A Fun One from the Vietnamese IMO
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- čas přidán 25. 06. 2024
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Today we cover a fun little problem from the Vietnamese international maths olympiad. We have given that a+b+c=2022 and 1/a+1/b+1/c=1/2022 and we have to find a certain exponentiated sum. Enjoy! =D
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these questions can actually sometimes be found in highschool entrance exams. crazy how we’ve learned all this.
It wouldn't be strange if it appeared in an international math olympiad question. But you know what's really crazy? Vietnamese students have to learn a lot about integrals, derivatives, all kinds of trigonometric formulas, but most of us cannot apply them in our daily work.
@@khanhphaminh1175 I'd say people should really stop bemoaning over the application of mathematics in real life. There are so many other reasons to blame for why students cannot perform well in real life situation, not the maths itself. Mathematics is always one of the most affordable approaches to teach students problem solving skills. Students have to come up with a beautiful solution to a hard piece of math problem with only the most basic materials (integrals, derivatives, trigonometric formulas,... as you've said). Isn't it the fundamental of problem solving? If you can introduce a new method of teaching problem solving skills that can be applied to a wide range of students in an affordable way, feel free to make a statement.
@@khanhphaminh1175 Not only Vietnamese. Many countries also teach students about integrals, derivatives... Stop complaining about that.
@@huyangquang1711 I just mentioned a fact. I don't understand why students have to spend time and money to learn a subject that is so difficult and low application
(for most professions), while still having to study dozens of other subjects in the same year. Regarding what you say "helping students think logically in real-life situations", I think it would be more effective to create a whole new subject called "How to think logically and solve most of common problems" (something that students in Law school definitely must learn, and books about this subject are sold in every bookstore - oh, we have the textbooks already!). A new approach like that is both more direct and effective than whirling students with cliched formulas, giving them no meaning of what this subject is use for, but unrealistically expecting that they will find out the meaning of learning that by themselves. I still remember an article, where a 3rd-grade high school student in Japan (equivalent to grade 12 in Vietnam) looked at the 10th grade entrance exam papers of Vietnamese students, and they exclaimed that these things they only study in college. If only a small percentage of students say that the current way of learning Math is meaningless, then you can say that they are learning the wrong way. But if the vast majority of students (except for those in the advanced math class) say it makes no sense, then the fault belongs to the education ministry's erroneous way of working.
@@dramafanpage8502 I'm 28 years old, have a job that has nothing to do with math, and still make good money, so what I'm saying isn't “complaining.” I'm feeling sorry for the money and time of hundreds and thousands of students who are still struggling with what they consider meaningless knowledge. What about many countries that teach those subjects? It's not what made those countries successful. Look, Vietnamese students have been taught that subject in so many years, and what is in return? A Vietnamese only earns an average of 3000 USD per year. In successful and prosperous countries, they teach students to think logically directly, to learn to be critical, to appreciate the arts, to be creative, to solve practical problems, but they are not successful by some math formulas
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.
A more concrete answer would be 1/393793903937577106479301743414185865462150020261045958348688864012331731304516123198845761433324488438219593808570961852064612014780620624672175445445623382895917883727710633016494081643796401279080059029404279015592123445442355194726188606869002467332145960773819828320146098500719543708842828909672937331352454838564658760130686616706686521039026472870725835633911635732397771739426937534540098162234792530262277400952960260275149179804966675094176984508713046317788443292024044223680238386586329406803944254718916260871815935375111696289754445942035497936545125868564313977047665770022714319377213978160622697269487278873119845809917406721550071458221311050676531396693901681843064092887169936333415826855426013543139044970705738911781672691422364080471999492019809282504071796204154517495882790498393867199563541419863517221565959261739686501178959417759309568516456104708920229992395230228903953063833065986161681444648022574918434659891914833112053428143196925629879411752075401355532544017417476788308017152966790564260548928897478386312037770428673468613582763168614277948609278870006347055324532210630733323281118380011252615931646701004690182075032123767325437226004873702559515067964070718560797194154998855092813980238968034648305431279385781862205048865127192139055595268009315185126800025295469023394604199061086359254433212349799318521040427532666989236907747701171802346780593669488132594120937585143570489494597412015258973495937532371308552027673665862452462496996340353575649988424249535811860120802497285563561335755572798634944409538335137253766015557723887443982767984207525075962600447534996987404419602791575303516970031347030489168581235574850237385515774445711182351283425539377557090153111789115365503761779089069170411660638323033599411217957676143335096505791910934533669664902763885344507187253341645042322357702417368822373002645986021656797833144415930392123612634034531334528128279264226280389919144186148073433090356229629574487001609945588996700009857931190244582787781981877820785356558637277173725635202063621762415552708860330373067798892396051070149086816524225341526084130866750847634080515977399218344052153044129811988865625708895172247970818244717511475389102792431715060679839217756732955249425619632592548104964162574028939416200191589900040276284996423183580269537829301856916783456589435696218814439711066029936512328512051652967106102752720230021501058568859576165644361305619324532969927181691887792263541733558947644497455771801727310607803753021989435677497179468397332031842874581636660316250318856415094613696232537028140098672438652578876667706317339702327770587307597082164866745269924314810790854857409922099931888208876920474910109408204469258308185663826674345827477443684328744712685776923850147437077008409008570138828158346805197980277364705951140387226569743797152383910274271531809109053141501420405008364306215728862473281280982903067496264707023676950228041397641708182387456387192125261922049660728334893434721711554690113649751953762093719649544106259506282234544368417844344916848902517657155685153361558834810291149212214391780357471530032013708218274165571182212666485358769490050426743923310358555482637184144991176745266924632888615386855771520624921929239357960686134080168753708870543782562454540905771134388971688342011522148286951383041405906792468715329987262184997340193718566532878070026244749756460332336440365671247018928365095159372452799133428334342520366099906761182526083567108201141288588724104232238357914686143195372407715037434209977040415079022412854911606517805279357734165915680428312491934938472712791319580956496831212171879918029774311776051369728473165255068483483066931996186347817761123025174218896915546274705502111123668819605957959687155582900219489982983060785692794814528133977551149181329523978870287909448589934972634913413696109945669390906497298702110302827788158757732929471952188976645780904858003262302119529067867009734725419656542228580307526339149805610503254400913020517140751943563039426792010722261237757210999688897482051826499691470761972612659923515867214526010081917608017819635760125117598408088791374527767798380362721651923772243995626328857066137238047500869683279361993297990285234019363341025302107780021780342462141320104632035676056177623067294477786045233140202827641212364911735918289045162519723315587505292207851118406684560762538232991887799490330991049955318927640825822145387807221109421203765448093540126843802557973601665027839307238700114797369398404860298797664371224359632764205544939799250187322420566038334231005428039078708951716390191762973818257167203166588555881429752420253511610115654895377758332885282660436756527741558431433851666754892799862780042186668220619278028300460620808670210170226049422822267856889193656382801651197151367708914970663241547420235743423123679241376599515853093965541673470565769082214045565719402743347871856207715349252073940433454197876400318235192833670976078217422087070910648435845672844035095784780021928512973373711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My eyes.
Ya i got same answer
@@ananysingh5216 use fermat :)
Wow. It actually is a 6010-digit number
I mean 6070
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.
well, advance math teaching usually fastfoward some of the bits so it can be really hard to understand sometimes, espeacially this kind of problems. To be honest, the explaination Flam gave here can be shorten from 2 board full of dogmatic ritual symbols to a page (maybe not even a full page) of an A4 notebook, that is how fast and compact the advance math students in highschool has to be
When the trees start speaking algebruh
Lol 😆
😂😂😂
Haha
LMAO
bruh
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
Nice
Sorry, but how can we form the cubic equation?
@@davarni116 The cubic polynomial p(x) = (x-a)(x-b)(x-c) clearly has the roots a, b, c. Multiplying it out, we get p(x) = x³ - (a+b+c)x² + (ab+bc+ac)x - abc.
Same
@@tomkerruish2982 Thanks!
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.
they don't care about practicality, they just want their student can climb to the top of their subjects
@@pinospin9588 sadly, that is exactly it. In my school, teachers were competing each other not out of pure sportsmanship but a ranking system of homeroom classes that forces them to do so. The homeroom teachers then went hard every week so their students could do better so their class could rise to the top of the board. Any week their class was in the bottom, they'd lash out. Pretty much just bullshit. Some teachers didn't give two fucks about ranking though lol.
Vietnamese education system basically nonsense
Integral? Derivative? Complex number?
What for?
Not everyone become engineer or scientist, teaching these in high school simply make no sense at all.
@@pinospin9588 many have had oppositions too. Even some of my teachers. But then the education ministry just doesn't care. I think one of the reasons is that they don't want to lose to, say, America or Japan. China, even. So they subjected students to whacky levels of education just to put up a front to the other countries.
@@pinospin9588 maths class are optional in France :) (in high school)
In Vietnam, these basic solutions you will encounter very often in high school, but they are not as scary as geometry.
Yé :)))
Vietnamese Geometry is real pain :(
that's quite gk3 imo tbh =))
Yé
Yeah, fuck geometry, I agree
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.
This question is easy compared to others where I don't even understand the problem
@@user-pv7wk5pd1r Yeah the geometry problems of Vietnamese IMO are brutal
The first question of first ever imo was the easiest i guess
Is that an IMO question? I remember I saw this one on lidsky problems in elementary mathematics
It is not an IMO problem from Vietnam. THE CHANNEL TOLD FAKE NEWS. This problem is just an exercise for the majority of grade-8-students (aged 14) in Vietnam.
Everybody gansgta until you need to calculate the numeric value of 1/2022^2023 by hand.
Thanks for posting this fun one!
glad to see some representation for us in Vietnam 🙏🙏🙏🙏
sick pfp
@@cthulhu5707 i know.
glad to see some Vietnamese and Mizuki fellows
2 mn
damn
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!
Dude this is so hard for me, how is this the easiest in your country? What do they teach you in highschool?
@@devd_rx Khang is making fun of indians
@@namename8986 sad reality, within every 3 minutes, a student suicides in India.
@@namename8986 didn't quite get you?
@@devd_rx we Indians like to brag about our entrance exams because they are a bit hard.
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
Thanks for the nice problem and such a beautiful solution!
Very interesting question! Thank your teaching!
Ich hab dich so lang gefolgt und jetzt kam endlich ein Video über Mathe aus meiner Heimat. Yayyy.
:))
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
The fact this can be a last question for an exam for grade 7th in Vietnamese is funny
Me who’s an 8th grade Vietnamese: at least I made it*
Edit:*at least
According to what I remember it's usually those extra mark questions for the smart lads so I usually just skipped these questions lol
@@duckyymomo5714 exactly. In a grade scale of 10, the question only worths 0.5.
that one last question is always a pass ticket for me :( 7 in math is too enough
@@anhph4746 that's enough?i'd be happy to get a 5,i always messed up the vi-et exercises
I got stuck at factoring when trying this, I sometimes forget that you can even factor expressions like b^2+ab+bc+ca
How? I guess i forgot too lol
Awesome! Keep it up lad!
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.
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'm a Vietnamese, I'm proud when I can't solve this problem
Haha, tự nhiên nghe thấy từ việt nam nên vào xem
Toán lớp 9 cũng có r :v
Một bài điểm 10 thi đại học thông thường mà, ai được học sẽ làm được thôi
@@trunghieu8922 đối với tôi 1 điểm coi như bỏ 😅
@@hongphuc2222 à thì nó dành cho mấy ông kiểu "còn 1 tiếng nữa làm gì bây giờ nhỉ" đó. Chẳng biết giờ trắc nghiệm 100% thì còn mấy dạng này không nữa
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.
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
That chalk has a great sound on that board
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.
Great video
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%.
mình tưởng nó thường là câu bdt chứ nhỉ
(TL: I thought that those last questions are inequality problems?)
Bro what kind of junior high are you talking about 💀
@@OakQueso the average Vietnamese junior high? these questions are definitely too common in 8th or 9th grade exams as the final quesiton worth 0.5 point out of 10.
@@tosyl_chloride How do students deal with these types of problems at such a young age? Are many of the students miserable and hopeless because of it? Do they spend all their time studying? Genuine questions. I would assume many people are unable to handle this type of rigor, at least without dramatically affecting their social lives and competency. I would assume it also destroys the ambitions the people who are unable to handle it.
I'm from America, where most people never get to problems this difficult. America is not stupid though; Students are largely allowed to go at an advanced pace if they are ambitious. We let people pursue their own interests. I believe letting people go at their own pace and not making them miserable is the best approach to learning, but you could let me know if I'm wrong.
@@OakQueso As I said, this question belongs to the category of "final questions" on each exam that is worth 0.5 out of 10 points, or 5%. Most students would just forgo this question altogether and focus on the other more manageable ones in order to maximize their points - getting a 7.0 or 8.0/10 is a common goal anyway. Only those tryhards who aim to join municipality- or region-level Olympiads, or are deathly fixated on fighting for a spot in entering top-tier public high schools would study about these questions.
im a vietnamese and damn these problems hit close to home
I remember this when studying for entrance exam into high school for the gifted in Vietnam
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
that's a wucking awesome problem
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.
Mid March I think! =)
whats wuck
I really like this exam so much !
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.
Hi Jessie. Thanks for sharing. If you are interested in math competitions, please consider
czcams.com/video/l5ef8BNduDs/video.html and other videos in the Olympiad playlist. Cheers
(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.
He used the wrong brackets but after that he said it was "or" lol
Also possible to solve with recurrence
I'd solved this kind of question few years ago. So it was a piece of cake for me 🔥
You deserve to get many more geeky subscribers!
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... ; )
Can we do it by taking natural log on both sides of the eqn….?
Hi, Thank you, It was cool 👌🙏🌹. please make more videos like this 🙏🌹. good luck 🖐
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)
When you realize this is the question on a test paper of a grade 8 student in Vietnam
As a Vietnamese, looking at this already made me writhed in pain
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
This solution is so elegant!
Even easier, as there are more variables than equations, just assume a = -1, b=1 and c = 2022.
where is the wuck ad smh.
also i hope the cold gets better. i got covid last week and i finally recovered lol.
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.
Need more Vietnamese IMO questions 🇻🇳🇻🇳🇻🇳
need more pho and banh mi questiions
How can i find the solution to that Gauss trick question in the intro of brilliant please?
3:08 giving us b + c divided by the set of complex numbers ;)
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.
It's not the only solution. The problem is to show all solutions gives the same final answer
"das is cool" the german coming back with exitement
What's the reference of your blackboard please ?
Yaaaas competition maths!
I find the product of sums together that adds b + c to be zero to be somewhat suspicious
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
Freaking genius, noice👏
damn
Flammy what is your opinion on umbral calculus? Would you ever make a video explaining these edgy math bois?
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
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.
Somebody busts this one out almost every year I feel
thanks
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
Thôi ông bớt xạo :D mấy bài bonus này dùng cho lớp 8 9 và thi chuyên thôi tiểu học đâu ra
@@duyphan803 thế tại lớp ông học kém quá rồi nha
@@duyphan803 tiểu học bài này là bài cuối và dùng để ôn hsg các cấp nhé. Lên 6 là làm suốt để ôn vô đội tuyển nhé. Còn lên 8 lên 9 là bài 45p nào cũng có. Còn chưa nói bọn ôn chuyên thì bài này không khác gì bài thưởng điểm với chúng nó
cái tầm của bài toán VMO như này nó phải khác với những bài phân loại đại trà. Bài này mà dành cho học sinh tiểu học thì tầm VMO của Vnam nó chỉ đến thế thôi à? học dốt cái xạo quần!
Ah yes. I remember this question in my highschool entrance exam. Damm, I didn't think it was this high level mathematics !
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
As a vietnamese i suffered from this thing TOO MUCH
in Vietnam always give the time value in each year.
IMO 2022 is not hosted in Vietnam? Which contest does this problem come from?
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.
wha does Vietnamese IMO mean? This year the IMO is in Oslo and it hasn't happened yet
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) 🙂
These are the formulas for the series and parallel association of resistors (capacitors) :P
Just how do they make these questions...
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 😅
this question can be solved by the "wouldn't be nice" theorem of blackpenredpen
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
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
Papa flammy is uploading more.
Not complaining 😏
I still remember this exact question in my 7th grade end of term exam 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! 😝🤪😏😎🤓
:^)
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
I have a question:
When youve already factored (b+c)[that monster] = 0, couldnt you assume b+c =0 from there already?
Or do we still need to factorize the monster to make sure (b+c) wont be cancelled out?
Sorry if this is a dumb question 🤣
You don't know whether (b+c) or [that monster] is the one that equal to 0 or not yet hench why you have to break them down like that
t.Vietnamese math tryhard
Nice maht problem papa 😘
Well Math is The nightmare of me since 6th grade and now + Algebra from 8th grade = Yeah Math is now 2x nightmare mode
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???
You should consider doing a Vietnamese geometry problem.
Don't
Don't
can't belive that i had to solve this since grade 9
at the second step, just put 1, -1, 2022 into the equation and you got the answer.
Yeah, 2022^2023 in the way. No thanks
Oh, that math problem in my good student math test grade 8
I solved this one like this.
After some mumbo jumbo, I got 2 sets of equations
1) a=-b
2) c^2 = -(ab+bc+ac)
Now, I had assumed that as a+b+c = 2022 , this the amount of positives must exceed the amount of negatives.
This, c^2 comes out to be negative (not allowed).
Thus, a=-b, and finally I too got
the expression as 2022^-2023.
Is this right??
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.
where is your tshirt from?
from my personal teespring shop, link in the description :)
some 20 yrs ago I solved this kind of problem using only 4 lines, now I will use 2 lines, which only differs from the numbers, 2000 is the key last time.