Romanian Mathematics Olympiad

This problem could easily be solved by approximating each radical but approximation was the only thing you were not allowed to do..

[BartBuzz]

Your combination of logic and math is always educational. Thanks for challenging me.

[flowerflour]

he speaks as if he's whispering secrets as he solves this, , , thing, , which feels right. It's one of those things you'd think you aren't meant to know as a child

[jackrubin]

Thank you. At 68 years old I learnt something new today.

[UncleHo]

As a Romanian graduating looong time ago, when the specialized mathematics gazzete called (guess the name) Gazeta Matematică was the nightmare of every student from 6th grade until end of 12th grade (high school) I never thought that I will see a day when a problem from that journal will make it to the whole world.

[dansheppard]

Best math problem channel on YouTube. Interesting problems with really clearly presented solutions.

[matei]

As a romanian that has the olympiad next week this video is really what i need to see. Keep up the good work 🫶🏼

[jay_]

Since sqrt(x) is a strictly concave function, you can use Jensen's inequality to establish that sqrt((5+6+...+13)/9) = sqrt(9) = 3 > (sqrt(5) + sqrt(6) + ... + sqrt(13))/9 = S/9, i.e. 3 > S/9.
Therefore S < 27.

[Sourabh-U]

I like your
''Never stop LEARNING,
Those who STOP, STOP LIVING''

[RADARTechie]

Man, it's nice looking at the comment section of a video and seeing smart people.

[RexxSchneider]

When I saw that the sequence was centred about √9, my initial reaction was to pair up the terms as you did and think of S-3 as:
( √(9-4)+√(9+4) ) + ( √(9-3)+√(9+3) ) + ( √(9-2)+√(9+2) ) + ( √(9-1)+√(9+1) ) then take a √9 = 3 out of each term to give:
3( √(1-4/9)+√(1+4/9) ) + 3( √(1-3/9)+√(1+3/9) ) + 3( √(1-2/9)+√(1+2/9) ) + 3( √(1-1/9)+√(1+1/9) )
So (S-3)/3 = ( (1-4/9)^(1/2)+(1+4/9)^^(1/2) ) + ( (1-3/9)^(1/2)+(1+3/9)^(1/2) ) + ( (1-2/9)^(1/2)+(1+2/9)^(1/2) ) + ( (1-1/9)^(1/2)+(1+1/9)^(1/2) )
Then I could take the binomial expansion of ( (1-x)^n + (1+x)^n ) where all the odd powers cancel out making each of the 4 paired terms very slightly less than 2.
That would make (S-3)/3 slightly less than 8, which leads to S being slightly less than 27, hence its floor would be 26.
The snag is that I'd have to do an estimate on what the "very slightly" was to ensure that 12 times the estimate is less than 1 to make sure S > 26.
Kudos to you; your method is cleaner (and clearer) than mine.

[romulusmilea]

Amazingly beautiful handwriting, Sir ! Congratulations ! This makes a Math teacher even better !

[gladysqueen]

I love how these problems turn from 'how the hell am I supposed to solve this' to 'it's just numbers, man' with the use of some shrewd applied logic. I don't think any channel makes me want to learn more about this than this one.

[somepianoguy]

Just found this channel. I love the passion and excitement you show for the problem. I like how you show a bit of thinking through the process at a pace that invites me to play around with the ideas myself. Subscribed!

[zbynekba]

I enjoyed the tricky part—if you can’t calculate the exact value, aim for the boundaries. Thank you for sharing with us. Great job!

[hrothgarrrrr]

I really appreciate how careful you are with language and notation! It's rare to see a teacher who has such precision.

[MutuAdrian]

I like that you show the process of thinking to get the solution, not just cut off videos that show everything at once and make it seem like you should know the answer instantly.

[AgassiUKR]

what a positive, joyful and openhearted fellow. Glad I know he exists.

[dlcipher]

I love that you asked "why does the series start at sqrt(5) and end at sqrt (13)". Indeed, these numbers were very deliberately and elegantly chosen. And once you pair them up, and you notice that 5x13 is exactly 1 greater than 8 squared, and that 8x10 is exactly 1 less than 9 squared, you're basically halfway to the solution.

[dannyo.r.r]

I think this is my new favorite channel

[bookert]

Draw the √x function, put a bar of unit length to the right of each integer (eg, the bar at 5 starts at 5, has height √5 and has a base of 1, etc). Looking at your result, you can see S < ∫√× dx from 5 to 14, which is 27.47. Repeat, but now use bars to the left & above the √x function, this time the integral goes from 4 to 13 and equals 25.81. So 25.91 < S < 27.47. This inequality is not quite close enough, so modify it by setting the √9=3, and doing the integrals on both sides of 9, then adding those two results to the 3 ; this tightens the inequality. The integrals are easy so these additional steps are quick.