Clearly the number of sides is even, let it be 2n. Suppose that AB and CD are two consecutive vertical sides (i.e. BC is a non vertical side), then m(ABC)=m(BCD). Note that one of ABC and BCD is an internal angle, hence the sum of the two consecutive internal angles are 360. Therefore the sum of internal angles is 360n. If no sides are intersecting, then the sum of internal angles is (2n-2)360. Hence n=2n-2, and n=2. It is trival to check the case n=2.

Sorry I made a small mistake, the equation should be 360n=180(2n-2), which is impossible for any n.

Problem: Let $\mathcal{P} = P_1 P_2 . . . P_{2n}$ be a polygon, possibly concave but not self-intersecting. Suppose there exists a line $\ell$ such that for each $k = 1 , 2, \ldots , n$, ray $P_{2k-1}P_{2k}$ intersects $\ell$ and is perpendicular to it. Prove that $\ell$ intersects $\mathcal{P}$. Draw the odd-labeled points first, that is $P_1, P_3, \ldots , P_{2n-1}$. Notice that all lines of the form $P_{2i-1}P_{2i}$ are parallel to each other. Draw lines $\ell_1, \ell_3, \ldots , \ell_{2n-1}$ through $P_1, P_3, \ldots , P_{2n-1}$ all parallel to each other. Each point $P_{2i}$ for all $i$ from $1 \to n$ must lie on line $\ell_{2i-1}$ by the parallel conditions. Assume that $\ell$ does not intersect $\mathcal{P}$. We rotate the polygon so that lines $\ell_i$ are horizontal. Consider the rightmost and leftmost odd-labeled points, $P_{2a-1}$ and $P_{2b-1}$, respectively. We see that $\ell$ must be to the right of $P_{2a-1}$ or left of $P_{2b-1}$ if it does not intersect $\mathcal{P}$. WLOG $\ell$ is far to the right. Since $\ell$ must also hit all rays $P_{2i-1}P_{2i}$, it follows that for all positive integers $i$, point $P_{2i}$ is to the right of $P_{2i-1}$. Now consider polygon $P_1P_2P_{2n-1}P_{2n}$. Note that $P_2$ is to the right of $P_1$ and $P_{2n-1}$ is to the left of $P_{2n}$. If $\mathcal{P}$ is to be not self-intersecting, then all other points must be on one side of $P_1P_{2n}$, but this does not hold, and we have our contradiction. $\blacksquare$ Remark: Please correct me if there is something wrong.

This problem feels a bit vacuous. Suppose otherwise. I claim that all rays $P_{2k-1}P_{2k}$ must point in the same direction. Otherwise, if there are two oppositely oriented rays $P_{2i-1}P_{2i}$ and $P_{2j-1}P_{2j}$, $\ell$ must pass between $P_{2i-1}$ and $P_{2j}$. But it follows that there exist points on both sides of $\ell$, which means that there exists a segment $P_iP_{i+1}$ that is intersected by $\ell$, thus $\ell$ intersects the interior. On the other hand, the quadrilateral $P_1P_2P_{2n-1}P_{2n}$ is self-intersecting now, contradiction!

Consider the leftmost and rightmost vertical sides. We can see that these are the leftmost and rightmost extents of the polygon, as each vertex must be part of a vertical side and any vertex further left or right would cause a contradiction. Thus the entire polygon is contained within the two lines given by these two sides. However, let $L_1L_2$ be the left edge and $R_1R_2$ be the right edge, with $L_1$ below $L_2$ and $R_1$ below $R_2.$ The edges of the polygon must connect $L_1$ to $R_2$ and $R_1$ to $L_2,$ and these paths must intersect by our containment.

Can someone please check and tell me if this solution actually works. I had solved this a few centuries back and after being reincarnating, I was too lazy to check if the solution actually works and just copied it down directly from an ancient manuscript. Suppose FTSOC it is possible. So all teh sides $P_{2k-1}P_{2k}$ must be on one side of $\ell$ and all $\overrightarrow{P_{2k-1}P_{2k}}$ must be directed perpendicularly towards $\ell$. Now if three adjacent vertices are collinear, then we can induct down. Otherwise all three consecutive vertices are non-collinear. We try to induct down. $n=2$ fails. Now consider $P_{2n-5},P_{2n-4},P_{2n-3},P_{2n-2},P_{2n-1},P_{2n}$. FTSOC this works. remove $P_{2n-3},P_{2n-2}$ and join $P_{2n-4} - P_{2n-1}$. THis does not work due to induction. So the segment $P_{2n-4}P_{2n-1}$ intersects $P_{2n} P_1$. Call $P_{2n}P_1$ as $\mathcal L$. Note that due to continuity, as $P_{2n-4}P_{2n-1}$ intersects $P_{2n}P_1$, we get any continuous curve from $P_{2n-4}$ to $P_{2n-1}$ intersects $\mathcal L$, and thus, $P_{2n-4}P_{2n-3}P_{2n-2}P_{2n-1}$ intersects it somewhere, contradiction.

jj_ca888 wrote: Problem: Let $\mathcal{P} = P_1 P_2 . . . P_{2n}$ be a polygon, possibly concave but not self-intersecting. Suppose there exists a line $\ell$ such that for each $k = 1 , 2, \ldots , n$, ray $P_{2k-1}P_{2k}$ intersects $\ell$ and is perpendicular to it. Prove that $\ell$ intersects $\mathcal{P}$. Assume not and let $\ell$ be a horizontal line below $\mathcal P$. Look at leftmost and rightmost vertical segment, which we denote $\overrightarrow{P_{2i-1} P_{2i}}$ and $\overrightarrow{P_{2j-1} P_{2j}}$, as shown below. [asy][asy] pair A = (-4,3); pair B = (-4,1); pair C = (2,6); pair D = (2,1.7); draw(A--B, blue+1.3, EndArrow(TeXHead), Margins); draw(C--D, blue+1.3, EndArrow(TeXHead), Margins); dot("$P_{2i-1}$", A, dir(180), blue); dot("$P_{2i}$", B, dir(180), blue); dot("$P_{2j-1}$", C, dir(0), blue); dot("$P_{2j}$", D, dir(0), blue); draw(A..(0,3)..(1,2)..D, dashed + deepgreen, Margins); draw(B..(0,2.3)..(0.1,3.2)..C, dashed + deepgreen, Margins); draw((-5.5,0)--(3.5,0), grey, Arrows); label("$\ell$", (1,0), dir(-90)); [/asy][/asy] Consider the rest of the polygonal path from $P_{2j}$ to $P_{2i-1}$ (i.e.\ the trajectory $P_{2j} \to P_{2j+1} \to \dots \to P_{2i-1}$), and from $P_{2i}$ to $P_{2j-1}$, which are shown as green paths in the cartoon above. They must intersect for continuity reasons --- contradiction.

HamstPan38825 wrote: On the other hand, the quadrilateral $P_1P_2P_{2n-1}P_{2n}$ is self-intersecting now, contradiction! why is that a contradiction? the entire polygon is given to be non-self-intersecting, but $P_2P_{2n-1}$ is not even an edge of the original polygon, so it intersecting $P_1P_{2n}$ should not be a contradiction no?