I'm sure I'm not the only one who has bought a solution before (or even gotten a question right and went to the solution page to see how other people did it) and was amazed by the supposedly obscure theorems that people cite for their answers (most recently, this happened with Lucas' theorem).

So I ask, what are some more obscure theorems and such that come in use? What kinds of problems would they be applied to? (try to stay away from mentioning theorems that most people already know, e.g. stars and bars, vieta's formulas, or what have you)

One theorem I've found personally that comes in handy is the Sophie-Germain identity. It's not too obscure, but at the same time, not that many people know it. The identity is:

\(a^4 + 4b^4 = ((a+b)^2 + b^2)((a-b)^2 + b^2) = (a^2 + 2ab + 2b^2)(a^2 - 2ab + 2b^2)\)

And it can be applied to problems such as these (credit: AOPS)

For what integer values of \(n\) is \(n^4 + 4^n\) composite?

Find the largest prime divisor of \(25^2 + 72^2\).

And there was a problem from \(1987 AIME\) and a similar one was featured on brilliant on the past. The problem was to compute \(\frac{10^4 + 324)(22^4 + 324)(34^4 + 324)(46^4 + 324)(58^4+324)}{4^4 + 324)(16^4 + 324)(28^4 + 324)(40^4 + 324)(52^4 + 324)}\)

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TopNewestOne of the things that I have learned from solving problems on Brilliant is that you don't have to know what a theorem is called to use it. I have used the Sophie-Germain identity without knowing what it's called for a really long time. I just used it as a factorization tool that helped me solve a lot of problems.

And it is not true that most people know things like 'stars & bars' and 'Vieta's formulas'. I, for one, had to learn what these are from solutions and posts on Brilliant. Our textbooks cover 'Vieta's formulas' but our books never mention what these formulas are called. So naturally I had never heard of Vieta's formulas before even though I knew what they were. Thus on Brilliant, I started to learn the names of things I already knew in addition to learning new things.

That being said, it is actually useful to know names of certain formulas and theorems. It helps you understand what people are saying and it reduces a lot of work. If I say, "because \(ABCD\) is a tangential quadrilateral, from the Pitot theorem [another example of something I'd known without knowing what it's called], we can say that \(AB+CD=AD+BC\)...", it's hard for someone to understand what I'm trying to stay if they don't know what the Pitot theorem is. So 'knowledge is power!'

And the Lucas' theorem thing happened to me too.

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Can you give me link to the problem that used Lucas theorem ?

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Not yet; it's still a live problem, and so for people who do have this problem (which may be you) they'd immediately know the solution.

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Well, I haven't really applied some of these to brilliant problems, but I suppose it will be good to keep in mind the following non-exhausted list of convoluted theorems:

Hölder's Inequality (If you aren't familiar with inequalities)

Jensen's Inequality (same comments as above)

Schur's Inequality (when used along with Hölder, can be quite sharp and hence powerful)

Desargue's Theorem (extremely useful in projective geometry)

Monge D'Alembert's Theorem (same comments as above)

Hensel's Lemma (Has quite some usage in algebra-like number theory problems)

The last few aren't really for brilliant level problems - they are geared towards usage in IMO-standard problems.

EDIT: Well, I think Cauchy isn't that obscure. (Maybe my judgement is very biased because I don't think Hölder of Jensen is obscure either)Log in to reply

And to add to the list of inequalities, of course there is Cauchy-Schwarz.

Though you're right, these don't really tend to be useful in brilliant problems. The only applications of inequalities I've really seen is AM-GM and then using that to find equality and there's no geometry section currently.

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Another theorem I thought of was Stewart's Theorem. Useful for when you have arbitrary cevians in a triangle.

Given a triangle ABC, draw a cevian from A to BC. Call the intersection D. Then, let \(AB = b, AC = c, BD = n, BC = m, AD = d\), and the following equality holds:

\(amn + ad^2 = b^2m + c^2n\)

And if you think you'll have a problem memorizing this, a handy mnemonic is \(man + dad = bmb + cnc\) -- "a man and a dad puts a bomb in the sink"

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You should get the book The Secrets of Triangles. Hundreds of awesome theorems and crazy properties.

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Where it is available?Please tell me !!!

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http://www.flipkart.com/secrets-triangles-mathematical-journey/p/itmd5agqdfycfwgq

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hey mursalin me too from Notre Dame College. Is there anyway we can contact through phone??I guess you could help me get through some problems as you are much experienced and better than me :')

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Trust me, I'm

notexperienced. I've always been in love with math but my interest in problem solving began no more than 6-7 months ago. In the beginning, I used to stare at problems doing absolutely nothing! And then things started to work out. I began solving problems!I'm still a novice problem solver and I'm still learning.

I don't want to post my phone number here. You can send me a private message in the BdMO online forum. My username is Mursalin.

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Another one that hasn't seemed to be mentioned yet is Pascal's Theorem, which states that for a cyclic hexagon \(ABCDEF\), the points determined by the intersections \(AB\cap DE\), \(BC\cap EF\), and \(CD\cap FA\) are collinear. Note that this also applies to degenerate hexagons as well; when there are double letters, as in degenerate hexagon \(AABCDE\), the line \(AA\) is meant to be the tangent to the circumcircle at point \(A\). An example of a problem that uses the degenerate form would be the following:

"Let \(ABCD\) be a convex cyclic quadrilateral with circumcircle \(\gamma\), and let \(P\) denote the intersection of its diagonals. Let \(X\) be the intersection of the tangents to \(\gamma\) from \(A\) and \(B\), and let \(Y\) be the intersection of the tangents to \(\gamma\) from points \(C\) and \(D\). Prove that line \(XY\) passes through \(P\)."

Try it!

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Jensen's inequality is great.

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