which means you will be able to write 20 two digit numbers + the first digit of the next two digit numbers.

10 to 29 is the first 20 two digit numbers so the next digit 3 is the answer. (first digit of the two digit number 30.)

Solution II: (50 - 9 ) / 2 = 20. 5 ; 20.5 + 9 = 29.5, so 29 pages + the first digit of the next two digit numbers, which is 3, the answer.


#5: What is the sum if you add up all the digits from 1 to 100 inclusive?

00  10  20  30  40  50  60  70  80  90

01  11  21  31  41  51  61  71  81  91

02  12  22  32  42  52  62  72  82  92

03  13  23  33  43  53  63  73  83  93

04  14  24  34  44  54  64  74  84  94

05  15  25  35  45  55  65  75  85  95

06  16  26  36  46  56  66  76  86  96

07  17  27  37  47  57  67  77  87  97

08  18  28  38  48  58  68  78  88  98

09  19  29  39  49  59  69  79  89  99

Solution I:
Do you see the pattern?  From 00 to 99 if you just look at the unit digits.
There are 10 sets of ( 1+ 2 + 3 ... + 9) , which gives you the sum of 10 * 45 = 450
How about the tenth digits? There are another 10 sets of (1 + 2 + 3 + ...9) so another 450
Add them up and you have 450 * 2 = 900 digits from 1 to 99 inclusive.
900 + 1 ( for the "1" in the extra number 100) = 901 

Solution II:
If you add the digits on each column, you have an arithmetic sequence, which is
45 + 55 + 65 ... + 135  To find the sum, you use average * the terms (how many numbers)
\(\dfrac {45+135} {2} * \left( \dfrac {135-45} {10}+1\right)\) =900
900 + 1 = 901

Solution III :
2*45*10¹ + 1 = 901


Problems to practice: Answers below.

#1: A book has 213 pages, how many digits are there?

#2: A book has 1012 pages, how many digits are there?

#3: If you write down all the digits starting with 1 and in the end there are a: 100, b: 501 and c: 1196 digits, what is the last digit you write down for each question?

#4: What is the sum of all the digits counting from 1 to 123? 

Answers: 

#1: 531 digits. 

#2: 2941 digits.

#3: a. 5, b. 3, c. 3

#4: 1038