A Fun One from the Vietnamese IMO 

these questions can actually sometimes be found in highschool entrance exams. crazy how we’ve learned all this.

A more concrete answer would be 1/393793903937577106479301743414185865462150020261045958348688864012331731304516123198845761433324488438219593808570961852064612014780620624672175445445623382895917883727710633016494081643796401279080059029404279015592123445442355194726188606869002467332145960773819828320146098500719543708842828909672937331352454838564658760130686616706686521039026472870725835633911635732397771739426937534540098162234792530262277400952960260275149179804966675094176984508713046317788443292024044223680238386586329406803944254718916260871815935375111696289754445942035497936545125868564313977047665770022714319377213978160622697269487278873119845809917406721550071458221311050676531396693901681843064092887169936333415826855426013543139044970705738911781672691422364080471999492019809282504071796204154517495882790498393867199563541419863517221565959261739686501178959417759309568516456104708920229992395230228903953063833065986161681444648022574918434659891914833112053428143196925629879411752075401355532544017417476788308017152966790564260548928897478386312037770428673468613582763168614277948609278870006347055324532210630733323281118380011252615931646701004690182075032123767325437226004873702559515067964070718560797194154998855092813980238968034648305431279385781862205048865127192139055595268009315185126800025295469023394604199061086359254433212349799318521040427532666989236907747701171802346780593669488132594120937585143570489494597412015258973495937532371308552027673665862452462496996340353575649988424249535811860120802497285563561335755572798634944409538335137253766015557723887443982767984207525075962600447534996987404419602791575303516970031347030489168581235574850237385515774445711182351283425539377557090153111789115365503761779089069170411660638323033599411217957676143335096505791910934533669664902763885344507187253341645042322357702417368822373002645986021656797833144415930392123612634034531334528128279264226280389919144186148073433090356229629574487001609945588996700009857931190244582787781981877820785356558637277173725635202063621762415552708860330373067798892396051070149086816524225341526084130866750847634080515977399218344052153044129811988865625708895172247970818244717511475389102792431715060679839217756732955249425619632592548104964162574028939416200191589900040276284996423183580269537829301856916783456589435696218814439711066029936512328512051652967106102752720230021501058568859576165644361305619324532969927181691887792263541733558947644497455771801727310607803753021989435677497179468397332031842874581636660316250318856415094613696232537028140098672438652578876667706317339702327770587307597082164866745269924314810790854857409922099931888208876920474910109408204469258308185663826674345827477443684328744712685776923850147437077008409008570138828158346805197980277364705951140387226569743797152383910274271531809109053141501420405008364306215728862473281280982903067496264707023676950228041397641708182387456387192125261922049660728334893434721711554690113649751953762093719649544106259506282234544368417844344916848902517657155685153361558834810291149212214391780357471530032013708218274165571182212666485358769490050426743923310358555482637184144991176745266924632888615386855771520624921929239357960686134080168753708870543782562454540905771134388971688342011522148286951383041405906792468715329987262184997340193718566532878070026244749756460332336440365671247018928365095159372452799133428334342520366099906761182526083567108201141288588724104232238357914686143195372407715037434209977040415079022412854911606517805279357734165915680428312491934938472712791319580956496831212171879918029774311776051369728473165255068483483066931996186347817761123025174218896915546274705502111123668819605957959687155582900219489982983060785692794814528133977551149181329523978870287909448589934972634913413696109945669390906497298702110302827788158757732929471952188976645780904858003262302119529067867009734725419656542228580307526339149805610503254400913020517140751943563039426792010722261237757210999688897482051826499691470761972612659923515867214526010081917608017819635760125117598408088791374527767798380362721651923772243995626328857066137238047500869683279361993297990285234019363341025302107780021780342462141320104632035676056177623067294477786045233140202827641212364911735918289045162519723315587505292207851118406684560762538232991887799490330991049955318927640825822145387807221109421203765448093540126843802557973601665027839307238700114797369398404860298797664371224359632764205544939799250187322420566038334231005428039078708951716390191762973818257167203166588555881429752420253511610115654895377758332885282660436756527741558431433851666754892799862780042186668220619278028300460620808670210170226049422822267856889193656382801651197151367708914970663241547420235743423123679241376599515853093965541673470565769082214045565719402743347871856207715349252073940433454197876400318235192833670976078217422087070910648435845672844035095784780021928512973373711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These old days, I have learnt from Vietnamese teacher, but it seems your explanation gets more impressive, I can get all the ideas easily. Thanks a lot. The way of education is inspirational.

I started trying to solve this algebraically, but once I realized that one of the three variables has to be negative the solution becomes fairly intuitive.

Ah yes. Vietnamese here. Problems like this made me hate math. I had to solve stuff like this when I was in 11th or 12th grade. My math teachers kept telling us these aren't that hard. That they always have some sort of trick (a cheat trick) to solving them. By the time I was having finals, I had a shit ton of tricks to solve different problems shoved into my head, but understood extremely little about why anything is solved in that certain way. At present, my teachers' impressions remain still and I don't know if this sort of math problem is normal for highschoolers around the world or not. I never truly understood anything. All I received were fast like the wind explanations which I took as scooping the surface and threw it onto my plate. Asking anymore didn't help. None of my teachers helped but I had to pretend to understand. Otherwise, they'd never get off my back. All throughout middle school and highschool, all I was was a machine using formulas without much thinking. That's it. I envy those who do have passionate and understanding teachers who actually teach.

The only Imo question i have managed to solve. It is quiet straightforward that 1/a+ 1/b+1/c = 1/(a+b+c) only when two of the letters are opposite numbers.

Good solution there is one more way to solve this from second equation we can get (ab+bc+ca) /abc = (1/2022). Let's assume ab+bc+ca=k, hence abc= 2022k. Hence we can form a cubic equation as below. x3- 2022x2+kx-2022k=0 whose roots would be a, b and c. Solving above quation we get x= 2022(a),+i√k(b), -i√k(c) Now putting the value we can cancel out the conjugate terms as i^2023 will be -i and (-i) ^2023 is +i and k^2023 remains common factor hence we will be left with (1/2022) ^2023

When the trees start speaking algebruh

In Vietnam, these basic solutions you will encounter very often in high school, but they are not as scary as geometry.

Thanks for posting this fun one!

Everybody gansgta until you need to calculate the numeric value of 1/2022^2023 by hand.

Father flammable mathematics I express my sincerest gratitude towards you for this here marvellous solution

Although I had known how to solve this math problem, I am impressed by your method of addressing it. Keep it up sir

glad to see some representation for us in Vietnam 🙏🙏🙏🙏

Thanks for the nice problem and such a beautiful solution!

Ich hab dich so lang gefolgt und jetzt kam endlich ein Video über Mathe aus meiner Heimat. Yayyy.

Ah yes, the easiest math question in my country's primary school entrance exam. I did not answer it correctly and brought shame to my family. All jokes aside, this was one hell of an interesting problem, for all of its intricacy. Thank you Papa Flammy!

I wasn't going to post a comment on Math Olympiad, but personally these, I think are perfect for Robotics or extracurricular activities and G/T classes where students have mastered the States' objectives and need something to do. My experiences as a teacher I found out that when I brought these students who could perform and solve these problems to sit the State Exam, they failed miserably since they overthink the concepts. I run into these problems with NCTM magazines, but I usually read and looked for activities and word problems to be used in class, rather than these types of problems. Just personal pref.

Very interesting question! Thank your teaching!

Awesome! Keep it up lad!

Very nice job. One point to note: it is a symmetric equation, and once you get - (b+c)/(a(a+b+c))=(b+c)/(bc) a solution is b+c=0 and a=2022. Being symmetric, it follows that b+a=0 and c=2022 and that a+c=0 and b=2022.

I bookmarked this video. Love it!

I did it without solving for a b c specifically(I still found S first by setting a b c to some specific numbers since it's much easier to solve if you know the desired equality). This is probably the most straightforward solution I've seen so far. Multiplying the first 2 equation give (a+b+c)[(ab+bc+ac)/abc]=1 Then multiply out ab(a+b)+bc(b+c)+ac(a+c)=-2abc =>3abc+ab(a+b)+bc(b+c)+ac(a+c)=abc =>(ab+bc+ac)(a+b+c)=abc Raise the first equation to the 2023 then multiply with third equation give S×2022^2023=[(a+b+c)(ab+bc+ac)/abc]^2023 Substitute the equality above and done

Thanks for the content this is nice !

I couldn't prove that S is constant by myself, but from the expression of this problem I can tell it's a constant value, so I just found one set of a, b, c that satisfies the conditions (a=2022, b=1, c=-1) and got the answer.

met this type of question in my friend's notebook when he was in 10th grade in a vietnamese gifted school in math major, but it might be suitable for high school entrance exam for the mark-10 question

In fact, mathematicians or even just mathematics lecturers in Viet Nam were truly unsung geniuses. Considering the field of Maths, attention drawn to them must be intensified. Maybe just because of the wars that deteriorate the improvement of education in Vietnam, many people in our countries couldn't stand a chance to reach further contemporary knowledge and invention. Otherwise, the world would probably have had a different view towards the Vietnamese. I really hope that the contributions and dedication of us to Maths will derive respect and reconsideration from the rest of the world.

I got stuck at factoring when trying this, I sometimes forget that you can even factor expressions like b^2+ab+bc+ca

Ah yes The usage of Algebra-chan Great one, Papa Flammy!

pretty interesting sum with a surprisingly easy answer honestly. liked the sum a lot

Great video

The fact this can be a last question for an exam for grade 7th in Vietnamese is funny

That chalk has a great sound on that board

I managed to get into Asia Pacific Mathematical Olympiad and that was the day I decided not to major in mathematics. Still love maths though

im a vietnamese and damn these problems hit close to home

Alternatively, teacher can do 1 easy thing... By the relation 1/a + 1/b = 1/2022 - 1/c By cross multiplying, we get ab + 2022c = 0 Then abc = 2022c² By the symmetry, a² = b² = c² By observing this n the given eqn a + b + c = 2022 We get a = -b and c = 2022... ; )

Here's another way to solve the problem. Since b+c=2022-a, and 2022(ab+bc+ca)=abc, we can solve simultaneously to obtain b=-2022a/c. Then, we sub this expression for b into b+c=2022-a, and solve the resulting quadratic equation, giving us a=-c or c=2022. But we can easily check that both solutions lead to the same value of S=1/2022^2023.

I really like this exam so much !

(a+b)(b+c)(a+c)=0 means a+b=0 or a+c=0 or b+c=0. which means a = -b or a = -c or b = -c. then we can get that c = 2022 or b = 2022 or c = 2022. and in all, the result still 1/2022^2023.

I'd solved this kind of question few years ago. So it was a piece of cake for me 🔥

In olympiades there are often shortcuts you can take. For example here, the question implies that for every a,b,c you get the same value. So just try a simple a,b and c. For example 1, -1, and 2022. Then 1/1-1/1+1/2022=1/2022 So 1/1²⁰²³-1/1²⁰²³+1/2022²⁰²³= 1/2022²⁰²³

Flammy, do you by chance have an estimate on the release date of Wuck? 🧐 I want to spend a few weeks wucking with my fellow wuckers.

I'm a Vietnamese, I'm proud when I can't solve this problem

You deserve to get many more geeky subscribers!

Hi, Thank you, It was cool 👌🙏🌹. please make more videos like this 🙏🌹. good luck 🖐

Really I have always been terrible at math and this is certainly not helping however I do find these videos fascinating. These problems seem to always be either 0 or 1? So great job I think and continued success. Physics is on a level that alludes me also. And I can't stop watching those videos either.

A less rigorous way to do this is to just say that since there are two equations and three unknowns, just assume one of the unknowns is a free variable. (1) a + b + c = 2022 and (2)1/a + 1/b + 1/c = 1/2022 S = 1/a^2023 + 1/b^2023 + 1/c^2023 Lcd(2): (bc + ac + ab)/abc = 1/2022 (c(b + a) + ab)/abc = 1/2022 Let b = free variable = -a (2) becomes: (ab)/abc = 1/2022 1/c = 1/2022 c = 2022 substitute b = -a and c = 2022 to (1) a - a + 2022 = 2022 2022 = 2022 (Tautology) (1/c)^2023 = (1/2022)^2023 = 1/(2022)^2023 and a = -b implies 1/a = -1/b 1/a^2023 = -1/b^2023 Substituting the above to S: S = 1/a^2023 - 1/a^2023 + 1/2022^2023 S = 1/2022^2023

that's a wucking awesome problem

I remember this when studying for entrance exam into high school for the gifted in Vietnam

Ah yes, the kind of question frequently given out on the end of Vietnamese junior high school math exams, taking up 5% of the score to gate off students without a math orientation from achieving a perfect 100%.

Also possible to solve with recurrence

Flammy what is your opinion on umbral calculus? Would you ever make a video explaining these edgy math bois?

Can we do it by taking natural log on both sides of the eqn….?

I got it through a bit more thought in a manner that made it much faster and easier to solve. I first noted that for 1/a+1/b+1/c=1/2022 if a,b,c are all positive then a>2022,b>2022, and c>2022 which makes the a+b+c>2022 thus I concluded that at least 1 number must be negative. And once I realized that it needed negative numbers I realized that additive inverses work for multiplicative inverses as well thus I got the solution of a=2022,b=2022,c=-2022 which satisfies both initial equations thus since they are raised to a odd power the additive inverses still apply thus 1/(2022)^2023 = S.

AM >= GM implies that only one term can be there since (a+b+c)(1/a+1/b+1/c) >=9. Hence either a=-b or b = -c or a = -c.

I used the following 2 homogeneous symmetric identities a^2(b+c)+b^2(c+a)+c^2(a+b)+2abc = (a+b)(b+c)(c+a) and a^2(b+c)+b^2(c+a)+c^2(a+b)+3abc = (a+b++c)(ab+bc+ca) 🙂

3:08 giving us b + c divided by the set of complex numbers ;)

you can also use vieta's formula for this. For the sake of brevity, replace a with 1/a(and vice versa), do it also with other variables. We have (ab+bc+ac)/(abc)=1/a+1/b+1/c=2022 ab+bc+ac=2022abc Let f(x)=x³-αx²+βx-γ be a polynomial with roots a,b,c. Then, α=a+b+c=¹/₂₀₂₂ γ=abc β=ab+bc+ac=2022abc=2022γ Let χ be any of the roots, then 0=f(χ)=χ³-¹/₂₀₂₂χ²+2022γ(χ-¹/₂₀₂₂) (χ-¹/₂₀₂₂)(χ²+2022γ)=0 Without loss of generality, a=¹/₂₀₂₂ Let b and c be the solutions of χ²+2022γ=0. Either γ0, we have b=-c ⇒ b²⁰²³+c²⁰²³=0 a²⁰²³+b²⁰²³+c²⁰²³ = (¹/₂₀₂₂)²⁰²³

As a Vietnamese, looking at this already made me writhed in pain

Need more Vietnamese IMO questions 🇻🇳🇻🇳🇻🇳

in equation min 4:19 we can say (b+c) = 0 so b = - c , then a= 2022 ; thiway you can go to minute 8:22 and save precious time .... anyway nice and good job

Solution proposed by : Abdul Mujib Lets Assume a,b,c be roots of the equation f(x) Then, a +b+c = 2022 (ab +bc+ca )/abc = 1/2022 Form a cubic polynomial, f(x) = x³- 2022x² +x - 2022 By factorising , (x²+1)(x-2022) = 0 (a,b,c ) = iota , negative iota , 2022 To find [(ab)²⁰²³ + (bc)²⁰²³+ (ca)²⁰²³]/ (abc)²⁰²³ Substituting the value you will get final answer as S = 1/(2022^2023)

One out of a,b,c has to be negative so just consider: a = 2022 b = -2022 c = 2022 and it should work because its an equation and any random number that fits the equations should give you the answer.

Somebody busts this one out almost every year I feel

When you realize this is the question on a test paper of a grade 8 student in Vietnam

Yaaaas competition maths!

"das is cool" the german coming back with exitement

where is the wuck ad smh. also i hope the cold gets better. i got covid last week and i finally recovered lol.

Papa, this is one that Wolfram Alpha didn't solve. At least not on the free version. It said I had to go Pro. I guess this proves you are Pro! 😝🤪😏😎🤓

Ah yes. I remember this question in my highschool entrance exam. Damm, I didn't think it was this high level mathematics !

I find the product of sums together that adds b + c to be zero to be somewhat suspicious

Ahhh, i remember i did this math 7 years ago when i was in high school. Still dont know how to do this and wonder how this help my life

I still remember this exact question in my 7th grade end of term exam lol

assuming the answer is unique, just try a= 1, b=-1 and c = 2022 and you can get the answer in 3 seconds But I think the most interesting part is how he proves it. It's still a nice video

Nice maht problem papa 😘

Thanks teacher for video ❤️📐

Lol i'm from Vietnam and I can confirm we see this kind of question for bonus mark at primary school and it appear more often around grade 7-9

this question can be solved by the "wouldn't be nice" theorem of blackpenredpen

How can i find the solution to that Gauss trick question in the intro of brilliant please?

These are the formulas for the series and parallel association of resistors (capacitors) :P

Well Math is The nightmare of me since 6th grade and now + Algebra from 8th grade = Yeah Math is now 2x nightmare mode

Papa flammy is uploading more. Not complaining 😏

Thank you for your solution, I did it in a different way. If you take the second equation 1/a + 1/b + 1/c = 1/2022 and simple add it we get (bc + ac + ab)/(abc) = 1/2022, lookint at this expression and the first one a + b + c = 2022 reminds me the Girard relations, so I adopted a + b + c = 2022, bc + ac + ab = 1 and abc=2022. I know I make something specific here. Then I wrote a polinomial equations using this values xˆ3 - 2022xˆ2 + x - 2022 = 0 now doing some factorization xˆ2(x - 2022) + (x - 2022) = 0 and (x - 2022)(xˆ2 + 1) = 0 we have here 3 solutions for x (2022 ; i ; -i). If we assume a = 2022, b = i and c = -i it is strange but this is a special case and we see that two of this values should be opposite in that case b and c, i and -i. If you replace this values to calculate what is asked you get the same result. Of course is strange getting imaginary values. Just a different approach to be discussed. Thank you

Ha, this kind of exam usually appears in the last question of a Maths Test(secondary school) in Vietnam, it's 0.5/10 pts 😅

As a vietnamese i suffered from this thing TOO MUCH

i just saw that you divide the first equation by 1^2022 to get the second equation so i divided 1/2022 (2022^-1) by that to get 2022^-2023 which is 1/2022^2023???

thanks

the proof process is great! I have a short version answer. For both a+b+c=2022 and 1/a+1/b+1/c=1/2022 to be true at the same time, two of the three unknowns have to be the same but with opposite signs. i.e., the sum of two out of three has to be zero. That is, a+b=0 or a+c=0 or b+c=0, only then can a=2022 and 1/a = 1/2022 both be true..

Fastest solution: we can easily see that a=b=2022, c=-2022 satisfy the conditions. Then calculation is simple. Such cheap tricks would not work in IMO, but sometimes finding the solution first helps though.

Let's put, u=1/a, w=1/b, v=1/c. Denote uw+uv+wv=K. Consider polynomial W(x)= x^3 -1/2022x^2+K*x-K/2022. u,w,v are roots of this polynomial, but also W(x)=(x^2 +K)(x-1/2022). Therefore (u,w,v)=(sqrt(-K),-sqrt(-K),1/2022) or permutations. This immediately gives us u^2023+w^2023+v^2023=(1/2022)^2023

IMO 2022 is not hosted in Vietnam? Which contest does this problem come from?

My solution: combining fractions we get abc = 2022(ab + bc + ca) so let ab + bc + ca = k, where k is some constant. By Vieta's we get that the cubic with a, b, c as roots is x^3 - 2022x^2 + kx - 2022k. Factor it into (x^2 + k)(x - 2022). Therefore clearly one of the roots is 2022, WLOG let this be a, and the other 2 roots are each other's inverses, therefore it is proven that WLOG a = 2022 and b = -c, so the b and c terms in the final evaluation cancel and we get (1/2022)^2023

I'm still crying on how I continuously skipped this kind of questions during my whole school life until now ( I'm a 10th grader btw ) as a Vietnamese *Now I gotta question my ability*

This was definitely one of the easiest IMO problems ever.

Oh, that math problem in my good student math test grade 8

can't belive that i had to solve this since grade 9

ELİNİZE SAĞLIK HOCAM

Just how do they make these questions...

Ở Việt Nam, bài này dành cho những bạn học trung học cơ sở hoặc trung học phổ thông, nhưng tui lên cấp 3 rùi mà vẫn không biết làm:”))))) (một bài toán đơn giản cho những đứa trâu bò)

1/2022^2022 is smaller than 1/2000^2000 which is 0.00..... (at least 6600 zeros, and then some). I'd say 0 is close enough as the answer.