Does 85 make AMC 12 the easier choice?

Not automatically. For the 2026–27 cycle, MAA lists AIME qualification thresholds of 100 on AMC 10 and 85 on AMC 12, subject to eligibility. AMC 12 also includes broader high-school content. A lower threshold and a broader paper can pull your decision in opposite directions. Source: current AIME thresholds.

The useful question is: on which contest can you produce a reliable qualifying performance with the time and knowledge available to you? A numerical comparison cannot tell you that. A student comfortable with trigonometry but weak at contest counting may need a different plan from a younger student with strong elementary problem-solving and no exposure to logarithms.

Nor should “qualifying” become the only reason to enter. Choose a contest that provides a worthwhile mathematical challenge and a preparation workload you can sustain. A sensible route remains useful even if one result falls short of a target.

Compare eligibility and content first

FeatureAMC 10AMC 12
Grade ceilingGrade 10Grade 12
Age ceiling in current policy17.5 years or younger on contest day19.5 years or younger on contest day
Standard format25 questions; 75 minutes25 questions; 75 minutes
Mathematical scopeElementary algebra, geometry, number theory and probability; excludes trigonometry and advanced algebra/geometry.Broader high-school mathematics, including trigonometry and advanced algebra/geometry; excludes calculus.
2027 AIME score threshold10085

The age wording above follows the official policy updated 30 October 2026, which says that it governs conflicting information. Both grade and age conditions apply; they are not alternatives. Other participation requirements also apply. Read the current AMC policy.

For the US/Canada programme, the policy includes school-attendance requirements. International participants should use the MAA international programme and their country or region’s International Group Leader. Being eligible by grade alone is not confirmation of a particular registration route.

Compare the score arithmetic without confusing it with difficulty

Both competitions use the same formula, 6C + 1.5B. Fourteen correct and eleven blank answers score 100.5. Eleven correct and fourteen blanks score 87. Those examples show how the thresholds can be met; they do not show how easy it is to obtain those correct answers.

A different distribution changes the result. Fourteen correct, ten blanks and one wrong scores 99. Eleven correct and fourteen blanks score 87, but eleven correct, twelve blanks and two wrong score 84. A plan expressed only as “get eleven questions right” hides the cost of additional incorrect attempts.

Use the AMC score calculator and AIME explanation to explore your own combination. Keep practice performance separate from official qualification, and do not add an A score to a B score.

Try six original skill checks

The first three questions illustrate useful shared problem-solving skills. The next three introduce extension material relevant to broader high-school preparation. They are deliberately short illustrations, not a representative miniature of either competition and not a validated selection test.

Attempt each before reading the solution. Record whether the obstacle is an unfamiliar fact, failure to recognise a method, or a calculation error. That distinction is more useful than turning six answers into a supposed predicted score.

Try it here

Shared skills and extension skills

Six original examples to inspect your mathematical preparation. There is no validated score threshold for choosing a contest from this set.

Question 01 · Algebraic identityShared-skill check A. Positive real numbers a and b satisfy a + b = 10 and ab = 17. Find (a − b)².
Need a hint?
Relate the square of the difference to the square of the sum.
Worked solution

(a − b)² = (a + b)² − 4ab = 100 − 68 = 32. The problem asks for a symmetric expression, so solving for a and b individually adds unnecessary work. Positive solutions do exist because the quadratic t² − 10t + 17 has two positive roots.

Question 02 · Counting and parityShared-skill check B. Two distinct integers are selected uniformly as an unordered pair from 1 through 9. What is the probability that their product is odd?
Need a hint?
A product is odd exactly when both selected integers are odd.
Worked solution

There are five odd integers, giving 5 × 4 / 2 = 10 favourable pairs. The total is 9 × 8 / 2 = 36 equally likely pairs. The probability is 10/36 = 5/18. Squaring 5/9 would treat the selections as independent draws with replacement.

Question 03 · Coordinate geometryShared-skill check C. A right triangle has perpendicular sides 6 cm and 8 cm. What is the distance from its right-angle vertex to the midpoint of the hypotenuse?
Need a hint?
Place the right-angle vertex at (0,0) and the other vertices on the axes.
Worked solution

Use vertices (0,0), (6,0) and (0,8). The hypotenuse midpoint is (3,4), whose distance from (0,0) is √(3² + 4²) = 5 cm. This also explains, in this case, the familiar result that the hypotenuse midpoint is equally distant from all three vertices.

Question 04 · Logarithms and domainExtension-skill check A. For real x > 2, log2x + log2(x − 2) = 3. Find x.
Need a hint?
Combine the logarithms, then check the original domain.
Worked solution

The logarithm rule gives log₂[x(x − 2)] = 3, so x(x − 2) = 8. Thus x² − 2x − 8 = (x − 4)(x + 2) = 0. The algebraic candidates are 4 and −2; the condition x > 2 leaves x = 4. The logarithm rule depends on positive arguments, so the rejected root cannot be restored.

Question 05 · Trigonometric identityExtension-skill check B. An acute angle θ satisfies sin θ + cos θ = 7/5. What is sin(2θ)?
Need a hint?
Square the given equation and use sin²θ + cos²θ = 1.
Worked solution

Squaring gives 1 + 2 sin θ cos θ = 49/25. Therefore 2 sin θ cos θ = 24/25. The double-angle identity gives sin(2θ) = 24/25. The given sum is between 1 and √2, so it is compatible with an acute angle; no inverse-trigonometric calculation is needed.

Question 06 · Complex-number powersExtension-skill check C. Let i² = −1 and z = 1 + i. What is the real part of z⁶?
Need a hint?
Square first, then reuse the simpler result.
Worked solution

z² = (1 + i)² = 1 + 2i + i² = 2i. Hence z⁶ = (z²)³ = (2i)³ = 8i³ = −8i. Its real part is 0. Expanding six factors is possible, but repeated use of a simple power is shorter and easier to check.

Read the pattern of strengths and gaps. Six illustrative questions cannot establish your likely AMC result.

If the extension questions contain unfamiliar notation, that tells you what to learn; it does not prove that AMC 12 is permanently unsuitable. If all six feel comfortable, test your reasoning with complete papers, where mixed topics, harder questions and time pressure change the task.

Choose a route using your present strengths

Try it here

Which description fits your situation?

Choose a profile for a suggested next task. These are teaching suggestions, not automatic eligibility decisions.

New to contest mathematics

Build the shared problem-solving skills

If you are eligible for AMC 10, begin with accessible AMC 10-style sets and learn how contest questions differ from routine exercises. Concentrate on factorisation, organised counting, diagrams and precise reading. A full paper can reveal your starting point, but follow it with targeted teaching rather than repeated low-information attempts.

Strong elementary contest skills

Compare the extra content deliberately

If you are eligible for both contests, inspect the AMC 12 topics you have already studied and those you have not. Try a suitable full paper and review whether missed questions reflect new content or solvable reasoning gaps. Choose from that evidence; the lower threshold alone is insufficient.

Comfortable with advanced school maths

Test contest fluency under time pressure

Knowledge of logarithms or trigonometry does not automatically produce an AMC 12 score. Use mixed problems without topic labels, practise switching methods and review early mistakes. If you are no longer grade-eligible for AMC 10, direct your contest preparation to an eligible route rather than comparing unavailable options.

Use official policy and your host to confirm eligibility and registration.

For a fairer comparison, use fresh full papers under the correct conditions and review them in the same way. Record the score, the topics you did not know and the number of errors you can now repair. One unusually easy or difficult paper should not carry the whole decision.

A small sequence of comparable attempts is more informative than one headline result, but do not exhaust every available paper just to make a choice. Once the evidence points clearly to a suitable route, spend the remaining time learning the mathematics that route requires.

Can you take both levels?

Eligible students may take an A competition and a B competition, including a permitted mix such as AMC 10A and AMC 12B. They cannot take AMC 10 and AMC 12 on the same day. Registration and availability must be arranged through the appropriate host; this is not an automatic option available at any time. Source: MAA registration policy.

If you use different levels on the two dates, make the preparation choice deliberately. Learning unfamiliar advanced material in a rush can consume time you might use to make existing skills more reliable. Conversely, a student who already knows the additional topics may find the second level a worthwhile challenge.

Whichever route you choose, keep an error log based on methods rather than paper labels. Factorisation, organised counting, precise diagrams and clear casework remain valuable throughout contest mathematics. The AMC portal offers level-specific practice so you can work at the appropriate depth.

Before you go

Common questions

Is AMC 12 always harder than AMC 10?
AMC 12 covers broader content, but a particular question or a student’s personal experience can differ. The useful decision depends on eligibility, mathematical preparation and performance on suitable full papers.
Does AMC 12 include calculus?
No. MAA excludes calculus from AMC 12. It does include broader high-school topics such as trigonometry and advanced algebra.
Can a grade 9 student choose AMC 12?
A student below grade 12 can meet the grade condition, but must also satisfy the age and other participation rules. Check the current policy and host; grade alone does not settle eligibility.
Can I take AMC 10A and AMC 12B?
The current policy permits an eligible student to take that combination. You cannot take AMC 10 and AMC 12 on the same day. Registration and local availability still need to be arranged.
Should I choose from my score on the six questions here?
No. They illustrate selected skills, not the full topic mix or difficulty of either contest. Use their explanations to identify gaps, then compare appropriate full-paper evidence.

Sources & further reading

Official rules and programme details can change. Use the linked source for your exact course, competition or examination session.

  1. MAA current AMC rules and policiesAuthoritative age and grade conditions, format, content, scoring and A/B combinations.
  2. MAA AIME information and FAQs2026–27 qualification thresholds.
  3. MAA international programmeInternational registration route.

Prepared by Thriving Scholars. Original teaching examples and illustrations in this guide are practice material, not official examination questions or predictions. Source checks: 10 November 2026. Published: 12 November 2026. Exam organisations and resource providers named here do not endorse this article.