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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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Silly Geometry Solutions

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(Drew here.) I've done a post on each of A, C and N, and decided it's finally time to finish off with G. My style of geometry is very unconventional, as people have pointed out throughout the years. If I had a nickel for every time I used inversion after I failed to perform an angle chase... I'd probably have over half a dollar by now. So here's a compilation of some of my most unique and/or elegant geometry solutions, and comments on my style. 2022 IMO Mock 2 P1 Let $APBCQ$ be a cyclic pentagon. A point $M$ inside triangle $ABC$ is such that $\angle MAB=\angle MCA$, $\angle MAC=\angle MBA$, and $\angle PMB=\angle QMC=90^\circ$. Prove that $AM$, $BP$, and $CQ$ concur. Being in my inversion addiction phase, after trivial angle chasing didn't work, I immediately attempted inversion. Point $M$ is where all the angle conditions are, so why not invert at $M$? I'll leave you to check that this just returns the same problem statement. At times even that could produce ...
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Polynomials and Newton Polygons

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(Guest author Andrew here!) A good exercise for Olympiad students is to prove the following: A real polynomial $f(x)\in \R[x]$ which is everywhere non-negative ( psd , short for positive semi-definite) must be a sum of squares ( sos ).  This is in some sense an algebraic witness to the analytic property of being non-negative, and is an example of what is known as a Positivstellensatz . One might reasonably ask whether this extends to the case of multiple variables, and it turns out the answer is no .  Consider the two variable polynomial $g(x,y)=x^4y^2+x^2y^4-3x^2y^2+1$ (the Motzkin polynomial ). Non-negativity follows from the AM-GM inequality. But how can we show that it isn't a sum of squares? We now introduce an object known as the Newton polytope. Given a polynomial $f$,  To every monomial with non-zero coefficient, assign a point whose $i^{th}$ coordinate is the degree of the $i^{th}$ variable in that monomial, e.g. assign to $x_1^n$ the point $(n, 0)$. The Newton p...
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EGMO 2025 ??% Speedrun

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(It's Glen again.) Last week, I wrote about my experience test-solving EGMO 2025/3 . As it turned out, I had some free time at EGMO (the night before Day 1, i.e. after we made the mark scheme for Problem 3, and the morning of Day 1 itself) to try the other problems, so I figured I'd write about them as well. I will try to be a little more brief than last week, so the post doesn't turn into a saga. Problem 1 ( EGMO 2025/1 ) For a positive integer $N$, let $c_1 < c_2 < \cdots < c_m$ be all positive integers smaller than $N$ that are coprime to $N$. Find all $N \geqslant 3$ such that $$\gcd( N, c_i + c_{i+1}) \neq 1$$ for all $1 \leqslant i \leqslant m-1$ Disclaimer: this was a pretty embarrassing case of wrong reasoning leading to the right answer. Try to spot where I went wrong! (Hint: there are many such places.) Let's try small cases: $N=3,4$ work. $N=5$ fails. $N=6$ works. If $N$ is odd, then the sequence starts with $N=1,2,\ldots$, so $3|N$. $N=3^k$ works: a...
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