Does every closed, smooth, non-self-intersecting loop have at least one point in the interior from which two perpendicular lines can be drawn with each intersecting the loop at 90° at all four points of intersection?
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It does not seem true to me. If you don't demand smoothness, I think a rectangular triangle is a counterexample. And I think a rectangular triangle with 'smoothed' edges might be good enough for the smooth case. – secavara Aug 28 '18 at 23:19
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I can somewhat imagine a method for finding a point in a right triangle with rounded corners. Draw an altitude from the hypotenuse to the right angled corner. Move it along the hypotenuse until it intersects the corner at a right angle. Then, construct a perpendicular to this altitude. Move it along the altitude until it intersects one of the other corners at 90°. I have no proof, but intuitively, I feel like the other corner is hit at 90°. If this is not the case, can anyone formalize a rule for which curves satisfy the conditions of the question as stated in the original question above. – William Grannis Aug 29 '18 at 00:44
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As Marco's answer shows, this is impossible. Just by counting degrees of freedom you could have expected this to be unlikely: you have three degrees of freedom (the point where the lines intersect, and the slope of one of the lines) but require four conditions to be satisfied (the angles at the four intersections with the curve). – Aug 29 '18 at 07:13
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Let $\gamma_1$ be a upper semicircle in the plane connecting $(1,0)$ to $(-1,0)$. Let $\gamma_2$ be defined by $r=f(\theta)$, $\pi \leq \theta \leq 2\pi$, where $f$ is such that $f(\pi)=f(2\pi)=1$, $f(\theta)\geq 1$ in the range, and $f$ is concave down with a unique maximum at $t\neq 3\pi/4$.
There are exactly two lines only that make a perpendicular angle with $\gamma_1 \cup \gamma_2$: the $x$-axis and the line given by $\theta=t$. These two lines were arranged to be not perpendicular to each other. So the answer to the question is negative.
However, one can prove it is possible for three of the angles to be right angles: Take two points on the curve that are farthest apart. The line $l$ through them is perpendicular to the curve at those two points. Now find a point $C$ on the curve that is farthest away from $l$. The line perpendicular to $l$ going through $C$ makes a right angle with the curve.
Marco
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I don't think it's possible for $f$ to be concave down and still have $f'(\pi)=0=f'(2\pi)$, which you need for smoothness. But concavity is too strong a condition anyway: you just need $f$ to be strictly increasing on $[\pi,t]$ and strictly decreasing on $[t,2\pi]$ (or vice versa). – Aug 29 '18 at 06:57
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Also, your last paragraph proves that it is always possible for three of the angles to be right angles for any smooth curve, not just for the one constructed in the previous two paragraphs, right? – Aug 29 '18 at 06:58
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