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little fermat
United Arab Emirates
Приєднався 26 лют 2021
My name is Muhammad Alhafi. I am a former IMO medalist and currently a math Olympiad coach. The goal of this channel is to help students preparing for math Olympiad contests.
British Math Olympiad | BMO1 2023: Problem 4
#IMO #BMO1 #NationalOlympiad #UKMT #MathOlympiad
In this video we solve the fourth problem of British Math Olympiad Round 1(BMO 1) of last year 2023 and which is a nice standard number theory problem.
IMO Past Problems: ua-cam.com/play/PL1qInO9wTp3kBq7mFRtDauNbuQnddGd6H.html
Functional Equations playlist: ua-cam.com/play/PL1qInO9wTp3lh0uGN9W6JufoBk7g0BDCL.html
Ultimate Geometry playlist: ua-cam.com/play/PL1qInO9wTp3m2wpdHSucdyAEKHKAqGuq1.html
Inequalities Tutorial: ua-cam.com/play/PL1qInO9wTp3mgEn4_X2niePNxrX26OJBl.html
Shorts: ua-cam.com/play/PL1qInO9wTp3m388GGbyWbw85LZRhn3dQw.html
__________________________________________________________________________________________
I upload videos concerning math Olympiad contests to help students preparing for math Olympiad and all people who share the love for some interesting math problems.
Don't forget to subscribe to the channel!
__________________________________________________________________________________________
Email me your thoughts! (littlefermat0@gmail.com)
Donate: www.paypal.com/paypalme/hommeida?locale.x=en_US
In this video we solve the fourth problem of British Math Olympiad Round 1(BMO 1) of last year 2023 and which is a nice standard number theory problem.
IMO Past Problems: ua-cam.com/play/PL1qInO9wTp3kBq7mFRtDauNbuQnddGd6H.html
Functional Equations playlist: ua-cam.com/play/PL1qInO9wTp3lh0uGN9W6JufoBk7g0BDCL.html
Ultimate Geometry playlist: ua-cam.com/play/PL1qInO9wTp3m2wpdHSucdyAEKHKAqGuq1.html
Inequalities Tutorial: ua-cam.com/play/PL1qInO9wTp3mgEn4_X2niePNxrX26OJBl.html
Shorts: ua-cam.com/play/PL1qInO9wTp3m388GGbyWbw85LZRhn3dQw.html
__________________________________________________________________________________________
I upload videos concerning math Olympiad contests to help students preparing for math Olympiad and all people who share the love for some interesting math problems.
Don't forget to subscribe to the channel!
__________________________________________________________________________________________
Email me your thoughts! (littlefermat0@gmail.com)
Donate: www.paypal.com/paypalme/hommeida?locale.x=en_US
Переглядів: 1 209
Відео
British Math Olympiad | BMO1 2023: Problem 3
Переглядів 3742 місяці тому
#IMO #BMO1 #NationalOlympiad #UKMT #MathOlympiad In this video we solve the third problem of British Math Olympiad Round 1(BMO 1) of last year 2023 and which is a standard geometry problem. IMO Past Problems: ua-cam.com/play/PL1qInO9wTp3kBq7mFRtDauNbuQnddGd6H.html Functional Equations playlist: ua-cam.com/play/PL1qInO9wTp3lh0uGN9W6JufoBk7g0BDCL.html Ultimate Geometry playlist: ua-cam.com/pla...
British Math Olympiad | BMO1 2023: Problem 2
Переглядів 3702 місяці тому
#IMO #BMO1 #NationalOlympiad #UKMT #MathOlympiad In this video we solve the second problem of British Math Olympiad Round 1(BMO 1) of last year 2023 and which is an easy number theory problem. IMO Past Problems: ua-cam.com/play/PL1qInO9wTp3kBq7mFRtDauNbuQnddGd6H.html Functional Equations playlist: ua-cam.com/play/PL1qInO9wTp3lh0uGN9W6JufoBk7g0BDCL.html Ultimate Geometry playlist: ua-cam.com/...
British Math Olympiad | BMO1 2023: Problem 1
Переглядів 4932 місяці тому
#IMO #BMO1 #NationalOlympiad #UKMT #MathOlympiad In this video we solve the first problem of British Math Olympiad Round 1(BMO 1) of last year 2023 and which is a nice but tricky counting problem. IMO Past Problems: ua-cam.com/play/PL1qInO9wTp3kBq7mFRtDauNbuQnddGd6H.html Functional Equations playlist: ua-cam.com/play/PL1qInO9wTp3lh0uGN9W6JufoBk7g0BDCL.html Ultimate Geometry playlist: ua-cam....
Inequalities Tutorial #7 -Wrapping up Muirhead
Переглядів 2992 місяці тому
#IMO #Algebra #Inequalities #MathOlympiad In this video we wrap up Muirhead's inequality by discussing an IMO shortlist problem that can't be solved directly by Muirhead, but rather some application of AM-GM is required first. Inequalities playlist: ua-cam.com/play/PL1qInO9wTp3mgEn4_X2niePNxrX26OJBl.html&si=ELV8IYRYz5cyq0sG IMO Past Problems: ua-cam.com/play/PL1qInO9wTp3kBq7mFRtDauNbuQnddGd6...
Inequalities Tutorial #6 - Muirhead vs IMO/P3
Переглядів 3723 місяці тому
#IMO #Algebra #Inequalities #MathOlympiad In this video we solve two inequality problems. The first is the classical Nesbitt's inequality and the second is the famous IMO 2005 Problem 3! Inequalities playlist: ua-cam.com/play/PL1qInO9wTp3mgEn4_X2niePNxrX26OJBl.html&si=ELV8IYRYz5cyq0sG IMO Past Problems: ua-cam.com/play/PL1qInO9wTp3kBq7mFRtDauNbuQnddGd6H.html Functional Equations playlist: ua...
Inequalities Tutorial #5 - Intro to Muirhead
Переглядів 4793 місяці тому
#IMO #Algebra #Inequalities #MathOlympiad In this video we move on to discuss a new very important inequality, which is a kind of generalization of AM-GM, but will save lots of time and effort compared to it and that is the Muirhead's inequality. Inequalities playlist: ua-cam.com/play/PL1qInO9wTp3mgEn4_X2niePNxrX26OJBl.html&si=ELV8IYRYz5cyq0sG IMO Past Problems: ua-cam.com/play/PL1qInO9wTp3k...
Inequalities Tutorial #4 - Wrapping up AM-GM
Переглядів 4954 місяці тому
#IMO #Algebra #Inequalities #MathOlympiad In this video we wrap up our discussion for the AM-GM inequality finishing the most important ideas and tricks needed in math Olympiad inequality problems. Inequalities playlist: ua-cam.com/play/PL1qInO9wTp3mgEn4_X2niePNxrX26OJBl.html&si=ELV8IYRYz5cyq0sG IMO Past Problems: ua-cam.com/play/PL1qInO9wTp3kBq7mFRtDauNbuQnddGd6H.html Functional Equations p...
Inequalities Tutorial #3 - More AM-GM ideas
Переглядів 5414 місяці тому
#IMO #Algebra #Inequalities #MathOlympiad In this video we continue with more ideas on AM-GM inequality, the ideas discussed include (Min/Max, Homogenization, important factorization formulae). Inequalities playlist: ua-cam.com/play/PL1qInO9wTp3mgEn4_X2niePNxrX26OJBl.html&si=ELV8IYRYz5cyq0sG IMO Past Problems: ua-cam.com/play/PL1qInO9wTp3kBq7mFRtDauNbuQnddGd6H.html Functional Equations playl...
Inequalities Tutorial #2 - Starting Examples on AM-GM
Переглядів 7034 місяці тому
#IMO #Algebra #Inequalities #MathOlympiad In this video we discuss Starting examples on AM-GM inequality which illustrate many techniques including (Pairing, Reverse AM-GM). Inequalities playlist: ua-cam.com/play/PL1qInO9wTp3mgEn4_X2niePNxrX26OJBl.html&si=ELV8IYRYz5cyq0sG IMO Past Problems: ua-cam.com/play/PL1qInO9wTp3kBq7mFRtDauNbuQnddGd6H.html Functional Equations playlist: ua-cam.com/play...
Inequalities Tutorial #1 - Intro to AM-GM
Переглядів 1,4 тис.5 місяців тому
#IMO #Algebra #Inequalities #MathOlympiad In this video we discuss an intro to math Olympiad inequalities beginning with the AM-GM inequality. Inequalities playlist: ua-cam.com/play/PL1qInO9wTp3mgEn4_X2niePNxrX26OJBl.html&si=ELV8IYRYz5cyq0sG IMO Past Problems: ua-cam.com/play/PL1qInO9wTp3kBq7mFRtDauNbuQnddGd6H.html Functional Equations playlist: ua-cam.com/play/PL1qInO9wTp3lh0uGN9W6JufoBk7g0...
UKMT Senior Mathematical Challenge (SMC) 2023
Переглядів 7629 місяців тому
#IMO #SMC #RegionalOlympiad #UKMT #MathOlympiad In this video we solve the Senior Mathematical Challenge (SMC) of this year 2023 and show the common tricks and how one approaches the paper. IMO Past Problems: ua-cam.com/play/PL1qInO9wTp3kBq7mFRtDauNbuQnddGd6H.html Functional Equations playlist: ua-cam.com/play/PL1qInO9wTp3lh0uGN9W6JufoBk7g0BDCL.html Ultimate Geometry playlist: ua-cam.com/pla...
Tangents of Circles| Angle Chasing| The Ultimate Geo Course
Переглядів 35410 місяців тому
#IMO #Geometry #AngleChasing #MathOlympiad In this video we discuss the basics of tangents one needs to keep in mind to handle geo problems involving them. IMO Past Problems: ua-cam.com/play/PL1qInO9wTp3kBq7mFRtDauNbuQnddGd6H.html Functional Equations playlist: ua-cam.com/play/PL1qInO9wTp3lh0uGN9W6JufoBk7g0BDCL.html Ultimate Geometry playlist: ua-cam.com/play/PL1qInO9wTp3m2wpdHSucdyAEKHKAqGu...
The Golden Point| Angle Chasing| The Ultimate Geo Course
Переглядів 40810 місяців тому
#IMO #Geometry #AngleChasing #MathOlympiad In this video we discuss the Incenter-Excenter lemma and we define what I like calling the golden point! IMO Past Problems: ua-cam.com/play/PL1qInO9wTp3kBq7mFRtDauNbuQnddGd6H.html Functional Equations playlist: ua-cam.com/play/PL1qInO9wTp3lh0uGN9W6JufoBk7g0BDCL.html Ultimate Geometry playlist: ua-cam.com/play/PL1qInO9wTp3m2wpdHSucdyAEKHKAqGuq1.html ...
Reflecting Orthocenter| Angle Chasing| The Ultimate Geo Course
Переглядів 28310 місяців тому
#IMO #Geometry #AngleChasing #MathOlympiad In this video we discover a secret weapon that blows up many hard problems involving the Orthocenter. IMO Past Problems: ua-cam.com/play/PL1qInO9wTp3kBq7mFRtDauNbuQnddGd6H.html Functional Equations playlist: ua-cam.com/play/PL1qInO9wTp3lh0uGN9W6JufoBk7g0BDCL.html Ultimate Geometry playlist: ua-cam.com/play/PL1qInO9wTp3m2wpdHSucdyAEKHKAqGuq1.html Sho...
The Orthic Triangle| Angle Chasing| The Ultimate Geo Course
Переглядів 38010 місяців тому
The Orthic Triangle| Angle Chasing| The Ultimate Geo Course
IMO 2023 P2 Review (Japan)| A sweet geometry
Переглядів 1 тис.11 місяців тому
IMO 2023 P2 Review (Japan)| A sweet geometry
IMO 2022 P4 Review (Norway)| Don't fear complex diagrams?
Переглядів 59811 місяців тому
IMO 2022 P4 Review (Norway)| Don't fear complex diagrams?
Cyclic Quadrilaterals| Angle Chasing| The Ultimate Geo Course
Переглядів 35911 місяців тому
Cyclic Quadrilaterals| Angle Chasing| The Ultimate Geo Course
A magical reflection| Angle Chasing| The Ultimate Geo Course
Переглядів 409Рік тому
A magical reflection| Angle Chasing| The Ultimate Geo Course
IMO 2023 P1 Review (Japan)| A divisibility problem
Переглядів 1,4 тис.Рік тому
IMO 2023 P1 Review (Japan)| A divisibility problem
IMO 2022 P3 Review (Norway)| A Hard Number Theory?
Переглядів 2,2 тис.Рік тому
IMO 2022 P3 Review (Norway)| A Hard Number Theory?
Orthocenter| Angle Chasing| The Ultimate Geo Course
Переглядів 611Рік тому
Orthocenter| Angle Chasing| The Ultimate Geo Course
IMO 2022 P2 Review (Norway)| A Functional Inequality?
Переглядів 4,8 тис.Рік тому
IMO 2022 P2 Review (Norway)| A Functional Inequality?
An Awesome System Of Equations| IMO 1961 P1
Переглядів 8432 роки тому
An Awesome System Of Equations| IMO 1961 P1
Wonderful bro, where are you from
can we substitute (0,y)after we prove f is surjective and get f(x)=x+f(0)
I have a question,can we substitute(0,y)into equation after we prove f is surjective and get f(x)=x+f(0)?
Fermat = little fermat
Im definitely going to have to rewatch this 😭
im a beginner at olympiad maths and at 15:20 you expanded the product very quickly and it would really help if you explained in more detail what is the logic behind such expansions, so that one can do this on the spot. Thank you for the nice videos!
This comes from experience. However if you try to think what terms should you get and always use the idea of symmetry for example if you get a then you should get b and c, same for ab will also have bc, and ca. If you think like that (let your mind expand not your hand, you get the grasp of it soon)
If we could show that f is surjection and find f(0) and g(0) problem become not messy anymore
we need to noticde that f(x) is a injection, then using f(f(x))=f(x*x) leads to the solution.
cool one, keep going man
good solution <3
Sir I am from India Do you know about JEE Advanced exam of India
if f(f(y))=y and f is surjective does that also mean f(y)=y?
@@Marlow998 If f(f(y)) =y then f is already surjective as f takes all values. Of course that doesn't mean f(y) =y. Take f(y) =1-y as a counter example
@@littlefermat Alright thanks for the quick reply also when we assume a number alpha such that f(alpha)= smth can we assume an another numbeer beta f(beta)= smth2 ?
Where are you from?
Oh dear Lord, your channel is EXACTLY what I needed to find. Thank you so much for this, I'll be bingingnthis series and taking extensive notes prepping for math olympiads. Hope to see y'all at the IMO!
May I ask what country you are from?
@@Sparkles.08 yeah sure but for what purpose? You look like a bot if anything tbf
@@kaiserquasar3178 wow, hold up. Just wanna be friends that's it. You're preparing for an IMO if I'm not mistaken and I'm taking part in a math competition too.... that's it, I'm no harm.
@@Sparkles.08 Fr? Well then, I'm from Lithuania. You?
@@kaiserquasar3178 I'm from Namibia
In the step where you conclude that 1=f(f(y))+f(y). Would it be possible to make a substitution such that u=f(y)? Which would make the equation 1=f(u)+u. This can then be rearranged to become f(u)=1-u.
How can we know that this is the only solution?
Hello sir, please do you offer 1 to 1 online lessons ?
@@Samuel-zs9gw Contact me on email at littlefermat0@gmail.com
Is there another word for the circle method? I have tried to search for it on Google but I didn’t get any results
@@user-bi3oc2jt4t Nah, I just came up with the word as it is literally like a circle!
@@littlefermat But where did you learn about it? I want to learn more about this tool
THANK YOU!!! All the other videos on here only talked about how to prove a function is surjective, and not how to use it’s properties to solve problems. This really helped me
@@user-bi3oc2jt4t glad you liked the video, the whole playlist is discussing techniques to solve functional equations.
@@littlefermatAwesome
Thank you!
@@fahrenheitwastaken you are welcome!
Very fun and easy problem!
thank you!
@@erapesmobile4216 you are welcome!
Great🎉
hi sir please can you solve this functional equation i try for more times but i can't solve it f(xf(y)-f(x))=2f(x)+xy
Hey sir, I was wondering what I should do in order to prepare for math Olympiad. Im currently year 10 (gcse) and can u please recommend a few textbooks for the preparation.
Solution 3: Let H be the orthocentre of ABC. Then HBX=HBA=HCA=HCX, so HBCX concyclic. This gives BXC=BHC=180-BAC=180-BZC=BYC. Also XCB=YCB and XB=YB, so triangles XCB and YCB are congruent. That gives BC perpendicular to XY, so BC perpendicular to ZX. Combining this with BXC=180-BZC implies that X is orthocenter of BCZ, so BZ perpendicular to CX
After BXHC concyclic we can also prove BC perpendicular to XY by using that reflecting (BHC) in BC gives (BAC), so reflection X’ of X in BC is the point in (ABC) with BX’=BX=BA which is Y. So XY perpendicular to BC
Beautiful solution
Another solution: B is on perp bisector of AY, so on angle bisector of AZY. So B is on angle bisector of AZX and on perp bisector of AX. But obviously ABXZ aren’t concyclic since X!=C. This means that the angle bisector of AZX is the same line as the perp bisector of AX. So Z is on perp bisector of AX. Since B is also onnthat perp bisector we have BZ perpendicular to AC
Solved in 1 minute, solution: angle chase BZX+ZXA=BZY+180-AXY=BAY+ABY/2=90, so BZ is perpendicular to AC
Well I came up with a difference solution using both "bad points". So let D be the intersection of |AY| and |BZ|. Notice that since |BY|=|BA| we have that B is the mid point of the arc AY so BZ is the angle bisector of [angle AZY] and used this with the fact that ABCZY is cyclic we'll have [angle AZB]=[angle YZB]=[angle BYA]=[angle BAY]. Now notice that ∆BYD~∆BZY because [angle BYD]=[angle BZY] and they have a common angle at B therefore [angle BDY]=[angle BYZ] but the ∆BXY is isosceles so [angle BYZ]=[angle BYX]=[angle BXY]=[angle BDY] so the quadrilateral BYXD is cylic thefore [angle DYX]=[angle XBD]. With this we have [angle AYZ]=[angle ABZ]=[angle XBD] so BZ is the angle bisector of [angle XBA ] and since ∆BXA is isosceles it is also it's perpendicular bisector so that finishes the problem.
Great job! To be honest, this problem is not that hard, that whatever angle you pick you can find a way to handle it. The strategy of reducing the number of bad points however becomes very important in much harder problems. I will upload a video explaining more with a much more suitable example.
@@littlefermatlooking forward to it.
@@littlefermatthanks
Thanks for the video. Here a solution involving the bad points X and Y: As |AB| = |BY|, one has that [angle AZB] = [angle ACB] = [angle BZY] = [angle BCY]. In the circle with center at B and passing through A, X and Y, one has that [angle AXY] + [angle ABY]/2 = 180°, which means that [angle AXZ] = [angle ABY]/2. As the points B and Z lie on opposite sides of AY, one has that [angle ABY] + [angle AZY] = 180° = 2*[angle AXZ] + 2*[angle BZY], i.e. [angle AXZ] + [angle BZY] = 90°, which means that AC and BZ are perpendicular.
Nice! Indeed there are many ways to tackle this problem regardless of considering angles with bad points. However, this strategy is affective when handling angles in harder problems. I will try to post a video where I illustrate the whole principle with a really complicated angle chasing problem.
Nice solution
@@littlefermatPlease do it :).
Lets go!!!
I am going to this year's IMO any advices for the contest?
congrats mate
Good luck my friend. Make sure to relax and have fun and don't forget to check the marking scheme with your leader in case you have written something valuable on draft during the competition. Also make sure you spend two weeks revising before you go there. Good luck again!
this was fairly easy problem
Indeed
Its a good day when little fermat uploads
I found the answer graphically in less than 1 minutes😭 1) the function is odd and therefore its graph is symmetrical about the origin 2) for all values of x, when we increase X by 1 f(x) is also increased by 1, therefore it had to be a line of gradient/slope 1, only the graph of y=X could hold 3) just checked if f(x)=X was true for the last one, And bam!, I know it just worked for that particular case but at that instant of my life, I felt so smart😭😭 It gives me hope Thanks so much for your videos ❤️!
We had this as P2 last year during nationals with the word KLAUSUR(German for exam)
Oh thanks for the info! and good luck to you
discontinuing inequalities?
😢
Nope but I was having fun with this bmo1 paper so I decided to post my solutions. I will surely continue the inequalities series once this is uploaded
@@littlefermat alr, do what u think is fun
Great video. In fact, the identities are valid for x different from 0 and 1. You can check that for all real numbers a, the piecewise function defined by f(x) = {x if x<>0 / a if x=0 / a+1 if x=1 satisfies the functional equation.
Thanks for the awesome video! I always love your content. Do you happen to offer any online coaching for the IMO? Love from Syria 🤍
How do you write pie -2x , plz explain
Next inequality? Cauchy?
Yep
Thnx❤
we deserve a projective geometry course bro. give that as an early Christmas gift.
Thanks bro