# 2010 Euclid Contest Discussion

As you might know, the 2010 Euclid Contest was officially taken yesterday. So lets discuss!

I thought it wasnt too bad. #10 was hard though (the triangle one). Here was the question:

For each positive integer n, let T(n) be the number of triangles with integer side lengths, positive area, and perimeter n. For example, T(6) = 1 since only such triangle with a perimeter of 6 has side lengths 2, 2 and 2.
(b) If m is a positive integer with m >=(greater than or equal to) 3, prove that T(2m) = T(2m-3).
(c) Determine the smallest positive integer n such that T(n) > 2010.

Also, #9 was hard:

(b) In triangle ABC, BC = a, AC = b, AB = c, and a < .5(b+c).
Prove that angle BAC < .5 (angle ABC + angle ACB).

Can you please help me with those problems?

Also, if anyone wants to share how they solved #7 and 8, that would be appreciated :)

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## Answers and Replies

2010 Euclid Contest , very easy. just 10(b)(c) i cannot do it.

9(b):
use this formula:A/SINA=B/SINB=C/SINC

7A:too easy .cannot remember the answer
7B:i just remenbered there are 2 points.
8A:cannot remember
8b:110
9A(i):too easy
9A(ii):120

oh ****. 9A(ii)should be 4. i misunderstand the queston.
but anyway, i can get at least 80:)

Does the fact that $${\sin}BAC<\frac{1}{2}({\sin}ABC+{\sin}BCA)$$ imply the desired result in problem 9b?

have you not got the answer 9(B)? very very easy!!!!

here is my answer:
A/SINA=B/SINB=C/SINC
so 2SINA<SINB+SINC

because this is permanent.
we know SINB+SINC is bigger or equal 2(root SINB*SINC) only when SINB=SINC , it can be equal.
so 2SINA<2(root SINB*SINC)
SINCE SINB=SINC
2SINA<2SINB=2SINC
SINA<SINB=SINC
SINCE A+B+C=180
SO WHEN SINA=SINB=SINC,A=B=C=60, THEN A MUST LESS THAN 60
SO 0<A<60
120<B+C<180

A<0.5(B+C)

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Doesn't that proof lose generality when you assume that $${\sin}B={\sin}C$$?

oh yes
i should rethink it

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