Show your work, or, how my math abilities started to decline
I
think it's problematic the way schools teach Algebra. Those meaningless
show-your-work approaches, without knowing what Algebra is truly about.
The overuse of calculators and the piecemeal way of teaching without
the unification of the math concepts are detrimental to our children's
ability to think critically and logically.
Of course
eventually, it would be beneficial to students if they show their work
with the much more challenging word problems (harder Mathcounts state
team round, counting and probability questions, etc...), but it's totally different from what some schools ask of our capable students.
How
do you improve problem solving skills with tons of worksheets by going
through 50 to 100 problems all look very much the same? It's called busy
work.
Quote: "Insanity: doing the same thing over and over again and expecting different results."
Quotes from Richard Feynman, the famous late Nobel-laureate physicist. Feynman relates his cousin's unhappy experience with algebra:
My
cousin at that time—who was three years older—was in high school and
was having considerable difficulty with his algebra. I was allowed to
sit in the corner while the tutor tried to teach my cousin algebra. I
said to my cousin then, "What are you trying to do?" I hear him talking
about x, you know."Well, you know, 2x + 7 is equal to 15," he said, "and
I'm trying to figure out what x is," and I say, "You mean 4." He says,
"Yeah, but you did it by arithmetic. You have to do it by algebra."And
that's why my cousin was never able to do algebra, because he didn't
understand how he was supposed to do it. I learned
algebra, fortunately, by—not going to school—by knowing the whole idea
was to find out what x was and it didn't make any difference how you did
it. There's no such a thing as, you know, do it by arithmetic, or you do it by algebra. It was a false thing that they had invented in
school, so that the children who have to study algebra can all pass it.
They had invented a set of rules, which if you followed them without
thinking, could produce the answer. Subtract 7 from both sides. If you
have a multiplier, divide both sides by the multiplier. And so on. A
series of steps by which you could get the answer if you didn't
understand what you were trying to do.
So I was lucky. I always learnt things by myself.
Showing posts with label problem solving. Show all posts
Showing posts with label problem solving. Show all posts
Sunday, May 25, 2025
Tuesday, November 12, 2024
A Skill for the 21st Century: Problem Solving by Richard Rusczyk
Does our approach to teaching math fail even the smartest kids ?
Quotes from that article "According to research from the University of California, Los Angeles, as many as 60 percent of all college students who intend to study a STEM (science, technology, engineering, math) subject end up transferring out. In an era when politicians and educators are beside themselves with worry over American students’ lagging math and science scores compared to the whiz kids of Shanghai and Japan, this attrition trend so troubles experts it has spawned an entire field of research on “STEM drop-out,” citing reasons from gender and race to GPAs and peer relationships."
A Skill for the 21st Century: Problem Solving by Richard Rusczyk, founder of "Art of Problem Solving".
Top 10 Skills We Wish Were Taught at School, But Usually Aren't
from Lifehacker
Quotes from that article "According to research from the University of California, Los Angeles, as many as 60 percent of all college students who intend to study a STEM (science, technology, engineering, math) subject end up transferring out. In an era when politicians and educators are beside themselves with worry over American students’ lagging math and science scores compared to the whiz kids of Shanghai and Japan, this attrition trend so troubles experts it has spawned an entire field of research on “STEM drop-out,” citing reasons from gender and race to GPAs and peer relationships."
A Skill for the 21st Century: Problem Solving by Richard Rusczyk, founder of "Art of Problem Solving".
Top 10 Skills We Wish Were Taught at School, But Usually Aren't
from Lifehacker
Saturday, October 15, 2022
16 17 Mathcounts handbook more interesting questions that have nicer solutions
Thanks to Achuth for trying out these questions and time them as an actual Mathcounts test. :)
First week : warm up 1, 4, 7. (time for 40 mins. like sprint)
Second week : warm up 2, 5, 8.
third week : workout 3 --> all right. (pair 1 to 6, 2 to 7, each time for 6 mins. as
target)
fourth week: workout 4 --> #95, then self correct.
At lesson: workout 5 and other harder problems.
These are nice questions that have various solutions, so it’s better to slow down and try them as puzzles.
Less is more and slow is fast.
If you are new to problem solving, one nice strategy is to make the question much simpler and explore ideas that come to your mind.
Answer key down below.
#105
When fully matured, a grape contains 80% water. After the drying process, called dehydration, the resulting raisin is only 20% water. What fraction of the original water in the grape remains after dehydration? Express your answer as a common fraction.
#112: Cora has five balls—two red, two blue and one yellow—which are indistinguishable except for their color. She has two containers—one red and one green. If the balls are randomly distributed between the two containers, what is the probability that the two red balls will be alone in the red container? Express your answer as a common fraction?
#116: A 12-foot by 12-foot square bathroom needs to be tiled with 1-foot square tiles. Two of the tiles are the wrong color. If the tiles are placed randomly, what is the probability that the two wrong-colored tiles share an edge? Express your answer as a common fraction.
#66: 4851
#105: 1/16
#112: 1/32
#116: 1/39
Sunday, April 15, 2018
Original Problem from a Student Problem Solver on Similar Triangles
An original problem on similar triangles from Varun (from FL)
Tuesday, February 14, 2017
2013 Mathcounts State Harder Problems
You can download this year's Mathcounts state competition questions here.
Trickier 2013 Mathcounts State Sprint Round questions :
Sprint #14:
From Varun:
Assume the term "everything" refers to all terms in the given set.
1 is a divisor of everything, so it must be first.
Everything is a divisor of 12, so it must be last.
The remaining numbers left are 2, 3, 4, and 6.
2 and 3 must come before 6, and 2 must come before 4.
Therefore, we can list out the possibilities for the middle four digits:
2,3,4,6
2,3,6,4
3,2,4,6
3,2,6,4
2,4,3,6
There are 5 ways--therefore 5 is the answer.
From Vinjai:
Here's how I did #14:
First, notice that 1 must be the first element of the set and 12 must be the last one.
So that leaves only 2,3,4,6 to arrange.
We can quickly list them out.
The restrictions are that 2 must be before 4, 3 must be before 6, and 2 must be before 6:
2,3,4,6 3,2,4,6 4,2,3,6 6,2,3,4
2,3,6,4 3,2,6,4 4,2,6,3 6,2,4,3
2,4,3,6 3,4,2,6 4,3,2,6 6,3,2,4
2,4,6,3 3,4,6,2 4,3,6,2 6,3,4,2
2,6,3,4 3,6,2,4 4,6,2,3 6,4,2,3
2,6,4,3 3,6,4,2 4,6,3,2 6,4,3,2
Only the bold ones work. So, the answer is 5.
#17: Common dimensional change problem
\(\overline {ZY}:\overline {WV}=5:8\) -- line ratio
The volume ratio of the smaller cone to the larger cone is thus \(5^{3}: 8^{3}\).
The volume of the frustum is the volume of the larger cone minus the volume of the smaller cone
= \(\dfrac {8^{3}-5^{3}} {8^{3}}\times \dfrac {1} {3}\times 8^{2}\times 32\times \pi\) = 516\(\pi\)
More problems to practice from Mathcounts Mini
#24: The answer is \(\dfrac {1} {21}\).
#28: Hats off to students who can get this in time !! Wow!!
From Vinjai:
Trickier 2013 Mathcounts State Sprint Round questions :
Sprint #14:
From Varun:
Assume the term "everything" refers to all terms in the given set.
1 is a divisor of everything, so it must be first.
Everything is a divisor of 12, so it must be last.
The remaining numbers left are 2, 3, 4, and 6.
2 and 3 must come before 6, and 2 must come before 4.
Therefore, we can list out the possibilities for the middle four digits:
2,3,4,6
2,3,6,4
3,2,4,6
3,2,6,4
2,4,3,6
There are 5 ways--therefore 5 is the answer.
From Vinjai:
Here's how I did #14:
First, notice that 1 must be the first element of the set and 12 must be the last one.
So that leaves only 2,3,4,6 to arrange.
We can quickly list them out.
The restrictions are that 2 must be before 4, 3 must be before 6, and 2 must be before 6:
2,3,4,6 3,2,4,6 4,2,3,6 6,2,3,4
2,3,6,4 3,2,6,4 4,2,6,3 6,2,4,3
2,4,3,6 3,4,2,6 4,3,2,6 6,3,2,4
2,4,6,3 3,4,6,2 4,3,6,2 6,3,4,2
2,6,3,4 3,6,2,4 4,6,2,3 6,4,2,3
2,6,4,3 3,6,4,2 4,6,3,2 6,4,3,2
Only the bold ones work. So, the answer is 5.
#17: Common dimensional change problem
\(\overline {ZY}:\overline {WV}=5:8\) -- line ratio
The volume ratio of the smaller cone to the larger cone is thus \(5^{3}: 8^{3}\).
The volume of the frustum is the volume of the larger cone minus the volume of the smaller cone
= \(\dfrac {8^{3}-5^{3}} {8^{3}}\times \dfrac {1} {3}\times 8^{2}\times 32\times \pi\) = 516\(\pi\)
More problems to practice from Mathcounts Mini
#24: The answer is \(\dfrac {1} {21}\).
#28: Hats off to students who can get this in time !! Wow!!
From Vinjai:
For
#28, there might be a nicer way but here's how I did it when I took the sprint
round:
#
4's # 3's # 2's # 1's
# ways
1 2 0
0 3
1 1 1
1 24
1 1 0
3 20
1 0 3
0 4
1 0 2
2 30
1 0 1
4 30
0 2 2
0 6
0 2 1
2 30
0 2 0
4 15
0 1 3
1 20
0 1 2
3 60
0 0 3
4 35
TOTAL:
277
#29:
\(\Delta ADE\) is similar to \(\Delta ABC\)
Let the two sides of the rectangle be x and y (see image on the left)
\(\dfrac {x} {21}=\dfrac {8-y} {8}\)
x =\(\dfrac {21\left( 8-y\right) } {8}\)
xy = \(\dfrac {21\left( 8-y\right) } {8}\) * y = \(\dfrac {-21y\left( y-8\right) } {8}\) =
\(\dfrac {-2l\left( y-4\right) ^{2}+21\times 16} {8}\)
From the previous equation you know when y = 4, the area \(\dfrac {21\times 16} {8}\)is the largest. The answer is 42.
Here is a proof to demonstrate that the largest area of a rectangle inscribed in a triangle is
half of the area of that triangle.
#30:
Solution I :
If (x, y) are the coordinates of the center of rotational points, it will be equal distance from A and A' as well as from B and B'.
Use distance formula, consolidate/simplify and solve the two equations, you'll get the answer (4, 1).
Solution II:
How to find the center of rotation from Youtube.
From AoPS using the same question
To sum up:
First, connect the corresponding points, in this case A to A' and B to B'.
Second, find the equation of the perpendicular bisector of line \(\overline {AA'}\), which is
y = - x + 5
and \(\overline {BB'}\), which is y = 5x - 19
The interception of the two lines is the center of rotation.
The answer is (4, 1).
2013 Mathcounts Target :
#3:
RT = D, unit conversions and different rates are tested here:
Make Joy's rate (speed) uphill be x m/s, his downhill speed be 2x m/s.
It takes Greg 3000 seconds (time) to reach the starting point and that is also what it takes Joy to
ride up to the hill and down to the same point.
\(\dfrac {7000} {x}+\dfrac {10000} {2x}=3000\) \(\rightarrow\) x = 4 m/s
#8:
Using "finding the height to the hypotenuse".( click to review)
you get \(\overline {CD}=\dfrac {7\times 24} {25}\).
Using similar triangles ACB and ADC, you get \(\overline {AD}=\dfrac {576} {25}\).
[\(\dfrac {24} {x}=\dfrac {25} {24}\)]
Using angle bisector (click to review),
you have \(\overline {AC}:\overline {AD}=\overline {CE}:\overline {ED}= 24: \dfrac {576} {25}\) = 600 : 576 = 25 : 24
\(\rightarrow\)\(\overline {ED}= \overline {CD}\times \dfrac {24} {24+25}\) = \(\dfrac {7\times 24} {25}\times \dfrac {24} {24+25}\) = \(\dfrac {576} {175}\)
\(\dfrac {x} {21}=\dfrac {8-y} {8}\)
x =\(\dfrac {21\left( 8-y\right) } {8}\)
xy = \(\dfrac {21\left( 8-y\right) } {8}\) * y = \(\dfrac {-21y\left( y-8\right) } {8}\) =
\(\dfrac {-2l\left( y-4\right) ^{2}+21\times 16} {8}\)
From the previous equation you know when y = 4, the area \(\dfrac {21\times 16} {8}\)is the largest. The answer is 42.
Here is a proof to demonstrate that the largest area of a rectangle inscribed in a triangle is
half of the area of that triangle.
#30:
Solution I :
If (x, y) are the coordinates of the center of rotational points, it will be equal distance from A and A' as well as from B and B'.
Use distance formula, consolidate/simplify and solve the two equations, you'll get the answer (4, 1).
Solution II:
How to find the center of rotation from Youtube.
From AoPS using the same question
To sum up:
First, connect the corresponding points, in this case A to A' and B to B'.
Second, find the equation of the perpendicular bisector of line \(\overline {AA'}\), which is
y = - x + 5
and \(\overline {BB'}\), which is y = 5x - 19
The interception of the two lines is the center of rotation.
The answer is (4, 1).
2013 Mathcounts Target :
#3:
RT = D, unit conversions and different rates are tested here:
Make Joy's rate (speed) uphill be x m/s, his downhill speed be 2x m/s.
It takes Greg 3000 seconds (time) to reach the starting point and that is also what it takes Joy to
ride up to the hill and down to the same point.
\(\dfrac {7000} {x}+\dfrac {10000} {2x}=3000\) \(\rightarrow\) x = 4 m/s
#8:
you get \(\overline {CD}=\dfrac {7\times 24} {25}\).
Using similar triangles ACB and ADC, you get \(\overline {AD}=\dfrac {576} {25}\).
[\(\dfrac {24} {x}=\dfrac {25} {24}\)]
Using angle bisector (click to review),
you have \(\overline {AC}:\overline {AD}=\overline {CE}:\overline {ED}= 24: \dfrac {576} {25}\) = 600 : 576 = 25 : 24
\(\rightarrow\)\(\overline {ED}= \overline {CD}\times \dfrac {24} {24+25}\) = \(\dfrac {7\times 24} {25}\times \dfrac {24} {24+25}\) = \(\dfrac {576} {175}\)
Thursday, January 5, 2017
Thursday, December 29, 2016
2013 Mathcounts School and Chapter Harder Problems
You can now download and discuss with your friends this year's school and chapter problems.
Here is the link to the official Mathcounts website.
Some more challenging problems from this year's Mathcounts school/or chapter problems.
2013 school team #10 : Three concepts are testing here :
Hint:
a. If you get rid of the remainder, the numbers will be evenly divided into 192, so you are looking at
those factors of 192 - 12 = 180
b. To leave a remainder of 12, those factors of 180 that are included in the Set must be smaller than 12, otherwise, you can further divide it.
c. To find the median, make sure to line up the numbers from the smallest to the largest and find the middle numbers. If there are even numbers of factors larger than 12, average the middle two. Otherwise, the middle number is the answer.
\(180=2^{2}\times 3^{2}\times 5 \) so there are (2 + 1) (2 + 1) (1 + 1) = 18 factors
The list on the left side gives you the first 9 and if you times those numbers with "5", you get 9 other factors,which are 5, 10, 20, 15, 18, 60, 45, 90 and 180.
Discard the factors that are smaller or equal to 12 and list all the other factors in order and find the median.
The answer is "36".
2013 Chapter Sprint:
#21: Dimensional change problem : The height of the top pyramid is \(\dfrac {2} {3}\) of the larger
pyramid so its volume is \(\left( \dfrac {2} {3}\right) ^{3}\) of the larger pyramid.
\(\left( \dfrac {2} {3}\right) ^{3}\times \dfrac {1} {3}\times \left( \dfrac {36} {4}\right) ^{2}\times 12
= \) \(96 cm^{3}\)
# 24: According to the given: \(xyz=720\) and \(2( xy+yz+zx)= 484 \) so
\(( xy+yz+zx )= 242\)
Since x, y and z are all integers, you factor 720 and see if it will come up with the same x, y and z values
for the second condition.
Problem writer(s) are very smart using this number because the numbers "6", "10", "12" would give you
a surface area of 252. (not right)
The three corrrect numbers are "8", "9", and "10" so the answer is \(\sqrt {8^{2}+9^{2}+10^{2}}=\) \(7\sqrt {5}\)
#25: Geometric probability: Explanations to similar questions and more practices below.
Probability with geometry representations form Aops.
Geometric probability from "Cut the Knots".
#26: This one is similar to 2002 AMC-10B #21, so try that question to get more practices.
2002 AMC-10B #21 link
#27:
\(\dfrac {1} {A}+\dfrac {1} {B}=\dfrac {1} {2}\)
\(\dfrac {1} {B}+\dfrac {1} {C}=\dfrac {1} {3}\)
\(\dfrac {1} {C}+\dfrac {1} {A}=\dfrac {1} {4}\)
Add them up and you have \(2 * (\dfrac {1} {A}+\dfrac {1} {B}+ \dfrac {1} {C})=\dfrac {13} {12}\)
\((\dfrac {1} {A}+\dfrac {1} {B}+ \dfrac {1} {C})=\dfrac {13} {24}\)
\(\dfrac {1} {\dfrac {1} {A}+\dfrac {1} {B}+\dfrac {1} {C}} = \)\(\dfrac {{24}} {13}\) hours
#28: Hint : the nth triangular number is the sum of the first "n" natural numbers and \(\dfrac {n\left( n+1\right) } {2}\) is how you use to find the sum.
From there, you should be able to find how many numbers will be evenly divided by "7".
#29 : Circle questions are very tricky so make sure to find more problems to practice accuracy.
#30 :
Solution I:
Read the solution that is provided by Mathcounts.org here.
Solution II:
Case 1 : \(x-1 > 0\rightarrow x > 1\) Times ( x - 1) on both sides and you have
\(x^{2}-1>8\) so x > 3 or x < -3 (discard)
Case 2: \(x-1 < 0\) so \(x < 1\) \(\rightarrow x^{2}-1 < 8\) [You need to change the sign since it's negative.]-3 < x < 3. Combined with x < 1 you have the range as -3 < x < 1
The answer is 60%.
2013 Mathcounts Target #7 and 8:
Target question #8 is very similar to 2011 chapter team #10.
It just asks differently.
Read the explanations provided on the Mathcounts official website.
They are explained very well.
Let me know if there are other easier ways to tackle those problems.
Hope this is helpful !! Thanks a lot !! Good luck on Mathcounts state.
Here is the link to the official Mathcounts website.
Some more challenging problems from this year's Mathcounts school/or chapter problems.
2013 school team #10 : Three concepts are testing here :
Hint:
a. If you get rid of the remainder, the numbers will be evenly divided into 192, so you are looking at
those factors of 192 - 12 = 180
b. To leave a remainder of 12, those factors of 180 that are included in the Set must be smaller than 12, otherwise, you can further divide it.
c. To find the median, make sure to line up the numbers from the smallest to the largest and find the middle numbers. If there are even numbers of factors larger than 12, average the middle two. Otherwise, the middle number is the answer.
\(180=2^{2}\times 3^{2}\times 5 \) so there are (2 + 1) (2 + 1) (1 + 1) = 18 factors
The list on the left side gives you the first 9 and if you times those numbers with "5", you get 9 other factors,which are 5, 10, 20, 15, 18, 60, 45, 90 and 180.
Discard the factors that are smaller or equal to 12 and list all the other factors in order and find the median.
The answer is "36".
2013 Chapter Sprint:
#21: Dimensional change problem : The height of the top pyramid is \(\dfrac {2} {3}\) of the larger
pyramid so its volume is \(\left( \dfrac {2} {3}\right) ^{3}\) of the larger pyramid.
\(\left( \dfrac {2} {3}\right) ^{3}\times \dfrac {1} {3}\times \left( \dfrac {36} {4}\right) ^{2}\times 12
= \) \(96 cm^{3}\)
# 24: According to the given: \(xyz=720\) and \(2( xy+yz+zx)= 484 \) so
\(( xy+yz+zx )= 242\)
Since x, y and z are all integers, you factor 720 and see if it will come up with the same x, y and z values
for the second condition.
Problem writer(s) are very smart using this number because the numbers "6", "10", "12" would give you
a surface area of 252. (not right)
The three corrrect numbers are "8", "9", and "10" so the answer is \(\sqrt {8^{2}+9^{2}+10^{2}}=\) \(7\sqrt {5}\)
#25: Geometric probability: Explanations to similar questions and more practices below.
Probability with geometry representations form Aops.
Geometric probability from "Cut the Knots".
#26: This one is similar to 2002 AMC-10B #21, so try that question to get more practices.
2002 AMC-10B #21 link
#27:
\(\dfrac {1} {A}+\dfrac {1} {B}=\dfrac {1} {2}\)
\(\dfrac {1} {B}+\dfrac {1} {C}=\dfrac {1} {3}\)
\(\dfrac {1} {C}+\dfrac {1} {A}=\dfrac {1} {4}\)
Add them up and you have \(2 * (\dfrac {1} {A}+\dfrac {1} {B}+ \dfrac {1} {C})=\dfrac {13} {12}\)
\((\dfrac {1} {A}+\dfrac {1} {B}+ \dfrac {1} {C})=\dfrac {13} {24}\)
\(\dfrac {1} {\dfrac {1} {A}+\dfrac {1} {B}+\dfrac {1} {C}} = \)\(\dfrac {{24}} {13}\) hours
#28: Hint : the nth triangular number is the sum of the first "n" natural numbers and \(\dfrac {n\left( n+1\right) } {2}\) is how you use to find the sum.
From there, you should be able to find how many numbers will be evenly divided by "7".
#29 : Circle questions are very tricky so make sure to find more problems to practice accuracy.
#30 :
Solution I:
Read the solution that is provided by Mathcounts.org here.
Solution II:
Case 1 : \(x-1 > 0\rightarrow x > 1\) Times ( x - 1) on both sides and you have
\(x^{2}-1>8\) so x > 3 or x < -3 (discard)
Case 2: \(x-1 < 0\) so \(x < 1\) \(\rightarrow x^{2}-1 < 8\) [You need to change the sign since it's negative.]-3 < x < 3. Combined with x < 1 you have the range as -3 < x < 1
The answer is 60%.
2013 Mathcounts Target #7 and 8:
Target question #8 is very similar to 2011 chapter team #10.
It just asks differently.
Read the explanations provided on the Mathcounts official website.
They are explained very well.
Let me know if there are other easier ways to tackle those problems.
Hope this is helpful !! Thanks a lot !! Good luck on Mathcounts state.
Tuesday, November 29, 2016
Geometry : Harder Chapter Level Quesitons
Question #1 : The area ratio of two equilateral triangles are 4 to 9 and the sum of their perimeter is 30 √ 3 . What is the area of the a. smaller triangle b. larger triangle?
Solution:
If the area ratio of two similar polygons is 4 to 9, their corresponding line ratio would be √ 4 to √ 9
or 2 to 3.[Make sure you know why.]
The perimeter of the two equilateral triangles is 30 √ 3 so the smaller triangle has a perimeter of
2/5 * 30√ 3 or 12 √ 3. One side is 4√ 3 . Using the formula of finding the area of an equilateral triangle \(\dfrac{\sqrt3*s^2}{4}\) , you get the area to be 12√ 3.
Use the same method to get the area of the larger triangle as 27√ 3.
You can also use ratio relationship to get the area of the larger triangle by
multiply 12√ 3 by 9/4.
2007 Mathcounts Chapter Sprint #30: In parallelogram ABCD, AB = 16 cm, DA = 32 cm, and sides AB and DA form a 45-degree interior angle. In isosceles trapezoid WXYZ with WX ≠ YZ, segment WX is the longer parallel side and has length 16 cm, and two interior angles each have a measure of 45 degrees. Trapezoid WXYZ has the same area as parallelogram ABCD. What is the length of segment YZ?
Make sure you know how to get the unknown leg fast. The height of the parallelogram is 8√2, so the area of the parallelogram is 48 square units. [Check out the special right triangle section here if you can't get the height fast.]
Let YZ of the trapezoid be x and draw the height. Using 45-45-90 degree angle ratio, you'll get the height. (See image above.)
Area of the trapezoid is average of the two bases time height. WX = 16 (given)
\(\dfrac{(16+x)* (16-x)}{4}\) = 48 ; 256 - x2 = 192 ; - x2 = - 64; x = 8 = YZ
Solution II:
Make the y be the height of the trapezoid. YZ = 16 - 2y. \(\dfrac{(16-2y + 16)}{2}\) * y = 48
\({(16 -y)* y = 48}\)\(\rightarrow\) \({16y -y^2 = 48}\) \(\rightarrow\) \({y^2 - 16y + 48 = 0}\) \(\rightarrow\) \({(y -12)(y -4)=0}\) \(\rightarrow\) \({y = 4}\) or \({y = 12}\)(doesn't work)
YZ = 16 - 2y. Plug in y = 4 and you have YZ = 8
Solution III: Let the height be y and you have \(\dfrac{(\overline{YZ}+ 16)* y}{2}"\) = 48 ; ( YZ + 16) * y = 96
When there are some numbers multiply together equal another number, it's a factoring question.
32 * 3 = 96, YZ = 16 (doesn't work)
24 * 4 = 96, YZ =8
Wednesday, July 6, 2016
Mathcounts Prep : Algebra Manipulation
Note that:
\(\left( x+y\right) ^{2}-2xy= x^{2}+y^{2}\)
\(\left( x-y\right) ^{2}+2xy= x^{2}+y^{2}\)
\(\left(x-y\right) ^{3}+3xy\left( x-y\right) =x^{3}-y^{3}\)
\(\left( x+y\right) ^{3}-3xy\left( x+y\right) =x^3 + y^{3}\)
\(\left( x+y+z\right) ^{2}-2\left( xy+yz+xz\right) =x^{2}+y^{2}+z^{2}\)
Applicable questions:
Question 1: If x + y = a and xy = b, what is the sum of the reciprocals of x and y?
Solution:
\(\dfrac {1} {x }+\dfrac {1} {y}=\dfrac {x +y} {xy}\)= \(\dfrac {a} {b}\)
Question 2: If \(x^{2}+y^{2}=153\) and x + y = 15, what is xy?
Solution:
\(\left( x+y\right) ^{2}-2xy= x^{2}+y^{2}\)
\(15^{2}-2xy=153\)\(\rightarrow xy=36\)
Question 3: If \(\left( x+y\right) ^{2}=1024\) , \(x^{2}+y^{2}\) = 530 and x > y , what is x - y?
Solution:
\(\left( x+y\right) ^{2}-2xy=x^{2}+y^{2}\)
1024 - 2xy = 530\(\rightarrow 2xy=494\)
\(\left( x-y\right) ^{2}+2xy=x^{2}+y^{2}\)
\(\left( x-y\right) ^{2}=36\)
x - y = 6
Question 4: x + y = 3 and \(x^{2}+y^{2}=89\), what is \(x^{3}+y^{3}\)?
Solution:
\(\left( x +y\right) ^{2}-2xy=x^{2}+y^{2}\)
9 - 2xy = 89 \(\rightarrow -2xy=80\) so xy = -40
\(\left( x+y\right) ^{3}-3xy\left( x+y\right) =27 - 3(-40)* 3 = 27 + 3*40*3 = x ^{3}+y^{3}\)
\(x ^{3}+y^{3}\)= 387
Question #5: If \(x+\dfrac {1} {x}=5\), what is \(x^{3}+\dfrac {1} {x ^{3}}\)?
Solution:
\(\left( x+\dfrac {1} {x}\right) ^{3}=x^{3}+3x^{2}.\dfrac {1} {x}+3x.\dfrac {1} {x^{2}}+\dfrac {1} {x^{3}}\)
\(5^{3}=x^{3}+3\left( x+\dfrac {1} {x}\right) +\dfrac {1} {x^{3}}\)
125 - 3*5 = \(x^{3}+\dfrac {1} {x ^{3}}\)
The answer is 110.
Question #6 : 2011 Mathcounts state sprint #24 : x + y + z = 7 and \(x^{2}+y^{2}+z^{2}=19\), what is the arithmetic mean of the three product xy + yz + xz?
Solution:
\(\left( x+y+z\right) ^{2}-2\left( xy+yz+xz\right) =x^{2}+y^{2}+z^{2}\)
\(7^{2}-2\left( xy+yz+xz\right) =19\)
xy + yz + xz = 15 so their mean is \(\dfrac {15} {3}=5\)
More practice problems (answer key below):
#1:If x + y = 5 and xy = 3, find the value of \(\dfrac {1} {x^{2}}+\dfrac {1} {y^{2}}\).
#2: If x + y = 3 and \(x^{2}+y^{2}=6\), what is the value of \(x^{3}+y^{3}\)?
#3: The sum of two numbers is 2. The product of the same two numbers is 5.
Find the sum of the reciprocals of these two numbers, and express it in simplest form.
#4:If \(x-\dfrac {6} {x}\) = 11, find the value of \(x^{3}-\dfrac {216} {x^{3}}\)?
#5: If \(x+\dfrac {3} {x} = 9\), find the value of \(x^{3}+\dfrac {27} {x^{3}}\)?
#6:If \(x+\dfrac {1} {x} = 8\), what is \(x^{3}+\dfrac {1} {x ^{3}}\)?
Answers:
#1 :\(\dfrac {19} {9}\)
#2: 13.5
#3: \(\dfrac {2} {5}\)
#4: 1529 [ \(11^{3}\)+ 3 x 6 x 11 =1529]
#5: 648 [\(9^{3}\)-3 x 3 x 9 = 648]
#6: 488 [ \(8^{3}\)– 3 x 8 = 488]
\(\left( x+y\right) ^{2}-2xy= x^{2}+y^{2}\)
\(\left( x-y\right) ^{2}+2xy= x^{2}+y^{2}\)
\(\left(x-y\right) ^{3}+3xy\left( x-y\right) =x^{3}-y^{3}\)
\(\left( x+y\right) ^{3}-3xy\left( x+y\right) =x^3 + y^{3}\)
\(\left( x+y+z\right) ^{2}-2\left( xy+yz+xz\right) =x^{2}+y^{2}+z^{2}\)
Applicable questions:
Question 1: If x + y = a and xy = b, what is the sum of the reciprocals of x and y?
Solution:
\(\dfrac {1} {x }+\dfrac {1} {y}=\dfrac {x +y} {xy}\)= \(\dfrac {a} {b}\)
Question 2: If \(x^{2}+y^{2}=153\) and x + y = 15, what is xy?
Solution:
\(\left( x+y\right) ^{2}-2xy= x^{2}+y^{2}\)
\(15^{2}-2xy=153\)\(\rightarrow xy=36\)
Question 3: If \(\left( x+y\right) ^{2}=1024\) , \(x^{2}+y^{2}\) = 530 and x > y , what is x - y?
Solution:
\(\left( x+y\right) ^{2}-2xy=x^{2}+y^{2}\)
1024 - 2xy = 530\(\rightarrow 2xy=494\)
\(\left( x-y\right) ^{2}+2xy=x^{2}+y^{2}\)
\(\left( x-y\right) ^{2}=36\)
x - y = 6
Question 4: x + y = 3 and \(x^{2}+y^{2}=89\), what is \(x^{3}+y^{3}\)?
Solution:
\(\left( x +y\right) ^{2}-2xy=x^{2}+y^{2}\)
9 - 2xy = 89 \(\rightarrow -2xy=80\) so xy = -40
\(\left( x+y\right) ^{3}-3xy\left( x+y\right) =27 - 3(-40)* 3 = 27 + 3*40*3 = x ^{3}+y^{3}\)
\(x ^{3}+y^{3}\)= 387
Question #5: If \(x+\dfrac {1} {x}=5\), what is \(x^{3}+\dfrac {1} {x ^{3}}\)?
Solution:
\(\left( x+\dfrac {1} {x}\right) ^{3}=x^{3}+3x^{2}.\dfrac {1} {x}+3x.\dfrac {1} {x^{2}}+\dfrac {1} {x^{3}}\)
\(5^{3}=x^{3}+3\left( x+\dfrac {1} {x}\right) +\dfrac {1} {x^{3}}\)
125 - 3*5 = \(x^{3}+\dfrac {1} {x ^{3}}\)
The answer is 110.
Question #6 : 2011 Mathcounts state sprint #24 : x + y + z = 7 and \(x^{2}+y^{2}+z^{2}=19\), what is the arithmetic mean of the three product xy + yz + xz?
Solution:
\(\left( x+y+z\right) ^{2}-2\left( xy+yz+xz\right) =x^{2}+y^{2}+z^{2}\)
\(7^{2}-2\left( xy+yz+xz\right) =19\)
xy + yz + xz = 15 so their mean is \(\dfrac {15} {3}=5\)
More practice problems (answer key below):
#1:If x + y = 5 and xy = 3, find the value of \(\dfrac {1} {x^{2}}+\dfrac {1} {y^{2}}\).
#2: If x + y = 3 and \(x^{2}+y^{2}=6\), what is the value of \(x^{3}+y^{3}\)?
#3: The sum of two numbers is 2. The product of the same two numbers is 5.
Find the sum of the reciprocals of these two numbers, and express it in simplest form.
#4:If \(x-\dfrac {6} {x}\) = 11, find the value of \(x^{3}-\dfrac {216} {x^{3}}\)?
#5: If \(x+\dfrac {3} {x} = 9\), find the value of \(x^{3}+\dfrac {27} {x^{3}}\)?
#6:If \(x+\dfrac {1} {x} = 8\), what is \(x^{3}+\dfrac {1} {x ^{3}}\)?
Answers:
#1 :\(\dfrac {19} {9}\)
#2: 13.5
#3: \(\dfrac {2} {5}\)
#4: 1529 [ \(11^{3}\)+ 3 x 6 x 11 =1529]
#5: 648 [\(9^{3}\)-3 x 3 x 9 = 648]
#6: 488 [ \(8^{3}\)– 3 x 8 = 488]
Tuesday, September 1, 2015
Testimonials for my services. So far, all through words of mouth locally or chance meets online and it's great.
Testimonials from my students/parents :)
I just did not get the opportunity earlier to thank you for all your help. _____ was able to make to the National MathCounts largely because of your excellent guidance and coaching. I do not know how to thank you. He had a great once-in-a-lifetime experience there and he really loved the competition as well as meeting other people.
If you can please provide me your mailing address, he wants to send you a gift as a token of thanks for your guidance and tutoring.
Sincerely,
Dear Mrs. Lin,
I have been meaning to tell you, but just didn't get the time. He just 'LOVES' your sessions. He said he is learning so much and gets to do lots of problems and he likes all the tips/shortcuts you are teaching. He looks forward to your session - he was so upset when we couldn't get back on time from _____ because he didn't want to miss your session. Honestly we came back Tue night only because he cried so much that he didn't want to miss your class:)
In school, his teacher focuses more on details like, put all the steps, write neatly, don't disturb the class by asking unnecessary questions, don't ask for more work, behave properly etc....so he is not too happy with math in school.
Thank you so much for making such a big difference in his life. He enjoys doing the homework you are assigning him and has not complained at all. He said "Mrs Lin is so smart and I love her classes...wish she lived next to our house...so I can go to her and have live classes' [disclaimer : I'm not ; my students are much smarter than I and I learn along with them and it's exhilarating ] Thank you so much!!
One thing he said was it would be helpful if he can have targeted practice worksheets on the tips covered during the class, after that class, so he can practice those shortcuts/tips more.
Hello Mrs. Lin,
This is ______ 's mom. We greatly appreciate your help in working with
______, keeping him motivated and providing him wth constant
encouragement. We are fortunate to have great mentors like you who
This is ______ 's mom. We greatly appreciate your help in working with
______, keeping him motivated and providing him wth constant
encouragement. We are fortunate to have great mentors like you who
are so selfless in their services to our younger generation.
We have a request, can you please share your address. ______ wanted
to send you a card to express his thanks.
BTW, _______ has his chapter level competition tomorrow.
Warm Regards
We have a request, can you please share your address. ______ wanted
to send you a card to express his thanks.
BTW, _______ has his chapter level competition tomorrow.
Warm Regards
Mrs. Lin,
Good Afternoon.
Good Afternoon.
Me and my son, ___ and I, have used your blog pages and got a great insight into several of the techniques that you use to solve the problems. I am glad to inform you that _____ was placed 14th in the State Mathcounts for ____. Your blog information has helped us a lot in this preparation and we really want to appreciate all that you do in sharing the information.
_____ is completing Geometry this year and he hopes to have a exciting next year for AMC 8 and Mathcounts. We hope to learn a lot more from your blog pages.
Regards,
Regards,
I just finished The One World School House by Salman Khan and I have been thinking of you and the online community you created. Even though _____ and ____ have taken web based courses before, what makes your class different is your inspiration and enthusiasm - you really care about them as unique individuals and you sincerely expect the best from them, more than they (and I sometimes) think they can achieve. Thank you for encouraging them to become a more responsible and self driven learner.
Dear Mrs.Lin,
I went to NSF tests today and got first place in the Math Bee III. The problems didn't seem that bad although I'm waiting to see my score online. I will most likely be going to nationals which will be held in Ohio this year so that's convenient. Lastly I would like to thank you for all you have done for me. I believe I've grown much more as a student under you and really appreciate everything you have shown me and taught me. I wouldn't be succeeding now if it wasn't for you. :)
Sincerely,
He could not participate in one regional bec of conflict with MathCounts chapter. We also want to share with you that he got admission into a private school for 9th gr with 100% scholarship. [more than 50 students in that high school are national merit semifinalists, so highly competitive]
Thank you,
An elementary whiz kid, national winners at Math Kangaroo and Math Olympiad.
_____ will be off to summer sleep away camp where he is not allowed to have any electronics starting _____.
_________ will be his last class with you for over a month. He really has been enjoying it and just so you know I never have to ask him twice to do the work- crazy because everything else I ask him to do takes at least 5 tries :)
He will be back online with you at _______ .
To be continued ...
I'm quite busy these days with resuming our Math Circle + many other projects (my students don't just excel at math, but many other areas equally fun and challenging + most Asian students have much bigger problems with critical reading/not to mention writing, taking initiatives and being strong leaders, and those are my other projects.
Work-Life balance is utmost important. Less is more.
Tuesday, June 16, 2015
Problem Solving Strategies : Complementary Counting
Check out Mathcounts here, the best competition math program for middle school students.
Download this year's Mathcounts handbook here.
Video to watch on complementary counting from "Art of Problem Solving"
Part 1
Part 2
Question: How many two-digit numbers contain at least one 9?
At beginning level, kids start to write down all the numbers that contain 9. However, this turns into impossible task if it's a three-digit or four-digit number. So let's try other ways to do it.
9 _ , if 9 is placed as the tens digit, unit digit can be chosen from 0 -9, altogether 10 numbers.
A lot of kids think there are only 9. Be careful. To find how many consecutive terms from number a to b, you do (b-a) +1
_ 9, there will be 9 choices (1-9) this time as the tens digit.(Why?) Zero can't be placed other than unit digit.
So total you have 10 + 9 - 1 (you've counted 99 twice) = 18
This way is better than the first one, but once the numbers become large, you will easily lose track of those double-counting, triple-counting numbers and over count your answers.
Here is a better way to tackle this type of problem: Think about the case of numbers that contain no 9s, and subtracting this value from the total number of two-digit numbers will give you the answer.
There are 99-10 +1 =90 total two digit numbers. There are 8 (the tens digit) x 9 (unit digit) = 72 numbers that contain no 9s. So 90-72 = 18 gives you the answer.
Try this question:
How many three-digit numbers contain at least one 9?
There are 999 - 100 + 1 or 999 - 99 = 900 three-digit numbers.
Or 9 x 10 (you can use the digit "0" now" x 10 = 900 three-digit numbers
900 - 8 x 9 x 9 = 252 numbers
This is called "complementary counting" and there are numerous problems that you can use this strategies to simplify the reasoning.
Here is another harder problem from 2003 Mathcounts Chapter Sprint Round #29:
Each day, two out of the three teams in a class are randomly selected to participate in a MATHCOUNTS trial competition. What is the probability that Team A is selected on at least two of the next three days? Express your answer as a common fraction.
Solution:
Use complementary counting.
If each day two of the team will be chosen, there will be 3C2 = 3 ways to choose the team -- AB, BC, or AC, so 1/3 of the chance that team A won't be chosen and 2/3 of the chance that team A will be chosen.
Case 1: Team A is not chosen on any of the three days. The probability is (1/3) 3= 1/27.
Case 2: Team A is chosen on one of the three days : The probability is (2/3) times (1/3)2 times 3C1 = 6/27 (A - -, - A - or - - A, which is 3C1 = 3 ways)
Total possibilities - none - at least 1 time = at least two times Team A will be chosen
so the answer is 1 - 1/27 - 6/27 = 20/27
Other applicable problems: (answer key below)
#1: 2006 AMC10 A: How many four-digit positive integers have at least one digit that is a 2 or 3?
#2: What is the probability that when tossing two dice, at least one dice will come up a "3"?
#3: If {x,y} is a subset of S={1,2,3,....50}. What is the probability that xy is even?
Answer key:
#1: 9000 - 7 x 8 x 8 x 8 = 5416
#2: The probability of the dice not coming up with a "3" is 5/6.
1 - (5/6)2 = 11/36
2. 1 - (25/50) (24/49) -- only odd times odd will give you odd product, the others will all render even product, so the answer is 37/49.
Download this year's Mathcounts handbook here.
Video to watch on complementary counting from "Art of Problem Solving"
Part 1
Part 2
Question: How many two-digit numbers contain at least one 9?
At beginning level, kids start to write down all the numbers that contain 9. However, this turns into impossible task if it's a three-digit or four-digit number. So let's try other ways to do it.
9 _ , if 9 is placed as the tens digit, unit digit can be chosen from 0 -9, altogether 10 numbers.
A lot of kids think there are only 9. Be careful. To find how many consecutive terms from number a to b, you do (b-a) +1
_ 9, there will be 9 choices (1-9) this time as the tens digit.(Why?) Zero can't be placed other than unit digit.
So total you have 10 + 9 - 1 (you've counted 99 twice) = 18
This way is better than the first one, but once the numbers become large, you will easily lose track of those double-counting, triple-counting numbers and over count your answers.
Here is a better way to tackle this type of problem: Think about the case of numbers that contain no 9s, and subtracting this value from the total number of two-digit numbers will give you the answer.
There are 99-10 +1 =90 total two digit numbers. There are 8 (the tens digit) x 9 (unit digit) = 72 numbers that contain no 9s. So 90-72 = 18 gives you the answer.
Try this question:
How many three-digit numbers contain at least one 9?
There are 999 - 100 + 1 or 999 - 99 = 900 three-digit numbers.
Or 9 x 10 (you can use the digit "0" now" x 10 = 900 three-digit numbers
900 - 8 x 9 x 9 = 252 numbers
This is called "complementary counting" and there are numerous problems that you can use this strategies to simplify the reasoning.
Here is another harder problem from 2003 Mathcounts Chapter Sprint Round #29:
Each day, two out of the three teams in a class are randomly selected to participate in a MATHCOUNTS trial competition. What is the probability that Team A is selected on at least two of the next three days? Express your answer as a common fraction.
Solution:
Use complementary counting.
If each day two of the team will be chosen, there will be 3C2 = 3 ways to choose the team -- AB, BC, or AC, so 1/3 of the chance that team A won't be chosen and 2/3 of the chance that team A will be chosen.
Case 1: Team A is not chosen on any of the three days. The probability is (1/3) 3= 1/27.
Case 2: Team A is chosen on one of the three days : The probability is (2/3) times (1/3)2 times 3C1 = 6/27 (A - -, - A - or - - A, which is 3C1 = 3 ways)
Total possibilities - none - at least 1 time = at least two times Team A will be chosen
so the answer is 1 - 1/27 - 6/27 = 20/27
Other applicable problems: (answer key below)
#1: 2006 AMC10 A: How many four-digit positive integers have at least one digit that is a 2 or 3?
#2: What is the probability that when tossing two dice, at least one dice will come up a "3"?
#3: If {x,y} is a subset of S={1,2,3,....50}. What is the probability that xy is even?
Answer key:
#1: 9000 - 7 x 8 x 8 x 8 = 5416
#2: The probability of the dice not coming up with a "3" is 5/6.
1 - (5/6)2 = 11/36
2. 1 - (25/50) (24/49) -- only odd times odd will give you odd product, the others will all render even product, so the answer is 37/49.
Tuesday, May 6, 2014
This Week's Work : Week 49 - for Inquisitive Young Mathletes
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 !!
Monday, November 4, 2013
Find the area of the petal, or the football shape.
Find the area of the football shape, or the petal shape.
The below Mathcounts mini presents two methods.
Circle and area revisited from Mathcounts mini
The first question is exactly the same as this one.
Besides the two methods on the videos, you can also use the following methods.
Solution III:
You can also look at this as a Venn Diagram question.
One quarter circle is A and the other is B, and both are congruent. (center at opposite corner vertexes)
The overlapping part is C.
A + B - C = 6^2 so C = A + B - 36 or 18pi - 36
Solution IV:
If you use the area of the rectangle,
which is 6 x 12 minus, the area of the half circle with a radius 6, you get the two white spots that are un-shaded.
Use the area of the square minus that will again give you the answer.
\(6^{2}-\left( 6*12-\dfrac {6^{2}\pi } {2}\right)\)
= 18pi - 36
Similar triangles and triangles that share the same vertexes/or/and trapezoid
Another link from my blog
Similar triangles, dimensional change questions are all over the place so make sure you really
understand them.
Take care and happy problem solving !!
The below Mathcounts mini presents two methods.
Circle and area revisited from Mathcounts mini
The first question is exactly the same as this one.
Besides the two methods on the videos, you can also use the following methods.
Solution III:
You can also look at this as a Venn Diagram question.
One quarter circle is A and the other is B, and both are congruent. (center at opposite corner vertexes)
The overlapping part is C.
A + B - C = 6^2 so C = A + B - 36 or 18pi - 36
Solution IV: which is 6 x 12 minus, the area of the half circle with a radius 6, you get the two white spots that are un-shaded.
Use the area of the square minus that will again give you the answer.
\(6^{2}-\left( 6*12-\dfrac {6^{2}\pi } {2}\right)\)
= 18pi - 36
Similar triangles and triangles that share the same vertexes/or/and trapezoid
Another link from my blog
Similar triangles, dimensional change questions are all over the place so make sure you really
understand them.
Take care and happy problem solving !!
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
Wednesday, May 29, 2013
This Week's Work : Week 14 - for Inquisitive Young Mathletes
Venn Diagram explained from AoPS : learn the two set and three set examples
Principle of Inclusion-Exclusion from AoPS: again, two set and three set examples
Video on Venn Diagrams with Two categories from AoPS
Video on Venn Diagrams with Three categories from AoPS
Word problems on Venn diagrams medium level (instant feedback)
Word problems on Venn diagrams medium hard level (instant feedback)
The common error for most students (especially boys, oh oh) is arithmetic error.
Make sure you line the category with specific number given right and check your math.
Chicken, rabbit algebra speed test : aim for less than 5 minutes. No calculator, please !!
Motivation, Not IQ, Matters Most For Learning New Math Skills from Time Magazine
Power of Ten Video
Principle of Inclusion-Exclusion from AoPS: again, two set and three set examples
Video on Venn Diagrams with Two categories from AoPS
Video on Venn Diagrams with Three categories from AoPS
Word problems on Venn diagrams medium level (instant feedback)
Word problems on Venn diagrams medium hard level (instant feedback)
The common error for most students (especially boys, oh oh) is arithmetic error.
Make sure you line the category with specific number given right and check your math.
Chicken, rabbit algebra speed test : aim for less than 5 minutes. No calculator, please !!
Motivation, Not IQ, Matters Most For Learning New Math Skills from Time Magazine
Power of Ten Video
Tuesday, April 23, 2013
This Week's Work : Week 9 -- for Inquisitive Young Mathletes
Part 1:
See below for links:
They are all related to dimensional change and similar polygons.
Dimensional change questions I
Dimensional change questions II
Dimensional change questions III : Similar Triangles
Par II:
Tangent Segments and Similar Triangles from Mathcounts Mini
If you have more time, download the extra word problems to see if you can solve them at
reasonable speed and accuracy.
Online timed test and problem of the week will be sent out through e-mail.
Time: 40 minutes without a calculator.
The Monty Hall Problem explained
See below for links:
They are all related to dimensional change and similar polygons.
Dimensional change questions I
Dimensional change questions II
Dimensional change questions III : Similar Triangles
Par II:
Tangent Segments and Similar Triangles from Mathcounts Mini
If you have more time, download the extra word problems to see if you can solve them at
reasonable speed and accuracy.
Online timed test and problem of the week will be sent out through e-mail.
Time: 40 minutes without a calculator.
The Monty Hall Problem explained
Tuesday, April 16, 2013
This Week's Work : Week 6 and 7 Review -- for Inquisitive Young Mathletes
Watch Joint Proportion from Art of Problem Solving
Spend some time pondering on "Work" word problems from Purple Math.
These are some very standard word problems you'll encounter in competition math.
Review --
dimensional change and probability links from previous weeks.
More practices on inverse and direct relation
From Regents Exam Prep
Link I
Link II
New concepts:
Height to the hypotenuse
How many ways to arrange the word "banana"? (with elements repeating)
Probability that two of the 3 friends were born on the same week day.
Question : If you can earn 0, 1, 3, 7 or 10 points with each shot and each person has three chances, how many scores can't be made?
Note:
Percentage increase (don't forget to minus the original 100% or 1) is very different from at what percent will it return to the original size or what is the size compared to the original.
Spend some time pondering on "Work" word problems from Purple Math.
These are some very standard word problems you'll encounter in competition math.
Review --
dimensional change and probability links from previous weeks.
More practices on inverse and direct relation
From Regents Exam Prep
Link I
Link II
New concepts:
Height to the hypotenuse
How many ways to arrange the word "banana"? (with elements repeating)
Probability that two of the 3 friends were born on the same week day.
Question : If you can earn 0, 1, 3, 7 or 10 points with each shot and each person has three chances, how many scores can't be made?
Note:
Percentage increase (don't forget to minus the original 100% or 1) is very different from at what percent will it return to the original size or what is the size compared to the original.
Subscribe to:
Posts (Atom)
