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Draw the triangle, and then draw it again rotated so that B is in the same corner that C was. Now since ICA < IBA and ICB < IBC, ICA + ICB < IBA + IBC. Since this is true, the angle at C is less than the angle at B. I forget the name of the proof, but if you have such a triangle, the side with a greater angle has a smaller diagonal (BB' and CC'). So Since angle C < B, CC' > BB'.
the problem were unsolve in along time ago !!!!!!!!!
next week , i will give the proof to my teacher ,please , !!!!!!!
very thanks to ganesh !!!!
Let white=0, black=1;
Maybe ganesh could post these one at a time under "Ganesh's Puzzles".
never fear!!! Flowers4Carlos is here to save the day!!!!
thanks . I will be waiting for you
Yes, that's right ... I said it wasn't really a proof.
your proff is wrong . please help me in problem B,C
No, not yet. Proofs are very difficult.