For AIME or USAMO veterans, what off the top of your head is a neat trick? I'm learning about AM-GM and Cauchy Schwartz and I'd like to know what other secrets are out there. I don't know many except basic stars and bars or PIE. What have you seen recurring many times over the AIME? Is there a recognizable pattern? Thank you doctors.

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TopNewestI was just going through some of the past ARML problems and I remembered a couple tricks.

Geometry: Whenever you have something involving inscribed circles in a triangle, but it's not entirely inscribed in a triangle (say two sides), just complete the triangle and you will have the answer. For example, Individual Problems #2: by symmetry, if you reflect the triangle about AC, the circle will be completely inscribed in it. In this case, \(s=\frac{2(12+13)}2=25\) and \(K=2\left(\frac12(5)(12)\right)=60\), so the answer is immediately \(r=\frac{K}s=\frac{60}{25}=\frac{12}5\). Try USAMTS 2013 Round 2 #2 as a tough practice.

Algebra: Whenever you're dealing with consecutive numbers, pick the central number to work with so that there are some cancellations. For example, Individual Problems #3 has \(\frac{a}{r^2}\frac{a}r(a)(ar)(ar^2)=32\) so \(a\) immediately is \(2\) because the \(r\)s cancelled. Try Dutch Mathematical Olympiad 1998 #5 as a tough practice.

Number Theory: When in doubt, use \(N=p_1^{e_1}p_2^{e_2}\dots p_n^{a_n}\). For example, Individual Problems #6 immediately has the only possible prime factors of \(a,b,c,d\) being \(2,3,5\). Hence, \(a=2^{a_1}3^{a_2}5^{a_3}\) and similar. Now the system is a linear system of equations: \(abc=2^{a_1+b_1+c_1}3^{a_2+b_2+c_2}5^{a_3+b_3+c_3}=2^53^15^3\) is turned into \(a_1+a_2+a_3=5\) and similar.

Combinatorics: Hope that you were strong enough in the other three to get you to the individual finals!

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Give an example of \(11\) consecutive positive integers the sum of whose squares is a perfect square. – Cody Johnson · 3 years ago

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– Cody Johnson · 3 years ago

Another one from ARML (2012, individuals #5): The four zeros of the polynomial \(x^4+jx^2+kx+225\) are distinct real numbers in arithmetic progression. Compute the value of \(j\).Log in to reply

Geometry: Stewarts, Menelaus, Ceva, Ptolemy, Symmedians, Euler Line, Trig (Homothety, Spiral Similarity, Cartesian/ Complex/ Barycentric Coordinates (if you're fine with bashing), and projective geometry for USAMO)

Combinatorics: Stars and bars, PIE, and Coloring will get you far

Algebra: Factoring (duh), Complex Numbers, Cauchy, AM-GM, Substitution (If abc=1, try letting \(a=\frac{x}{y}, b=\frac{y}{z}\), and \( c=\frac{z}{x}\) or use trig) (Chinese Dumbass Notation, Schur, Muirhead, Sum of Squares (S.O.S Method), Fudging for USAMO)

Number Theory: Euler Theorem (totient), Euclidean Algorithm, Chinese Remainder Theorem, (Lifting the Exponent, Quadratic Residues for USAMO) – Srnth <3 · 3 years ago

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– Spock Weakhypercharge · 3 years ago

This is the best possible post I could ask for. There are many new theorems I've never seen before, I look forward to it.Log in to reply

– Narahari Bharadwaj · 2 years, 11 months ago

When you say symmedians, is there a specific theorem you have in mind or just properties of symmedians in general?Log in to reply

– Srnth <3 · 2 years, 11 months ago

Sorry I was on a vacation and didn't really get to respond to this till now. I meant the properties of symmedians in general. There is a lecture by Yufei Zhao (Lemmas in Euclidean Geometry) that you can find that contains some properties and applications of symmedians. It also contains many other geometric properties. You should definitely check it out if you want to learn more geo :DLog in to reply

– Narahari Bharadwaj · 2 years, 11 months ago

Thanks for the tip. Great topics, learned a lot of new stuff!Log in to reply

Try a lot of trig, and complex numbers probably. and master not making stupid mistakes. – David Lee · 3 years ago

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When you multiply/divide complex numbers, you add/subtract the arguments. Roots of complex numbers are spread out equally in a circle. Real numbers are complex numbers, too.

Ex: 2014 AIME II Question 10 http://www.artofproblemsolving.com/Wiki/index.php/2014

AIMEIIProblems/Problem10Because 2014 is a complex number, assume that z=2014, and the problem becomes a Roots of Unity problem(multiplied out by 2014).

Also, coordinate bashing is not advised(in USA(J)MO it may lose a few points for elegance), but it can be useful with the aid of some other non-coordinate theorems.

Ex: 2014 AIME II Question 14 http://www.artofproblemsolving.com/Wiki/index.php/2014

AIMEIIProblems/Problem14Start with knowledge of 30-60-90 triangles, 45-45-90 triangles, and 15-75-90 triangles to get coordinates for a triangle with a vertex at the origin, then use distance formula for a few things, take the triangle off the grid, use median and angle-bisector theorems, put back on the grid, then use distance formula to get the final result.

I hope this helped! – Tristan Shin · 3 years ago

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– Alexander Sludds · 3 years ago

You might want to fix your links.Log in to reply

– Tristan Shin · 3 years ago

Sorry, I'm not a good writer. How do you add links?Log in to reply

In geometry problems, if they ask for a very general condition, simplify the problem by assuming, say, right triangles or collinearities. These can tackle problems very quickly. Example: AIME I 2009/4 – Cody Johnson · 3 years ago

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