2020 Mathcounts State Prep: Simon's Favorite Factoring Trick

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Download this year's Mathcounts handbook here.

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

I:  $xy+x+y+1=(x+1)(y+1)$

II:  $xy-x-y+1=(x-1)(y-1)$

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 : ${\displaystyle \frac{6\times 5\times 4\times 1\times 4C2}{6^4}}$ = ${\displaystyle \frac{5}{9}}$

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 ${\displaystyle \frac{4!}{2!}}$ =12ways to arrange the chosen 4 numbers.[same method when you arrange AABC]

So the answer is${\displaystyle \frac{20\ast 3\ast 12}{6^4}}$ = ${\displaystyle \frac{5}{9}}$

Hints/links or Solutions to 2014 Harder Mathcounts State Sprint and Target question

Links, notes, Hints or/and solutions to 2014 Mathcounts state harder problems.

2014, 2015 Mathcounts state are harder 

Sprint round:

#14 :
Solution I :

(7 + 8 + 9)  + (x + y + z)  is divisible by 9, so the sum of the three variables could be 3, 12, or 21.
789120 (sum of 3 for the last three digits) works for 8 but not for 7.
21 is too big to distribute among x, y and z (all numbers are district),
thus only x + y + z = 12 works and z is an even number
__ __ 0 does't work (can't have 6 6 0 and the other pairs all have 7, 8 or 9)
264 works (789264 is the number)

Solution II : 
789000 divided by the LCM of 7, 8 and 9, which is 504 = 1565.47...
Try 504 * 1566 = 789264 (it works)
The answer is 264.

#18:
Watch this video from Mathcounts mini and use the same method for the first question,
you'll be able to get the answer. It's still tricky, though.

#23 : Drop the heights of the two isosceles triangles and use similar triangles to get the length of FC.
Then solve.

#24: 
The key is to see 2¹⁰ is 1024 or about 10³

2³⁰ = ( 2¹⁰ )³  or about (10³  )³about 10⁹ so the answer is 10 digit.

#25:
As you can see, there are two Pythagorean Triples : 9-12-15 and 9-40-41.
Base (40-12) = 28 gives you the smallest area.
The answer is 28 * 18 = 504

#26 : Let there be A, B, C three winners. There are 4 cases to distribute the prizes.
A     B    C
1      1     5    There are 7C1 * 6C1 * ${\displaystyle \frac{3!}{2!}}$ = 126 ways -- [you can skip the last part for C
because it's 5C5 = 1]

1       2    4    There are 7C1* 6C2 * 3! = 630

1      3     3    There are 7C1 * 6C3 * ${\displaystyle \frac{3!}{2!}}$ = 420

2      2    3     There are 7C2 * 5C2 * 3 (same as above)

Add them up and the answer is 1806.

If you can't see why it's ${\displaystyle \frac{3!}{2!}}$ when there is one repeat, try using easier case to help you understand.

What about A, B two winners and 4 prizes ?
There are 2 cases, 1 3 or 2 2, and you'll see how it's done.

#27 : Read this and you'll be able to solve this question at ease, just be careful with the sign change.
Vieta's Formula and the Identity Theory

#28: There are various methods to solve this question.
I use binomial expansion :
${11}^{12}={(13-2)}^{12}=12C0\ast {13}^{12}$+ $12C1\ast {13}^{11}\ast 2^1$+... $12C11\ast {13}^1\ast 2^{11}$+ $12C12\ast 2^{12}$ Most of the terms will be evenly divided by 13 except the last term, which is $2^{12}$ or 4096, which, when divided by 13, leaves a remainder of 1.

Solution II :
$11\equiv -2\left(mod13\right)$ ; $(-2{)}^{12}\equiv 4096\equiv 1\left(mod13\right)$

Solution III :  
Or use Fermat's Little Theorem (Thanks, Spencer !!)
${11}^{13-1}\equiv {11}^{12}\equiv 1(mod13)$

Target Round : 

#3: Lune of Hippocrates : in seconds solved question.
^__^

#6: This question is very similar to this Mathcounts Mini.
My students should get a virtual bump if they got this question wrong.

#8: Solution I : by TMM (Thanks a bunch !!)
Using similar triangles and Pythagorean Theorem.

The height of the cone, which can be found using the Pythagorean  is . 
Usingthediagram below, let  be the radius of the top cone and let  be the height of the topcone. 
Let  be the slant height of the top cone.


Drawing the radius as shown in the diagram, we have two right triangles. Since the bases of the top cone and the original cone are parallel, the two right triangles are similar. So we have the proportionCross multiplying yields This is what we need.

Next, the volume of the original cone is simply . 
The volume of the top cone is .
From the given information, we know that We simply substitute the value of  from above to yield We will leave it as is for now so the decimals don't get messy.

We get  and .

The lateral surface area of the frustum is equal to the lateral surface area of the original cone minus the lateral surface area of the top cone. The surface area of the original cone is simply 
. 
The surface area of the top cone is . 
So our lateral surface area is 

All we have left is to add the two bases. The total area of thebases is . So our final answer is 

Solution II : 
Using dimensional change and ratio, proportion.

Cut the cone and observe the shape.

The circumference of the larger circle is 20pi (10 is the radius) and the base of
the cone circle circumference is 10pi (5 is the radius), which means that the cut-off cone shape is a half circle because it's ${\displaystyle \frac{10\pi }{20\pi }}$ or ${\displaystyle \frac{1}{2}}$ of the larger circle. (180 degrees)

To find the part that is the area of the frustum not including the top and bottom circles,
you use the area of the half circle minus the area of the smaller half circle.

Since the volume ratio of the smaller cone to larger cone = 2 to 3, the side ratio of the
two radius is ${\displaystyle \frac{\sqrt[3]{2}}{\sqrt[3]{3}}}$.

Using this ratio, we can get the radius of the smaller circle as 10 * ${\displaystyle \frac{\sqrt[3]{2}}{\sqrt[3]{3}}}$ and the radius of the top circle of the frustum as 5 * ${\displaystyle \frac{\sqrt[3]{2}}{\sqrt[3]{3}}}$.

Now we can solve this :
 ${\displaystyle \frac{1}{2}}$$[{10}^2\pi -{(10\times {\displaystyle \frac{\sqrt[3]{2}}{\sqrt[3]{3}}})}^2\pi ]$ + $5^2\pi +{(5\times {\displaystyle \frac{\sqrt[3]{2}}{\sqrt{3}}})}^2\pi$ = about 176 (after you round up)ional change and ratio, proportion.

Cut the cone and observe the shape.

The circumference of the larger circle is 20pi (10 is the radius) and the base of
the cone circle circumference is 10pi (5 is the radius), which means that the cut-off cone shape is a half circle because it's ${\displaystyle \frac{10\pi }{20\pi }}$ or ${\displaystyle \frac{1}{2}}$ of the larger circle. (180 degrees)

To find the part that is the area of the frustum not including the top and bottom circles,
you use the area of the half circle minus the area of the smaller half circle.

Since the volume ratio of the smaller cone to larger cone = 2 to 3, the side ratio of the
two radius is ${\displaystyle \frac{\sqrt[3]{2}}{\sqrt[3]{3}}}$.

Using this ratio, we can get the radius of the smaller circle as 10 * ${\displaystyle \frac{\sqrt[3]{2}}{\sqrt[3]{3}}}$ and the radius of the top circle of the frustum as 5 * ${\displaystyle \frac{\sqrt[3]{2}}{\sqrt[3]{3}}}$.

Now we can solve this :
 ${\displaystyle \frac{1}{2}}$$[{10}^2\pi -{(10\times {\displaystyle \frac{\sqrt[3]{2}}{\sqrt[3]{3}}})}^2\pi ]$ + $5^2\pi +{(5\times {\displaystyle \frac{\sqrt[3]{2}}{\sqrt{3}}})}^2\pi$ = about 176 (after you round up)

Solution III : Another way to find the surface area of the Frustum is : 
median of the two half circle [same as median of the two bases] * the height [difference of the two radius]
${\displaystyle \frac{1}{2}}(2\times 10\pi +2\times 10\times {\displaystyle \frac{\sqrt[3]{2}}{\sqrt[3]{3}}}\pi )$* $(10-10\times {\displaystyle \frac{\sqrt[3]{2}}{\sqrt[3]{3}}})$

Notes to 2018 Mathcounts chapter more interesting problems

2018 Mathcounts Chapter Spring problems : solutions down below 
Thanks to a boy mathlete who tried these problems and e-mail me for feedback. 

Please try these problems first before reading the explanations. :D

#25 : Three employees split a bonus valued at some number of dollars. Arman first receives $10morethanonethirdofthetotalamount.Bernardothenreceives$3 more than one half of what was left. Carson receives the remaining $25. What is the total dollar value of the bonus?

#27: For a particular list of four distinct integers the mean, median and range have the same value. If the least integer in the list is 10, what is the greatest value for an integer in the list?

#29: There are two values of x such that $|{\displaystyle \frac{x-2018}{x-2019}}|={\displaystyle \frac{1}{6}}$. . What is the absolute difference between these two values of x? Express your answer as a common fraction.

Target #8 : Four congruent circles of radius 2 cm intersect with their centers at intersection points as shown. What is the area of the shaded region? Express your answer in terms of π.

#25 : It's easier if you go backward and use inverse operations to solve this question. 
(25 + 3) *2 = 56 and 56 + 10 = 66, which is ${\displaystyle \frac{2}{3}}$ of the original bonus value, or 
what is left after ${\displaystyle \frac{1}{3}}$ was given out. 

${\displaystyle \frac{2}{3}}$ of bonus is 66 dollars, so the answer is 99. 

#27: Let the average be x and the three other numbers be a, b, c and $a<b<c$.
The least number is 10 (given), so 
$10+a+b+c$ = $4x$---> equation 1 
${\displaystyle \frac{a+b}{2}}$ = $x$ (how to find the median), so $a+b$ = $2x$ --- (2)
$c-10=x$ (given because it's the range), $c=x+10$ ---(3)
Substitute (2) and (3) to equation one and you have $10+2x+x+10=4x$, so $x=20$
$C=20+10=30$, the answer 

#28 : Let $x-2018=y$ , then $x-2019=y-1$
We then have either 
$|{\displaystyle \frac{y}{y-1}}|={\displaystyle \frac{1}{6}}$  $\to$ $y$ = ${\displaystyle \frac{-1}{5}}$

or $|{\displaystyle \frac{y}{y-1}}|={\displaystyle \frac{-1}{6}}$ $\to$ $y$ = ${\displaystyle \frac{1}{7}}$

Their positive difference is ${\displaystyle \frac{12}{35}}$ , the answer. 


Target #8 : Thanks to a 5th grader girl mathlete's solution: 

If you move parts around, you'll see the answer is exactly a semi-circle with a radius 2, so the answer is $2\pi$, the answer. 






or check out another solution from me: