Bay Area Math Olympiad

by pythag011, Feb 25, 2009, 4:34 PM

I didn't take, but I saw 5,6,7

Hmm...

Problem 5:

A set is called magic if, for every 2 distinct i,j in the set, i+j/gcd(i,j) is in the set. Find all magic sets.

Solution:

First, we prove the set can only have two elements. Assume it has more than 2 elements. Consider the two smallest elements. Let the smallest be a and the second smallest be b. Two cases:

They are relatively prime. Then a+b is in the set, and a+b and a are also relatively prime, so 2a+b is in the set, and ka+b is in the set for all positive k, so the set is infinite, which is impossible.

They are not relatively prime. Then, gcd(a,b) is at least 2, so $ \frac{a+b}{gcd(a,b)} \le \frac{a+b}{2} <  b,$ and its in the set, so it must equal a. This immediately implies that b = (gcd(a,b)-1)a, so b is a multiple of a, and gcd(a,b) = a, which implies that b = (a-1)a.

Similarly, we get an expression for c, and it gives a contradiction, yay.

Problem 6 is too long to post here... but you can do some stuff with triangular coordinates mod 3...

Problem 7: cyclic sum of $ \frac{cosA}{cos(B-C)}$

Well, you cosA=cos(180-B-C)=-cos(B+C), and then you get some expressions, divide numerator and denominator by cosBcosC, and then you get something in terms of tangents, then multiply numerator and denominator by tanA and use tanAtanBtanC=tanA+tanB+tanC, and then it reduces to the inequality cyclic sum x+y/x+y+2z is at least 3/2, which is easy normalization, adding 3 to both sides, and cauchy-schwarz lemma, qed.

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Are u "copying" my solution, except the last step?? :P . I wish I'm as good at NT+Geometry as I'm "good at" inequality!

by ghjk, Feb 26, 2009, 1:10 AM

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I didn't copy it. I came up with it independently.

by pythag011, Feb 26, 2009, 1:41 AM

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WAIT #5 LOOKS LIKE APMO LOLOL

by not_trig, Mar 1, 2009, 5:31 PM

Click here if you did not make white MOP (Fake copy of pythag's blog)

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