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.

Tuesday, December 6, 2016

Harder Mathcounts State Question

2012 Mathcounts State Sprint #30: In rectangle ABCD, shown here, point M is the midpoint of side BC, and point N lies on CD such that DN:NC = 1:4. Segment BN intersects AM and AC at points R and S, respectively. If NS:SR:RB = x:y:z, where x, y and z are positive integers, what is the minimum possible value of x + y + z? 

Solution I :


\(\overline {AB}:\overline {NC}=5:4\) [given]

Triangle ASB is similar to triangle CSN (AAA)

\(\overline {NS}:\overline {SB}= 4 : 5\)

Let \(\overline {NS}= 4a,  \overline {SB}= 5a.\)






Draw a parallel line to \(\overline {NC}\) from M and mark the interception to \(\overline {BN}\)as T.

 \(\overline {MT}: \overline {NC}\) = 1 to 2. [\(\Delta BMT\) and \(\Delta BCN\) are similar triangles ]

\(\overline {NT} = \overline {TB}= \dfrac {4a+5a} {2}=4.5a\)

\(\overline {ST} = 0.5a\)

 \(\overline {MT} :  \overline {AB}\) = 2 to 5
[Previously we know  \(\overline {MT}: \overline {NC}\) = 1 to 2 or 2 to 4 and  \(\overline {NC}:\overline {AB}= 4 : 5\) so the ratio of the two lines  \(\overline {MT} :  \overline {AB}\) is 2 to 5.]


\(\overline {TB} = 4.5 a\)  [from previous conclusion]

Using 5 to 2 line ratio [similar triangles \(\Delta ARB\) and \(\Delta MRT\) , you get \(\overline {BR} =\dfrac {5} {7}\times 4.5a =\dfrac {22.5a} {7}\) and \(\overline {RT} =\dfrac {2} {7}\times 4.5a =\dfrac {9a} {7}\)

Thus, x : y : z = 4a : \( \dfrac {1} {2}a + \dfrac {9a} {7}\) : \(\dfrac {22.5a} {7}\) = 56 : 25 : 45

x + y + z = 126

Solution II : 
From Mathcounts Mini: Similar Triangles and Proportional Reasoning

Solution III: 
Using similar triangles ARB and CRN , you have \(\dfrac {x} {y+z}=\dfrac {5} {9}\).
9x = 5y + 5z ---- equation I

Using similar triangles ASB and CSN and you have \( \dfrac {x+y} {z}=\dfrac {5} {4}\).
4x + 4y = 5z  ---- equation II

Plug in (4x + 4y) for 5z on equation I and you have 9x = 5y + (4x + 4y) ; 5x = 9y ; x = \(\dfrac {9} {5}y\)
Plug in x = \(\dfrac {9} {5}y\) to equation II and you have z  =  \( \dfrac {56} {25}y\)

x : y : z = \(\dfrac {9} {5}y\)  : y  :  \( \dfrac {56} {25}y\) =  45 y :  25y  :  56y

45 + 25 + 56 = 126


























Solution IV : Yes, there is another way that I've found even faster, saved for my private students.

Solution V : from Abhinav, one of my students solving another similar question : 

2016 AMC 10 A, #19 





2016 AMC 10 B #19 : Solution from Abhinav 


Thursday, December 1, 2016

Mass Points Geometry

Some of the harder/hardest questions at Mathcounts can be tackled at ease using mass point geometry
so spend some time understanding it.

Mass Point Geometry : This one is easier to understand with questions that have detailed solutions.

Mass Point Geometry : another link for basics

2014-15 Mathcounts handbook Mass Point Geometry Stretch
from page 39 to page 40

(Talking about motivation, yes, there are students already almost finish
this year's Mathcounts' handbook harder problems.)

From Wikipedia

From AoPS

Mass Point Geometry by Tom Rike

Videos on Mass Point :

Basics 

Mass Point Geometry Part I 

Mass Point Geometry : Split Masses Part II 

Mass Point Geometry : Part III 


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?

Solution I:
Make sure you know how to get the unknown leg fast. The height of the parallelogram is 82, 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

Monday, November 14, 2016

2017 Mathcounts State Prep: Some Counting and Probability Questions on Dot Grids

Check out Mathcounts here, the best competition math program for middle school students.
Download this year's Mathcounts handbook here. (It's free.)

#5 1993 Mathcounts National Target : Find the probability that four randomly selected points on the geoboard below will be the vertices of a square? Express your answer as a common fraction.













#5 2004 AMC 10A: A set of three points is chosen randomly from the grid shown. Each three-point (same image as the below question) set has the same probability of being chosen. What is the probability that the points lie on the same straight line?
















2007 Mathcounts Chapter Sprint #29 : The points of this 3-by-3 grid are equally spaced
horizontally and vertically. How many different sets of three points of this grid can be the three
vertices of an isosceles triangle?












Solution:
#5 National Target: There are 16C4 = \(\dfrac {16\times 15\times 14\times 13} {4\times 3\times 2\times 1}\)= 1820 ways to select 4 points on the geoboard.

There are 3 x 3 = 9  one by one squares and 2 x 2 = 4 two by two squares and 1 x 1 = 1 three by three squares. (Do you see the pattern?)
                                                         




There are 4 other squares that have side length of √ 2
and 2 other larger squares that have side length of 5.

9 + 4 + 1 + 4 + 2 = 20 and \(\dfrac {20} {1820}=\dfrac {1} {91}\)






#5: Solution:
AMC-10A: There are 9C3 = \(\dfrac {9\times 8\times 7} {3\times 2\times 1}\)= 84 ways to chose the three dots and 8 of the lines connecting the three dots will form straight lines. (Three verticals, three horizontals and two diagonals.) so 
\(\dfrac {8} {84}=\dfrac {2} {21}\)

#29: Solution: 
Use the length of the two congruent legs to solve this problem systematically. 



 There are 16   1 - 1 - \(\sqrt {2}\)    isosceles triangles.
There are 8    \(\sqrt {2}\)  by  \(\sqrt {2}\) by 2 isosceles triangles. (See that ?)







There are 4     2 - 2 - \(2\sqrt {2}\)  isosceles triangles.
There are 4    \(\sqrt {5}\)  by  \(\sqrt {5}\) by 2 isosceles triangles.
Finally, there are \(\sqrt {5}\)  by  \(\sqrt {5}\) by \(\sqrt {2}\) isosceles triangles.
Add them up and the answer is 36.  




Monday, November 7, 2016

How Many Zeros?

Problems: (Solutions below.)

#1. 2003 Chapter Team # 7--How many zeros are at the end of (100!)(200!)(300!) when multiplied out?

#2. How many zeros are at the end of 2013!? 

#3. How many zeros are at the end of 10! *9!*8!*7!*6!*5!*4!*3!*2!*1!*0!?

#4. What is the unit digit of 10! + 9! + 8! + 7! + 6! + 5! + 4! + 3! + 2! + 1! + 0!?

#5. The number \(3^{4}\times 4^{5}\times 5^{6}\) written out in full. How many zeros are there are at the end of the number?

#6. How many zeros are at the end of (31!)/(16!*8!*4!*2!*1!)

#7. 2009 National Sprint #18-- What is the largest integer n such that \(3^{n}\) is a factor of 1×3×5×…×97×99?












Solutions:
#1.For 100!, there are -- 100/5 = 20 , 20/5 = 4 (Stop when the quotient is not divisible by 5 and then add up all the quotients.), or 20 + 4 = 24 zeros.
For 200!, there are 200/5 =40 , 40/5 = 8, and  8/5 = 1, or total 40 + 8 + 1 = 49 zeros.
For 300!, there are 300/5 = 60, 60/5 = 12, and 12/5 = 2, or total 60 + 12 + 2 = 74 zeros.
Add all the quotients and you get 147 zeros. 

#2. Use the same method as #1 and the answer is 501 zeros.

#3:Starting at 5!, you have one "0", the same goes with 6!, 7!, 8!, and 9!
10! will give you 2 extra "0"s. Thus total 7 zeros.

#4: Since starting with 5! you have "0" for the unit digit, you only need to check 4! + 3! + 2! + 1! + 0!.
24 + 6 + 2 + 1 + 1 = 34 so the unit digit is 4.

#5: Make sure to prime factorize all the given number sequences, in this case, it's \(3^{4}\times 2^{10}\times 5^{6}\) after you do that.
2 * 5 = 10 will give you a zero since there are fewer 5s than 2s so the answer is 6 zeros.

#6: You need the same number of 2 and 5 multiple together to get a "0".
31! gives you 30/5 = 6, 6/5 = 1 or 6 + 1 = 7 multiples of 5
31! gives you 10//2 = 15...15/2 = 7...7/2 = 3...3/2 = 1 or 15 + 7 + 3 + 1 = 26 multiples of 2. 16!*8!*4!*2!*1! gives you 4 multiples of 5 and 8 + 4 + 2 + 1 (16!) + 4 + 2 + 1 (8!) + 2 + 1 (4!) + 1(2!)
= 26 multiples of 2.
Thus the multiples of 2s all cancel out, the answer is "0" zeros. 

#7: There are 3*1, 3*3, 3*5...3*33 or \(\dfrac {33-1} {2}+1=17\) multiples of 3.
There are 9*1, 9* 3...9*11 or \(\dfrac {11-1} {2}+1 = 6\) extra multiples of 3.
There are 27*1, 27*3 or 2 extra multiples of 3.
There is 81*1 or 1 extra multiples of 3.
Add them up and the answer is 26.

Wednesday, October 26, 2016

2014 AMC 8 Results , Problems, Answer key and Detailed Solutions.

E-mail me if you want to join my group lessons for Mathcounts/AMCs prep.
Different groups based on your current level of proficiency.

My students are amazing. They NEVER stop learning and they don't just do math.

Please don't contact me if you just want state/national questions since I'm extremely busy these days with the preps.
You can purchase Mathcounts National tests and other prep materials at Mathcounts store.

You can also use my online blog contents. If you really understand those concepts, I'm sure you can be placed in your state's top 10 in the above average states [Every year I'd received some online students'/parents' e-mails about their (their kids') state results], not the most competitive states, which are crapshoot for even the USAJMO qualifiers, that I know.

Take care and best of luck.

Have fun problem solving and good luck.

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 !!



ACT worksheets

ACT worksheets

Let me know if you have any other questions. Best and take good care, Mrs. Lin 



worksheet : 


Similar triangles : 



worksheet : 

Monday, October 10, 2016

2012 Harder Mathcounts State Target Questions

Check out Mathcounts here -- the best competition math program for middle schoolers up to the 
state and national level. 

# 6: A semicircle and a circle are placed inside a square with sides of length 4 cm, as shown. The circle is tangent to two adjacent sides of the square and to the semicircle. The diameter of the semicircle is a side of the square. In centimeters, what is the radius of the circle? Express your answer as a decimal to the nearest hundredth. [2012 Mathcounts State Target #6]


#6:  Solution:
Using Pythagorean theory: (2 + r)2 = (4-r)2 + ( 2- r)2
4 + 4r + r2 = 16 - 8r + r2 + 4 - 4r + r2
 r2 - 16 r + 16 = 0
Using the quadratic formula You have 8 ± 4√ 3
Only 8 - 4 3 = 1.07 works

There is a Mathcounts Mini #34 on the same question. Check that out !!

The above question looks very similar to this year's AMC-10 B #22, so try that one.
(cover the answer choices so it's more like Mathcounts)

2014 AMC-10 B problem #22 

#8: In one roll of four standard, six-sided dice, what is the probability of rolling exactly three different numbers? Express your answer as a common fraction. [2012 Mathcounts State Target #8]

Solution I : Permutation method
If order matters, there are 6 * 5 * 4 * 1 ways to choose the number, 1 being the same number as one of the previous one.
Let's say if you choose 3 1 4 1.

Now for the placement of those 4 numbers on the 4 different dice. There are 4C2 ways to place where the two "1" will
be positioned so the answer is : \(\dfrac {6\times 5\times 4\times 1\times 4C2} {6^{4}}\) = \(\dfrac{5}{9}\)
Solution II:  Combination method

There are 6C3 = 20 ways to choose the three numbers.

There are 3 ways that the number can be repeated. [For example: If you choose 1, 2, and 3, the fourth number could be 1, 2 or 3.]

There are \(\dfrac {4!} {2!}\) =12
ways to arrange the chosen 4 numbers.[same method when you arrange AABC]
So the answer is\(\dfrac{20* 3 *12}{6^4}\) = \(\dfrac{5}{9}\)

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.

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 \(4\sqrt {3}\).



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.  :) 

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. 
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.


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 !!


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 

#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

#6: Answer: 136 square units

#7: Answer: 98 square units

#8: Answer: 66 square units 

Tuesday, August 16, 2016

Pathfinder

From Mathcounts Mini :

Counting/Paths Along a Grid

From Art of Problem Solving

Counting Paths on a Grid 

Math Principles : Paths on a Grid : Two Approaches 


Question #1: How many ways to move the dominoes on a 6 by 6 checker board if you can only move the dominoes to the right or to the bottom starting from the upper left and you can't move the dominoes diagonally? 

Solution :
You can move the dominoes 5 times to the right at most and 5 down to
the bottom at most, so the answer is \(\dfrac {\left( 5+5\right) !} {5! \times 5!}\) = 252 ways







Question # 2: How many ways can you  move from A to B if you can only move downward and to right? 

Solution : There are \(\dfrac {\left( 4+4\right) !} {4!\times 4!}\) * 2 * \(\dfrac {\left( 4+4\right) !} {4!\times 4!}\) = 9800 ways from A to B

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]

Sunday, July 3, 2016

Sum of All the Possible Arrangements of Some Numbers

Check out Mathcounts, the best middle school competition math program up to the national level.

 Questions to ponder: (detailed solutions below)

#1: Camy made a list of every possible distinct four-digit positive integer that can be formed using each of the digits 1, 2 , 3 and 4 exactly once in each integer. What is the sum of the integers on Camy's list?

#2: Camy made a list of every possible distinct five-digit positive even integer that can be formed using each of the digits 1, 3, 4, 5 and 9 exactly once in each integer. What is the sum of the integers on Camy's list? (2004 Mathcounts Chapter Sprint #29)













Solutions:
#1:  
Solution I: 
There are 4! = 24 ways to arrange the four digits. Since each digit appears evenly so each number will appear 24 / 4 = 6 times.
1 + 2 + 3 + 4 = 10 and 10 times 6 = 60 ; 60 (1000 +100 +10 + 1) = 60 x 1111 = 66660, which is the answer.

Solution II: 
The median of four numbers 1, 2, 3, 4 is (1 + 2 + 3 + 4) / 4 = 2.5 and there are 4! = 24 ways to arrange
the four numbers. 
2.5 (1000 + 100 + 10 + 1) x 24 = 66660 

#2:  
Solution I: 
Since this time Camy wants five-digit even integer, which means that the number "4" has to be at the unit digit and only 1, 3, 5, 9 can be moved freely.
Again there are 4! = 24 ways to arrange the four numbers. 1 + 3 + 5 + 9 = 18 and 18 x 6 = 108 (Each number that can be moved freely appears 6 times evenly.)108 x 11110 + 4 x 24 = 1199976

Solution II: 

Since this time Camy wants five-digit even integer, which means that the number "4" has to be at the unit digit and only 1, 3, 5, 9 can be moved freely. 
There will be 4! = 24 times the even number 4 will be used so 4 x 24 = 96
As for the remaining 4 numbers, their average (or mean) is \(\dfrac{1 + 3+ 5 + 9} {4} = 4.5\)                               
4.5 * ( 10000 + 1000 + 100 + 10) * 24 (arrangements)  + 96 = 4.5 * 11110 * 24 + 96 = 1199976

Other applicable problems: (answers below)

#1: What is the sum of all the four-digit positive integers that can be written with the digits 1, 2, 3, 4 if each digit must be used exactly once in each four-digit positive integer? (2003 Mathcounts Sprint #30)

#2: What is the average (mean) of all 5-digit numbers that can be formed by using each of the digits 1, 3, 5, 7, and 8 exactly once? (You can use a calculator for this question.) (2005 AMC-10 B)

#3: What is the sum of all the four-digit positive integers that can be written with the digits 2, 4, 6, 8 if each digit must be used exactly once in each four-digit positive integer?

#4: What is the sum of all the 5-digit positive odd integers that can be written with the digits 2, 4, 6, 8, and 3 if each digit must be used exactly once in each five-digit positive integer?  

#5:What is the sum of all the four-digit positive integers that can be written with the digits 2, 3, 4, 5 if each digit must be used exactly once in each four-digit positive integer?






  





Answer key: 
#1: 66660
#2: \(\dfrac{1 + 3 + 5 + 7 + 8}{5} = 4.8\)
4.8 * 11111 =\(\color{red}{53332.8}\) 
#3: 133320
#4: 1333272
#5: 93324