Can a figure inside a circle be seen at right angle from any point on the circle?

Question

A convex, closed figure lies inside a given circle. The figure is seen from every point of the circumference at a right angle (that is, the two rays drawn from the point and supporting the convex figure are perpendicular). Prove that the center of the circle is a center of symmetry of the figure.

Firstly does the question mean for the figure to be two-dimensional? Otherwise do there exist such figures in $3$D (in other words if the $3$D figure doesn't intersect the circle)? In $2$D, the only such object I can think of is the diameter of a circle, but I was wondering if there are more.

Answer 1

It does imply that everything is in the plane, otherwise it wouldn't say "figure" or talk about two rays "supporting" it, it would be a cone in 3D. Diameter is a good suggestion due to the Thales' theorem but it won't work because the condition is violated at the endpoints of the diameter. One obvious solution is a concentric circle of smaller radius. Inscribe a square into the original circle, and then inscribe the smaller circle into the square. Rotating the square around the smaller circle shows that it will be seen at $90^\circ$ angle from any point on the original one. By the Pythagorean theorem the radius of the smaller circle is $1/\sqrt{2}$ of the larger one's for this to happen.

This construction may generalize. At first, it seems that the inscribed circle is the only such figure, but it also at first seems that the circle is the only curve of constant width, while in fact there are infinitely many of them. In this case the set of points from which a compact convex figure is seen at a constant angle $\nu$ is called its $\nu$-isoptic. The $\pi/2$-isoptic is called orthoptic. The OP question is equivalent to asking if a circle is the only convex figure having a circular orthoptic. According to Kurusa's Is a Convex Plane Body Determined by an Isoptic?

"In 1950 Green proved that if the $\nu$-isoptic of a convex body D is a circle, then D must be a disc provided $1-\nu/\pi$ is an irrational or rational number with even numerator in its lowest terms. If the numerator is odd, then there are continuum many bodies not even similar to each other with that circle as their $\nu$-isoptic".

Since for $\nu=\pi/2$ we have $1-\nu/\pi=1/2$, this is the case of a continuum of non-similar solutions. According to Green, Ernst Strauss noticed earlier that a family of them is easy to construct classically: inscribe a rectangle into a circle, and inscribe an ellipse into that rectangle with its axes equal and parallel to the sides of the rectangle. Then it can be proved that the circle is the orthoptic of the ellipse. These ellipses interpolate between the insquared circle described above, and the circle's diameters, which are their degenerate limits, and "almost" orthoptics. Green constructs more general solutions by using support functions of convex figures.

Answer 2

There are infinitely many figures like that.

Let $p(\theta)$ be any smooth periodic function of period $2\pi$. It is known that the necessary and sufficient condition for such a function $p(\theta)$ to be the support function of a convex set $K$ is $$p(\theta) + p''(\theta) > 0 \quad\text{ for all }\;\theta$$

The boundary of the convex set $K$ is given by the parametrization:

$$\begin{cases} x &= p(\theta)\cos\theta - p'(\theta)\sin\theta,\\ y &= p(\theta)\sin\theta + p'(\theta) \cos\theta \end{cases} $$ If you pick a function $p(\theta)$ which satisfies an additional constraint $$p(\theta)^2 + p\left(\theta + \frac{\pi}{2}\right)^2 = 1$$

then for every $\theta$, the pair of tangent lines at $\theta$ and $\theta + \frac{\pi}{2}$ intersect at some point on the unit circle. Since they are perpendicular to each other, the angle subtended by $K$ at that intersection will be a right angle.

As an example, pick any $\alpha \in [0,1)$ and consider the function $\displaystyle\;p(\theta) = \sqrt{\frac{1+\alpha \cos(2\theta)}{2}}$.
It is easy to check $\displaystyle\;p(\theta) + p''(\theta) = \frac{1-\alpha^2}{\sqrt{2}(1 + \alpha\cos(2\theta))^{3/2}} > 0$ for all $\theta$. This give us a convex set bounded by the curve $$\begin{cases} x(t) &= \frac{1+\alpha}{\sqrt{2}}\frac{\cos(t)}{\sqrt{1 + \alpha\cos(2\theta)}}\\ y(t) &= \frac{1-\alpha}{\sqrt{2}}\frac{\sin(t)}{\sqrt{1 + \alpha\cos(2\theta)}} \end{cases}$$ With help of an CAS, one can simplify this to the ellipse $\displaystyle\;\frac{x^2}{1+\alpha} + \frac{y^2}{1-\alpha} = \frac12$. This is equivalent to the director circle mentioned by others in comment.

For a more complicated example, one can try something like $\displaystyle p(\theta) = \sqrt{\frac{1+\beta\cos(6\theta)}{2}}$ where $\beta \in [0,\frac{1}{17})$ instead. This will give you a figure that looks like a hexagon with the edge curved outward.

Answer 3

Let the given outside circle pass through the origin. Then power of a circle from origin to a circle

$ (x- h)^2 + y ^2 = R^2 $ is:

$ h^2 - R^2 = T^2$

And the subtended angle at pole is:

$ 2 \tan^{-1} \left (\dfrac{R}{T} \right)$

If the above should be $\pi/2,$ then $ R= T, h = \sqrt 2 R $

Thus in the plane the radius of inside circle locus equals length of tangent, inside circle has radius $ R= h/ \sqrt 2,$ which exists as a unique envelope between the perpendicular tangents.

In 3D,likewise by rotational symmetry, the required surface locus is a sphere of $ R= h/ \sqrt 2.$ where horizons for an observer on the given circle is the set of small circles of latitude $45^{0}$.