USAMO2019

You can also do this without making a common demoninator. If the numbers were increased to i.e something like 15234/15235 and 15676/15677, finding the common denominator would be reduntant and be extremely time confusing. Instead, we can notice that both the fractions are 1/n away from 1. 1/5 is clearly greater than 1/7, as spliting the same thing into 5 things compared to 7 things, is clearly greater. Thus, we are subtracting 1/5 from 4/5, SOMETHING THAT is greater than subtracting 1/7, so therefore 6/7 is greater .

n*A+n*B suffices because angle B is represented as the interior angle in each vertex

Sol for ASHME Q20 1992:

Since it is an n-pointed star, we have a 2n-polygon. As we know, the sum of angles in a polygon formula is (n-2)(180).

Thus, the sum of angles in our n-pointed star is 180(2n-2). If the angle at each vertex A(n) is some number A(i) (Note that the () represent subscripts) and the angle at each vertex B(n) equals B(i), we have the following equation:

n*A+n*B=180(2n-2). Note that the acute angle at B(i) is 360-B, so therefore A(i) is simply 360-B-10=350-B Hence, A+B=350. Therefore we can set up the following equation:

350*n=180(2n-2)

After we solve for n, we get that n=36!

Hello!

I have qualified for USJMO twice, and USAMO once. Currently I am in 12th grade. I would say that AoPS certainly helps, but I never really used Alcumus, as the questions there are quite easy. Currently, there is around 1 more month left 'till the amc10a, and if you want to make AIME, you should start spamming tests as of now. I would say 5 amc10 tests MINIMUM a day should prepare you fine. However, you should make sure you understand the concepts to solving the problems. Furthermore, I would say you should get some AoPS books, the introduction to alegebra being the most important. If you want to make USJMO and above, I would say you'd need the precalculus book, which has much information in it.

Now, often times there is time pressure, along with the difficulty of the problems. I would suggest "ftw" on AoPS to help increase your speed in solving and skimming the problems. Note that amc10 questions 1-10 should take each maximum of 2 mins per question. Questions 10-20 take some time, and usually for me, require around 4-5 min to figure out 17-20. I suggest taking some of the newest tests, to get an idea of the difficulty of the current tests. If you can score a 120 and above consistently on amc10 tests, i would say you are fine to make AIME.

Some important concepts for the AMC tests include but are not limited to:

AM-GM

Cauchy-Schwartz Inequality

Triangluar Inequality

Pythagorean Theorem

Shoelace formula

Menelaus Theorem

Stewarts Theorem

Angle-Bisector Theorem

Basic trig (Sum of sins, cosines, law of sines, etc)

Veita's formula (extended)

Geometric and Arithemetic series

Difference and Sum of squares

Binomial Theorem

Telescoping

Minimum/Maximum value

Basic Log

Abs Value

Floor and Ceiling

Recursion

Advanced Factorial Concepts

Combination & Purmutation

Hockey Stick Identity

Mass point geometry

Stars and Bars

Descartes Circle theorem

Roots of Unity

That is all of I can think of now. These concepts are extremely useful in doing well in the AMC tests. Good luck! :_)

Hello!

Notice our equations 2 and 3 are equivalent.

xy-y=yz-x

y(z+1)=x(z+1)

x=y

Now we can rewrite our 2nd equation as x(z-1)=0

First we note that x cannot =0, as x,y, and z are all positive. Thus, z=1.

We can plug this in, giving us x^2-1=15. Hence, x=4 (x cannot be -4)

Therefore, x=4, y=4, and z=1. Thus, our answer is 4*4*1= 16.

Hello! I like the letters of the problem :-).

Notice that "USAMO" is one of the last numbers in the five letter words. If we let A=1, M=2, O=3, S=4, and U=5, we can write the numbers in term of "12345." Clearly, USAMO is only of the last letters in the sequence, as it starts with U. Note that "USAMO" is equal to 54123. Now, we can simply count backwords! The last number is 54321, followed by 54312, 54231, 54213,54132, and 54123. Thus, we can clearly see that "USAMO" is 5 away from the last number. There are 5! ways to order the numbers. The 5th to the last is 120-5= 115.

Uhhh, expanding we get 4x^4+12x^2+4. Thus, the coefficients are 4, 12, and 4. 4^2+12^2+4^2=176. (Tell me if I am misinterpreting the question. I am unsure if this is what you are looking for.)

I am done with calculus :-)

Hello!

First note that the first equation is clearly a circle. We can complete the square on the first equation to get (x+2)^2+(y+3)^2=25 This is a circle with a center (-2,-3) and radius 5. The second equation can be expressed as y=4x/5 +13/5. After doing a quick graph, we can clearly see that there is 2 intersections.

Hello!

We can express the number of cheesecakes people have in terms of x. Let the number of cheesecakes Juan has be x.

Thus, Clarence has 3x-5 cheesecakes. Darlene has 1/2(3x-5)+2.

Therefore, we can simply combine these expressions. x+3x-5+3x/2-5/2+2=11x/2 -11/2

Hello!

We can look for a pattern to count these dots. Note that the 1st grid is 1x2, the second one is 2x3, the third one is 3x4, and so on. This means that the 11th grid is 11*12. Now, we simply have to add these numbers. 2+6+12+20+30+42+56+72+90+110+132. After adding, we get 572, as our answer.