Combo problem about regular polygons

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Combo problem about regular polygons

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Source: 1st TASIMO, Day1 Problem3

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NJAX

#1 h May 18, 2024, 9:06 AM • 2 Y

Y by dmusurmonov, Seungjun_Lee

$Abdulqodir$ cut out $2024$ congruent regular $n-$ gons from a sheet of paper and placed these $n-$ gons on the table such that some parts of each of these $n-$ gons may be covered by others. We say that a vertex of one of the afore-mentioned $n-$ gons is $visible$ if it is not in the interior of another $n-$ gon that is placed on top of it. For any $n>2$ determine the minimum possible number of visible vertices.

Proposed by David Hrushka, Slovakia

This post has been edited 3 times. Last edited by NJAX, May 18, 2024, 9:41 AM

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#2 h Apr 27, 2025, 5:49 PM • 3 Y

Y by B1t, sami1618, Pomegranat

Suppose first that $n \geq 3$ is even, write $n = 2k$ , $k \geq 2$ , and consider an $n$ -gon $X$ which is not covered by any other (there must clearly be at least one, take any if there are more than one) and an $n$ -gon $Y$ which has this property after removing $X$ from the table. Since any diagonal connecting a pair of opposite vertices in $Y$ is the longest among all segments contained in $Y$ , and hence also in $X$ , no pair of opposite vertices of $Y$ can be covered by the interior of $X$ .

It follows that there are at least $k$ visible vertices of $Y$ which together with $2k$ visible vertices from $X$ make a lower bound $3k$ .

If $n$ is odd, i.e., $n = 2k + 1$ for some $k \geq 1$ , we choose $X$ and $Y$ in the same way and moreover pick $Z$ being ``on top'' after both $X$ and $Y$ are removed. Then we argue analogously as above (using the fact that the diagonal connecting a vertex $V$ of an $(2k+1)$ -gon $G$ with any endpoint of the opposite side to $V$ in $G$ realizes the longest segment contained in $G$ ) to conclude that at least $k + 1$ vertices from $Y$ are visible and at least one vertex of $Z$ must be visible. Thus, we get a lower bound of
$(2k+1) + (k+1) + 1 = 3k+3.$
The following constructions prove that those lower bounds are achievable for any $n \geq 3$ .

First, consider the case $n = 2k$ again. Take three copies $P, Q, R$ of a regular $2k$ -gon, place $R$ with its side $AB$ going horizontally and being the bottom one, $Q$ precisely on top of $R$ , and $P$ precisely on top of $Q$ . Then rotate $Q$ around $A$ and $P$ around $B$ , one clockwise, the other counterclockwise, both by equally large angles so that the segment $AB$ is not covered by $P$ and $Q$ anymore and the opposite vertex of $A$ in $Q$ and the opposite vertex of $B$ in $P$ meet.

It is easy to see that only $A$ and $B$ are now visible among the vertices of $R$ . Shifting finally $P$ and $Q$ down by distances $d$ and $2d$ for some $d>0$ small enough makes no vertices of $R$ visible and precisely $k$ vertices of $Q$ visible. The remaining $2019$ $2k$ -gons can be placed identically as $R$ (under it) which makes in total $2k+k=3k$ visible vertices.

Secondly, let $n = 2k+1$ . Analogously, place four copies of a $(2k+1)$ -gon on the table, one precisely on top of the previous ones, one common side lying horizontally and below. Then observe that shifting the top layer to the right, the second one to the left, then the first one down, then the second one also down (but less), and finally the third one up, with all distances chosen properly (and after hiding $2018$ more copies of our $(2k+1)$ -gon under the fourth layer) we get $(2k+1) + (k+1) + 1 = 3k+3$ visible vertices.

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