Відео

How does the bound from above @6:08 work? Why is there a fixed c such that for every large enough n the inequality holds?

No way, I have actually compeated on this MEMO, It took us (Croatia) to solve this one 10 minutes but it was the only one on the team competition that We aced.

Awesome funcional equation and clever solution!

3:38 Why was it enough to show that AF=OF?

Nice

Thank you so much 🙌🏾

Thanks for the video. I am unsure as to why you motivate the expected value as a linear function that must be equal to the probability of the event for the case of an indicator. Why is that more fundamental or why not just observe that from the definition?

This is really well done, thank you 🙏

Thank You

This is really great, better than A LOT of measure theoretic so called "Introduction" textbook! Please make more :)

Great video. A Second video on this topic would be highly appreciated.

Funny intro. These type of videos are amazing. More of then would be amazing.

Sooo gooodd

Hi there ,there's a small request pls bring series that teaches mathematics from basic to Olympiad level if you do that then it is going to very helpful for students ❤ by the way your knowledge is amazing appreciate your work man

Can you do some videos on Grid combinatorics problems and Combinatorics problems from Russia?

from Morocco thank you son...despite my foccusing and attempt to understand...i couldnot...surely i m so old for that🤣😂 i shared your link on my facebook page

amazing!

Very very cool! Keep up the great work

Amazing

Amazing vid/!!

Nice stuff

MATHA SIZE THORAA BARAA KARO...

It all seems very complex, it would be great if you could make a video starting from grass root level of maths and explaining all concepts irrespective of it being long otherwise great video🫡

Guys, if you start organising contests you probably will have a very good community.

Recently I solved a problem to find all finite sets that satisfy: if a and b distinct elements of S then (a+b) / gcd(a,b) is also in S. The funny thing is, the solution is: exactly the same, just characterizing the 3 smallest elements.

Amazing explanation

4:55 why does that imply f is injective and bijective?

Thank you! We want more. Keep going!

Why dony you use triagular inequality

Sir, which book will be great for learning that level of number ? please can you tell me

5:00 Unless you’re not allowed to use calculus, it seems like it would be easier to show the square root function is concave by showing its second derivative is always negative. If f(x) = x ¹/² then f’(x) = x ⁻¹/² /2 and f’’(x) = - x ⁻³/² / 4. Note that x ⁻³/² is always positive (it’s the reciprocal of the cube of a positive square root), so the second derivative is always negative (since it’s negative that number divided by 4).

. On a natural number n you are allowed two operations: (1) multiply n by 2 or (2) subtract 3 from n. For example starting with 8 you can reach 13 as follows: 8 → 16 → 13. You need two steps and you cannot do in less than two steps. Starting from 11, what is the least number of steps required to reach 121? Is 10 correct or there can be least number of steps which are less than 10

Wow what a solution ❤

Wild 💪🏼

But what’s the thinking process behind for coming up the inequality a_(n+2) >= a_n + 3?

Your videos are just mind blowing 🤘

I solved it with mod 16. The hardest part is to see that you should take mod 16/mod 4 than it's easy.

Only Geo in this years imo

I want to get connected with you in any social media you want. I want to discuss maths. I want you and viewers of this channel to be my friends . Reply all please

Is your black pen also a Pilot V Board Master? Black line looks thicker that blue and red lines.

I just press like button to appreciate your effort, but I never watch it, since I know I wouldn't get it even if I watch it hundred times

Sehr schönes Video!

why did assume alpha is even in the first line?

The description says RMM Shortlist, not IMO

I don’t even understand the question…

So, in other words you're saying that triangles TNC and TNC' are equal, which further implies that TC = TC'. But how are these traingles equal? We only have a common side TN, NC = NC' and <NTC=<NTC'. Is that really enough to conclude the equation?

Continue like that !!

There is another proof to this beautiful problem: Let O be the center and r be the radius of (AFE). So: BO² - r² = BF. BA = BD.BC CO² - r² = CE. CA = CD.CB We can substract them: BO² - CO² = BD² - CD² That means OD is perpendecular to BC. So: BO² = BD² + OD² We can use first equation: BD² + OD² - r² = BD.BC Hence: OD² - r² = BD.DC Also if we define T as the second intersection of AD and (AFE): OD² - r² = DT.DA Hence: DP.DA = DB.DC = DT.DA That means DP = DT. Also we know that the symmetry of T whit respect to OD is on the (AFE) again. Let's name that point Q'. By symmetry, DP = DT = DQ'. That means <TQ'P = 90°. So BC || TQ' hence OD perpendecular BC and OD perpendecular TQ'. And BC bisects TP. Hence BC is perpendecular bisector of Q'P. So Q' = Q and we are done.

12:47 I don't think p(c) is unique. It seems easy to find a counterexample for n=3.

For n=6, m=7 is not >=11?