SIMO Retiree Blog

Glen & Sheldon here. At last, our schedules aligned enough for us to try this year's SMO Open together. Here's an outline of our thought process while solving the problems, featuring some screenshots from our Cocreate whiteboard. Problem 1 In the triangle $ABC$, $\angle B>90^\circ$, the incircle touches the sides $BC$ and $CA$ at $D$ and $E$, respectively. The lines $ED$ and $AB$ intersect at $P$. The incircle of the triangle $AEP$ touches the sides $PE$ and $AP$ at $D_1$ and $E_1$, respectively. The lines $E_1D_1$ and $AE$ intersect at $P_1$. Suppose $P,C,E,B$ are concyclic. Prove that $BE$ is parallel to $PP_1$. Weird-looking problem. The concyclic condition means $\angle ABC = \angle AED$. Maybe it could be less weird if we write it in terms of the angles of the triangle, $\angle ABC = 90^\circ + \frac{\angle ACB}2$. Actually, the equal angles imply $\triangle ABC \sim \triangle AEP$. The parallel condition is equivalent to $\frac{AB}{AE} = \frac{AP}{AP_1}$, and the s...

(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...

 (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 $MA=MB$...

(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...