Monday, December 15, 2014

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

From AoPS videos :

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

Sunday, December 7, 2014

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

#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?

# 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

Wednesday, November 12, 2014

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

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

#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?

#1: 531 digits.
#2: 2941 digits.
#3: a. 5, b. 3, c. 3
#4: 1038

Tuesday, September 23, 2014

Similar triangles, Trapezoids and Triangles that Share the Same Vertices

Check out Mathcounts here, the best competition math program for middle school students.

This is an interesting question that requires understanding of dimensional changes. (They are everywhere.)

Question: If D and E are midpoints of AC and AB respectively and the area of ΔBFC = 20, what is the
a. area of Δ DFB?
b. area of Δ EFC?
c. area of Δ DFE?
e. area of trapezoid DECB?
f. area of Δ ABC?

Solution:

DE is half the length of BC (D and E are midpoints so DE : BC AD  : AB = 1 : 2

Δ DFE and ΔCFB are similar and their area ratio is 12 : 22  = 1 : 4  (If you are not sure about this part, read this link on similar triangles.)

so the area of Δ DFE = (1/4) of ΔBFC = 20 = 5 square units.

The area of Δ DFB = the area of Δ EFC = 5 x 2 = 10 square units because Δ DFE and Δ DFB,
Δ DFE and ΔEFC share the same vertexs D and E respectively, so the heights are the same.
Thus the area ratio is still 1 to 2.

Δ ADE and ΔDEF share the same base and their height ratio is 3 to 1, so the area of
Δ ADE is 5 x 3 = 15 square units.

[DE break the height into two equal length and the height ratio of Δ DFE and ΔCFB is 1 to 2 (due to similar triangles) so the height ratio of Δ ADE and ΔDEF is 3 to 1.]

The area of trapezoid DEBC is 45 square units.

The area of Δ ABC is 60 square units.

Extra problems to practice (answer below):
The ratio of   AD and AB is 2 to 3,  DE//BC and the area of Δ BFC is 126, what is the area of

a. Δ DFE ?

b. Δ DFB ?

c.  Δ EFC ?

e. How many multiples is it of Δ ABC to ΔBFC?

a. 56 square units
b. 84 square units
c. 84 square units
d. 280 square units
e. 5 times multiples.

Thursday, September 4, 2014

Sep. 4th, 2014 AMC-8 and Mathcounts State/National prep

To prepare for AMC-8 and Mathcounts simultaneously, it's a good idea to delete AMC-8 multiple choice options to make the test more like Mathcounts problems.

There are some tricky questions for AMC-8 test, so if you are at the Mathcounts state level in above average states, you might get better scores on AMC-10 tests than on AMC-8. (Sigh...)

That's what happened to quite a few of my students because their level is way up so they might not as focused as they worked on the more challenging, more interesting questions. Oh dear !!

Review the following very frequently tested concepts. You really don't need a lot of tools (formulas) but deeper understanding, tenacity and the love of thinking outside the box.

Review similar triangles

Review counting and probability

I'll put solutions to some of the "Mass Points" questions soon.

We all love "Mass Points".

Have fun problem solving. Cheers, Mrs. Lin

Wednesday, August 20, 2014

Notes to Sunday Nights' Problem Solving Group Lessons

This week's work : Please review the last 8 hardest AMC-8 problems from 2005 to 2011 + finish the last 8 hardest AMC-8 problems from 2012 and 2013 if you haven't done that.

Some notes and questions.

Question #1 : How many ways can you climb up a ten-step staircase if you climb only one or two steps at a time ?

Solutions I :
Make a chart, starting with a smaller case.
one-step staircase -- 1 way, which is 1.
two-step staircase -- 2 ways, which is 1, 1 or 2.
three-step staircase -- 3 ways, which is 1 1 1, 1 2 or 2 1.
four-step staircase -- 5 ways, which is 1111, 211, 121, 112, 2222.

Notice the pattern - - 1 , 2, 3, 5, 8, 13, 21, 34, 55, 89, which is the answer; it also happens that it's
part of the Fibonacci numbers.

Why does it work that way ?

Well, if you climb the 3-step staircase, there are two cases :
You either take one step at first, and there are 2-step left, which leaves you two ways to climb the remaining staircase.
Or you take two step at first, and there are 1 step left, which leaves you one way to climb the remaining staircase.
If you climb the 4-step staircase, again, there are two similar cases :
You either take one step at first, and  then there are 3-step left, which leaves you 3 ways to climb the remaining staircase.
Or you take two step at first, and there are 2 step left, which leaves you 2 way to climb the remaining staircase.

Thus, it's always the sum of the previous two terms for the next staircase steps.
This concept is called recursion.

Now try another question :
2010 AMC-8 #25 : Every day at school, Jo climbs a flight of 6 stairs. Joe can take the stairs, 1, 2 or 3 at a time. For example, Jo could climb, 3, then 1, then 2. In how many ways can Jo climb the stairs ?

Solution I :
List out all the possible ways.
111111  -- 1 way
21111 -- 5C1 = 5 ways to arrange the steps
2211 -- 4!/2! x 2! = 6 ways to arrange the steps
222-- 1 way
3111 -- 4C1 or 4 ways to arrange the steps
123 -- 3! or 6 ways
33 -- 1 way
Sum them up and the answer is 24.

Solution II :
Using recursion, starting with the smallest case.
1 step -- 1 way
2 steps-- 2 ways (11, or 2)
3 steps -- 4 ways (111, 21, 12, or 3)
4 steps -- 1 + 2 + 4 = 7 ways (why??)
5 steps -- 2 + 4 + 7 = 13 ways.
6 steps -- 4 + 7 + 13 = 24 ways, which is the answer.

Besides these, please review the following questions.

Q #1  1988 AMC-8 #25 : A palindrome is a whole number that reads the same forwards as backwards. If one neglects the colon, certain times displayed on a digital watch are palindromes. Three examples are 1: 01, 4: 44 and 12: 21. How many times during a 12-hour period will be palindromes?

Q#2  2004 Mathcounts sprint  #21: If |-2a + 1| < 13, what is the sum of the distinct possible integer values of a?

Q#3  2004 Mathcounts sprint #30 : A particular right square-based pyramid has a volume of 63,960 cubic meters and a height of 30 meters. What is the number of meters in the length of the lateral height (AB) of the pyramid? Express your answer to the nearest whole number.

#1 : 57

#2 : 6

#3 : 50

Thursday, August 14, 2014

Dimensional Change Questions II

Dimensional change questions II:   Answer key below.
If you've found you are not solid yet with these problems,

1a. There is a regular cylinder, which has a height equal to its radius. If the radius and height are both increased by 20%, by what % does the total volume of the cylinder increase?

1b. If the radius and height are both decreased by 20%, by what % does the total volume of the cylinder decrease?

1c. If the radius is increased by 50% and the height is decreased by 25%, what % of the volume of the original cylinder does the volume of the new cylinder represent?

1d. If the radius is increased by 25% and the height is decreased by 50%, what % of the volume of the original cylinder does the volume of the new cylinder represent?

1e. If the height is increased by 300%, what % does the radius need to be decreased by for the volume to remain the same?

2. If the side of a cube is increased by 30%, by what % does the total surface area of the cube increase? By what % does the volume increase?

3a. If the volume of a cube increases by 174.4%, by what % does the total surface area of the cube increase?

3b. By what % did the side length of the cube increase?

Answer key to dimensional change questions II:

1a. There is a regular cylinder, which has a height equal to its radius. If the radius and height are both increased by 20%, by what % does the total volume of the cylinder increase?

72.8%

1b. If the radius and height are both decreased by 20%, by what % does the total volume of the cylinder decrease?

48.8% (Only 0.83 = 0.512 = 51.2% of the original percentage left and 100% - 51.2% = 48.8%.)

1c. If the radius is increased by 50% and the height is decreased by 25%, what % of the volume of the original cylinder does the volume of the new cylinder represent?

168.75%

1d. If the radius is increased by 25% and the height is decreased by 50%, what % of the volume of the original cylinder does the volume of the new cylinder represent?

78.125%

1e. If the height is increased by 300%, what % does the radius need to be decreased by for the volume to remain the same?

50%

2. If the side of a cube is increased by 30%, by what % does the total surface area of the cube increase? By what % does the volume increase?

The surface area will increase 69% and the volume will increase 119.7%

3a. If the volume of a cube increases by 174.4%, by what % does the total surface area of the cube increase?

96%

3b. By what % did the side length of the cube increase?

40%

Wednesday, August 6, 2014

Dimensional Change questions I:

Questions written by Willie, a volunteer.  Answer key and detailed solutions below.

1a. There is a regular cylinder, which has a height equal to its radius. If the radius and height are both increased by 50%, by what % does the total volume of the cylinder increase?

1b. If the radius and height are both decreased by 10%, by what % does the total volume of the cylinder decrease?

1c. If the radius is increased by 20% and the height is decreased by 40%, what % of the volume of the original cylinder does the volume of the new cylinder represent?

1d. If the radius is increased by 40% and the height is decreased by 20%, what % of the volume of the original cylinder does the volume of the new cylinder represent?

1e. If the height is increased by 125%, what % does the radius need to be decreased by for the volume to remain the same?

2. If the side of a cube is increased by 50%, by what % does the total surface area of the cube increase?

3a. If the volume of a cube increases by 72.8%, by what % does the total surface area of the cube increase?

3b. By what % did the side length of the cube increase?

4. You have a collection of cylinders, all having a radius of 5. The first cylinder has a height of 2, the second has a height of 4, the third a height of 6, etc. The last cylinder has a height of 50. What is the sum of the volumes of all the cylinders (express your answer in terms of pi)?

Answer key: (Each question should not take you more than 30 seconds to solve if you really understand the concepts involved.)

1a.  The volume of a cylinder is πr2x h (height). The radius itself will be squared and the height stays at constant ratio. The volume will increased thus (1.5)3 - 13 -- the original 100% of the volume = 2.375
=237.5%

1b.  Like the previous question: 13 - 0.93 [when it's discount/percentage decrease, you use the 100% or 1 - the discount/decrease percentage] = 0.271 =  27.1% decrease

1c.  1.22 [100% + 20% increase = 1.2] x 0.6 [100% -40% = 0.6] = 0.864  or
86.4% of the original volume

1d.  1.42 [100% + 40% increase = 1.4] x 0.8 [100% -20% = 0.8] = 1.568 = 156.8% of the original volume

1e.  When the height of a cylinder is increased 125%, the total volume is is 225% of the original cylinder, or 9/4.
Since the radius is used two times (or squared), it has to decrease 4/91/2 = 2/3 for the new cylinder to have the same volume as the old one. [9/4 times 4/9 = 1 or the original volume.]
1 - (2/3) = 1/3 = 0.3 = 33.3%

2. Surface area is 2-D so 1.52 - 1 = 1.25 = 125% increase

3a. If a volume of a cube is increased by 72.8 percent, it's 172.8% or 1.728 of the original volume. Now you are going from 3-D (volume) to 2-D (surface area). 1.7282/3 = 1.44 or 44% increase. [Don't forget to minus 1 (the original volume) since it is asking you the percentage increase.]

3b. From surface area, you can get the side increase by using 1.441/2 = 1.2, so 20% increase.
Or you can also use 1.7281/3 = 1.2;  1.2 - 1 = 20%

4. The volume of a cylinder is πr2x h . (2 + 4 + 6 + ...50) x 52π = (25 x 26) x 25π =16250π

Tuesday, July 15, 2014

Analytical Geometry : Circle Equations

Circle Equations from Math is Fun

How to Find Equation of a Circle Passing 3 Given Points

7 methods included ; Amazing !!

Practice finding the equation of a Circle given 3 points --

Q #1 : (1, 3), (7, 3) and (1, -3)

Answer : (x -4)2 + y2 = 18
Q #2  : (3, 4), (3, -4), (0, 5)
Answer : x2 + y2 = 25
Q #3 : A (1, 1), B (2, 4), C (5, 3)
Answer : (x-3)2 + (y -2)2 = 5
Solution :
The midpoint of line AB on the Cartesian plane is $$(\frac{3}{2}, \frac{5}{2})$$ and the slope is $$(\frac{3}{1})$$ so the slope of the perpendicular bisector of line AB is $$(\frac{-1}{3})$$.
The equation of the line bisect line AB and perpendicular to line AB is thus :
y - $$(\frac{5}{2}$$) =$$\frac{-1}{3}$$ [x - $$(\frac{3}{2})$$] --- equation 1
The midpoint of line BC on the Cartesian plane is $$(\frac{7}{2}, \frac{7}{2})$$ and the slope is $$(\frac{-1}{3})$$ so the slope of the perpendicular bisector of line BC is 3.
And the equation of the line bisect line BC and perpendicular to line BC is
y - $$(\frac{7}{2})$$ = 3 [x - $$(\frac{7}{2})$$] --- equation 2
Solve the two equations for x and y and you have the center of the circle being (3, 2)
Use distance formula from the center circle to any point to get the radius =
$$\sqrt{5}$$.
The answer is : (x - 3)2 + (x - 2)2 = 5

More practices on similar questions :  (Answers below for self check)

Q #1 : A (2, 5) , B (2, 13) ,  and C (-6, 5 )

Q #2 : A (0, 7), B ( 6, 5 ), and C (-6, -11 )

Q #3 : A (3, -5) , B (-4, 2) and C (1, 7 )

#1 :  (x - 2) 2 + ( y - 9 ) 2 = 32

#2 :   x2 + (y + 3)2 = 100

#3 :  (x - 2) 2 + ( y -1) 2 = 37

Tuesday, July 8, 2014

This Week's Work : Week 53 for Inquisitive Young Mathletes

For this week's work, review 1991 Mathcounts National sprint and target round questions you got wrong or not fast enough. Please try once more to see if you now can solve them at ease.

For this week's review, try the following Mathcounts Mini :

Similar Triangles and Proportional Reasoning

Try the follow-up problems

Detailed solutions

I'll send you notes/solutions/links to some of the hardest problems later.

Take care and happy problem solving !!

Thursday, June 19, 2014

This Week's Work : Week 52 for Inquisitive Young Mathletes

Finish 2000 Mathcounts National sprint and target rounds according to the rules.

E-mail me your scores + what questions you skip, couldn't solve fast, preferably with
the wrong answers put down so I can tell if you have some ideas.

Constructive Counting from Mathcounts Mini

Try some questions from the activity sheet till you fully understand the concepts.

Activity sheet solutions

Monday, June 9, 2014

This Week's Work : Week 51 for Inquisitive Young Mathletes

For this week's work :

Try 2013 Mathworks Math Contest problems from Texas State University

Answer key and statistics can be viewed here.

I'll send you detailed solutions once we finish all the problems.

Practice more "at least" problems :

2014 AMC-10 B problem 16

Solution

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

Counting the Number of Subsets of a Set

Constructive Counting

More Constructive Counting

Probability and Counting

Tuesday, June 3, 2014

This Week's Work : Week 50 - for Inquisitive Young Mathletes

Counting Restrictions from Glyia

Very nice introductory level counting questions with restrictions.

from Mathcounts Mini : Counting the Number of Subsets of a Set

from AoPS :

Counting with Restrictions Part 3  : This one is tricky. Make sure you understand the concepts well.

Try this question from 2011 AMC-8 # 23 :

How many 4-digit positive integers have four different digits, where the leading digit is not zero, the integer is a multiple of 5, and 5 is the largest digit?

Solutions from AoPS

Counting with Combinations Part 3

At our group lessons, we'll continue working on harder Mathcounts concepts, not just the accuracy, but speed.

Take care and happy problem solving !!

Tuesday, April 29, 2014

This Week's Work : Week 48 - for Inquisitive Young Mathletes

 Tackling Problems with Vieta Section 10.4: Vieta for Quadratics Part 1 Section 10.4: Vieta for Quadratics Part 2 Section 10.4: Tackling Problems with Vieta Practicing Problems using Vieta's formula The Binomial Theorem
Casework Counting Part I

Casework Counting Part II

Tuesday, April 15, 2014

This Week's Work : Week 47 - for Inquisitive Young Mathletes

 This week, we'll continue discussing this year's Mathcounts state harder problems. Please try these problems before our lesson. Review sprint round questions we went over at last Sunday's lesson, especially #18 to see if you can get that question right and fast the first try. The technique used for this question is very similar to the first explanation from this episode of Mathcounts mini -- more construction counting. Review Binomial Theorem and see if you can use the following concepts solving this year's state sprint # 28. (You can also use modular arithmetic, or mod.) Chapter 14: The Binomial Theorem
Have a calculator ready for target and team round questions.

Take care and happy problem solving !!

Tuesday, April 8, 2014

This Week's Work : Week 46 - for Inquisitive Young Mathletes

Finally, this year's Mathcounts state problems are clear for discussions.

Check that out here !! We'll talk about those harder problems this week at our group lessons.

Please let me know how you did/what scores you got if you have followed this blog.
So far, there were a few states' top 10, 20, etc.. and two students I met online will go to the Nationals.

Some e-mails make my day and definitely make me more productive.

For this week's work :

If you haven't watched this video, do so this week.

Joint Proportion : In the video, three methods are discussed and as you can see, the last method is much, much
faster so try that to see if you fully understand the relationship now and can implement it on new, seemingly hard
problems.

We'll continue working on some rate, proportion questions (the harder ones) at our lessons besides reviewing and
practicing speed to solve other harder, counting/probability and geometry questions.

New Mathcounts Mini : Recognizing Squares and Solving a Simpler Problem

From TED Talk : The Magic of Fibonacci Numbers presented by Arthur Benjamin

Wednesday, April 2, 2014

This Week's Work : Week 45 - for Inquisitive Young Mathletes

See if you can use "sticks and stones" to solve the following questions (it requires a little twist).

Tossing 3 dice, what is the probability that the sum is 10 ?

Tossing 4 dice, what is the probability that the sum is 10 ?

Tossing 4 dice, what is the probability that the sum is 18 ?

If you want to try harder problems, check out this year's AIME II, 1, 3 and 4. (related to what we've learned recently)

This week's new links, working together, direct and inverse proportion, rate, time distance.

Take care and happy problem solving.
Cheers, Mrs. Lin

Saturday, March 29, 2014

Mathcounts Prep -- Number Sense

Check out Mathcounts: the best competition math program up to the national level.

Problems: (Solutions below)
#1: 2005 Chapter Team-- A standard deck of playing cards with 26 red cards and 26 black cards is split into two piles, each having at least one card. In pile A there are six times as many black cards as red cards. In pile B, the number of red cards is a multiple of the number of black cards. How many red cards are in pile B?

#2:  2000 State sprint #30. Joe bought a pumpkin that cost $10$ cents more per pound than his sister's. Together, the two pumpkins weighed $20$ pounds, but Joe's pumpkin was heavier. Joe paid $\ 3.52$ dollars and his sister paid $48$ cents. How many pounds did Joe's pumpkin weigh?

Solutions :

#1: You know the total cards in pile A is a multiple of 7 because there are six times as many black cards as the red cards. (given)

6 Black, 1 Red on pile A gives you 20 Black and 25 Red cards on pile B. (doesn't work)
12 Black and 2 Red cards on pile A gives you 14 Black and 24 Red cards on pile B. (doesn't work)
18 Black and 3 Red cards on pile A gives you 8 Black and 23 Red cards on pile B. (doesn't work)
24 Black and 4 Red cards on pile A gives you 2 Black and 22 Red cards on pile B. Yes!!
The answer is 22 Red cards.

#2:
Solution I :
Let x dollars be the cost per pound for Joe's sister's pumpkin and x + .1 dollars are the cost per pound for Joe's pumpkin. Since the pounds of each pumpkin is the cost $\div$ cost per pound, we have

Solution II:  Make a list:
Joe's sister        Joe
1 lb.                 19 lb.       (doesn't work since 19 x 58 cents are too much)
2 lbs.                18 lbs      (doesn't work)
3 lbs.                17 lbs      (No)17 x (48/3 + 10) = 442 (still too much)
4 lbs                 16 lbs      16 x (48/4 + 10)= 352 (yes)

Tuesday, March 25, 2014

This Week's Work : Week 44 - for Inquisitive Young Mathletes

From Mathcounts Mini : More Constructive Counting

Also try the warm-up, retry the problems in the video on your own and spend some time
pondering on those follow-up, harder problems.

#7 and 9 are more challenging.

Review 2014 harder school round problems, especially the "Harvey" method.

Try the first 8 2012 AMC-10 B questions again and see if you have the speed.

Have fun problem solving !! Mrs. Lin

Monday, March 17, 2014

This Week's Work : Week 43 - for Inquisitive Young Mathletes

From Mathcounts Mini : Using Similarity to Solve Geometry Problems

Also try the warm-up, retry the problems in the video on your own and spend some time
pondering on those follow-up, harder problems.

#4, 5, 6 are state level problems. #7 is challenging.

From Mathcounts Mini : Tangent Segments and Similar Triangles

#8 is state/national level.

Have fun problem solving !! Mrs. Lin

Thursday, February 20, 2014

Problem Solving is Fun

There is a goat, says the question. The goat is tethered to one edge of a barn, on a rope 10 meters long. If the barn is 9 meters long and 5 meters wide, and the goat cannot enter the barn, what is the total area that the goat can cover?

To my eleven-year-old mind, this question brought up countless other questions: what, for example, would inspire the goat to traverse as much distance as possible in the first place? Would it not be more sensible for the goat to double 'round and chew off the rope, then escape the tyranny of the barn and traipse off to whatever happy meadows were certainly waiting in store? What exactly is the farmer hiding in the barn that is so important that he will not even allow an innocent goat to see it?

The problem with the stereotypical math questions posed in high school is that they all look something like this: If x^2 - 6x + 5 = 0, what is x? Such a question gives no incentive to solve it. There is no sympathy for 'x', no tears shed over the tragedy of a beleaguered animal. All the question does is to continue to drill factoring and quadratics into brains already well-tired of such things.

Stemming from this is my proposition on how to make math 'cool'. Rather than the continuous drilling of formulas and equations, problems in math classes should be set up like the problems offered on standard math competitions such as the MathCounts programs and the AMCs. The goat question is a perfect example - it provides the blending of a possible real-life situation (if any mathematicians are striving to become farmers with curious goats) with the concept of sectors of circles. Schools offer the basics; almost all high school-students know that the area of a circle is "pi-r-squared". However, I have offered the goat question - a standard on several MathCounts state tests - to a roomful of honors-level high-schoolers during my time as MathLeague coach, and been rewarded with a bunch of blank stares.

"We don't know how to do that," say my teammates. "Why is there a goat? The goat is confusing us!"

And here's the problem - schools force the memorization of formulas, but do nothing to teach applied math. In order to really understand math, students should be offered questions which challenge their imagination and understanding. It's not hard, either - after explaining the goat question once, all of the MathLeague students present at the lecture could solve any similar questions; all they needed was a moment's exposure to competitional math.

Additionally, several of the - well, granted, not-very-mature - students suddenly caught a glimpse of how I, in all my madness, could call math 'fun'. "This is kind of cool," one boy admitted, after a ten-minute discussion on how we could find the number of diagonals that a convex pentagon - whimsically named DUCKS - had. "I like ducks."

So it's easy, really, to make math fun. Amuse the students - put math in situations they've never seen before, with not only goats, but ducks, and cows, and even once, to the chagrin of our MathLeague supervisor who had already been losing faith in the human race, a llama. Offer new ways to play with probability - probability as used in gambling, for example, which tends to amuse students to no end.

And, finally, challenge the kids, because that's the only way they're going to go from memorizing formulas to applying them. After all, my teammates can now say quite confidently each time they see a question dealing with circles, "I've got your goat."

Monday, February 17, 2014

2014 Mathcounts State Prep : Inscribed Circle Radius and Circumscribed Circle Radius of an equilateral triangle

#1: P is the interior of equilateral triangle ABC, such that perpendicular segments from P to each of the sides of triangle ABC measure 2 inches, 9 inches and 13 inches. Find the number of square inches in the area of triangle ABC, and express your answer in simplest radical form.

#2: AMC 2007-B: Point P is inside equilateral  ABC. Points Q, R, and S are the feet of the
perpendiculars from P to AB, BC, and CA, respectively. Given that PQ = 1,PR = 2, and PS = 3, what is AB ?
#3:

This is an equilateral triangle. If the side is "S", the length of the in-radius would be$$\dfrac {\sqrt {3}} {6}$$ of  S (or $$\dfrac {1} {3}$$of the height) and the length of the circum-radius would be$$\dfrac {\sqrt {3}} {3}$$ of  S (or $$\dfrac {2} {3}$$of the height).

You can use 30-60-90 degree special right triangle angle ratio to get the length of each side as well as the height.

Solution I: Let the side be "s" and break the triangle into three smaller triangles.

$$\dfrac {9+13+2} {2}$$= 12s (base times height divided by 2)= $$\dfrac {\sqrt {3}} {4}\times s^{2}$$

s = 16 3

Area of the triangle = 192 3

Solution II: Let the side be "s" and the height of the equilateral triangle be "h"
24* s (by adding 9, 2 and 13 since they are the height of each smaller triangle)  = s*h
(Omit the divided by 2 part on either side since it cancels each other out.)
h = 24
Using 30-60-90 degree angle ratio, you get $$\dfrac {1} {2}$$ s = 8 3  so  s = 16 3
Area of the equilateral triangle = $$\dfrac {24\times 16\sqrt {3}} {2}$$ = $$192\sqrt {3}$$

#2: This one is similar to #1, the answer is 4 3

Thursday, January 23, 2014

Three Pole Problems : Similar triangles

Question: If you know the length of x and y, and the whole length of $$\overline {AB}$$,

A: what is the ratio of a to b and

B: what is the length of z.

Solution for question A:
$$\Delta$$ABC and $$\Delta$$AFE are similar so $$\dfrac {z} {x}=\dfrac {b} {a+b}$$. -- equation 1
Cross multiply and you have z ( a + b ) = bx

$$\Delta$$BAD and $$\Delta$$BFE are similar so $$\dfrac {z} {y}=\dfrac {a} {a+b}$$. -- equation 2
Cross multiply and you have z ( a + b ) = ay

bx = ay so $$\dfrac {x} {y}=\dfrac {a} {b}$$  same ratio

Solution for question B:
Continue with the previous two equations, if you add equation 1 and equation 2, you have:
$$\dfrac {z} {x}+\dfrac {z} {y}=\dfrac {b} {a+b}+\dfrac {a} {a+b}$$
$$\dfrac {zy+zx } {xy}=1$$ $$\rightarrow$$ z = $$\dfrac {xy} {x+y}$$

Applicable question:

$$\overline {CD}=15$$ and you know $$\overline {DB}:\overline {BC}=20:30=2:3$$
so  $$\overline {DB}=6$$ and $$\overline {BC}=9$$

$$\overline {AB}=\dfrac {20\times 30} {\left( 20+30\right) }$$ = 12

Monday, January 6, 2014

2014 Mathcounts State/National/AMC Prep : More Counting and Probability Problems from Mathcounts Mini / AoPS

From Mathcounts Mini : Video tutorials on counting and probability for Mathcounts state prep

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.