Is the figure the circumference of a unit circle?

Question

A friend of mine taught me the following question. I've never heard such a strange and interesting question!

Qustion: Supposing that a figure $S$, which is constituted by points, satisfies the following four conditions, can we say that $S$ is the circumference of a unit circle?

1. $S$ crosses at two points every line which passes through the origin.

2. $S$ crosses at one point every tangent of the unit circle whose center is the origin.

3. $S$ crosses at two points every line $L_x$ which satisfies the following two: (i) $L_x$ is parallel to $x$-axis. (2) The distance $d_x$ between $L_x$ and $x$-axis satisfies $d_x\lt1$.

4. $S$ crosses at two points every line $L_y$ which satisfies the following two: (i) $L_y$ is parallel to $y$-axis. (2) The distance $d_y$ between $L_y$ and $y$-axis satisfies $d_y\lt1$.

I've tried to solve this, but I'm facing difficulty. He said the answer is NO without his memory of the very figure. Can we find a special counterexample?

Edit : I'm sorry. The Question 2 is not an appropriate question, so I deleted it.

Answer 1

I've just got the following figure.

The answer for Question is No.

$S$ has eight line segments $AB, CD, EF, GH, MN, OP, QR, ST$ and four points $I, J, K, L$ without twelve points $A, B, C, D, E, F, G, H, M, P, R, T$ where $$A(0,1), B(-\frac12, \frac12), C(-1,0), D(-\frac12, -\frac12), E(0,-1), F(\frac12, -\frac12), G(1,0), H(\frac12, \frac12), I(0,\frac12), J(-\frac12, 0), K(0,-\frac12), L(\frac12,0), M(0,\sqrt{10}), N(\frac{\sqrt{10}}{2},\frac{\sqrt{10}}{2}), O(-\frac{\sqrt{10}}{2},\frac{\sqrt{10}}{2}), P(-\sqrt{10},0), Q(-\frac{\sqrt{10}}{2},-\frac{\sqrt{10}}{2}), R(0,-\sqrt{10}), S(\frac{\sqrt{10}}{2},-\frac{\sqrt{10}}{2}), T(\sqrt{10},0).$$

I think this figure is one of the counterexamples.

Answer 2

EDIT: This figure is not quite correct as a counterexample.

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I managed to construct this counterexample. It is very, very "hackish" and contrived. It is probably not what your friend had in mind. But it is a counterexample!

Anything in $\color{blue}{\text{blue}}$ is not part of the actual figure.

1. Define (only this is in polar coordinates) $R(\varphi):=(\sqrt{2}, \varphi)$. Then the arc in first quadrant is defined as the set of points $\{R(\varphi):0 < \varphi < \pi/2\}$. Whereas the arc in third quadrant is defined as the set of points $\{R(\varphi):\pi < \varphi < 3\pi/2\}$.
2. $A = (0, A_y)$ such that $A_y < 1$ is infinitesimally close to $1$.
3. $B = (B_x,0)$ such that $B_x > -1$ is infinitesimally close to $-1$.
4. $C = (C_x, 0)$ such that $C_x < 1$ is infinitesimally close to $1$.
5. $D = (0, D_x)$ such that $D_x > -1$ is infinitesimally close to $-1$.
6. $P_1 = (-\sqrt{2}/2, \sqrt{2}/2)$.
7. $P_2 = (\sqrt{2}/2, -\sqrt{2}/2)$.

Please critique this figure.

Some comments:

1. You might be worried that the lines $y = 0$ and $x = 0$ would violate property 1, 3 and 4. They do not, because the arcs do not actually touch the axes.
2. Would the tangent lines $y = x \pm \sqrt{2}$ violate property 2? No, because, again, the arcs do not touch the axes and those tangent lines touch one and only one of the points $P_1$ and $P_2$.
3. Each of the tangent lines $y = \pm 1$ crosses only at one point. They pass through neither point $A$ nor $D$.