Question

Given that there exists a triangle whose sides are a,b,c. Then prove that there exists a triangle whose `sqrta,sqrtb,sqrtc`?

Answer 1

Given that there exists a triangle whose sides are a,b,c.

So we have following 3 inequalities satisfied.

• `a+b>c`

• `b+c>a`

• `c+a>b`

Considering the first one

`a+b>c`

`=>(sqrta+sqrtb)^2-2sqrt(ab)>c`

`=>(sqrta+sqrtb)^2>c +2sqrt(ab)`

`=>sqrta+sqrtb>sqrtc`

Similarly from 2nd inequality we get

`sqrtb+sqrtc>sqrta`

And from 3rd inequality we get

`sqrtc+sqrta>sqrtb`

So we can say if there exists a triangle having sides `a,bandc` then there exists a triangle having sides `sqrta,sqrtbandsqrtc`

Answer 2

My interest in the problem:

Explanation:

Choosing `a = 2, b = 4 and c = 3`, and using scale and compass

only, four conjoined `triangles DEF`,

with sides `sqrta = sqrt2, sqrt b = 2 and sqrt c = sqrt3` are constructed.

The graph shows one pair over the base `EF = sqrt2`,

with vertices.

`D ( 1/sqrt 8, +-sqrt (23/8) ), E ( 0, 0 ) and F ( sqrt2, 0 )`

There are three such pairs, and all have the central

common `triangle DEF`

graph{(x^2+y^2-3)((x-sqrt2)^2+y^2-4)(x^2+y^2-0.01)((x-sqrt2)^2+y^2-0.01)((x-sqrt(1/8))^2+(y-sqrt((23)/8))^2-0.01)((x-sqrt(1/8))^2+(y+sqrt((23)/8))^2-0.01)=0[-4 4 -2 2]}