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Olympiads math help >.<"

ok i will copy the question word for word

" Triangle ABC is inscribed in circle O of radius one. Angle A measures 30 degrees and angle B measures 70 degrees. M and N are points on the circle O such that M is closer to C than to B, and AM and AN trisects Angle A. The area of quadrilateral BNMC is "P sin20degrees - square root of Q", where P and Q are positive rational numbers in simplest form. Find 16(P+Q)."

some one pls help, i have no idea even after drawing it ...
thanks in advance

Comments

coolbho3000
After drawing it, I have no idea either

Angle A has been trisected.. so the individual portions (angles NAM, BAN (lol, ban), and CAM) of it are 10 degrees each. And angle C has to be 80 degrees. Other than that, no idea. Need to think more.
 
Joey90
I don't understand how angle A can be trisected by MN, surely that means that either M or N lies on top of A...

Also, what olympiad are you referring to?

EDIT, wait, AN and AM trisect it! ok I'll have a look :P

EDIT2, nothing seems immediately obvious, have you worked out the side lengths of the triangle? And if I know what you are talking about, these questions should take hours :|
 
L
seems to be practice for comc, canadian open math challenge i think, but should be wayy past that contests level
 
Joey90
Well, BC is length 1, so OBC is an equilateral triangle (side 1)

And that's all I can be bothered to work out this morning.

You can work out the other two sides, but without a calculator, they don't work out nicely... (if you care, they are just under 2 and just under 1.9)
 

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