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The following question was asked as part of the Canadian Mathematical Olympiad Qualifying Repechage in 2016.

Consider a convex polygon $P$ with $n$ sides and perimeter $P_0$. Let the polygon $Q$, whose vertices are the midpoints of the sides of $P$, have perimeter $P_1$. Prove that $P_1 \geq \frac{P_0}{2}$.

The official solution is as follows.

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I can't make any sense of the second paragraph. (Isn't $w$ just $v_i$?)

First, am I missing something here? Second, if there really is, as it seems to me, an error in this solution, then is the problem even correct? What would a correct solution be?

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    That's a terribly, terribly worded "proof". You are right, $w \equiv v_i$ and $x \equiv v_{i+2}$ if taken at face value. – 2017-02-05

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The problem is correct, but the "official solution" is completely bungled.

What they likely had in mind is: let $w_i = v_iv_{i+2} \cap v_{i-1}v_{i+1}$ and $x_i = w_{i+1}=v_iv_{i+2} \cap v_{i+1}v_{i+3}$. Then by triangle inequalities:

$$ \begin{cases} \begin{align} v_iw_i+w_iv_{i+1} & \ge v_iv_{i+1} \\ v_{i+1}x_i+x_iv_{i+2} & \ge v_{i+1}v_{1+2} \end{align} \end{cases} $$

Adding up those inequalities for all vertices and using that $v_iw_i+x_iv_{i+2} \le v_iv_{i+2}$ does indeed prove $\sum v_iv_{i+2} \ge \sum v_iv_{i+1}\,$, which in turn proves the proposed $2 P_1 \ge P_0$.