AMC 10 AMC 12 Guide

What is AMC 10 or 12? How to Prepare for them?

American Mathematics Contest 10/12 (AMC 10/12) is the 2nd stage of Math Olympiad Contest in the US after AMC 8. The contest is in multiple choice format and aims to develop problem solving abilities. The difficulty of the problems dynamically varies and are based on important mathematical principles. These contests have a lasting educational value.

Eligibility for AMC 10/12:

Students in grade 10 or below and under 17.5 years of age on the day of the contest can take the AMC 10. Students in grade 12 or below and under 19.5 years of age on the day of the contest can take the AMC 12.

AMC 10/12 Exam Syllabus:

The syllabus for AMC 10 include:

  • Elementary Algebra
  • Basic Geometry Knowledge like Pythagorean Theorem, area and volume formulas
  • Elementary Number Theory
  • Elementary Probability

The syllabus for AMC 10 excludes:

  • Trigonometry
  • Advanced Algebra
  • Advanced Geometry

The syllabus of AMC 12 includes high school mathematics curriculum and including all the topics mentioned above. Only calculus is excluded in the advanced topics.

AMC 10 and AMC 12 Test Format:

As mentioned earlier, the format of the tests are multiple choice type. It is a 75 minute test with 25 questions. The exams are held on two dates under the names AMC 10A and 10B for AMC 10 and AMC 12A and 12B for AMC 12.

Upcoming AMC 10 and AMC dates:

For AMC 10/12A:

Early bird Registration: through Dec. 2, 2020

Regular Registration: Dec. 3 – 19, 2020

Late Registration: Dec. 20, 2020 – Jan. 9, 2021

Exam Days: Thursday, February 4, 2021
For AMC 10/12B:

Early bird Registration: through Dec. 8, 2020

Regular Registration: Dec. 9 – Jan 9, 2020

Late Registration: Jan. 10, 2020 – 13, 2021

Exam Days: Wednesday, February 10, 2021

Books for Preparation:

Cheenta believes “It is important to completely solve at least one book per topic before going into the harder ones. You do not need to solve hard books for all topics.”

So, we prefer, one should keep “Challenges and Thrills of Pre College Mathematics” as a Common Minimum Requirement for your preparation.

How to Prepare for AMC 10 or AMC 12?

  • Utilize the resources: Here are the resources required to help you perform better in AMC 10 or AMC 12 Preparation:
  • Solve Past Papers and similar questions from similar contests like SMO (Singapore), UKMT (Junior, Senior), University of Waterloo BMC, Math Kangaroo which may serve as practice problems for AMC 10/12.
  • Take tests and Track your Tests: Tests are essential for knowing your weaknesses and strengths. Attempting quizzes and mock tests helps the students to evaluate themselves.
  • Get a coach: A coach can see and analyze your performance and guide you in the best way possible. He/She can help you strengthen your weaknesses by letting you work in the right direction.
AMC 10 AMC 12 Guide

Upcoming USA Math Competition

In this post, we are going to discuss about the upcoming USA Math Contests that would be useful for you to consider if you are looking to participate in Mathematics Competitions.

1. American Mathematics Competitions (AMC)

American Mathematics Competition (AMC) are the mathematics contests for United States students that leads them to International Mathematics Olympiad (IMO).


These competitions are divided into 3 levels:

  1. AMC 8, for classes 8 or below
  2. AMC 10, for classes 10 or below
  3. AMC 12, for classes 12 or below

Higher scores in AMC 10 or 12 lead the stuedents to American Invitational Mathematics Examination (AIME) or USAMO/USAJMO.

Important Dates:

For AMC 8:

Early bird Registration: through October 1, 2020

Regular Registration: 2nd – 29th October 2020

Late Registration: 30th October – 3rd November 2020

Exam Days: Tuesday, November 10, 2020 through Monday, November 16, 2020

For AMC 10A/12A:

Early bird Registration: through Dec. 2, 2020

Regular Registration: Dec. 3 – 19, 2020

Late Registration: Dec. 20, 2020 – Jan. 9, 2021

Exam Days: Thursday, February 4, 2021
For AMC 10B/12B:

Early bird Registration: through Dec. 8, 2020

Regular Registration: Dec. 9 – Jan 9, 2020

Late Registration: Jan. 10, 2020 – 13, 2021

Exam Days: Wednesday, February 10, 2021


AIME I: Wednesday, March 10, 2021

AIME II: Thursday, March 18, 2021

For more info:

2. Perennial Math

Perennial Math has 3 types of competitions: Online Math contest, Onsite Tournament and Virtual Tournaments.


Online and Onsite Tournaments are for students of Grade 3-8 and Virtual Tournaments are for students of grade 3 -12.

Important dates:

The Online Math Contest usually occurs in 2 seasons. 4 tests are held in each season. Season 1 is November through February and Season 2 is January through April.

The On Site Tournament is online for this year due to COVID 19. Here are the Ucoming Events:

  1. Birmingham AL Tournament – 24th October 2020
  2. Ohio Tournament – 14th October 2020
  3. Douglasville, GA Tournament – 15th December 2020
  4. Memphis TN Tournament – 23rd January 2020

Virtual Tournaments are Video Conferencing Tournaments in a WebEx Chat Room with a Gameshow Host. The Upcoming Virtual Tournaments are listed below:

  1. Pumpkin pi Fest (for individuals) – 17th October 2020
  2. Mathoween (for individuals) – 31st October 2020
  3. Mathgiving (for individuals) – 21st November 2020
  4. Winterchill ( for individuals) – 12th December 2020
  5. Winterdash (Grade 3-8) Team Event (for teams) – 15th – 17th December 2020
  6. Top 10 Championships (for individuals) – 19th December 2020

For more info:

3. SCUDEM V 2020

This fall due to COVID 19, SCUDEM V 2020 will be held in a virtual approach.


SCUDEM V 2020 is an opportunity for students, in high school or undergraduate school, to engage in a differential equations modeling experience. One can register with a team of 3 or individually. A team will be formed for individuals. They can apply to participate as a student, coach or judge.

Important Dates:

  1. For Students: Registration closes 23rd October 2020
  2. For Coaches: Registration closes 15th October 2020
  3. For Judges: Registration closes 13th October 2020

For more info:

4. Rocket City Math League

The Rocket City Math League is a free, international math contest granted from Mu Alpha Theta.


It  is open to all middle, high school, and two-year college students enrolled in pre-algebra through precalculus and above math courses.

Important Dates:

  1. Interschool Test: Distributed Nov 10, 2020
  2. Round One: January 12, 2021
  3. Round Two: February 16, 2021
  4. Round Three: March 9, 2021

For more info:

5. Purple Comet! Math Meet

The Purple Comet! Math Meet is a free, annual, international, online, team, mathematics competition, where teams from 1 – 6 students compete by submitting solutions to a list of mathematics problems.


It is designed for middle and high school students.

Important Dates:

Next Contest: 6 – 15th April 2021

For more info:

6. National Internet Math Olympiad

National Internet Math Olympiad is an online Math Contest partnered with Online Math Open (OMO). Teams of upto 4 participate in the contest who are given more than a week to answer 30 short answer questions, ranging from the early AMC to late IMO.


High school students are eligible for this.

Important Dates:

Fall Contests occurs in October to November and Spring contests in March and April.

For more info:

7. M3 Challenge

MathWorks Math Modeling (M3) Challenge helps students to to work as a team to tackle a real-world problem under time and resource constraints, akin to those faced by professional mathematicians working in industry.


Students of age 16 – 19 in US participate in this competition.

Important Dates:

Registration opens in November for M3 Challenge 2021.

For more info:

8. MathCON

MathCON is an innovative mathematics platform, including a competition series and educational resources and opportunities for students and educators, developed to enhance mathematics education.


MathCON is open to all students in grades 4-12, including public, charter, private and home schooled students.

Important Dates:

Registration going on for 2021.

For more info:

AMC 12 USA Math Olympiad

Area of Trapezoid | AMC 10A, 2018 | Problem 9

Try this beautiful problem from AMC 10A, 2018 based on area of trapezoid.

Area of Trapezoid | AMC 10A

All of the triangles in the diagram below are similar to isosceles triangle ABC , in which AB = AC . Each of the 7 smallest triangles has area 1 and \(\triangle {ABC}\) has area 40.  What is the area of trapezoid DBCE ?

  • 24
  • 25
  • 26
  • 20

Key Concepts

2D – Geometry

Isosceles Triangle

Area of Triangle

Check the Answer

Answer: 24

American Mathematics Competition, 2018 Problem 9

Challenges and Thrills – Pre – College Mathematics

Try with Hints

We can try this problem using this hint :

Let the base length of the small triangle be  x :

Then, there is a triangle ADE encompassing the 7 small triangles and sharing the top angle with a base length of 4x.

Try the rest of the sum ………………………..

Continuing from the last hint:

Because the area is proportional to the square of the side, let the base BC be \(\sqrt {40}x\). Then triangle ADE  has an area of 16. So the area is 40 – 16 = 24(Answer).

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AMC 12 USA Math Olympiad

Triangle Inequality Problem – AMC 12B, 2014 – Problem 13

Try this beautiful problem from AMC 12 based on Triangle inequality problem.

Problem – Triangle Inequality

Real numbers a and b are chosen with 1 < a < b such that no triangle with positive area has side lengths 1,a and b or \(\frac {1}{b},\frac {1}{a}\) and 1. What is the smallest possible value of b?

  • \(\frac {3+\sqrt 3}{2}\)
  • \(\frac {5}{2}\)
  • \(\frac {3+\sqrt 5}{2}\)
  • \(\frac {3+\sqrt 6}{2}\)

Key Concepts

Triangle Inequality



Check the Answer

Answer: \(\frac {3+\sqrt 5}{2}\)

American Mathematics Competition – 12B ,2014, Problem Number – 13

Secrets in Inequalities.

Try with Hints

Here is the first hint where you can start this sum:

It is given \( 1 >\frac {1}{a} > \frac {1}{b } \) . Use Triangle Inequality here :



\(\frac {1}{a} + \frac {1}{b} >1 \)

If we want to find the lowest possible value of b , we create we try to create two degenerate triangles where the sum of the smallest two sides equals the largest side. Thus we get : a = b – 1

Now try to do the rest of the sum.………………….

We already know \(\frac {1}{a} + \frac {1}{b} = 1\)

After substituting we will get :

\(\frac {1}{b – 1} + \frac {1}{b} = \frac {b+b-1}{b(b-1)} = 1 \)

\(\frac {2b – 1}{b(b-1)} = 1 \)

Now do the rest of the calculation ………………………..

Here is the rest of steps to check your problem :

\( 2b – 1 = b^2 – b \)

Now Solving for b using the quadratic equation, we get

\(b^2 – 3b + 1 = 0 \)

\(b = \frac {3 + \sqrt 5}{2} \) (Answer)

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AMC 10 AMC 12 AMC 8 India Math Olympiad Math Olympiad PRMO Regional Mathematics Olympiads USA Math Olympiad

Can we prove that the length of any side of a triangle is not more than half of its perimeter?

Can we Prove that ……..

The length of any side of a triangle is not more than half of its perimeter

Key Concepts

Triangle Inequality



Check the Answer

Answer: Yes we can definitely prove that by Triangle Inequality

Mathematical Circles – Chapter 6 – Inequalities Problem 3

Mathematical Circles by Dmitri Fomin , Sergey Genkin , Llia Itenberg

Try with Hints

We can start this sum by using this picture below

The length of the three sides of this triangle are a,b and c. So if we apply triangle inequality which implies that the length of one side of a triangle is less than the sum of the lengths of the two sides of that triangle. In reference to the theorem

b + c > a

Proof based on triangle

So can you try to do the rest of the sum ????????

According to the question we have to find the perimeter at first

Perimeter is the sum of the length of all sides of the triangle = a + b + c

And the length of each side is a or b or c.

We have to prove : a + b + c > length of any one side

This can be one of the most important hint for this problem. Try to do the rest of the sum …………………………..

Here is the rest of the sum :

As stated above if we use triangle inequality :

b + c > a

Lets add a to both the sides

a + b + c > a + a

a + b + c > 2 a

The left hand side of the above inequality is the perimeter of this triangle.

perimeter > 2 a

So , \(\frac {perimeter}{2} > a \)

\(\frac {perimeter}{2} \) = semi perimeter

Hence this is proved that the length of one side of a triangle is less than half of its perimeter.

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