Magnitude of sum of vectors $\le2$

Question

Let $V_1V_2\cdots V_{2n}$ be a convex inscribed polygon of the unit circle. Let $\mathbf x=\sum\limits_{k=1}^n\overrightarrow{V_{2k-1}V_{2k}}$, prove that $|\mathbf x|\le2$.

I started by setting coordinates. Let $V_i~(\cos\theta_i,\sin\theta_i)$ for $1\le i\le2n-1$ and $V_{2n}~(1,0)$. WLOG, let $0<\theta_1$$<\theta_2$$<\cdots$$<\theta_{2n-1}$$<2\pi$. Let the unit vector with the same direction with $\mathbf x$ be $\mathbf j=(\cos\alpha,\sin\alpha)$. We need to prove that $\mathbf j\cdot\mathbf x\le2$.

We could get a trigonometry expression from this, but I don't know how to prove it.

Answer 1

Suppose $\ \exists\ V_i\ $ s.t. $\ \vert \mathbf x\vert > 2\ $. I could use complex notation (real, imaginary parts), but to keep in line with OP, I'll use the $\ \mathbf i\ $ and $\ \mathbf j\ $ vectors. I'll assume that the $\ V_i\ $ were originally labelled anti-clockwise around the unit circle.

1. Rotate all the vertices around the circle about the origin, each by the same angle $\ \theta,\ $ so that they are the same distance away from one another as before, but so that $\ \sum\limits_{k=1}^n\overrightarrow{V_{2k-1}V_{2k}} = -s \mathbf i,\ s>2.\ $ Let the vertices keep their original labels $\ V_{1},\ V_2,\ldots, V_{2k},\ $ but consider them from now in their new positions.
2. For each $\ j\in \{1,\ldots, 2n\},\ $ re-label $\ V_j\ $ as $\ {V_i}^{\ '}\ $ so that: $\ V_1^{\ '}\ $ is the furthest right point on the unit circle, that is, the point with the maximum value of $\ \overrightarrow{0\ {V_{j}}^{\ '} } \cdot \mathbf i,\ $ where $\ 0\ $ is the origin; $\ V_2^{\ '}\ $ is the next vertex counter-clockwise on the unit circle to $\ V_1^{\ '},\ $ etc.

The effect of (2) is that:

If $\ V_1^{\ '}\ $ had an odd subscript originally (for example, $\ V_3\ $), then $\ \sum\limits_{k=1}^n\overrightarrow{ \ {V_{2k-1}}^{\ '}\ {V_{2k}}^{\ '}}= \sum\limits_{k=1}^n\overrightarrow{ V_{2k-1} V_{2k}}.$

Else if $\ V_1^{\ '}\ $ had an even subscript originally (for example, $\ V_4\ $), then $\ \sum\limits_{k=1}^n\overrightarrow{ {V_{2k-1}}^{\ '} {V_{2k}}^{\ '} } = \underbrace{\sum\limits_{k=1}^{2n}\overrightarrow{V_{k}V_{k+1}} }_{=\mathbf 0} - \sum\limits_{k=1}^n\overrightarrow{ V_{2k-1} V_{2k} }.\ $ Here, $\ V_{2n+1}:= V_1.$

The point is that, in either case, $\left\lvert \sum\limits_{k=1}^n\overrightarrow{ {V_{2k-1}}^{\ '} {V_{2k}}^{\ '} } \right\rvert = \left\lvert \sum\limits_{k=1}^n\overrightarrow{ V_{2k-1} V_{2k} } \right\rvert = s > 2.\ $

Now define $ W:= \mathbf i + \mathbf 0 \mathbf j,\ Y:= -\mathbf i + \mathbf 0 \mathbf j,\ $ and let $\ 2Z\ $ be the greatest even number such that $\ 1\leq Z\leq n\ $ and that maximises $\ -\overrightarrow{0\ {V_{2Z}}^{\ '} } \cdot \mathbf i,\ $ in other words, either $\ {V_{2Z}}^{\ '}\ $ or $\ {V_{2Z-1}}^{\ '}\ $ is the furthest left point on the unit circle. Then:

$$ s = \left\lvert \left(\sum\limits_{k=1}^n\overrightarrow{ {V_{2k-1}}^{\ '} {V_{2k}}^{\ '}}\right) \cdot \mathbf i \right\rvert = \left\lvert \sum\limits_{k=1}^n \left(\overrightarrow{ {V_{2k-1}}^{\ '} {V_{2k}}^{\ '} } \cdot \mathbf i\right) \right\rvert \leq \left\lvert \sum\limits_{k=1}^Z \left(\overrightarrow{ {V_{2k-1}}^{\ '} {V_{2k}}^{\ '} } \cdot \mathbf i\right) \right\rvert $$

$$ \leq \left\lvert \overrightarrow{W {V_{2}}^{\ '} } \cdot \mathbf i + \sum\limits_{k=2}^{Z-1} \left(\overrightarrow{ {V_{2k-1}}^{\ '} {V_{2k}}^{\ '} } \cdot \mathbf i \right) + \overrightarrow{ {V_{2Z-1}}^{\ '} Y} \cdot \mathbf i \right\rvert \leq \left\lvert -2 \mathbf i \right \rvert \leq 2, $$

a contradiction.

I suspect some of my working is redundant (or perhaps lacking in detail in some bits), but I believe I have at least given a detailed sketch of a proof.

Answer 2

Observations towards a solution. Fill in the gaps as needed.

Let $ \vec{x} = \sum\overrightarrow{V_{2k-1} V_{2k}}$. Assume $\vec{x} \neq 0$, since otherwise there is nothing to prove.
Consider $ x_i = \vec{x} \cdot \overrightarrow{ V_i V_{i+1}}$.

• Claim: $\{ x_i \geq 0 \} = \{i, i+1, i+2, \ldots j\} $ and $ \{ x_i < 0 \} = \{ j+1, j+2, \ldots i+2n - 1 \}$, where the indices are interpreted modulo $2n$. Furthermore $i \neq j$ (because $\vec{x} \neq 0$.)
  • (The following observation is not required for the proof, but provides a one-line explanation for the claim.) In fact, by drawing $ \vec{x}$ such that it is normal to the unit circle, we can determine $i, j$ as vertices of the 2 edges that intersect with $\vec{x}$ (extended as needed). (It's easier to understand if you draw it out. See note below.)
• If $ x_{2k} > 0$, then we can increase $|\vec{x}|$ by setting $ V_{2k} = V_{2k+1}$,
• If $ x_{2k-1} < 0 $, then we can increase $|\vec{x} | $ by setting $ V_{2k-1} = V_{2k}$,
• With these in place, show that we get $ \vec{x} = \overline{ V_i V_j }$.
• Hence $|\vec{x}| \leq 2$, with equality when
  • $V_iV_j$ is a diameter
  • $V_{2k} = V_{2k+1}$ when $i \leq 2k \leq j $.
  • $V_{2k-1} = V_{2k}$ when $j \leq 2k-1 \leq i+2n$

Notes

• The claim that $ \vec{x} = \overline{V_i V_j}$ could be hypothesized by studying the 2 equality cases listed out in the comments, and realizing that we sometimes want $V_{2k-1} = V_{2k}$ but other times we want $V_{2k} = V_{2k+1}$, which basically boils down to $x_i = \vec{x} \cdot \overline{V_i V_{i+1} }$. This proof falls out from that consideration, though it remains to check the details to ensure that the claim is valid.
• The naive claim that $|x| \leq \max | V_i V_{i+1} |$ can be easily invalidated by having extra vertices. EG Modifying OP's equality case and let $V_1V_4$ be a diameter with $V_2 = V_3$, and then $V_5 = V_6, \ldots$.
  • This reinforces the importance of studying $x_i$.
  • It also suggests the $ |x| \leq \max |V_i V_j|$.
• Pictorial explanation of the observation that wasn't required. For an inscribed convex polygon $BCDEFG$ and ANY non-zero vector $HI$ drawn normal to the circle (hence passing through the center $A$), can you classify which of $HI \cdot BC, HI \cdot CD, HI \cdot DE, HI \cdot FG, HI \cdot GB$ are positive and negative? Why does the sign toggle when the edge cuts $HI$? Think about the trigonometric form of the dot product. Now, let $HI = \vec{x}$.