Відео

my bad. mis count.

why are there 26 syms.in the diag. but 24 in the calc.???

Great solution! thank you :)

why angle BI1A equal 90+theta/2?

i got totally stuck in a part

Do you do olympiad classes sir

Do you do olympiad classes sir ?

1. Proof it is a surjective function And use this property and get f(0)=1 2.then put y=0 and you will get a relation between f(x-1)and f(x) And then simply find the sequence and It's done ❤️❤️❤️😫

One of the easiest p6

who is there after egmo? XD

It is good solution

what it will be if we coloringin to 6 colors,how many ways will be

Thank you so much, this was a very nice explanation!

I am so wondering about a few things : 1. If we number them from 1 to n(n+1) Then we remove the n(n-1), then we make them closer all together, then we get the remaining players of order : a_1 < a_2 < ... < a_2n. Is it? 2. Does the main idea suggests that we need to prove that, out of n(n+1) we can obtain 2n players with height sequence is either increasing or decreasing?

Shouldn't the final result for c be y^2-b = 7^2-16 = 49-16 = 33 and not 49?

Amazing playlist sir. Really made me understand the incenter-excenter lemma well

angle PTR=TRA but PTR=TRA=SKR how?? someone please explain!!

There's a very simple method to solve this.

daily dose of dopamine :)

Searching for this explanation. Sera

Thanks. Your use of a concrete example of colouring of 2X2 chess board helped a lot. I am going to teach the derivation of finite crystallographic group of rotations in my course in crystallography. I will use this example.

Nice

clearly understand

How did you think that f(x) can be taken as alpha*n? Why not alpha*n + some constant say c?

Beautiful problem! Thanks. We never get tired of geometry aren't we ? 🙂

It looks like that KI pass through H in the drawing. Does it happen always or accidentally?

lovely!

I swear to god phantom points are one heck of a thing...

nice and neat

I did it in a different way

THIS IS SO BEAUTIFUL!!!!! Very nice solution!!!!!

Part (c) is actually easier than the video once you solve part (b). That is, since in part (b) you proved that f(m,n)≤½max(m,n), then you can make either m or n arbitrarily large and thus f(∞,n)<∞ which means that there is no constant upper bound C for which f(m,n)<C for all m,n

beautiful proof!!!

How do we prove that A, I and L are collinear

I feel like this was by far the hardest problem on the exam

نظرية جيدة . رسم واضح مرتب . شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا . تحياتنا لكم من غزة فلسطين .

4:17

نظرية جميلة جيدة . رسم واضح مرتب . شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا . تحياتنا لكم من غزة فلسطين .

Hi, I have worked on this problem on and off , for years. Thank you for this elegant solution, more elementary than the official solutions

Where are you 😢 plz upload more

Very good explanation

Please make a full course on math olympiad basic to advance .

Thank you!

very good. Thanks for the big help.

7 yrs And Still Learning outcomes 🎉 thnx

Fantastic ❤❤❤❤❤

where do u think it would be by today's sstandards

Nice❤❤❤❤

5:00 uhh just look at the cyclic quad with two feet and point A on it and see that the opposite angle is 180-A?

For some reason nobody did a circle centered at M with radius MK. Radius zero is nit very intuitive so 😝