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.

Basics 

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

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

From Wikipedia

From AoPS

Mass Point Geometry by Tom Rike

Another useful notes 

Videos on Mass Point :

Mass Points Geometry Part I 

Mass Points Geometry : Split Masses Part II 

Mass Points Geometry : Part III 

other videos from Youtube on Mass Points

It's much more important to fully understand how it works, the easier questions the weights align
very nicely.

The harder problems the weights are messier, not aligning nicely, so you need to find ways to may them integers (LCM) for easier solving.

Let me know if you have questions. I love to help (:D) if you've tried.

Face Diagonal and Space Diagonal of a Rectangular Prism

Face diagonal and space diagonal of a cube 

Ways to calculate face and space diagonal.

Each side of the cube is x units long.

Use  45-45-90 degree angle ratio
( 1 - 1 - √ 2  ) or Pythagorean theorem to get the face diagonal.

Using Pythagorean theorem twice and you'll get the space diagonal.







Face diagonal and Space diagonal of a rectangular prism.

Same way to figure out the face

diagonal of a rectangle as well as

space diagonal of a rectangular prism.

Use Pythagorean theorem or

30-60-90 degree angle ratio

(1 -√ 3  - 2) to figure out the face

diagonal and Pythagorean theorem

twice to figure out the space diagonal.

2020 Mathcounts State Prep: Simon's Favorite Factoring Trick

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

The most common cases of Simon's Favorite Factoring Trick are:

I:  \(xy+x+y+1=\left( x+1\right) \left( y+1\right)\)

II:  \(xy-x-y+1=\left( x-1\right) \left( y-1\right)\)

It's easy to learn. Here is the best tutorial online, by none other than Richard Rusczyk.
The method Rusczyk uses at the second half is very nifty. Thanks!!

Questions to ponder:(answer key below)
#1: Both x and y are positive integers and \(x>y\). Find all positive integer(s) that \(xy+x+y=13\) 
#2: Both x and y are positive integers and \(x>y\). Find all positive integer(s) that \(2xy+2x-3y=18\)
#3: Find the length and the width of a rectangle whose area is equal to its perimeter.
#4: Twice the area of a non-square rectangle equals triple it's perimeter, what is the area of the rectangle? 













Answer key:
#1:  x = 6 and y = 1
#2: ( x, y ) = (4, 2) 
#3: Don't forget square is a kind of rectangle (but not the other way around) so there are two answers: 
4 by 4 and 3 by 6 units. 
#4: One side is 4 units and the other 12 units so the answer is 4 x 12 or 48 square units. 
There is another one, 6 by 6 that would fit if the question doesn't specify non-square rectangle. 

2019 AMC 8 problems, solutions and some thoughts

2019 AMC 8 problems and solutions, for students, by students

A student's reflection on this year's test : 

Mrs. Lin,
I did the AMC 8 yesterday, and it was actually quite easier than last year. I was reviewing my answers, and I believe I only got the last two wrong. I used stars and bars for the last one, but did 21C2 instead of 20C2. I could’ve done number 24, because geometry is really my best subject in math. I realized that I should’ve used mass points later on. It’s fine, though, because it’s still a good score. I think that many people could solve this test pretty well because in many of the last questions you could plug in the multiple choice answers and get the right answer. Also, a lot of it was just plain algebra. Question 20 was actually just an equation, which anybody who knows basic algebra can solve. I thought that I would never say this, but I honestly wish that it was harder, because I was hoping for some interesting problems. Those are the problems that get people’s gears turning; this year the problems were quite basic. I think many people will get really good scores on this test, which, along with a good thing, is also not so good because it brings down the credibility of the test.

Thanks,

some links that you can review those very basic, but extremely useful strategies on this 
year's seemingly harder, but not really last two questions. 

mass points  learn together with triangles sharing the same vertex 

dimensional change / scaling 

balls and urns, stars and bars  (lots of variations or twists on this one, so 
you need to fully understand the concept so to use it well. Be patient !!!!!) 

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 \(\dfrac {4!} {2!}\) =12ways to arrange the chosen 4 numbers.[same method when you arrange AABC]

So the answer is\(\dfrac{20* 3 *12}{6^4}\) = \(\dfrac{5}{9}\)