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If I have a point K inside a euclidean disc what will be chord (chord which goes through the point K) with the smallest length. I think it will be chord which is perpendicular to the diameter, which goes through the point K, but I don't know is this true. I tried to prove this with formula of length of the chord $L=2R \sin (A/2)$, where $A$ is angle between two radii drawn to the ends of chord, but the only thing I deduced from this formula is that length would be the smallest when a is the smallest. So I'm stuck with this. Is my assumption true (if so can you help with the proof) and if not what chord will have the smallest length ?

Arnaud Mortier
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Юрій Ярош
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  • Yes, this is true, and just as important, $K$ will be the midpoint of this chord. – Paul Feb 16 '18 at 13:14
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    Draw the perpendicular from the centre $O$ to a chord through $K$. Note that the shorter the perpendicular distance the longer the chord. – almagest Feb 16 '18 at 13:14
  • @Arthur Kind of – Юрій Ярош Feb 16 '18 at 13:14
  • @ArnaudMortier I said that I'm talking about chord which goes through point K. – Юрій Ярош Feb 16 '18 at 13:15
  • @ArnaudMortier He said in the circle, not on the circle. Now that it's edited to "inside", it's even clearer. – Arthur Feb 16 '18 at 13:18
  • @ArnaudMortier OP said $K$ is in the circle. And it is a fixed point. – Jaideep Khare Feb 16 '18 at 13:18
  • @Arthur I see that now. Yet it doesn't make sense, you should say disc in that case. – Arnaud Mortier Feb 16 '18 at 13:18
  • @Arthur You beat me by 8 sec – Jaideep Khare Feb 16 '18 at 13:18
  • @ArnaudMortier No, I don't think so. I would personally either use in or inside a circle, or on a disc, and those would all mean the same to me. Now, on a circle means something different from these, I agree. – Arthur Feb 16 '18 at 13:20
  • @ArnaudMortier Inside the circle. – Юрій Ярош Feb 16 '18 at 13:22
  • @Arthur a circle is $1$-dimensional, if you say in a circle when you mean inside the bounded component of its complement, that's not proper speaking. – Arnaud Mortier Feb 16 '18 at 13:22
  • @ArnaudMortier If you required that people were that precice with their language all the time, writing anything would take ages, and reading it would take longer. Abuse of terminology is a thing for a reason, and I don't think many would have actual trouble understanding what "a point inside a circle" means, which is the entire point of speaking and writing in the first place: being understood. Overzealous pedantry actively hinders communication in some cases, and it is therefore something I oppose strongly. – Arthur Feb 16 '18 at 13:27
  • @Arthur I'm not being pedant, I honestly pictured a point on a circle when I first read the question. On the contrary, knowing the correct words to use is precisely a way to be understood, nothing more. I have to admit that "inside a circle" would have been clearer though - and I would not have found it necessary to intervene. – Arnaud Mortier Feb 16 '18 at 13:32
  • @ArnaudMortier Yeah, you're right. When it said "in" it was easy to confuse it, and avoiding confusion is also a big part of communication. Sorry I ranted off on you. – Arthur Feb 16 '18 at 13:38
  • @Arthur no worries. I also understand why you could think that I was just nitpicking. – Arnaud Mortier Feb 16 '18 at 13:49
  • @ArnaudMortier OK, sorry, you're right, I needed to be more presice. – Юрій Ярош Feb 16 '18 at 14:42

3 Answers3

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I am.afraid Arthur, Winter said it all .

Just for fun:

Point $K$ given, inside a circle .

For definiteness consider the chord $KM$ , where $M$ is the centre of the circle.

$K$ divides the chord $KM$ into $2$ segments of lengths $a$ and $b.$

Inersecting Chord Theorem:

$ab = xy$ where $x,y$ are the lengths of segments formed by any chord through $K.$

$2$ equations:

1)$xy=ab=C$ (constant , given)

2)$S:= x +y $.

Want to minimize $S$ with the constraint 1).

Use 1) to eliminate $y$:

$S = x + C/x $.

AM-GM :

$S = x+C/x \ge 2√C.$

Minimum attained for $x=C/x= √C$.

Then $y= (ab)/√C =√C$.

Hence $x=y$ , the perpendicular chord to chord $KM $.

Note: $KM$ is a symmetry axis, hence $x=y$ implies the chord through $K$ is perpendicular.

Peter Szilas
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Let's say we have a unit circle around the origin $O$. A secant $\ell$ having distance $d(\ell,O)$ to $O$ is of length $$L=2\sqrt{1-d(\ell,O)^2}.$$ We can minimize this by maximizing $d(\ell,O)$.

Now let's restrict to secants $\ell$ through $K$. We know that $d(\ell,O)$ is the length of a line segment between $O$ and $\ell$, which is perpendicular to $\ell$. The point where this segment meets $\ell$ is the closest point of $\ell$ to $O$. All other points of $\ell$ are further away from $O$. Because $\ell$ has to go through $K$, it cannot have a distance greater than $d(K,O)$ from $O$, but by choosing $K$ to be the closest point on $\ell$ to $O$, we also ensure that $d(\ell,O)$ is not smaller than $d(K,O)$. Therefore, this maximizes $d(\ell,O)=d(K,O)$ and minimizes the secants length to:

$$L_{\min}=2\sqrt{1-d(K,O)^2}.$$

Since the segment from $O$ to $K$ is a (part of the) radius of the circle, and meets $\ell$ perpendicularly in $K$, your intuition was right.

M. Winter
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Given any point $P$ and any circle $C$, the power of $P$ with respect to $C$ is defined as follows:

Take a line through $P$ that intersects $C$ twice, at $A$ and $B$. The power of $P$ with respect to $C$ is $|PA|\cdot |PB|$, and is independent of what line was chosen.

(According to some conventions the power is negative when $P$ is on the inside of $C$ and positive when $P$ is outside, but we will say it is positive. Until we have to compare powers of several different points it doesn't matter.)

Given a chord through $K$, the power of $K$ with respect to the circle is length of the two parts of chord on either sides of $K$. It is the same no matter which chord you draw. You want the sum of the two parts to be minimal.

Now note that if two quantities vary so that their product is the same (such as the lengths of the two parts of the chords), their sum is the least when the quantities are equal. (Replace "their product" with "the square root of their product" and "their sum" with "their average", and you've got exactly the AM-GM inequality, which proves this statement.)

Arthur
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