Edited by Richard Walker, Friday, 6 Sept 2024, 14:41
Someone asked on Quora how, given two concentric circles, we can construct a square having two vertices on one circle ands two on the other.
I think this was answered by the notable Alon Amit, but I deliberately haven't looked at his solution yet. Here is my solution.
Choose a point A on the outer circle c and rotate c through 90 degrees about A, to give circe c', shown fotted.
Because c' is the rotated image of c, any point P' on c' is the image on a point P on c, and any point P on c is the inverse image of a point P' on c'. Angle PAP' is a right angle, from the construction, and length PA = length P'A by the same reasoning. This defines the square PAP'Q with two vertices on c, shown shown shaded.
All we now need to do is choose P' to be the point where the dotted circle c' intersects the inner circle and we obtain the required square, since Q must lie on the inner circle by symmetry.
Of course if the inner circle is too small this doesn't work, because the dotted circle and the small one don't intersect. The limiting case is when the inner radius is ~ 0.414 of the outer one.
An Interesting Geometrical Construction
Someone asked on Quora how, given two concentric circles, we can construct a square having two vertices on one circle ands two on the other.
I think this was answered by the notable Alon Amit, but I deliberately haven't looked at his solution yet. Here is my solution.
Choose a point A on the outer circle c and rotate c through 90 degrees about A, to give circe c', shown fotted.
Because c' is the rotated image of c, any point P' on c' is the image on a point P on c, and any point P on c is the inverse image of a point P' on c'. Angle PAP' is a right angle, from the construction, and length PA = length P'A by the same reasoning. This defines the square PAP'Q with two vertices on c, shown shown shaded.
All we now need to do is choose P' to be the point where the dotted circle c' intersects the inner circle and we obtain the required square, since Q must lie on the inner circle by symmetry.
Of course if the inner circle is too small this doesn't work, because the dotted circle and the small one don't intersect. The limiting case is when the inner radius is ~ 0.414 of the outer one.