Math olympiad problem you can solve

Mathematical problem from Moscow math olympiad for the graduate class.

#Puzzle #Math #YuriNeuro

Timecodes:
0:00-Intro
0:51-Math Olympiad Problem
0:56- Parametrization
1:32- Logarithm base rule
2:15- Logarithm base (chain) rule derivation
2:54 -Log base rule application for the parametrization
3:32- Simplifing qubic equation
3:45-(x-a)(x-b)(x-c) factorization formula
3:51- Answer
4:51- New math olympiad puzzle to solve
5:17- Share your ideas and check neuroscience videos

Mathematical puzzle from Moscow school mathematical olympiad (2018 and 2021). To solve this math problem one should recall the knoledge about logarithm base (logarithm chain) rules and polynomial factorization. For instance, one should now how to expand (x-a)(x-b)(x-c). However, one should not memorize neither logarithm base change (logarithm switch) rule, nor
(x-a)(x-b)(x-c) formula and can derive them by applying the definition of logartihmic function and guesses about polynomial factoring.

At the end, another mathematical olympiad question from Russian school math olympiad is given for the further discussion.

If you are interested in this mathematical question, check the other videos in this topic.

Other puzzle videos:

If you are interested in science, you can check my recent videos about novel neuroscience research.

Sources:

[yurineuro]

Timecodes:
0:00-Intro
0:51-Math Olympiad Problem
0:56- Parametrization
1:32- Logarithm base rule
2:15- Logarithm base (chain) rule derivation
2:54 -Log base rule application for the parametrization
3:32- Simplifing qubic equation
3:45-(x-a)(x-b)(x-c) factorization formula
3:51- Answer
4:51- New math olympiad puzzle to solve
5:17- Share your ideas and check neuroscience videos

[theevilmathematician]

a(a-1/b) is a²-a/b. a(b-1) is simply ab-a. But b=a-a/b. a(b-1) becomes a(a-a/b-a). Hence, if (a, b)>0, a²-(a/b) > a- (a/b)-a.

[bruh-cvec]

Found this as an ad! Finally the algorithm figured the truth thing to do

[bencheesecake]

We have b= a(1-1/b), so putting the parentheses as one fraction: b = a(b-1)/b, and since b != 0, b^2 = a(b-1).

This looks a lot like our goal, but with b^2 instead of b. Can we compare b to b^2?

If b< 1 then b^2 < b, if b = 1, b^2 = b, and if b > 1, b^2 > b, so it looks like all three are possible, but can we put them back in the original equation?

If b < 1, then (b-1) < 0, so we have b = a(b-1)/b < 0, because both a, b > 0, but then b< 0 and b > 0, so contradiction.

If b = 1, then b = a(b-1)/b = a*0/b = 0, then b = 1 and b = 0, therefore:

If b > 1, b = a(b-1)/b > 0, so no contradiction, and therefore b < a(b-1)

[lovefootball]

b = a(1 - 1/b)
which becomes
b = a(b - 1)/b
So, b^2 = a(b - 1).

So, b < a(b - 1)

[АртурГолицын-тю]

Will you make more videos about math olympiads?

[ABc-esnv]

Also found this as an add. So happy I solved it. Shoutout to Michael Penn who taught me this method for solving cubics.

[yurineuro]

Dear viewers, thanks a lot for the active discussion and your comments! I do appreacite it. Every comment will receive a response, but it sometimes not exactly on the same day. Hope for your understanding!=)

[umeshjain]

Answer is b > a(b-1)

As the given equation becomes
b = √a(b-1)
and b is always > 0 as written
So b > a(b-1)

[NavneetSingh-wtjo]

It is very easy questions for an Indian guy who is preparing for jee advanced

[ddystopia]

b < a(b - 1) i think
// не судите строго, это просто на глаз

[Zaki-zglt]

Where can I learn calculus from A to Z

[keithmasumoto]

Yes, that was nice I-0pu, thank you.

[abhishekbhamare]

0:23 A beautiful mind
This movies so sad 😢

[kalpadipbhowmik]

Bro Where are you live in ?
Thanks for I like and love MATHS and OTHERS 👉❤👈

[mushroom_foot]

Make a video of classical physics book available in