Some of the harder/hardest questions at Mathcounts can be tackled at ease using mass point geometry
so spend some time understanding it.
Basics
2014-15 Mathcounts handbook Mass Point Geometry Stretch
from page 39 to page 40
(Talking about motivation, yes, there are students already almost finish
this year's Mathcounts' handbook harder problems.)
From Wikipedia
From AoPS
Mass Point Geometry by Tom Rike
Another useful notes
Videos on Mass Point :
Mass Points Geometry Part I
Mass Points Geometry : Split Masses Part II
Mass Points Geometry : Part III
other videos from Youtube on Mass Points
It's much more important to fully understand how it works, the easier questions the weights align
very nicely.
The harder problems the weights are messier, not aligning nicely, so you need to find ways to may them integers (LCM) for easier solving.
Let me know if you have questions. I love to help (:D) if you've tried.
Showing posts with label Mathcounts state. Show all posts
Showing posts with label Mathcounts state. Show all posts
Sunday, March 6, 2022
Tuesday, September 18, 2018
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)?
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
1d. 1.42 [100% + 40% increase = 1.4] x 0.8 [100% -20% = 0.8] = 1.568 = 156.8% of the original volume
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
Thursday, January 5, 2017
Wednesday, October 5, 2016
2017 Mathcounts State Prep: Volume of a Regular Tetrahedron and Its Relationship with the Cube it's Embedded
How to find volume of a tetrahedron (right pyramid) with side length one.
The above link gives you a visual interpretation of the relationship of a regular tetrahedron, its
relationship with the cube that it is embedded and the other kind of tetrahedron (right angle pyramid).
The side of the cube is \(\dfrac {S} {\sqrt {2}}\) so the volume of the regular embedded tetrahedron is
\(\dfrac {1} {3}\times \left( \dfrac {S} {\sqrt {2}}\right) ^{3}\)=\(\dfrac {1} {3}*\dfrac {s^{3}} {2\sqrt {2}}\)= \(\dfrac {\sqrt {2}S^{3}} {12}\).
You can also fine the height of the tetrahedron and then \(\dfrac {1} {3}\)*base*space height to get the volume.
Using Pythagorean theory, the hypotenuse S and one leg which is \(\dfrac {2} {3}\) of the height of the equilateral triangle base, you'll get the space height.
The above link gives you a visual interpretation of the relationship of a regular tetrahedron, its
relationship with the cube that it is embedded and the other kind of tetrahedron (right angle pyramid).
The side of the cube is \(\dfrac {S} {\sqrt {2}}\) so the volume of the regular embedded tetrahedron is
\(\dfrac {1} {3}\times \left( \dfrac {S} {\sqrt {2}}\right) ^{3}\)=\(\dfrac {1} {3}*\dfrac {s^{3}} {2\sqrt {2}}\)= \(\dfrac {\sqrt {2}S^{3}} {12}\).
You can also fine the height of the tetrahedron and then \(\dfrac {1} {3}\)*base*space height to get the volume.
Using Pythagorean theory, the hypotenuse S and one leg which is \(\dfrac {2} {3}\) of the height of the equilateral triangle base, you'll get the space height.
Monday, April 6, 2015
Rate, Time, and Distance Question
Question: Harder SAT question: Esther
drove to work in the morning at an average speed of 45 miles per hour.
She returned home in the evening along the same route and averaged 30
miles per hour. If Esther spent a total of one hour commuting to and
from work, how many miles did Esther drive to work in the morning?
Solutions I:
Let t be the time it took Esther to drive to work and (1-t ) be the time it took him to return home.
Solution II:
Let D be the distance from Esther's home to work.
Solutions I:
Let t be the time it took Esther to drive to work and (1-t ) be the time it took him to return home.
Since rate times time = distance, we can set the equation as 45 t = 30 (1- t), 75t = 30 so
t = \(\frac{\Large{2}}{\Large{5}}\) Plug in t to the previous equation : 45 * \(\frac{\Large{2}}{\Large{5}}\) = 18 miles ,which is the answer.
t = \(\frac{\Large{2}}{\Large{5}}\) Plug in t to the previous equation : 45 * \(\frac{\Large{2}}{\Large{5}}\) = 18 miles ,which is the answer.
Solution II:
Let D be the distance from Esther's home to work.
\(\frac{\Large{D}}{\Large{45}}\) + \(\frac{\Large{D}}{\Large{30}}\) = 1 (hour)
Times 90 both sides to get rid of the denominator and you have 2D + 3D = 90
D = 18 miles
Times 90 both sides to get rid of the denominator and you have 2D + 3D = 90
D = 18 miles
Solution III:
The rate ratio between driving to work and returning home is 45 : 30 or 3 : 2.
Since rate and time are inversely related (rt = d), the time ratio between the two is 2 : 3.
45*\(\frac{\Large{2}}{\Large{5}}\) * 1 (hour) = 18 miles
Solution IV: (from Varun in FL)
The average speed can be found by harmonic mean: \(\dfrac {2\times 45\times 30}{45+35}\) = 36 mph. She drives for a total of 1 hour, so she drives 36 miles. Since she drives the same distance both ways, the distance to work is \(\dfrac {36} {2}\)=18 miles.
Question 1991 Mathcounts National #28 : A man is running through a train tunnel. When he is \(\frac{\Large{2}}{\Large{5}}\) of the way through, he nears a train that is approaching the tunnel from behind him at a speed of 60 mph. Whether he runs ahead or runs back, he will reach an end of the tunnel at the same time the train reaches that end. At what rate, in miles per hour, is he running? (Assume he runs at a constant rate.)
Solution : When the man went back \(\frac{\Large{2}}{\Large{5}}\) of the length of the tunnel, the train is at the beginning of the tunnel.
Same thing happened if the men went ahead \(\frac{\Large{2}}{\Large{5}}\) of the length of the tunnel. The time it took the train to run through the length of the tunnel, the man could only run \(\frac{\Large{1}}{\Large{5}}\) of the same distance.
So the speed of the man is \(\frac{\Large{1}}{\Large{5}}\) of the train's speed, which is \(\frac{\Large{1}}{\Large{5}}\) * 60 or 12 mph.
Question #3: Sally drives to her aunt's house, which is 84 miles away, at 40 miles per hour and comes back home at 60 miles per hour. What is her average speed?
Solution I:
To find average speed, you use total distance over total time it takes Sally to drive.
It takes Sally \(\frac{\Large{84}}{\Large{40}}\) + \(\frac{\Large{84}}{\Large{60}}\), or total 3.5 hours
to drive 84 * 2 = 168 miles.
\(\frac{\Large{168}}{\Large{3.5}}\) = 48 mph
Solution II:
Similar to dimensional change question, as long as the segments are constant, the distance Sally drives is extraneous. Thus, it's much easier to use \(\frac{\Large{2}}{\frac{\Large{1}}{\Large{40}} - \frac{\Large{1}}{\Large{60}}}\)or \(\frac{\Large{2*40*60}}{\Large{40+60}}\) = 48 mph
Question #4: If Sally drives to her work at 40 miles per hour, she will be 15 minutes late. If she drives to her work at 60 miles per hour, she will be 15 minutes early, what speed will she arrive at her work on time? How far away is her office ?
Solution I:
Let t be the time (hpm) Sally drives to her work on time. According to the given, we can set up
the following equation:
40 (t + \(\frac{\Large{1}}{\Large{4}}\)) = 60 ( t- \(\frac{\Large{1}}{\Large{4}}\) )
40 t + 10 = 60 t - 15 ; 25 = 20t ; t = 1.25
Plug in to get the distance as 40 (1.25 + 0.25) = 60 miles
To get the average, use total distance over total time she spent \(\rightarrow\) \(\frac{D}{T}\) = \(\frac{\Large{120}}{\Large{1.5 + 1}}\) = 48 mph
Solution II :
Again, have you noticed that if both time (late or early) to arrive to the destination is the same, the easiest
method is to still use the harmonic mean. \(\rightarrow\) \(\frac{\Large{2*a * b}}{\Large{a + b}}\)
a and b are the rates. \(\frac{\Large{2*40*60}}{\Large{40+60}}\) = 48 mph
45*\(\frac{\Large{2}}{\Large{5}}\) * 1 (hour) = 18 miles
Solution IV: (from Varun in FL)
The average speed can be found by harmonic mean: \(\dfrac {2\times 45\times 30}{45+35}\) = 36 mph. She drives for a total of 1 hour, so she drives 36 miles. Since she drives the same distance both ways, the distance to work is \(\dfrac {36} {2}\)=18 miles.
Question 1991 Mathcounts National #28 : A man is running through a train tunnel. When he is \(\frac{\Large{2}}{\Large{5}}\) of the way through, he nears a train that is approaching the tunnel from behind him at a speed of 60 mph. Whether he runs ahead or runs back, he will reach an end of the tunnel at the same time the train reaches that end. At what rate, in miles per hour, is he running? (Assume he runs at a constant rate.)
![]() |
Solution : When the man went back \(\frac{\Large{2}}{\Large{5}}\) of the length of the tunnel, the train is at the beginning of the tunnel.
Same thing happened if the men went ahead \(\frac{\Large{2}}{\Large{5}}\) of the length of the tunnel. The time it took the train to run through the length of the tunnel, the man could only run \(\frac{\Large{1}}{\Large{5}}\) of the same distance.
So the speed of the man is \(\frac{\Large{1}}{\Large{5}}\) of the train's speed, which is \(\frac{\Large{1}}{\Large{5}}\) * 60 or 12 mph.
Question #3: Sally drives to her aunt's house, which is 84 miles away, at 40 miles per hour and comes back home at 60 miles per hour. What is her average speed?
Solution I:
To find average speed, you use total distance over total time it takes Sally to drive.
It takes Sally \(\frac{\Large{84}}{\Large{40}}\) + \(\frac{\Large{84}}{\Large{60}}\), or total 3.5 hours
to drive 84 * 2 = 168 miles.
\(\frac{\Large{168}}{\Large{3.5}}\) = 48 mph
Solution II:
Similar to dimensional change question, as long as the segments are constant, the distance Sally drives is extraneous. Thus, it's much easier to use \(\frac{\Large{2}}{\frac{\Large{1}}{\Large{40}} - \frac{\Large{1}}{\Large{60}}}\)or \(\frac{\Large{2*40*60}}{\Large{40+60}}\) = 48 mph
Question #4: If Sally drives to her work at 40 miles per hour, she will be 15 minutes late. If she drives to her work at 60 miles per hour, she will be 15 minutes early, what speed will she arrive at her work on time? How far away is her office ?
Solution I:
Let t be the time (hpm) Sally drives to her work on time. According to the given, we can set up
the following equation:
40 (t + \(\frac{\Large{1}}{\Large{4}}\)) = 60 ( t- \(\frac{\Large{1}}{\Large{4}}\) )
40 t + 10 = 60 t - 15 ; 25 = 20t ; t = 1.25
Plug in to get the distance as 40 (1.25 + 0.25) = 60 miles
To get the average, use total distance over total time she spent \(\rightarrow\) \(\frac{D}{T}\) = \(\frac{\Large{120}}{\Large{1.5 + 1}}\) = 48 mph
Solution II :
Again, have you noticed that if both time (late or early) to arrive to the destination is the same, the easiest
method is to still use the harmonic mean. \(\rightarrow\) \(\frac{\Large{2*a * b}}{\Large{a + b}}\)
a and b are the rates. \(\frac{\Large{2*40*60}}{\Large{40+60}}\) = 48 mph
Labels:
distance,
Mathcounts,
Mathcounts problems,
Mathcounts state,
Rate,
rt = d,
SAT harder math,
time,
word problems
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
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
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 !!
Q #1 : (1, 3), (7, 3) and (1, -3)
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 )
Answer key :
#1 : (x - 2) 2 + ( y - 9 ) 2 = 32
#2 : x2 + (y + 3)2 = 100
#3 : (x - 2) 2 + ( y -1) 2 = 37
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 --
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 =
\(\sqrt{5}\).More practices on similar questions : (Answers below for self check)
Q #1 : A (2, 5) , B (2, 13) , and C (-6, 5 )
Q #3 : A (3, -5) , B (-4, 2) and C (1, 7 )
Answer key :
#1 : (x - 2) 2 + ( y - 9 ) 2 = 32
#2 : x2 + (y + 3)2 = 100
#3 : (x - 2) 2 + ( y -1) 2 = 37
Tuesday, May 6, 2014
This Week's Work : Week 49 - for Inquisitive Young Mathletes
Simon's Favorite Factoring Trick from AoPS
Notes from my blog
Now try this year's team round #9 using SFFT.
Review all the team round questions.
From Mathcounts Mini : Maximum Areas of Inscribed Rectangles and Triangles
The boring algebra proof from my blog. (Oh, dear !!)
Notes from my blog
Now try this year's team round #9 using SFFT.
Review all the team round questions.
From Mathcounts Mini : Maximum Areas of Inscribed Rectangles and Triangles
The boring algebra proof from my blog. (Oh, dear !!)
Tuesday, April 29, 2014
This Week's Work : Week 48 - for Inquisitive Young Mathletes
Tackling Problems with Vieta
Practicing Problems using Vieta's formula The Binomial Theorem |
- Section 14.2: Introducing the Binomial Theorem
- Section 14.4: Using the Binomial Theorem Part 1
- Section 14.4: Using the Binomial Theorem Part 2
- Section 14.4: Using the Binomial Theorem Part 3
- Section 14.5: Using the Binomial Theorem Part 4
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 |
- Section 14.2: Introducing the Binomial Theorem
- Section 14.4: Using the Binomial Theorem Part 1
- Section 14.4: Using the Binomial Theorem Part 2
- Section 14.4: Using the Binomial Theorem Part 3
- Section 14.5: Using the Binomial Theorem Part 4
Take care and happy problem solving !!
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?Problems: (Solutions below)
#2: 2000 State sprint #30. Joe bought a pumpkin that cost
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 :
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)
The answer is 16 pounds.
Labels:
Mathcounts,
Mathcounts problems,
Mathcounts state
Friday, November 1, 2013
Counting I : Ways to Avoid Over Counting or Under Counting
Check out Mathcounts here, the best competition math program for middle school students.
Download this year's Mathcounts handbook here.
Question #1: How many ways to count a triple of natural numbers whose sum adds up to 12
if A. order doesn't matter? B. if order matters?
Solution I:
A. Find a systematic way to solve this type of problem in an organized manner.
We can start with the smallest natural number "1".
(1, 1, 10), (1, 2, 9), (1, 3, 8), (1, 4, 7), (1, 5, 6) Since (1, 5, 6) is the same as (1, 6, 5) so stop.
(2, 2, 8), (2, 3, 7), (2, 4, 6), (2, 5, 5)
(3, 3, 6), (3, 4, 5)
(4, 4, 4) so total there are 12 ways.
B. Since in this case order matters so if you add up all the possible ways, for example, there are "3" ways to arrange (1, 1, 10) -- 3!/2! = 3
This is similar to "how many ways to arrange the letters 'odd'.
If all letters/numbers are different, you have 3! ways to arrange them; however, in this case, the two numbers 1, 1 are indistinguishable and there are 2! ways to arrange them, thus 3!/2! = 3
There are 3! or 6 ways to arrange triples such as (1, 2, 9).
Add all the possible ways and there are total 55 ways.
Solution II:
B.There is a much easier way to tackle this question, using bars and stars or sticks and stones method.
There are 11 spaces between 12 stones. @__@__@__@__@__@__@__@__@__@__@__@
If you places the two sticks on two of the spaces, you'll split the stones into three groups.
For example, if you have @ | @@@@@ | @@@@@@
The triples are 1, 5, and 6.
If it's @@@@@@@@ | @@ | @@ , the triples are 8, 2, and 2.
So 11C2 = 55 ways
Let me know if you are still confused. Have fun problem solving !!
Download this year's Mathcounts handbook here.
Question #1: How many ways to count a triple of natural numbers whose sum adds up to 12
if A. order doesn't matter? B. if order matters?
Solution I:
A. Find a systematic way to solve this type of problem in an organized manner.
We can start with the smallest natural number "1".
(1, 1, 10), (1, 2, 9), (1, 3, 8), (1, 4, 7), (1, 5, 6) Since (1, 5, 6) is the same as (1, 6, 5) so stop.
(2, 2, 8), (2, 3, 7), (2, 4, 6), (2, 5, 5)
(3, 3, 6), (3, 4, 5)
(4, 4, 4) so total there are 12 ways.
B. Since in this case order matters so if you add up all the possible ways, for example, there are "3" ways to arrange (1, 1, 10) -- 3!/2! = 3
This is similar to "how many ways to arrange the letters 'odd'.
If all letters/numbers are different, you have 3! ways to arrange them; however, in this case, the two numbers 1, 1 are indistinguishable and there are 2! ways to arrange them, thus 3!/2! = 3
There are 3! or 6 ways to arrange triples such as (1, 2, 9).
Add all the possible ways and there are total 55 ways.
Solution II:
B.There is a much easier way to tackle this question, using bars and stars or sticks and stones method.
There are 11 spaces between 12 stones. @__@__@__@__@__@__@__@__@__@__@__@
If you places the two sticks on two of the spaces, you'll split the stones into three groups.
For example, if you have @ | @@@@@ | @@@@@@
The triples are 1, 5, and 6.
If it's @@@@@@@@ | @@ | @@ , the triples are 8, 2, and 2.
So 11C2 = 55 ways
Let me know if you are still confused. Have fun problem solving !!
Thursday, September 5, 2013
3 - D Geometry and harder AMCs, AIME links
For Mathcounts advanced group -- harder Mathcounts State/National problem on 3-D geometry.
From Mathcounts Mini :
Using Similarity to Solve Geometry Problems
Area and Volume
Relationship Between A Box's Dimensions, Volume and Surface Area
AMC, AIME videos from AoPS
Harder/Hardest AMC-10, 12 and AIME problems explained by Mrs. Rusczyk
From Mathcounts Mini :
Using Similarity to Solve Geometry Problems
Area and Volume
Relationship Between A Box's Dimensions, Volume and Surface Area
AMC, AIME videos from AoPS
Harder/Hardest AMC-10, 12 and AIME problems explained by Mrs. Rusczyk
Monday, December 24, 2012
2013 Mathcounts State Prep: Counting and Probability
Question #1: Rolling two dice, what is the probability that the product is a multiple of 3?
Solution I :
As long as one of the two numbers turn up as 3 or multiple of 3 (in this case, a "6"), the product of the two numbers will be a multiple of 3.
There are 6 * 6 = 36 ways to get 2 numbers. Out of the 36 pairs, you can have
3 - 1
3 - 2
3 - 3
3 - 4
3 - 5
3 - 6, 6 ways.
However, there are only 5 ways left if the other die has a 3 since 3 - 3 only counts as once.
1 - 3
2 - 3
4 - 3
5 - 3
6 - 3
Next we look at "at least one number is "6".
6 - 1
6 - 2
6 - 4 (We already used 6 - 3)
6 - 5
6 - 6 so 5 ways.
The other way around, we have
1 - 6
2 - 6
4 - 6
5 - 6 Total 4 ways, so the answer is \(\frac{\Large{( 6 + 5 + 5 + 4 )}}{\Large{36}}\) = \(\frac{\Large{ 20}}{\Large{36}}\) = \(\frac{\Large{5}}{\Large{9}}\).
Solution II:
The easiest way to solve this problem is to use complementary counting, which is 1 (100% or total possible way) - none of the the multiples of 3 showing up, so 1 - \(\frac{\Large{4}}{\Large{6}}\) *\(\frac{\Large{4}}{\Large{6}}\) = 1 - \(\frac{\Large{2}}{\Large{3}}\) * \(\frac{\Large{2}}{\Large{3}}\)= \(\frac{\Large{5}}{\Large{9}}\)
Question #2: [2002 AMC-12B #16] Juan rolls a fair regular eight-sided die. Then Amal rolls a fair regular six-sided die. What isthe probability that the product of the two rolls is a multiple of 3?
Solution:
Using complementary counting (see solution II of the previous question), 1 - \(\frac{\Large{6}}{\Large{8}}\) *\(\frac{\Large{4}}{\Large{6}}\) = 1 - \(\frac{\Large{3}}{\Large{4}}\) * \(\frac{\Large{2}}{\Large{3}}\)= \(\frac{\Large{1}}{\Large{2}}\)
Solution I :
As long as one of the two numbers turn up as 3 or multiple of 3 (in this case, a "6"), the product of the two numbers will be a multiple of 3.
There are 6 * 6 = 36 ways to get 2 numbers. Out of the 36 pairs, you can have
3 - 1
3 - 2
3 - 3
3 - 4
3 - 5
3 - 6, 6 ways.
However, there are only 5 ways left if the other die has a 3 since 3 - 3 only counts as once.
1 - 3
2 - 3
4 - 3
5 - 3
6 - 3
Next we look at "at least one number is "6".
6 - 1
6 - 2
6 - 4 (We already used 6 - 3)
6 - 5
6 - 6 so 5 ways.
The other way around, we have
1 - 6
2 - 6
4 - 6
5 - 6 Total 4 ways, so the answer is \(\frac{\Large{( 6 + 5 + 5 + 4 )}}{\Large{36}}\) = \(\frac{\Large{ 20}}{\Large{36}}\) = \(\frac{\Large{5}}{\Large{9}}\).
Solution II:
The easiest way to solve this problem is to use complementary counting, which is 1 (100% or total possible way) - none of the the multiples of 3 showing up, so 1 - \(\frac{\Large{4}}{\Large{6}}\) *\(\frac{\Large{4}}{\Large{6}}\) = 1 - \(\frac{\Large{2}}{\Large{3}}\) * \(\frac{\Large{2}}{\Large{3}}\)= \(\frac{\Large{5}}{\Large{9}}\)
Question #2: [2002 AMC-12B #16] Juan rolls a fair regular eight-sided die. Then Amal rolls a fair regular six-sided die. What isthe probability that the product of the two rolls is a multiple of 3?
Solution:
Using complementary counting (see solution II of the previous question), 1 - \(\frac{\Large{6}}{\Large{8}}\) *\(\frac{\Large{4}}{\Large{6}}\) = 1 - \(\frac{\Large{3}}{\Large{4}}\) * \(\frac{\Large{2}}{\Large{3}}\)= \(\frac{\Large{1}}{\Large{2}}\)
Friday, December 21, 2012
2013 Mathcounts State Prep: Similar Triangles and Height to the Hypotenuse

Question:
#1: Δ ABC is a 3-4-5 right triangle. What is the height to the hypotenuse?
Solution:
Use the area of a triangle to get the height to the hypotenuse.
Let the height to the hypotenuse be "h"
The area of Δ ABC is \(\Large\frac{3*4}{2}\)= \(\Large\frac{5*h}{2}\)
Both sides times 2 and consolidate: h = \(\Large\frac{3*4}{5}\) = \(\Large\frac{12}{5}\)
Practice: What is the height to the hypotenuse?
Question:
#2: How many similar triangles can you spot?
Solution:
There are 4 and most students have difficulty comparing the largest one with the other smaller ones.
Δ ABC is similar to Δ ADE, Δ FBD, ΔGEC. Make sure you really understand this and can apply this to numerous similar triangle questions.
Question:
#3: What is the area of □ DEGF if \(\overline{BF}\) = 9 and \(\overline{GC}\) = 4
Solution:
Using the two similar triangles Δ FBD and ΔGEC (I found using symbols to find the corresponding legs
to be much easier than using the lines.), you have \(\frac{\Large{\overline{BF}}}{\Large{\overline{FD}}}\) = \(\frac{\Large{\overline{GE}}}{\Large{\overline{GC}}}\).
s (side length of the square) = \({\overline{GE}}\) = \({\overline{FD}}\)
Plug in the given and you have 9 * 4 = s2 so the area of □ DEGF is 36 square units. (each side then is square root of 36 or 6)
Question:
#4: Δ ABC is a 9-12-15 right triangle. What is the side length of the square?
Solution :
The height to the hypotenuse is\(\frac{\Large{9*12}}{\Large{15}}\) = \(\frac{\Large{36}}{\Large{5}}\)
Δ ABC is similar to Δ ADE. Using base and height similarities, you have \(\frac{\Large{\overline{BC}}}{\Large{\overline{DE}}}\) = \(\frac{\Large{15}}{\Large{S}}\) = \(\frac{\frac{\Large{36}}{\Large{5}}}{\frac{\Large{36}}{\Large{5}} - \Large{S}}\)
Cross multiply and you have 108 - 15*S = \(\frac{\Large{36}}{\Large{5}}\) *S
S =\(\frac{\Large{180}}{\Large{37}}\)
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