IMO Math

Starting now, I will occasionally post Putnam problems on this blog. However, the main content will still be IMO Problems.  These problems will be under the tag #putnam.

Hello,  I have enabled ads on this blog as a fund for my future education. Please cooperate with this and if they annoy you, click the down arrow button next to the ad to hide it. Thank you!

 Hello,  I just noticed that the blog was private for a while. I fixed the problem now. Sorry for the inconvenience.

 From now on, I will only stream on Twitch if someone wants one-on-one detailed help on a solution they don't understand on my blog. 

This was the huge surprise I was talking about in the other post! Now, we have a site like this for Physics as well, which you can find at  thinkphysics1.blogspot.com .  I have combined these two to form my organization named Thonks, and its website can be found at  thonks123123.github.io/Thonks .  I hope you like it!

 Hi,  I'm sorry but I won't be able to stream today.  I have a little pain in neck and it hurts a bit when I talk. I will stream again next week (Friday, April 15th) at 7:00 PM

Now, we are officially spread all throughout the world! We have views from every continent (well other than Antarctica, of course).  We already had it spread in all continents other than Africa but now we have many views from Africa as well. Before, it was technically "spread" in Africa, but there were minimal views from there.  I'm really happy about this :) Thank you so much to everyone who views my blog :D I hope it continues to help you.

Hi,  I have decided that sometimes, I will pick problems that are not from the IMO shortlists but I believe that it is an IMO-level problem (in other words, I think that the problem could have appeared in the IMO as well). Because after all, there are a limited number of problems.  

In  this  post, Anonymous suggested that I should start a Twitch stream. Thanks for the suggestion! I will stream every Friday at 7:00 PM EST, starting March 25th. You can find my Twitch account here .  You can submit problems that you want me to solve during the stream at my email, imomath@imomath.xyz, or at my discord handle, math4l#4750.  If I get no problem requests, then I will pick a random problem to solve and explain. 

Today (March 18th) is my birthday! I was born on March 18th, 2009.  Thanks to my parents for supporting me in my journey so much :) I probably would've given up a long time back if it wasn't for them.  Now it's time for 2009 IMO SL #18, which is 2009 IMO SL #G3.  Let $ABC$ be a triangle. The incircle of $ABC$ touches the sides $AB$ and $AC$ at the points $Z$ and $Y$, respectively. Let $G$ be the point where the lines $BY$ and $CZ$ meet, and let $R$ and $S$ be points such that the two quadrilaterals $BCYR$ and $BCSZ$ are parallelogram. Prove that $GR=GS$. Let the incircle touch $\overline{BC}$ at $X$ and let the $A$-excircle $\omega_A$ touch $\overline{BC}$ at $X'$ and $\overline{AC}$ at $Y'$. Denote $\omega_R$ and $\omega_S$ as the circles centered at $R$ and $S$ respectively, both with radius $0$. Notice that $BR=CY=CX=BX'$ and $YR=BC=AY'-AY=YY'$, so we have that $\overline{BY}$ is the radical axis of $\omega_A$ and $\omega_R$. Similarly, $\overline{CZ}

Due to several emails requesting to close QOTD (people found them distracting to the main purpose of the blog), I will not be posting any more quotes. I have deleted the two quotes I posted as well. By the way, if you didn't know, you can contact me at imomath@imomath.xyz or at math4l#4750 on Discord for any questions or suggestions. 

This blog just reached 1 million views... I'm glad to know that my solutions are helping people! Keep visiting this blog, more interesting solutions are to come.  As of now, I have no plans of closing this blog, and there will usually be 2-3 posts a week.   Also, if you have any questions or suggestions for this blog, please contact me at imomath@imomath.xyz  

Hello,  I am making changes to this site today, so please visit after 6pm EST. The content might look messed up and weird before that.  Thank you!

$$ne^n = \sum_{k=0}^{\infty}\frac{n^k(n - k)^2}{k!}.$$   I dreamt of this identity yesterday night but I don't even know how to prove it lol

Due to the request of a few users, I got a custom domain at  imomath.xyz . The  math4l.blogspot.com  link will still work, but you can just go to  imomath.xyz  now. 

I.  Am.  Completely.  Shocked.  I couldn't even think, in my dreams, for this blog to blow up so much. I'm glad to know that my solutions are helping people! Keep visiting this blog, more solutions are to come! As of now, I have no plans of closing this blog, and I plan to continue posting daily.  Btw, I won't be posting much of these "X views" anymore, since they're getting so frequent lol. I'll only post at some major number of views, like 100K, 150K, 200K, etc.

I am speechless. Thank you so much for 10K views! I will try to keep posting interesting content every day. Keep visiting! By the way, I have recently added a new feature in this blog. If you have a specific contest math problem you would like me to solve, please use the "Problem/Advice request" form on the left side of the homepage. It doesn't have to be an olympiad problem, but as long as it is a contest problem you can submit it through the form. You can use the same form if you want to ask for any advice. I will make a blog post regarding your problem/advice request.

Thanks everyone for 3K views on this blog! I didn't expect it to blow up this much. Keep viewing it, more interesting content is to come!

At this point,  I have decided that I am only going to do IMO Shortlist problems.  Once I finish EGMO, I will keep doing this if I don't feel like the problems have started to repeat.  If I do feel that way, then I will then go towards research and most likely close this blog.  These plans might change in the future.

Since winter break is now over for me, although I will still be posting daily, I will post 3-6 problems instead of more.  However, the content will still be high quality! You won't ever run out of posts to view, and you won't ever be bored.