If (0,0),(a,11) and (b,37) are points of a equilateral triangle.....Find ab?
1057nopes dude.....i dun knw de correct answer but the options are 216,495,315 and 365...
36hmm.... question is a bit different one....
u can get the soln. here...
http://web2.uwindsor.ca/math/wlyee/Putnam/Fall07/UWUndergraduateMathContestSolutions.pdf
my approach was a bit different one but came up with -407
but there comes two values of a in the soln abve out of which -407 is unacceptable.. :P
1a much better way would be to use complex numbers,
let z1 = 0
z2 = a + 11i
z3 = b + 37i
we know that if z1 , z2 , z3 form equilateral triangle then ,
z12 + z22 + z32 = z1 z2 + z2 z3 + z3 z4
(by rotation formula...which actually follows from the fact tht angle between them is 60)
plug in the values and simplify and equate real and imaginary part to get,
a = 21 √3 and b = 5√3
.:. ab = 315
1its best to use the fromula for equilatreal trinagle which
is
x3=x1+x2+/-√3(y2-y1)/2 and similaarly for y cooridnate except instead of +- it is -+ it is very time saving and eliminates the need to do any computation
