Posts

SMO Open 2025 ??% Speedrun

-
Image
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...
Post a Comment
Read more

IOI 2025/1 was surprisingly mathy

-
Image
(Glen here.) Recently, there was some discussion about this year's IOI in a group chat, and Ker Yang remarked that the first task, " Souvenirs ", was basically a math question. Of course, I had to have a look. The original description is pretty long, so here's a rewriting in a more math olympiad style: Sheldon is buying french fries from a fast food restaurant, which sells fries in $N$ different sizes (with packagings labelled from largest to smallest from $0$ to $N-1$), with integer costs $P_0 > P_1 > \ldots > P_{N-1} > 0$. To buy fries, Sheldon can pay $\$M$ where $M$ is a positive integer, and then define a sequence $i_1,i_2,\ldots$ as follows: $i_1$ is the smallest integer such that $P_{i_1}\le M$, and in general, $i_k$ is the smallest integer $>i_{k-1}$ such that $P_{i_k} \le M - P_{i_1} - \cdots P_{i_{k-1}}$. If no such $i_k$ can be defined, the sequence terminates. Then, Sheldon receives the fries with labels $i_1,i_2,\ldots$ along with his chang...
Post a Comment
Read more

IMO 2025 Livesolve (Day 1)

-
Image
(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...
Post a Comment
Read more

SMO Senior 2025 ??% Speedrun

-
Image
 (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 $MA=MB$...
Post a Comment
Read more