This problem was proposed by Burii.

The answer is $n-3$ for all $n\geq 3$. Let $M(n)$ denote the minimal number of Bohemian vertices of a convex $n$-gon, also let $B_i$ be the reflection of $A_i$ with respect to the midpoint of $A_{i-1}A_{i+1}$. Claim: $M(3)=0$. Proof(without words): [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(5cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -10.44, xmax = 10.82, ymin = -10.48, ymax = 5.96; /* image dimensions */ draw((-4,-1)--(-2,-4)--(-6,-5)--cycle, linewidth(0.8)); /* draw figures */ draw((-4,-1)--(-2,-4), linewidth(0.8)); draw((-2,-4)--(-6,-5), linewidth(0.8)); draw((-6,-5)--(-4,-1), linewidth(0.8)); /* dots and labels */ dot((-4,-1),linewidth(2pt) + dotstyle); label("$A_1$", (-3.92,-0.92), NE * labelscalefactor); dot((-2,-4),linewidth(2pt) + dotstyle); label("$A_2$", (-1.92,-3.92), NE * labelscalefactor); dot((-6,-5),linewidth(2pt) + dotstyle); label("$A_3$", (-5.92,-4.92), NE * labelscalefactor); dot((-4,-8),linewidth(2pt) + dotstyle); label("$B_1$", (-3.92,-7.92), NE * labelscalefactor); dot((-8,-2),linewidth(2pt) + dotstyle); label("$B_2$", (-7.92,-1.92), NE * labelscalefactor); dot((0,0),linewidth(2pt) + dotstyle); label("$B_3$", (0.08,0.08), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Claim: $M(4)=1$. Proof: First note that $A_iA_{i+1}B_iA_{i-1}$ is a parallelogram. Assume that $A_1A_2A_3A_4$ is not a parallelogram, hence it has two obtuse angles, we divide now in two cases If the obtuse angles are at neighbour vertices, say at $A_1$ and $A_2$, let $h_1,h_2$ be the heights of $A_1,A_2$ onto $A_3A_4$, assume that $h_1<h_2$ then $A_1A_2B_1A_4$ is a parallelogram that lies entirely in $A_1A_2A_3A_4$, wich finishes this case. if the obtuse angles are at $A_1$ and $A_3$, then if $\angle A_2>180^{\circ}-\angle A_1$, let the parallel line through $A_4$ intersect $A_2A_4$ at $D$, if $A_3$ does not lie in $A_2D$ then $A_1A_2B_1A_4$ is a parallelogram lying in $A_1A_2A_3A_4$, and if $A_3$ lies on $A_2D$ then $A_2A_3B_2A_1$ is a parallelogram lying in $A_1A_2A_3A_4$, and the case $\angle A_2<180^{\circ}-\angle A_1$ is identical, hence there is always a Bohemian vertex on a convex quadrilateral. And for an example: [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(5cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.3, xmax = 16.58, ymin = -10.2, ymax = 6.24; /* image dimensions */ draw((2,1)--(0,-3)--(8,-4)--(7,0)--cycle, linewidth(0.8)); /* draw figures */ draw((2,1)--(0,-3), linewidth(0.8)); draw((0,-3)--(8,-4), linewidth(0.8)); draw((8,-4)--(7,0), linewidth(0.8)); draw((7,0)--(2,1), linewidth(0.8)); /* dots and labels */ dot((2,1),linewidth(2pt) + dotstyle); label("$A_1$", (2.08,1.08), NE * labelscalefactor); dot((0,-3),linewidth(2pt) + dotstyle); label("$A_4$", (0.08,-2.92), NE * labelscalefactor); dot((8,-4),linewidth(2pt) + dotstyle); label("$A_3$", (8.08,-3.92), NE * labelscalefactor); dot((7,0),linewidth(2pt) + dotstyle); label("$A_2$", (7.08,0.08), NE * labelscalefactor); dot((5,-4),linewidth(2pt) + dotstyle); label("$B_1$", (5.08,-3.92), NE * labelscalefactor); dot((-1,1),linewidth(2pt) + dotstyle); label("$B_3$", (-0.92,1.08), NE * labelscalefactor); dot((3,-3),linewidth(2pt) + dotstyle); label("$B_2$", (3.08,-2.92), NE * labelscalefactor); dot((10,0),linewidth(2pt) + dotstyle); label("$B_2$", (10.08,0.08), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Claim: $M(n)=n-3$ for $n\geq 5$. Proof: We proceed by induction on $n$, we have shown that for $n=4$ this is true. Assume now that it holds for all $k<n$, now, for the convex polygon $A_1A_2\dots A_n$ consider the convex polygon $A_1A_2\dots A_{n-1}$, by induction hypothesis there exists a vertex that is Bohemian, say $A_1$, then we remove $A_1$ and we consider $A_2A_3\dots A_{n-1}$, note that if $A_2,A_3$ are Bohemian in the convex polygon $A_2A_3\dots A_n$ they will also be Bohemian in the convex polygon $A_1A_2\dots A_n$, hence the minimal number of Bohemian vertices is at least $1+(n-1)-3=n-3$ by induction hypothesis. An example is [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(5cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.3, xmax = 16.58, ymin = -10.14, ymax = 6.3; /* image dimensions */ /* draw figures */ draw((4,-2)--(10,-2), linewidth(0.8)); draw((10,-2)--(10,-4), linewidth(0.8)); draw((4,-2)--(6,-7), linewidth(0.8)); draw((6,-7)--(8,-6), linewidth(0.8)); /* dots and labels */ dot((4,-2),linewidth(2pt) + dotstyle); label("$A_3$", (4.08,-1.92), NE * labelscalefactor); dot((10,-2),linewidth(2pt) + dotstyle); label("$A_2$", (10.08,-1.92), NE * labelscalefactor); dot((10,-4),linewidth(2pt) + dotstyle); label("$A_1$", (10.08,-3.92), NE * labelscalefactor); dot((6,-7),linewidth(2pt) + dotstyle); label("$A_4$", (6.08,-6.92), NE * labelscalefactor); dot((8,-6),linewidth(2pt) + dotstyle); label("$A_5$", (8.08,-5.92), NE * labelscalefactor); dot((12,-7),linewidth(2pt) + dotstyle); label("$B_3$", (12.08,-6.92), NE * labelscalefactor); dot((4,-4),linewidth(2pt) + dotstyle); label("$B_2$", (4.08,-3.92), NE * labelscalefactor); dot((6,-1),linewidth(2pt) + dotstyle); label("$B_4$", (6.08,-0.94), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] where for the case $n=5$ we directly join $A_1A_5$ and for $n\geq 6$ we break it till we get the desired number of vertices. $\blacksquare$

is there any relation to the fact that any convex n-gon can have at most 3 acute angles

@above While I was trying to solve this problem I found some relations.

We claim that the answer is $n-3$ for all $n\geq 3$. Difine a vertex $A_{i}$ ANTI-BOHEMIAN vertex if its reflection with respect to the midpoint of $A_{i-1}A_{i+1}$ lies outside the polygon. So we prove that there is at most $3$ ANTI-BOHEMIAN vertices. $\textbf{First Lemma.}$ $A_{i}$ is an ANTI-BOHEMIAN vertex $\Longleftrightarrow$ $\pi - \angle A_{i-1}A_{i}A_{i+1} > \min(\angle A_{i-2}A_{i-1}A_{i} , \angle A_{i}A_{i+1}A_{i+2})$ $\textbf{Proof:}$ By reflection of $A_{i}$ with respect to midpoint of $A_{i-1}A_{i+1}$, it makes a parallelogram. Use this fact and the rest is tirivial. $\textbf{Second Lemma.}$ If a vertex $A_{i}$ is an ANTI-BOHEMIAN vertex; then at least one of the vertices $A_{i-1}$ or $A_{i+1}$ is also ANTI-BOHEMIAN. $\textbf{Proof:}$ By using the first lemma, and WLOG assume that $\pi -\angle A_{i-1}A_{i}A_{i+1} > \angle A_{i}A_{i+1}A_{i+2}$ ;then we conclude that $\pi -\angle A_{i}A_{i+1}A_{i+2} > \angle A_{i-1}A_{i}A_{i+1}$. Done. Assume that $A_{i}$ and $A_{i+1}$ are a pair of ANTI-BOHEMIAN vertices. So; $$(\pi-\angle A_{i-1}A_{i}A_{i+1})+(\pi-\angle A_{i}A_{i+1}A_{i+2}) > (\angle A_{i}A_{i+1}A_{i+2})+(\pi-\angle A_{i}A_{i+1}A_{i+2})= \pi$$It means that sum of the exterior angles of these two vertices are more than $\pi$. Conclusion: Assume that there exist two separate pair of ANTI-BOHEMIAN vertices(with no common vertex). Then the sum of exterior angles is more than $2 \pi$. Contradiction. Hence because of second lemma, it doesn’t exist $4$ or more ANTI-BOHEMIAN vertices. So there is at most $3$ ANTI-BOHEMIAN vertices, and they’re consecutive. Now we prove that $3$ is the best bound for the number of ANTI-BOHEMIAN vertices. Assume a $n$-gon such that $\angle A_{n}A_{1}A_{2}$,$\angle A_{1}A_{2}A_{3}$ and $\angle A_{2}A_{3}A_{4}$ are acute. By using the first lemma, we get that $A_{1}$ , $A_{2}$ and $A_{3}$ are ANTI-BOHEMIAN. and now we’re done.

This is identical to Problem 4 in the 1964 Miklos Schweitzer competition.

Mary987 wrote: W... $\textbf{First Lemma.}$ $A_{i}$ is an ANTI-BOHEMIAN vertex $\Longleftrightarrow$ $\pi - \angle A_{i-1}A_{i}A_{i+1} > \min(\angle A_{i-2}A_{i-1}A_{i} , \angle A_{i}A_{i+1}A_{i+2})$ $\textbf{Proof:}$ By reflection of $A_{i}$ with respect to midpoint of $A_{i-1}A_{i+1}$, it makes a parallelogram. Use this fact and the rest is tirivial. ... The first Lemma seems incorrect as an example of the following situation (see attachments), so how to correct this? [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(5cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -5.455051543068859, xmax = 8.051115502927109, ymin = -3.4357300827649238, ymax = 3.317353440233073; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); draw(arc((-1.358745393538734,-0.18620556621259715),0.3175431123666764,-130.45071680675017,-41.077298668987304)--(-1.358745393538734,-0.18620556621259715)--cycle, linewidth(0.7) + qqwuqq); draw(arc((-0.24239404466581793,1.1231528300152562),0.3175431123666764,-130.45071680675014,-22.942392627077123)--(-0.24239404466581793,1.1231528300152562)--cycle, linewidth(0.7) + qqwuqq); draw(arc((-0.1934445214904088,-1.2019495208166893),0.3175431123666764,23.888977916496852,138.9227013310127)--(-0.1934445214904088,-1.2019495208166893)--cycle, linewidth(0.7) + qqwuqq); /* draw figures */ draw((-0.24239404466581793,1.1231528300152562)--(-1.358745393538734,-0.18620556621259715), linewidth(0.7)); draw((-1.358745393538734,-0.18620556621259715)--(-0.1934445214904088,-1.2019495208166893), linewidth(0.7)); draw((-0.1934445214904088,-1.2019495208166893)--(0.5465132806613243,-0.874215643007063), linewidth(0.7)); draw((0.5253437398368792,0.7981780821241024)--(-0.24239404466581793,1.1231528300152562), linewidth(0.7)); draw((-0.24239404466581793,1.1231528300152562)--(-0.1934445214904088,-1.2019495208166893), linewidth(0.7) + dotted); draw((-0.24239404466581793,1.1231528300152562)--(-2.021335920777582,-0.9633520004218044), linewidth(0.7) + dotted); /* dots and labels */ dot((-0.24239404466581793,1.1231528300152562),dotstyle); label("$A_1$", (-0.43787036767537246,1.2850775210863403), NE * labelscalefactor); dot((-1.358745393538734,-0.18620556621259715),dotstyle); label("$A_2$", (-1.5281017201342948,0.014905071619632494), NE * labelscalefactor); dot((-0.1934445214904088,-1.2019495208166893),dotstyle); label("$A_3$", (-0.3955312860264823,-1.4034541636181912), NE * labelscalefactor); dot((0.5465132806613243,-0.874215643007063),dotstyle); label("$A_4$", (0.6311914439591048,-0.9377242654803984), NE * labelscalefactor); dot((0.5253437398368792,0.7981780821241024),dotstyle); label("$A_n$", (0.5676828214857694,0.904025786246328), NE * labelscalefactor); dot((0.9280746350051485,-0.13806198666430491),dotstyle); label("$A'_2$", (0.9699040971502262,-0.027434010029257768), NE * labelscalefactor); label("$89.37^{\circ}$", (-1.5175169497220722,-0.7154440868237245), NE * labelscalefactor,qqwuqq); label("$107.51^{\circ}$", (-0.3531922043775921,0.6076522147040961), NE * labelscalefactor,qqwuqq); label("$115.03^{\circ}$", (-0.448455138087595,-0.8001222501215051), NE * labelscalefactor,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

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