Let me know if you have any other questions. Best and take good care, Mrs. Lin
Lesson Equations of Circles
Practice : Practice with Equations of Circles
worksheet :
Similar triangles :
worksheet :
Wednesday, October 26, 2016
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.
Tuesday, September 27, 2016
2016 AMC-8 Prep
Interesting articles on math for this week:
How to Fall in Love With Math by Manil Suri from the New York Times
The Simpsons' secret formula : it's written by math geeks by Simon Singh from the Guardian
For this week's self studies (part I work):
Review:
2009 #25 : Review using the Harvey method. :D
2008 #25 Don't use the method on the link. Use the much faster method we talked about at our lesson.
2008 #24 Make a chart. Slow down on similar question such as this one.
This type of problem is very easy to make mistake on under or overcounting.
Skip first and definitely slow down and double, triple check.
2007 #25 Read the solution if you don't get the method we talked about at our lesson.
It takes time to develop insights so you need to be patient.
If you understand the method, this question will be easy, right ?
Stay with this question longer.
2007 #24
Aayush's method is faster.
(To get the sum of three digits that is a multiple of 3, you either get rid of 1 or 4 [do you see it jumps by 3, why?] ) , so the answer is 1/2.
2006 #25
I've seen other problems (AMC-10s) using the oddest prime, which is "2", the only prime number that is even.
Thus, make sure you understand this question.
2006#24 Taking out the factor question.
Also, learn "1001 = 7 x 11 x 13"
"23 x 29 = 667"
2005 #25 Venn diagram is your friend.
2005 #24 Working backwards is the way to go.
For part II work :
This week, work on the last 5 problems from AMC-8 year 2010, 2011, 2012, 1999 and 1998.
Here is the link from AoPs.
Sam' original question:
David has a bag of 8 different-colored six-sided dice. Their colors are red, blue, yellow, green, purple, orange, black, and white. What is the probability that David takes out a red die, rolls a 6, then takes out a purple die, and rolls another 6 without replacement?
Solution:
The probability of rolling a 6 on a red die is 1/8 * 1/6 = 1/48. The probability of rolling a purple die and rolling a 6 after that, without replacement, is 1/7 * 1/6 = 1/42. Therefore, to get both events, 1/48 * 1/42 = 1/2016.
Evan's compiled question:
\(\sqrt {18+8\sqrt {2}}=a+b\sqrt {c}\)
a, b and c are positive integers. Find a + b + c.
Solution:
Square both sides and you have \(18+8\sqrt{2}\) = \( a^2 + 2ab\sqrt {2} + b^2c\)
You can see ab = 4 = 4 x 1 and c = 2
a = 4, b = 1 and c = 2 so the sum is 7.
Sounak's problem:
A rhombus with sides 4 is drawn. It has an angle of 60 degrees. What is length of the longer diagonal?
Solution:
Well first you have to draw the rhombus's height .The resulting triangle will be 30,60, 90 triangle.
We know the hypotenuse is 4 so now we know the rest of the sides are \(2\sqrt {3}\) and 2.
Now if we draw the diagonal we see that it makes another right angle triangle.
We know the legs of this triangle are the same as the previous lengths so then we know the diagonal is \(4\sqrt {3}\).
How to Fall in Love With Math by Manil Suri from the New York Times
The Simpsons' secret formula : it's written by math geeks by Simon Singh from the Guardian
For this week's self studies (part I work):
Review:
2009 #25 : Review using the Harvey method. :D
2008 #25 Don't use the method on the link. Use the much faster method we talked about at our lesson.
2008 #24 Make a chart. Slow down on similar question such as this one.
This type of problem is very easy to make mistake on under or overcounting.
Skip first and definitely slow down and double, triple check.
2007 #25 Read the solution if you don't get the method we talked about at our lesson.
It takes time to develop insights so you need to be patient.
If you understand the method, this question will be easy, right ?
Stay with this question longer.
2007 #24
Aayush's method is faster.
(To get the sum of three digits that is a multiple of 3, you either get rid of 1 or 4 [do you see it jumps by 3, why?] ) , so the answer is 1/2.
2006 #25
I've seen other problems (AMC-10s) using the oddest prime, which is "2", the only prime number that is even.
Thus, make sure you understand this question.
2006#24 Taking out the factor question.
Also, learn "1001 = 7 x 11 x 13"
"23 x 29 = 667"
2005 #25 Venn diagram is your friend.
2005 #24 Working backwards is the way to go.
For part II work :
This week, work on the last 5 problems from AMC-8 year 2010, 2011, 2012, 1999 and 1998.
Here is the link from AoPs.
Sam' original question:
David has a bag of 8 different-colored six-sided dice. Their colors are red, blue, yellow, green, purple, orange, black, and white. What is the probability that David takes out a red die, rolls a 6, then takes out a purple die, and rolls another 6 without replacement?
Solution:
The probability of rolling a 6 on a red die is 1/8 * 1/6 = 1/48. The probability of rolling a purple die and rolling a 6 after that, without replacement, is 1/7 * 1/6 = 1/42. Therefore, to get both events, 1/48 * 1/42 = 1/2016.
Evan's compiled question:
\(\sqrt {18+8\sqrt {2}}=a+b\sqrt {c}\)
a, b and c are positive integers. Find a + b + c.
Solution:
Square both sides and you have \(18+8\sqrt{2}\) = \( a^2 + 2ab\sqrt {2} + b^2c\)
You can see ab = 4 = 4 x 1 and c = 2
a = 4, b = 1 and c = 2 so the sum is 7.
Sounak's problem:
A rhombus with sides 4 is drawn. It has an angle of 60 degrees. What is length of the longer diagonal?
Solution:
Well first you have to draw the rhombus's height .The resulting triangle will be 30,60, 90 triangle.
We know the hypotenuse is 4 so now we know the rest of the sides are \(2\sqrt {3}\) and 2.
Now if we draw the diagonal we see that it makes another right angle triangle.
We know the legs of this triangle are the same as the previous lengths so then we know the diagonal is \(4\sqrt {3}\).
Friday, September 23, 2016
First week's review reminders
For 16-17 Mathcounts handbook :
Warm-up 1 : review #3, 7, 9.
Think of ways to get these questions in seconds.
Warm-up 2 : review 17,18,19,20 -- slow down on #18, similar types are very easy to
get wrong the first try
direct and inverse relationship
From AoPS videos :
Instructions for reviewing
2014 School Rounds link here
Sprint :
#18 (trial and error are faster), #20 (loose pennies out first)
#22, #26, #27, #28, #29, #30 -- Read the solutions and make sure to fully understand
what major concept(s) is (are) tested -- these are not hard problems.
Target :
#1 : Calendar questions could be tricky, so slow down a bit.
Check what stage it starts first.
#2 : WOW, this one is a killer, so tedious -- hard to get it right the first time.
# 3 and 4 are in seconds questions.
# 5: One line, two numbers and you are done.
With a calculator, you don't even need to write anything down, right? :)
#6 : more interesting question
#7 and 8 : standard questions, not hard, just need to be careful and thorough.
Team :
# 2, 3, 5, 6 (very tedious, I suggest skipping first)
8, 9 (more tedious than 8 and 10, so don't work on problems in order), and
#10 (very easy if you've noticed what is actually tested) It's in seconds question.
Agai, scan those questions and only try those that are your weak spots or marked red.
Please check the solution files I'd sent you to learn the better methods.
Whenever you have extra time, use the other links (individually based) to keep learning.
Keep me posted. Have fun at problem solving.
Don't just do math. :)
Warm-up 1 : review #3, 7, 9.
Think of ways to get these questions in seconds.
Warm-up 2 : review 17,18,19,20 -- slow down on #18, similar types are very easy to
get wrong the first try
direct and inverse relationship
From AoPS videos :
2014 School Rounds link here
Sprint :
#18 (trial and error are faster), #20 (loose pennies out first)
#22, #26, #27, #28, #29, #30 -- Read the solutions and make sure to fully understand
what major concept(s) is (are) tested -- these are not hard problems.
Target :
#1 : Calendar questions could be tricky, so slow down a bit.
Check what stage it starts first.
#2 : WOW, this one is a killer, so tedious -- hard to get it right the first time.
# 3 and 4 are in seconds questions.
# 5: One line, two numbers and you are done.
With a calculator, you don't even need to write anything down, right? :)
#6 : more interesting question
#7 and 8 : standard questions, not hard, just need to be careful and thorough.
Team :
# 2, 3, 5, 6 (very tedious, I suggest skipping first)
8, 9 (more tedious than 8 and 10, so don't work on problems in order), and
#10 (very easy if you've noticed what is actually tested) It's in seconds question.
Agai, scan those questions and only try those that are your weak spots or marked red.
Please check the solution files I'd sent you to learn the better methods.
Whenever you have extra time, use the other links (individually based) to keep learning.
Keep me posted. Have fun at problem solving.
Don't just do math. :)
Math related video : Making Stuff Faster,
which includes "The Travelling Salesman Problem" and a competition between an astrophysicist and a paleontologist on how to move passengers boarding the planes faster 
Good luck, from Mrs. Lin
Wednesday, September 21, 2016
2016 AMC-8 prep
Want to join our group for the up-coming AMC 8 test in Nov. ?
The problems are more complex, including many steps, occasionally not going in difficulty order, or/and there are troll/ trap questions, so it's GREAT to deter students to just memorize the formulas,
I do see some brilliant/ inquisitive students who are not good test takers, if you belong to that group, there are other ways to shine.
E-mail me on that. It's really much, much better to just sit back and enjoy the problems.
Check this awesome note from AoPS forum.
You know, the most amazing thing about various competitions is the energy, the pleasure, the spontaneity, the camaraderie and the kindred spirits.
Thanks a lot to those diligent, inquisitive boys and girls for their impromptu, collaborated efforts.
You are one of its kind. :)
2015 unofficial AMC 8 problems and detailed solutions from online whiz kids.
This year's AMC 8 official statistics is rolling in.
My online whiz kids NEVER stop learning because ___ is who they are and it doesn't have to be all math and science related.
E-mail me at thelinscorner@gmail.com if you want to join us who LOVE problem solving (and many other areas equally challenging and engaging.)
Unofficial , official problems, answer key and detailed solutions to 2014 AMC8 test + official statistics.
This year's (2013) AMC-8 results can be viewed here.
2013 AMC-8 problems in pdf format (easier to print out and work on it as a real test)
Try this if your school doesn't offer AMC-8 test.
40 minutes without a calculator.
If you want to use the test to prepare for Mathcounts, cover the multiple choice
options to make the questions harder unless you have to see the choices to answer
the question(s) asked.
2013 AMC-8 problems
2013 AMC-8 answer key
2013 AMC-8 problems with detailed (multiple) solutions
Trickier problems : #18, 21(mainly the wording) and maybe 25 (slightly tricky)
2014 AMC-8 Result Statistics can be viewed here.
2014 AMC-8 problems and solutions from AoPS wiki.
Comments :
#4 : We've been practicing similar problems to #4 so it should be a breeze if you see right away that the prime number "2" is involved. You'll get a virtual bump if you forgot about that again.
#10 : Almost every test has this type of problem, inclusive, exclusive, between, calendar, space, terms, stages... It's very easy to twist the questions in the hope of confusing students, so slow down on this type of question or for the trickier ones, skip it first. You can always go back to it if you have time left after you get the much easier-to-score points. (such as #12, 13, 18 -- if you were not trolled and others)
#11 : Similar questions appear at AMC-10, Mathcounts. For harder cases, complementary counting is easier.
This one, block walk is easier.
#12: 1/ 3!
How about if there are 4 celebrities ? What is the probability that all the baby photos match with the celebrities ? only 1 baby photo matches, only2 baby photos match, 3 baby photos match, or none matches ?
#13: number theory
For sum of odd and /or even, it's equally likely --
odd + odd = even ; even + even = even
odd + even = odd ; even + odd = odd
For product of odd and/or even, it's not equally likely --
odd * even = even ; even * even = even
odd * odd = odd
For product probability questions, complementary counting with total minus the probability of getting odd product (all odds multiply)is much faster.
SAT/ACT has similar type of problems.
#15 : Central angle and inscribed angles --> Don't forget radius is of the same length.
Learn the basics from Regents prep
#17: rate, time and distance could be tricky
Make sure to have the same units (hour, minutes or seconds) and it's a better idea writing down
R*T = D so you align the given infor. better.
Also, sometimes you can use direct/inverse relationship to solve seemingly harder problems in seconds.
Check out the notes from my blog and see for yourself.
#18 : Trolled question. Oh dear !!
1 4 6 4 1 , but it doesn't specify gender number(s) so
(4 + 4)/2^4 is the most likely.
#19 : more interesting painted cube problems --> one cube is completely hidden inside.
Painted cube animation from Fairy Math Tutors
Painted cube review Use Lego or other plops to help you visualize how it's done.
#20 : Use 3.14 for pi and if you understand what shape is asked, it's not too bad.
#21: You can cross out right away multiples of three or sum of multiples of 3 by first glance.
For example 1345AA, you can cross out right away 3 and 45 (because 4 + 5 = 9, a multiple of 3).
You don't need to keep adding those numbers up. It's easier this way.
#22 : To set up two-digit numbers, you do 10x + y.
To set up three-digit numbers, you do 100x + 10y + z
For those switching digits questions, sometimes faster way is to use random two or three digit numbers, not in this case, though.
#23 : This one is more like a comprehension question. Since it relates to birthday of the month, there are not many two digit primes you need to weight, so 11, 13 and 17. (19 + 17 would exceed any maximum days of the month). From there, read carefully and you should get the answer.
#24 : a more original question --
To maximize the median, which in this case is the average of the 50th and 51st term, you minimized the first 49 terms, so make them all 1s.
Don't forget the 51st term has to be equal or larger than the 50th term.
#25: The figure shown is just a partial highway image. 40 feet is the diameter and the driver's speed is 5 miles per hour, so units are not the same --> trap
I've found most students, when it comes to circular problems, tend to make careless mistakes because there are just too many variables.
Areas, circles, semi-circles, arch, wedge areas, and those Harvey like "think outside the box" fun problems.
Thus, it's a good idea to slow down for those circular questions. Easier said than remembered.
Happy Holiday !!
The problems are more complex, including many steps, occasionally not going in difficulty order, or/and there are troll/ trap questions, so it's GREAT to deter students to just memorize the formulas,
I do see some brilliant/ inquisitive students who are not good test takers, if you belong to that group, there are other ways to shine.
E-mail me on that. It's really much, much better to just sit back and enjoy the problems.
Check this awesome note from AoPS forum.
You know, the most amazing thing about various competitions is the energy, the pleasure, the spontaneity, the camaraderie and the kindred spirits.
Thanks a lot to those diligent, inquisitive boys and girls for their impromptu, collaborated efforts.
You are one of its kind. :)
2015 unofficial AMC 8 problems and detailed solutions from online whiz kids.
This year's AMC 8 official statistics is rolling in.
Yeah, my boys' and girls' names are there. :)
Move on to the most fun Mathcounts competition and of course, AMC 10 and AIME tests.
My online whiz kids NEVER stop learning because ___ is who they are and it doesn't have to be all math and science related.
E-mail me at thelinscorner@gmail.com if you want to join us who LOVE problem solving (and many other areas equally challenging and engaging.)
Unofficial , official problems, answer key and detailed solutions to 2014 AMC8 test + official statistics.
In Chinese : 《專題報導》亞裔尖子生 職場難出頭？
This year's (2013) AMC-8 results can be viewed here.
2013 AMC-8 problems in pdf format (easier to print out and work on it as a real test)
Try this if your school doesn't offer AMC-8 test.
40 minutes without a calculator.
If you want to use the test to prepare for Mathcounts, cover the multiple choice
options to make the questions harder unless you have to see the choices to answer
the question(s) asked.
2013 AMC-8 problems
2013 AMC-8 answer key
2013 AMC-8 problems with detailed (multiple) solutions
Trickier problems : #18, 21(mainly the wording) and maybe 25 (slightly tricky)
2014 AMC-8 Result Statistics can be viewed here.
2014 AMC-8 problems and solutions from AoPS wiki.
Comments :
#4 : We've been practicing similar problems to #4 so it should be a breeze if you see right away that the prime number "2" is involved. You'll get a virtual bump if you forgot about that again.
#10 : Almost every test has this type of problem, inclusive, exclusive, between, calendar, space, terms, stages... It's very easy to twist the questions in the hope of confusing students, so slow down on this type of question or for the trickier ones, skip it first. You can always go back to it if you have time left after you get the much easier-to-score points. (such as #12, 13, 18 -- if you were not trolled and others)
#11 : Similar questions appear at AMC-10, Mathcounts. For harder cases, complementary counting is easier.
This one, block walk is easier.
How about if there are 4 celebrities ? What is the probability that all the baby photos match with the celebrities ? only 1 baby photo matches, only2 baby photos match, 3 baby photos match, or none matches ?
#13: number theory
For sum of odd and /or even, it's equally likely --
odd + odd = even ; even + even = even
odd + even = odd ; even + odd = odd
For product of odd and/or even, it's not equally likely --
odd * even = even ; even * even = even
odd * odd = odd
For product probability questions, complementary counting with total minus the probability of getting odd product (all odds multiply)is much faster.
SAT/ACT has similar type of problems.
#15 : Central angle and inscribed angles --> Don't forget radius is of the same length.
Learn the basics from Regents prep
#17: rate, time and distance could be tricky
Make sure to have the same units (hour, minutes or seconds) and it's a better idea writing down
R*T = D so you align the given infor. better.
Also, sometimes you can use direct/inverse relationship to solve seemingly harder problems in seconds.
Check out the notes from my blog and see for yourself.
#18 : Trolled question. Oh dear !!
1 4 6 4 1 , but it doesn't specify gender number(s) so
(4 + 4)/2^4 is the most likely.
#19 : more interesting painted cube problems --> one cube is completely hidden inside.
Painted cube animation from Fairy Math Tutors
Painted cube review Use Lego or other plops to help you visualize how it's done.
#20 : Use 3.14 for pi and if you understand what shape is asked, it's not too bad.
#21: You can cross out right away multiples of three or sum of multiples of 3 by first glance.
For example 1345AA, you can cross out right away 3 and 45 (because 4 + 5 = 9, a multiple of 3).
You don't need to keep adding those numbers up. It's easier this way.
#22 : To set up two-digit numbers, you do 10x + y.
To set up three-digit numbers, you do 100x + 10y + z
For those switching digits questions, sometimes faster way is to use random two or three digit numbers, not in this case, though.
#23 : This one is more like a comprehension question. Since it relates to birthday of the month, there are not many two digit primes you need to weight, so 11, 13 and 17. (19 + 17 would exceed any maximum days of the month). From there, read carefully and you should get the answer.
#24 : a more original question --
To maximize the median, which in this case is the average of the 50th and 51st term, you minimized the first 49 terms, so make them all 1s.
Don't forget the 51st term has to be equal or larger than the 50th term.
#25: The figure shown is just a partial highway image. 40 feet is the diameter and the driver's speed is 5 miles per hour, so units are not the same --> trap
I've found most students, when it comes to circular problems, tend to make careless mistakes because there are just too many variables.
Areas, circles, semi-circles, arch, wedge areas, and those Harvey like "think outside the box" fun problems.
Thus, it's a good idea to slow down for those circular questions. Easier said than remembered.
Happy Holiday !!
Monday, August 29, 2016
Mathcounts Strategy: Shoestring (or Shoelace) method of finding the area of any polygon
Check out Mathcounts here, the best competition math program for middle school students.
Download this year's Mathcounts handbook here.
Shoelace formula from Wikipedia
More on Shoelace
Problems: Solutions below
#2 2007 Chapter Target Round: A quadrilateral in the plane has vertices at (1,3), (1,1), (2, 1) and (2006, 2007). What is the area of the quadrilateral?
#3: Find the area of a polygon with coordinates (1, 1), (3, -1), ( 4, 4), and (0.3)
#4: What is the number of square units in the area of the pentagon whose vertices are
Solution I: Draw a rectangle and use the area of the rectangle minus the four triangles to get the area of the quadrilateral polygon.
Use this link to practice finding the area of any irregular polygon. Keep in mind that a lot of the times you don't need to use shoestring method. Be flexible!! Scroll to the middle section.
#2 Answer: 3008 square units
#3: Answer: 10.5 square units
#4: Answer: 22 square units
#5: Answer: 45.5 square units
#6: Answer: 136 square units
#7: Answer: 98 square units
#8: Answer: 66 square units
Download this year's Mathcounts handbook here.
Shoelace formula from Wikipedia
More on Shoelace
Problems: Solutions below
#1: Find the area of a quadrilateral polygon given the four end points (3, 5), (11, 4), (7,0) and (9,8) in a Cartesian plane.
#2 2007 Chapter Target Round: A quadrilateral in the plane has vertices at (1,3), (1,1), (2, 1) and (2006, 2007). What is the area of the quadrilateral?
#3: Find the area of a polygon with coordinates (1, 1), (3, -1), ( 4, 4), and (0.3)
#4: What is the number of square units in the area of the pentagon whose vertices are
(1, 1 ), ( 3, -1), (6, 2), (5, 6), and (2, 5)?
#5: Find the area of a polygon with coordinates ( -6, 0), (0, 5), (3, -2), and (4, 7)
#6: Find the area of a polygon with coordinates (20, 0), (0, 12), (3, 0), (4, -4)
#7: Find the area of a polygon with coordinates (-8, 4), (2, 12), (3, -5), (4, -4)
#8: Find the area of a triangle with coordinate (-8, -4), (-3, 10), (5, 6)
Solution I: Draw a rectangle and use the area of the rectangle minus the four triangles to get the area of the quadrilateral polygon.
Solution
II: Using shoestring method. First, plug in the four points. Second,
choose one starting point and list the other points in order (either clockwise or counterclockwise) and at the
end, repeat the starting point. The answer is 33 square units.
Use this link to practice finding the area of any irregular polygon. Keep in mind that a lot of the times you don't need to use shoestring method. Be flexible!! Scroll to the middle section.
#2 Answer: 3008 square units
#3: Answer: 10.5 square units
#4: Answer: 22 square units
#5: Answer: 45.5 square units
#7: Answer: 98 square units
#8: Answer: 66 square units
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]
Sunday, March 6, 2016
2016 Mathcounts State Prep : How to Avoid Making Careless Mistakes
Why Can Some Kids Handle Pressure While Others Fall Apart? from the New York Times
Relax! You'll Be More Productive from the New York Times
First, please read Mathcounts "Forms of Answers" carefully for a few times and remind yourself not to write down the unallowable answers.
Stop Making Stupid Mistakes by Richard Rusczyk from Art of Problem Solving
How to Avoid Careless Mathematical Errors from Reddit
Common errors from all my students:
Number 1 issue with most of the boys/and a few girls who have strong intuition in math is their horrible handwriting since their minds work much faster than their hands can handle. (See!! I'm making excuses for them.)
I call some of my students "second try ___ [Put your name here.] or third try ____ because I don't need to teach them how to solve a problem but the normal pattern is that it has to take them twice or three times to get it finally right. Thus, they need to slow down and organize their thoughts.
Practice your numbers: 4 (it's not 21), 7, 9 (don't dislocate the circle on top), 2 (don't make it looks like a 7), etc...
35 cents is very different from 0.35 cents.
Unless stated, you write improper fraction as your answer. Make sure to simplify the answer.
Many mistakes happen when it involves fractions, negative numbers, parenthesis, and questions that
have radicals on the denominator and you need to simplify it, so make sure to double check your math.
Some problems takes many steps to reach the final solutions. Some involve lots of data, extreme long strings of information so if those are your weakness, skip them first and go back to them later.
Every point is the same so don't stay with a tedious question too long and panic later not able to finish the last few harder questions which might be much easier to solve arithmetic wise if you know how.
Circle questions trouble many students. Make sure you are aware of what's being tested.
Is it circumference or area, is it diameter or radius, is it linear to area (squared) or vice versa ?
Are the units the same (most frequently appeared errors)?
Don't forget to divide by 2 for the triangle, times 6 if you use area of an equilateral triangle to get the hexagon. For geometry questions or unit conversions, don't do busy work, write down the equivalent equations and cancel like crazy.
Same goes to probability questions, cancel, cancel and cancel those numbers. You don't need to practice mental math multiplication for most of the Mathcounts problems. Think Smart!!
The answer for probability questions can only be 0 to 1, inclusive.
For lots of algebra questions, manipulations are the way to go or number sense. Use the digit clue to help you narrow down the "trial and error" method.
Make sure you have the terms and space type questions right. Calendar questions, how many numbers (inclusive, exclusive, between, ...), Stage doesn't necessarily start with 1 or 0.
Make sure you don't over-count or under-count.
Mos students got counting questions wrong when it involves limit.
Case in point : How many non-congruent triangles are there if the perimeter is 15?
Answer: There are 7 of them. Try it !!
To be continue...
Relax! You'll Be More Productive from the New York Times
First, please read Mathcounts "Forms of Answers" carefully for a few times and remind yourself not to write down the unallowable answers.
Stop Making Stupid Mistakes by Richard Rusczyk from Art of Problem Solving
How to Avoid Careless Mathematical Errors from Reddit
Common errors from all my students:
Number 1 issue with most of the boys/and a few girls who have strong intuition in math is their horrible handwriting since their minds work much faster than their hands can handle. (See!! I'm making excuses for them.)
I call some of my students "second try ___ [Put your name here.] or third try ____ because I don't need to teach them how to solve a problem but the normal pattern is that it has to take them twice or three times to get it finally right. Thus, they need to slow down and organize their thoughts.
Practice your numbers: 4 (it's not 21), 7, 9 (don't dislocate the circle on top), 2 (don't make it looks like a 7), etc...
35 cents is very different from 0.35 cents.
Unless stated, you write improper fraction as your answer. Make sure to simplify the answer.
Many mistakes happen when it involves fractions, negative numbers, parenthesis, and questions that
have radicals on the denominator and you need to simplify it, so make sure to double check your math.
Some problems takes many steps to reach the final solutions. Some involve lots of data, extreme long strings of information so if those are your weakness, skip them first and go back to them later.
Every point is the same so don't stay with a tedious question too long and panic later not able to finish the last few harder questions which might be much easier to solve arithmetic wise if you know how.
Circle questions trouble many students. Make sure you are aware of what's being tested.
Is it circumference or area, is it diameter or radius, is it linear to area (squared) or vice versa ?
Are the units the same (most frequently appeared errors)?
Don't forget to divide by 2 for the triangle, times 6 if you use area of an equilateral triangle to get the hexagon. For geometry questions or unit conversions, don't do busy work, write down the equivalent equations and cancel like crazy.
Same goes to probability questions, cancel, cancel and cancel those numbers. You don't need to practice mental math multiplication for most of the Mathcounts problems. Think Smart!!
The answer for probability questions can only be 0 to 1, inclusive.
For lots of algebra questions, manipulations are the way to go or number sense. Use the digit clue to help you narrow down the "trial and error" method.
Make sure you have the terms and space type questions right. Calendar questions, how many numbers (inclusive, exclusive, between, ...), Stage doesn't necessarily start with 1 or 0.
Make sure you don't over-count or under-count.
Mos students got counting questions wrong when it involves limit.
Case in point : How many non-congruent triangles are there if the perimeter is 15?
Answer: There are 7 of them. Try it !!
To be continue...
Tuesday, November 10, 2015
Mathcounts : Geometry -- Medians ; Squares in Isosceles Right Triangle, Similar Triangles, Triangles share the same vertex
Check out Mathcounts here, the best competition math program for middle school students.
Download this year's Mathcounts handbook here.
This one appears at 91 Mathcounts National Target #4. It's interesting.
Q: triangle ABC is equilateral and G is the midpoint. BI is the same length as BC and the question asks about the area of quadrilateral polygon GDBC.
We'll also find the area of triangle ADG and triagle DBI here.
Solution I: The side length is 2 and it's an equilateral triangle so the area of triangle ABC is
\(\dfrac {\sqrt {3}} {4}\times 2^{2}=\sqrt {3}\).
Using similar triangles GCF and ACE, you get \(\overline {GF}\)= \(\dfrac {\sqrt {3}} {2}\)
From there, you get area of \(\Delta\)GIC = \(4*\dfrac {\sqrt {3}} {2}*\dfrac {1} {2}=\sqrt {3}\)
\(\Delta\)FCH is a 30-60-90 degree right triangle so \(\overline {DH}=\sqrt {3}*\overline {BH}\)
Using similar triangles IDH and IGF, you get \(\dfrac {2+\overline {BH}} {3.5}=\dfrac {\sqrt {3}\overline {BH}} {\dfrac {\sqrt {3}} {2}}\) = 2\(\overline {BH}\);
2 +\(\overline {BH}\)= 7\(\overline {BH}\); \(\overline {BH}\)= \(\dfrac {1} {3}\)
\(\overline {DH}=\dfrac {\sqrt {3}} {3}\) = the height
Area of \(\Delta\)DBI = 2 * \(\dfrac {\sqrt {3}} {3}\)*\(\dfrac {1} {2}\) = \(\dfrac {\sqrt {3}} {3}\)
Area of quadrilateral polygon GDBC = \(\sqrt {3}-\dfrac {\sqrt {3}} {3}\)= \(\dfrac {2\sqrt {3}} {3}\)
The area of both \(\Delta\)ABC and \(\Delta\)GIC is \(\sqrt {3}\) and both share the quadrilateral polygon GDBC (It's like a Venn diagram).
Thus, the area of both triangles are the same \(\dfrac {\sqrt {3}} {3}\).
Solution II:
There are some harder AMC-10 questions using the same technique.
Using triangles share the same vertex, you get the two same area "a"s and "b"s because the length of the base is the same.
From the area we've found at solution I, we have two equations:
2a + b = \(\sqrt {3}\)
2b + a = \(\sqrt {3}\)
It's obvious a = b so you can find the area of the quadrilateral being the \(\dfrac {2} {3}\)of the equilateral triangle so the answer is \(\dfrac {2\sqrt {3}} {3}\).
Vinjai's solution III:
Draw the two extra lines and you can see that the three lines are medians, which break the largest triangle into 6 equal parts. [You can proof this using triangles share the same vertices : notes later.]
Quadrilateral polygon FECB is \(\dfrac {2} {3}\)of the equilateral triangle so the answer is \(\dfrac {2\sqrt {3}} {3}\).
This is similar to 2005 Mathcounts National Target #4 questions. AMC might have similar ones.
Q: Both triangles are congruent and isosceles right triangles and each has a square embedded in it. If the area of the square on the left is 12 square unites, what is the area of the square on the right?
Solution I:
Solution II:
Download this year's Mathcounts handbook here.
This one appears at 91 Mathcounts National Target #4. It's interesting.
Q: triangle ABC is equilateral and G is the midpoint. BI is the same length as BC and the question asks about the area of quadrilateral polygon GDBC.
We'll also find the area of triangle ADG and triagle DBI here.
Solution I: The side length is 2 and it's an equilateral triangle so the area of triangle ABC is
\(\dfrac {\sqrt {3}} {4}\times 2^{2}=\sqrt {3}\).
Using similar triangles GCF and ACE, you get \(\overline {GF}\)= \(\dfrac {\sqrt {3}} {2}\)
From there, you get area of \(\Delta\)GIC = \(4*\dfrac {\sqrt {3}} {2}*\dfrac {1} {2}=\sqrt {3}\)
\(\Delta\)FCH is a 30-60-90 degree right triangle so \(\overline {DH}=\sqrt {3}*\overline {BH}\)
Using similar triangles IDH and IGF, you get \(\dfrac {2+\overline {BH}} {3.5}=\dfrac {\sqrt {3}\overline {BH}} {\dfrac {\sqrt {3}} {2}}\) = 2\(\overline {BH}\);
2 +\(\overline {BH}\)= 7\(\overline {BH}\); \(\overline {BH}\)= \(\dfrac {1} {3}\)
\(\overline {DH}=\dfrac {\sqrt {3}} {3}\) = the height
Area of \(\Delta\)DBI = 2 * \(\dfrac {\sqrt {3}} {3}\)*\(\dfrac {1} {2}\) = \(\dfrac {\sqrt {3}} {3}\)
Area of quadrilateral polygon GDBC = \(\sqrt {3}-\dfrac {\sqrt {3}} {3}\)= \(\dfrac {2\sqrt {3}} {3}\)
The area of both \(\Delta\)ABC and \(\Delta\)GIC is \(\sqrt {3}\) and both share the quadrilateral polygon GDBC (It's like a Venn diagram).
Thus, the area of both triangles are the same \(\dfrac {\sqrt {3}} {3}\).
Solution II:
There are some harder AMC-10 questions using the same technique.
Using triangles share the same vertex, you get the two same area "a"s and "b"s because the length of the base is the same.
From the area we've found at solution I, we have two equations:
2a + b = \(\sqrt {3}\)
2b + a = \(\sqrt {3}\)
It's obvious a = b so you can find the area of the quadrilateral being the \(\dfrac {2} {3}\)of the equilateral triangle so the answer is \(\dfrac {2\sqrt {3}} {3}\).
Vinjai's solution III:
Draw the two extra lines and you can see that the three lines are medians, which break the largest triangle into 6 equal parts. [You can proof this using triangles share the same vertices : notes later.]
Quadrilateral polygon FECB is \(\dfrac {2} {3}\)of the equilateral triangle so the answer is \(\dfrac {2\sqrt {3}} {3}\).
This is similar to 2005 Mathcounts National Target #4 questions. AMC might have similar ones.
Q: Both triangles are congruent and isosceles right triangles and each has a square embedded in it. If the area of the square on the left is 12 square unites, what is the area of the square on the right?
Solution I:
Solution II:
Monday, October 26, 2015
Triangular Numbers & Word Problems: Chapter Level
Triangular Numbers : From Math is Fun. 1, 3, 6, 10, 15, 21, 28, 36, 45...
Interesting Triangular Number Patterns: From Nrich
Another pattern: The sum of two consecutive triangular numbers is a square number.
What are triangular numbers? Let's exam the first 4 triangular numbers:
The 1st number is "1".
The 2nd number is "3" (1 + 2)
The 3rd number is "6" (1 + 2 + 3)
The 4th number is "10" (1 + 2 + 3 + 4)
.
.
The nth number is \(\frac{n(n+1)}{2}\)
It's the same as finding out the sum of the first "n" natural numbers.
Let's look at this question based on the song "On the Twelve Day of Christmas"
(You can listen to this on Youtube,)
On the Twelve Day of Christmas
On the first day of Christmas
my true love gave to me
a Partridge in a Pear Tree
On the second day of Christmas,
My true love gave to me,
Two Turtle Doves,
And a Partridge in a Pear Tree.
On the third day of Christmas,
My true love gave to me,
Three French Hens,
Two Turtle Doves,
And a Partridge in a Pear Tree
On the fourth day of Christmas,
My true love gave to me,
Four Calling Birds,
Three French Hens,
Two Turtle Doves,
And a Partridge in a Pear Tree.
The question is "How many gifts were given out on the Day of Christmas?"
Solution I"
1st Day: 1
2nd Day: 1 + 2
3rd Day 1 + 2 + 3
4th Day 1 + 2 + 3 + 4
.
.
12th Day 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12
Altogether, you'll have 12 * 1 + 11 * 2 + 10 * 3 + 9 * 4 + 8 * 5 + 7 * 6 + 6 * 7 + 5 * 8 + 4 * 9 + 3 * 10 + 2 * 11 + 1 * 12 = 12 + 22 + 30 + 36 + 40 + 42 + 42 + 40 + 36 + 30 + 22 + 12 = 364
Solution II: The sum of the "n" triangular number is a tetrahedral number.
To get the sum, you use \(\frac{n(n+1)(n+2)}{6}\)
n = 12 and \(\frac{12 (13)(14)}{6}= 364 \)
Here is a proof without words.
Applicable questions: (Answers and solutions below)
#1 Some numbers are both triangular as well as square numbers. What is the sum of the first three positive numbers that are both triangular numbers and square numbers?
#2 What is the 10th triangular number? What is the sum of the first 10 triangular numbers?
#3 What is the 20th triangular number?
#4 One chord can divide a circle into at most 2 regions, Two chords can divide a circle at most into 4 regions. Three chords can divide a circle into at most seven regions. What is the maximum number of regions that a circle can be divided into by 50 chords?
#5: Following the pattern, how many triangles are there in the 15th image?
Answers:
#1 1262 The first 3 positive square triangular numbers are: 1, 36 (n = 8) and 1225 (n = 49).
#2 55; 220 a. \(\frac{10*11}{2}=55\) b. \(\frac{10*11*12}{6}=220\)
#3 210 \(\frac{20*21}{2}=210\)
#4 1276
1 chord : 2 regions or \(\boxed{1}\) + 1
2 chords: 4 regions or \(\boxed{1}\) + 1 + 2
3 chords: 7 regions or \(\boxed{1}\) + 1 + 2 + 3
.
.
50 chords:\(\boxed{1}\) + 1 + 2 + 3 + ...+ 50 = 1 + \(\frac{50*51}{2}\) =1276
#5: 120
The first image has just one triangle, The second three triangles. The third 6 triangles total.
It follows the triangular number pattern. The 15th triangular is \(\frac{15*16}{2}\) =120
Interesting Triangular Number Patterns: From Nrich
Another pattern: The sum of two consecutive triangular numbers is a square number.
What are triangular numbers? Let's exam the first 4 triangular numbers:
The 1st number is "1".
The 2nd number is "3" (1 + 2)
The 3rd number is "6" (1 + 2 + 3)
The 4th number is "10" (1 + 2 + 3 + 4)
.
.
The nth number is \(\frac{n(n+1)}{2}\)
It's the same as finding out the sum of the first "n" natural numbers.
Let's look at this question based on the song "On the Twelve Day of Christmas"
(You can listen to this on Youtube,)
On the Twelve Day of Christmas
On the first day of Christmas
my true love gave to me
a Partridge in a Pear Tree
On the second day of Christmas,
My true love gave to me,
Two Turtle Doves,
And a Partridge in a Pear Tree.
On the third day of Christmas,
My true love gave to me,
Three French Hens,
Two Turtle Doves,
And a Partridge in a Pear Tree
On the fourth day of Christmas,
My true love gave to me,
Four Calling Birds,
Three French Hens,
Two Turtle Doves,
And a Partridge in a Pear Tree.
The question is "How many gifts were given out on the Day of Christmas?"
Solution I"
1st Day: 1
2nd Day: 1 + 2
3rd Day 1 + 2 + 3
4th Day 1 + 2 + 3 + 4
.
.
12th Day 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12
Altogether, you'll have 12 * 1 + 11 * 2 + 10 * 3 + 9 * 4 + 8 * 5 + 7 * 6 + 6 * 7 + 5 * 8 + 4 * 9 + 3 * 10 + 2 * 11 + 1 * 12 = 12 + 22 + 30 + 36 + 40 + 42 + 42 + 40 + 36 + 30 + 22 + 12 = 364
Solution II: The sum of the "n" triangular number is a tetrahedral number.
To get the sum, you use \(\frac{n(n+1)(n+2)}{6}\)
n = 12 and \(\frac{12 (13)(14)}{6}= 364 \)
Here is a proof without words.
Applicable questions: (Answers and solutions below)
#1 Some numbers are both triangular as well as square numbers. What is the sum of the first three positive numbers that are both triangular numbers and square numbers?
#2 What is the 10th triangular number? What is the sum of the first 10 triangular numbers?
#3 What is the 20th triangular number?
#4 One chord can divide a circle into at most 2 regions, Two chords can divide a circle at most into 4 regions. Three chords can divide a circle into at most seven regions. What is the maximum number of regions that a circle can be divided into by 50 chords?
#5: Following the pattern, how many triangles are there in the 15th image?
Answers:
#1 1262 The first 3 positive square triangular numbers are: 1, 36 (n = 8) and 1225 (n = 49).
#2 55; 220 a. \(\frac{10*11}{2}=55\) b. \(\frac{10*11*12}{6}=220\)
#3 210 \(\frac{20*21}{2}=210\)
#4 1276
1 chord : 2 regions or \(\boxed{1}\) + 1
2 chords: 4 regions or \(\boxed{1}\) + 1 + 2
3 chords: 7 regions or \(\boxed{1}\) + 1 + 2 + 3
.
.
50 chords:\(\boxed{1}\) + 1 + 2 + 3 + ...+ 50 = 1 + \(\frac{50*51}{2}\) =1276
#5: 120
The first image has just one triangle, The second three triangles. The third 6 triangles total.
It follows the triangular number pattern. The 15th triangular is \(\frac{15*16}{2}\) =120
Tuesday, October 13, 2015
2015 Mathcounts State Prep : Inscribed Cricle Radius and Similar Triangles
Question : \(\Delta\) ABC is an equilateral triangle. Circle "O" is the inscribed circle and it's radius is 15.
What is the length of the radius of the smaller circle p which is tangent to circle "O" and the two sides?
Here is the link to the basics of inscribed circle radius as well as circumscribed circle radius of an equilateral triangle.
Solution I :
The length of the radius of an inscribed circle of an equilateral triangle is \(\dfrac {1} {3}\) of the height so you know AO is \(\dfrac {2} {3}\) of the height or 30 (the height is 15 + 30 = 45 unit long)
\(\Delta\) AEP is similar to \(\Delta\) AFO \(\rightarrow\) \(\dfrac {r} {15}=\dfrac {AP} {30}\)
so \(\overline {AP}\) = 2r.
\(\overline {AP}+\overline {PO}=30\) \(\rightarrow\)2r + r + 15 = 30 \(\rightarrow\) 3r = 15 so r = 5
or \(\dfrac {1} {3}\) of the larger radius
Solution II:
\(\Delta\) APE is a 30-60-90 right triangle, so \(\overline {AP}\) = 2r
\(\overline {PO}\) = r + 15
\(\overline {AP}+\overline {PO}\) \(\rightarrow\) 2r + r + 15 = 30 \(\rightarrow\) 3r = 15 so r = 5
or \(\dfrac {1} {3}\) of the larger radius
This is an AMC-10 question.
\(\Delta\) ABC is an isosceles triangle.
The radius of the smaller circle is 1 and the radius of the larger circle is 2,
A: what is the length of \(\overline {AP}\) ?
B. what is the area of \(\Delta\) ABC?
Solution for question A:
\(\Delta\) AEP is similar to \(\Delta\) AFO \(\rightarrow\) \(\dfrac {1} {2}=\dfrac {AP} {AP +3}\)
2\(\overline {AP}\) = \(\overline {AP}\) + 3 \(\rightarrow\) AP = 3
Using Pythagorean theorem, you can get \(\overline {AE}\) = \(2\sqrt {2}\)
\(\Delta\) AEP is similar to \(\Delta\) ADC [This part is tricky. Make sure you see that !!]
\(\rightarrow\) \(\dfrac {1} {\overline {DC}}=\dfrac {AE} {AD}\) = \(\dfrac {2\sqrt {2}} {8}\)
\(\overline {DC}\) = \(2\sqrt {2}\) and \(\overline {BC}\) = 2 * \(2\sqrt {2}\) = \(4\sqrt {2}\)
The area of \(\Delta\) ABC = \(\dfrac{1}{2}\)*\(4\sqrt {2}\) * 8 = \(16\sqrt {2}\)
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
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.
Wednesday, July 1, 2015
Wednesday, June 24, 2015
Problem Solving Strategy: Complementary Counting
#: How many three digit numbers contain the digit "9" at least once?
Solution:
For the "at least" questions: a lot of the time, the easiest way to solve the problems is to use the total number of ways minus the number of ways which do not satisfy the criteria we're looking for.
There are 999 -100 + 1 = 900 three digit numbers.
There are 8 x 9 x 9 = 648 numbers that do not contain the digit "9" at all, so the answer is 900 - 648 = 252
#2: How many three digit numbers have at least two digits that are the same?
Solution:
Use complementary counting to find how many three digit numbers have no digit that are the same.
9 digit choices for the hundredth digit (no 0), 9 for the tenth digit (0 is allowed) and 8 digit left for the unit digit, so 9 * 9 * 8 = 648
900 (3 digit numbers) - 648 = 252
#3: How many of the natural numbers from 1 to 600, inclusive, contain the digit "5" at least once?
(The numbers 152 and 553 are two natural numbers that contain the digit 5 at least once, but 430 is not.)
Solution:
From 1 to 600 inclusive, there are 600 - 1 + 1 = 600 numbers.
___ ___ ___ If we're looking for all the three-digit numbers which do not contain any 5s: In the hundreds place, you can put 0, 1, 2, 3, or 4 (we'll get rid of '000' later). In the tens place, you can put 9 digits, and there are another 9 possible digits for the units digit, so 5 x 9 x 9 = 405.
405 - 1 (to get rid of '000') + 1 (to compensate for not counting the number '600') = 405, which is the total number of numbers which don't use the digit '5'.
600 - 405 =195, which is the answer.
#4: If you toss three coins, what is the probability that at least one coin lands heads up?
Solution:
There is a \(\dfrac {1} {2}\times \dfrac {1} {2}\times \dfrac {1} {2}=\dfrac {1} {8}\)chance to get all tails = no heads, so if we want at least one head, the answer is 1 - \(\dfrac {1} {8}\)= \(\dfrac {7} {8}\).
Other applicable problems (answers below):
#1: Amy tosses a nickel four times. What is the probability that she gets at least as many heads as tails ?
#2: What is the probability that the product of the top faces on 2 standard die is even when rolled?
#3: 3 numbers are selected at random, with replacement, from the set of integers from 1 to 600 inclusive. What is the probability that the product of the 3 numbers is even ? Express your answer as a common fraction in lowest terms.
#4: 9 fair coins are flipped. What is the probability that at least 4 are heads?
#5: How many 3 digit numbers does not contain the digit 1 but have at least one digit that is 5?
Answer: #1-\(\dfrac {11} {16}\) ; #2-\(\dfrac {3} {4}\) ; #3- \(\dfrac {7} {8}\) ; #4-\(\dfrac {191} {256}\) #5: 200
Solution:
For the "at least" questions: a lot of the time, the easiest way to solve the problems is to use the total number of ways minus the number of ways which do not satisfy the criteria we're looking for.
There are 999 -100 + 1 = 900 three digit numbers.
There are 8 x 9 x 9 = 648 numbers that do not contain the digit "9" at all, so the answer is 900 - 648 = 252
#2: How many three digit numbers have at least two digits that are the same?
Solution:
Use complementary counting to find how many three digit numbers have no digit that are the same.
9 digit choices for the hundredth digit (no 0), 9 for the tenth digit (0 is allowed) and 8 digit left for the unit digit, so 9 * 9 * 8 = 648
900 (3 digit numbers) - 648 = 252
#3: How many of the natural numbers from 1 to 600, inclusive, contain the digit "5" at least once?
(The numbers 152 and 553 are two natural numbers that contain the digit 5 at least once, but 430 is not.)
Solution:
From 1 to 600 inclusive, there are 600 - 1 + 1 = 600 numbers.
___ ___ ___ If we're looking for all the three-digit numbers which do not contain any 5s: In the hundreds place, you can put 0, 1, 2, 3, or 4 (we'll get rid of '000' later). In the tens place, you can put 9 digits, and there are another 9 possible digits for the units digit, so 5 x 9 x 9 = 405.
405 - 1 (to get rid of '000') + 1 (to compensate for not counting the number '600') = 405, which is the total number of numbers which don't use the digit '5'.
600 - 405 =195, which is the answer.
#4: If you toss three coins, what is the probability that at least one coin lands heads up?
Solution:
There is a \(\dfrac {1} {2}\times \dfrac {1} {2}\times \dfrac {1} {2}=\dfrac {1} {8}\)chance to get all tails = no heads, so if we want at least one head, the answer is 1 - \(\dfrac {1} {8}\)= \(\dfrac {7} {8}\).
Other applicable problems (answers below):
#1: Amy tosses a nickel four times. What is the probability that she gets at least as many heads as tails ?
#2: What is the probability that the product of the top faces on 2 standard die is even when rolled?
#3: 3 numbers are selected at random, with replacement, from the set of integers from 1 to 600 inclusive. What is the probability that the product of the 3 numbers is even ? Express your answer as a common fraction in lowest terms.
#4: 9 fair coins are flipped. What is the probability that at least 4 are heads?
#5: How many 3 digit numbers does not contain the digit 1 but have at least one digit that is 5?
Answer: #1-\(\dfrac {11} {16}\) ; #2-\(\dfrac {3} {4}\) ; #3- \(\dfrac {7} {8}\) ; #4-\(\dfrac {191} {256}\) #5: 200
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.
Thursday, May 14, 2015
Games: logical puzzle: the wolf, the sheep and the cabbages
Logical games:
How to cross the river without the wolf eating the sheep, the sheep eating the cabbages.
Here is the link to the puzzle "the wolf, the sheep and the cabbages".
Two versions of "Leap Frog Puzzle".
Leap Frog Puzzle
Frogs Game
How to cross the river without the wolf eating the sheep, the sheep eating the cabbages.
Here is the link to the puzzle "the wolf, the sheep and the cabbages".
Two versions of "Leap Frog Puzzle".
Leap Frog Puzzle
Frogs Game
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