Question

Inscribed Quadrilateral ?

Answer 1

The quadrilateral is a regular quadrilateral.

Explanation:

If the points lie on a circle with radius `1` then they can be represented as

`e^(i phi_1)+e^(i phi_2)+e^(i phi_3)+e^(i phi_4)=0`

or equivalently

`sin phi_1+sin phi_2+sin phi_3+sin phi_4 = 0`

and

`cos phi_1+cos phi_2+cos phi_3+cos phi_4 = 0`

(Here we used de Moivre's identity `e^(i phi) = cos phi + i sin phi` )

grouping and squaring both sides

`(sin phi_1+sin phi_2)^2=(-(sin phi_3+sin phi_4))^2`
`(cos phi_1+cos phi_2)^2=(-(cos phi_3+cos phi_4))^2`

and adding side by side we get at

`2+2(sin phi_1 sin phi_2+cos phi_1 cos phi_2) = 2 +2(sin phi_3 sin phi_4+ cos phi_3 cos phi_4)`

or

`cos(phi_2-phi_1) = cos(phi_4-phi_3)`

grouping now

`(sin phi_2+sin phi_3)^2=(-(sin phi_1+sin phi_4))^2`
`(cos phi_2+cos phi_3)^2=(-(cos phi_1+cos phi_4))^2`

we get at

`cos(phi_3-phi_2) = cos(phi_4-phi_1)`

So we can conclude that the quadrilateral is a regular quadrilateral.