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Spacetime, Special Relativity, and a Lot of Circles

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(It's Glen again!) This post is going to be long, and as the title suggests, contain very many circles. Relatedly, most of the diagrams in here are hand-drawn and scanned because no way am I going to figure out how draw all of these diagrams on computer. You have been warned.
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IMO 2025 Livesolve (Day 1)

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(Drew here.) As a retired contestant, I decided it would be fun to attempt the IMO 2025 paper and see how well I would do on it! Let's get started. Problem 1 (IMO 2025/1) A line in the plane is called sunny if it is not parallel to any of the $x$–axis, the $y$–axis, or the line $x+y=0$.  Let $n\ge3$ be a given integer. Determine all nonnegative integers $k$ such that there exist $n$ distinct lines in the plane satisfying both of the following: for all positive integers $a$ and $b$ with $a+b\le n+1$, the point $(a,b)$ lies on at least one of the lines; and exactly $k$ of the $n$ lines are sunny. To begin, I decided to try the $n=3$ case, as that's the smallest one. The points form a right triangle with $3$ points on each edge, and a sunny line is a line not parallel to any of the edges of the main triangle. We can shift the points a bit to instead form an equilateral triangle, and a sunny line is any line not parallel to any of the sides of the equilateral triangle. Important...
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SMO Senior 2025 ??% Speedrun

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 (Glen here.) The blog has been somewhat inactive because everyone has been busy, so sorry if we kept you waiting! I got sent this year's SMO Senior Round 2 paper this morning, so I thought I'd give it a shot. I don't have paper with me, so that's an extra challenge, I guess. Problem 1 Let $ABCD$ be a convex quadrilateral and $M$ be the intersection of its diagonals. Through $M$ draw a line meetin [sic] the side $AB$ at $P$ sand [sic] the side $CD$ at $Q$. Find all the quadrilaterals so that there exists the segment $PQ$ that divides the triangles $ABM$ and $CDM$ into $4$ similar triangles. ??? What on earth is this??? Here's a picture:  Ok, we probably need $\angle APM = \angle MPB = 90^\circ$? Yeah, if not, then each triangle has two distinct angles that add up to $180^\circ$, which is bad. This also tells us that $AB||DC$, as $PQ$ is perpendicular to both. If $\angle APM = 90^\circ$, then the only two possibilities for $APM$ and $BPM$ to be similar are if $MA=MB$...
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