How to demonstrate this?We have equilateral trinagle #DeltaOAB#:#O_(((0,0))),A_(((m,n)));m,ninNN;m,n!=0# and#B_(((x,y)));x,yin(0,oo)#.Demonstrate that #B# can not have both coordinates natural numbers.

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Answer 1 · 2017-04-15T10:02:09

See below.

Let #R(pi/3)=((Cos(pi/3), -Sin(pi/3)),(Sin(pi/3), Cos(pi/3)))= ((1/2,-sqrt3/2),(sqrt3/2,1/2))#

be the rotation matrix which rotates #pi/3# radians.

Having a vertice at the origin, If #(m,n)# is an equilateral triangle vertice then the other vertice is located at

#((x),(y)) = R(pmpi/3)cdot((m),(n))#

Taking the counterclockwise rotation to follow we have

#((x),(y)) = R(pi/3)cdot((m),(n))=((m/2 - (sqrt[3] n)/2), ((sqrt[3] m)/2 + n/2) )#

so clearly #((x),(y))# can not be represented as an integer couple.

Note that #nm ne 0#

Answer 2 · 2017-04-23T13:24:04

For Proof, refer to the Explanation.

We use Reductio Ad Absurdum to prove the Result.

For this, suppose, to the contrary, that, there exists an equilateral

#DeltaOAB,# having vertices #O(0,0), A(m,n), B(x,y),# such that,

#m,n,x,y in NN, and, m^2+n^2!=0.#

Knowing that, Area #A_(Delta)#of an equilateral #DeltaOAB# is given by,

#A_(Delta)=sqrt3/4*OA^2=sqrt3/4(m^2+n^2)..........(ast).#

On the other hand, from Co-ordinate Geometry,

#A_(Delta)=1/2|D|,# where,

#D=|(0,0,1),(m,n,1),(x,y,1)|=|my-nx|,#

#:. A_(Delta)=1/2|my-nx|............(ast').#

#(ast), &, (ast') rArr 2|my-nx|/(m^2+n^2)=sqrt3,# which is a

contradiction, because, with, #m,n,x,y in NN, &, m^2+n^2ne0,#

the L.H.S. is a Natural No., whrereas, the R.H.S., an Irrational No.

Therefore, our supposition that all the co-ordinates of an equilateral

triangle are Natural Nos. is False.

Henc, the Proof.