Germany - Math Olympiad Problem | Be Careful!
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- čas přidán 7. 06. 2024
- You should know this approach. Many goes WRONG! Solution
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Theres no way this is an olympiad problem. This is like algebra ii at most.
finding the fourth root makes it a maths olympiad problem
@@pieterkok7486 How? This is high school algebra, even in the US that has bad math instruction.
@@pieterkok7486u gotta be joking 😂
this cant possibly be a math olympiad question, thers no way
Maybe its math olympiad for school?
Euler's formula: e^iθ = cosθ + i sinθ
a^4=(a-1)^4
(a/(a-1))^4 = 1 = e^(2niπ) for n = 0, 1, 2, 3, ...
take fourth root
a/(a-1)= e^(iπn/2) = 1 , i, -1, -i
giving no solution for 1 (it is an eqn of order 3, after all) but the others are
a= i/(1-i) = (1+i)/2
a= 1/(1+1) = 1/2
a= -i/(1+i) = (1-i)/2
The e-trick is more of a guess, about all the results of forth root of 1
@@zyklos229 I have no idea why you would say that. Using Euler's formula is well established maths (and the easiest way to solve equations of the form (f(x)^n=g(x)^n for some order n).
Someone loved complex analysis
Almost right, you just forgot, at the beginning, to say that a can't be equal to 1, so you can divide by a-1.
the good way of solving any problem
Question: Solve for the value of a.
Answer: x = ...
Wait... what?!!! 😱🤔
lol
This means the problem is not solved.
Wo ist denn der mittelfinger wo ist denn der mittelfinger, hier bin ich , hier bin ich .....😂
@@Archik4let x=a
Just from looking the equation, imediately I knew that a = 1/2 should be one solution. That's because the only way a number to the fourth power to be equal to this number minus one to the fourth power is if the number minus one is the same number with switched sign, that is, x^4 = (-x)^4, therefore a-1 = - a, and a = 1/2. Also, I noticed that the fourth power terms would cancel out on both sides of the equation, and the remaining equation would be a cubic. The other two solutions should be irrational, because there is no other possible real solution to a^4 = (a-1)^4, as I said above. So the only calculation needed is to divide the polynomial "a^4 - (a-1)^4 " by (a-1/2) and solve the resulting quadratic equation. Dividing - 4a³ +6a² - 4a + 1 by a - 1/2 gives a² - a + 1/2, and finding the roots gives a = (1+- i)/2. Very simple problem.
I had a similar thought. I saw 1/2 the same way you did. I carried out (a-1)^4, subtracted a^4 and got a cubic. I used synthetic division to factor out x - 1/2 and found the complex zeros of the resulting quadratic.
You wrote "I saw" - while this may be obvious, I believe it's better to deduce this solution. For example, you could take the fourth root, leading to an equation like |a| = |a-1|, which has a single real solution, a = 1/2. Then, you could divide the polynomials (a-1/2 is also a polinomial) and proceed as you have.
@@Kirkemuswe dont have to take actually take forth root. as soon as we “saw” 1/2 is an answer, we can substitute a=1/2 and confirm that both sides equal. and that’s it
Simple? Well, so please explain why the approach of applying the 4th root on both sides does not work. If you do that, I will agree you truly understood the problem.
@@lucianofinardi7222 what you mean "does not work"? It works, it gets one of the solutions, a = 1/2, which is the only real solution. The other ones are obviously complex numbers, then you have to divide the solution by (a-1/2) and solve the remaining polynomial.
Far too easy for an Olympiad. Main problem I had was to convince myself it's actually just 3 solutions but to get the solutions were very easy.
Be Careful not to change the "a" to an "x" in your answer!
You can do that very quickly ! a^4 = (a-1)^4 means : a = a - 1 or a = -(a-1) or a = i(a - 1) or a = -i(a-1)
1st equation : no solution and the 3 others are very easy to resolve
Its like x²=y², x=-y or x=y, but for fourth [word]. Thanks, good to know.
What word should I write?
You can take fourth roots on both sides, but the catch is that the results are only equal to each other when multiplied with one of the fourth roots of unity. So suppose that a^4 = (a-1)^4, then a = (a-1)*z, for some z such that z^4 = 1. Then, a(1-z) = -z, so a = z/(z-1). The fourth roots of unity are 1, -1, i and -i, so just plugging in these values for z you obtain the answers. Note that z = 1 does not give a solution because you are dividing by 0.
Take the 4th root 9f both sides now a=a-1, yes
First expand the right hand side and cancel the a^4 term:
a^4 = (a-1)^4
a^4 = a^4 - 4*x^3 + 6*x^2 - 4*x + 1
x^3 - 1.5*x^2 + x - 0.25 = 0
Now recognize that x = 0.5 is a root of the original equation, by inspection. So we can divide out (x - 0.5), which leaves
a^2 - a + 2 = 0
The roots of that are 0.5 +/- j*0.5.
Not "j". Its "i".
j and k is quaternions
@@aristoferElectric Engineers denote √-1 as j (as i means current in eee) :p
@@user-iy7jk1yd6d ok, but this is math, not electric enginery
The oddity of this solution is a 4th-order polynomial with only 3 roots.
Try expanding it out. The quartic terms cancel.
@@chaosredefined3834 I realize that. It's just interesting that the equation "appears" it would have 4 roots.
Technically, I think it does have four roots, two of them being coincident roots at a=½
@@trueriver1950 no because at the end you have a cubic
A short way to the three solutions is as follows : you have the possible cases : i*x = x-1 ,-x= x-1 ,-i*x = x-1 ,which leads to the three solutions (1+i)/2 , (1 - i)/2, 1/2 .
a^4=(a-1)^4 so a = w*(a-1) where w is any 4-th root of unity. Then, a*(w-1)=w so a = 1+1/(w-1). Valid solutions: w=+-i, w=-1
Substitute a = b + 1/2, and solve for b. Answer falls out much more easily.
How did you get this substitution? It is obvious why it has to work, but...? Perfect intuition?
Lol, symmetry.
a⁴=(a-1)⁴
Obviously a=½ is a solution.
4a³-6a²+4a-1=0
By dividing it by (2a-1) we get
(2a-1)(2a²-2a+1)=0
D=4-8=-4
my direct impulse:
1. a1 = 0.5
2. a^4 cancels out -> polynomial 3rd order
3. polynomial division by (x-0.5)
4. abc (or pq) formula
(I'm German)
I just watched up to minute 2 and now I'm sad
I wouldn't expect complex solutions is what we are looking for.
So, I'd just got both sides to the power of 1/4 and get |a| = |a - 1|. This tells us, a must be in the interval of [0, 1], which gives us solution a = 1/2.
My method:
Since a number and its negative raised to an even power give the same solution, a = -(a-1) ==> a = 1/2
Expand the RHS
Cancel a^4
Factor out (a-1/2) using long division
Solve the other factor, a second order polynomial
Get answers a = (1 +/- i) / 2
The problem statement must include specification of a field in which solution are to be sought. There is no reason to silently assume it is C.
Has no roots in GF2. 😢
it is basically a-1= |a| (a has to be negative otherwise we’ll get -1 = 0), so it is a - 1 = -a, a = 1/2
I would suggest that two coincident real roots exist at a=1/2. There have to be four roots!
The fsktotization from the firdt step eorks for the 2nd step too. In the second term insert -i^2 before a
its just expanding (a-1)^4 , getting a trinomial. done, factor out solve, you will get 1/2 and (1+-i)/2
How should you factor it?
If 2 powers with the same exponent are equal they should have the same base and in this particular case the exponent is even it is possible that the bases are opposite numbers. a is never equal to a-1 but -a = a-1 when a=1/2
Not true in general, not even in this particular case if you're solving over the field of complex numbers (which wasn't stated in the problem, to be fair, but follows from his explanations). Here you have to consider all 4 possible values of the (complex) 4th root (considered as a multi-valued function).
Feels like a problem from the International Olympiad in Algebra 2
Gleichung vierten Grades,
also vier Nullstellen/Wurzeln ...
a = 1/2 ist als doppelte Nullstelle zu berücksichtigen.
Please use a for your answer at the end and not a. Also this can be trivialized by using complex 4th root of unity
The solution can be shorter. Let b = a - 1/2, so the equation will be (b+1/2)^4 = b-1/2)^4 , it can be transformed to b*(b^2 + 1/4) = 0
a basic school task, we clicked these very quickly
So we say in the hypothesis (a belongs to C - complex one).
Simpler method, take square root of both sides,
a^2 = ± (a-1)^2
expand RHS and simplify
+ve root gives a = 1/2
-ve root gives (1 ± i)/2
Dilly dalliy approach lengthening the presentation!
|a| = |a-1| -> a=a-1; a=-a+1 -> a=1/2
What happened to the 4th answer?
It's a cubic in disguise because when you multiply it out and set it to zero you get a³ as the highest power
The root 1/2 is repeated root.
Just take example of (x-1)^2=0 it has only one "solution" ie 1 but two roots 1 and 1 as it is a quadratic equation(necessarily 2roots)
Back to School!@@GooogleGoglee
Not really @@KG1_007
The forth answer has no solution
What is the 4th solution?
Almost correct explanation…
√-4 = √(-1×4) = √(i² × 4) = √i² × √4 = i × 2 = 2i.
Stating that √-1 = i is a sloppy shortcut ;)
The final answers are basically wrong because you are solving for a & not x-That's what my teacher would have said💔😭
1/2 is an obvious solution.
a⁴ = (a−1)⁴ ⇒ 4a³ - 6a² + 4a - 1 = 0
f(a) = 4a³ - 6a² + 4a - 1 is strictly increasing,
so a=1/2 is the only solution.
Why does a fourth degree equation have only three solutions? The basic theorem of algebra is violated, however.
Multiply both sides by 16 = 2^4.
So, (2a)^4 = (2(a - 1))^4.= (2a - 2)^4.
Let u be the arithmetic mean of 2a and 2a - 2; i.e., u = 2a - 1.
2a = u + 1.
2a - 2 = (2a - 1) - 1 = u - 1.
This, (u + 1)^4 = (u - 1)^4.
0 = (u + 1)^4 - (u - 1)^4
= ((u + 1)^2 + (u - 1)^2) * ((u + 1)^2 - (u - 1)^2)
= (u^2 + 2u + 1 + u^2 - 2u + 1) * ((u + 1) + (u - 1)) * ((u + 1) - (u - 1)
= (2u^2 + 2) * 2u * 2
= 4u * 2 * (u^2 + 1)
= 8u * (u + i)(u - i).
Divide both sides by 8.
u = 0 or u = ±i
2a - 1 = 0 or 2a - 1 = ±i.
2a = 1 or 2a = 1 ± i.
a = 1 / 2 or a = (1 ± i) / 2.
First get both sides four-rooted, you end with a=a-1. The answer is nope.
🤣🤣🤣
What is the name of the method you used? I thought negative numbers don't have square roots. Is there a reason the "i" is i ?
This is complex numbers. Try to learn about it.
На какой уровень образования составлена эта олимпиадная задача?
With guessing and test, i already found it would be 1/2, just 3 seconds 😂
4th order polynomial has 4 roots not 3. !
I mean immediately looking at this question I knew one answer was 1/2
a = +/- (a-1)
This task is solved by 2 steps with geometric considerations
I can't see anything wrong with: a = 1/2
a⁴ = 0.0625000
(a - 1)⁴ = 0.0625000
Is it an exercise of problem solving or problem making?
Trivial as fuck. Just "see" that a=1/2 is one (rational) solution, then take out the linear factor (a - 1/2) and you get a quadratic equation with rational coefficients which you can solve the standard way to get the two complex conjugate roots which are 1/2*(1 +/- i). Done.
The approach with substitution + binomial formula is nice if you absolutely want to avoid polynomial division, but is not really necessary.
Hard to believe this is a problem from math olympiad.
interesting that a 4th equation has 3 roots, not 4?
The a^4 cancels. It’s a cubic in disguise
(a-1)^4=a^4+4... etc. Therefore we have a^4 on both sides, so they cancel out leaving the highest term as -4a^3
|a| = |a-1| has the unique solution, a = ½
This should be solved within 1 minute, not 10 minutes
(a-1)^2=+/-a^2 then the only real solution is 1/2.
Would the easier answer would be a = infinity?
What the f-
that must be the fourth root
Cadê a quarta solução?
Force power?!?!?
"i" don't work as a solution.
what does this bracket sign mean? 8:23
Absolute value
@@edstudio6181 Thank you
go back to school
That's vertical bars not parentheses...
The bars indicate absolute Value. The absolute value is just a numbers distance from 0 on the number line (remember that distance as we know it is always a positive real number).
This means that |-3| = 3, because the distance of minus 3 to 0 is just 3.
For real numbers, we can also simplify the absolute value to turning any negative into the corresponding positive number. But that definition doesnt hold up for complex numbers.
a = 0.5?
Lol, why not give the whole task?
I have already passed complex analysis, but how shall I know that some German kids also know it?
Tbh I did this in my head.
Another 5s one. +/- 0.5 are immediately obvious answers. Might be other roots.
Only +0.5
-0.5 is an obvious answer. Wrong, but obvious.
i just did this in 3 methods ._.
a=.5
It's obviously 0.5 . Why would you want to spend 10 minutes solving it?
Also, you have to prove this is the only solution. That's why you can't just say the answer
Слишком просто для Олимпиады. Я решил уравнение и даже проверил устно, без бумаги, доски, и т.п. Решил чисто в уме. A1=0.5, A2,3=(1 +/- i)/2
Исходное представление намекает, что м.б. чётная степень съест знак. Было бы всё в левой части я мог бы не догадаться 😊
Итак abs(a)=abs(a-1), очевидно одного знака под модулем быть не может, имеем a=1-a, A1=0.5. Подставляем - первый корень есть.
Один множитель есть, а 4я степень сократится в итоге будет квадратный многочлен. -4а^3+6a^2-4a+1=0. Делим на -4, затем на (a-0.5). a^2-a+1/4=0. Отсюда ещё 2 иррациональных корня. Проверить тоже можно устно - там вовсе легко, т.к. 1*1 и i*i сокращаются.
Легкотня.
Зачем проверять устно, когда это можно сделать молча в уме?
@@eugnsp Не понял вопроса? Я делал именно молча, в уме. Термин "устно" использовал как противоположеность "письменному" решению или "у доски".
I don't believe this is a olympiad question.
1/2
It's a 8th grade problem 😅
This answer is incomplete, this equation should have 4 answers not just three
perhaps we have the same situation as with (x-1)² = 0. we have two solutions: x1 = 0, x2 = 0
No. This is actually a third degree equation, because the leading term cancels. Compare x^500 + x = x^500 + 1, which has only one solution.
Обычная задача для 8 или 9 класса (из 11).
trivial solution is a = -(a-1) > a = 0.5. Why waste any further time ?
a^4 = (a - 1)^4
a^2 = (a - 1)^2 or a^2 = -(a - 1)^2
a = a - 1 or a = 1-a or a = i(a-1) or a = -i(a-1)
Case 1: a = a - 1. No solution
Case 2: a = 1 - a. Therefore 2a = 1, so a = 1/2
Case 3: a = i(a - 1). Expanding gives a = ia - i, so a - ia = -i. Multiply by 1+i, 2a = 1-i, so a = (1-i)/2
Case 4: a = -i(a - 1). So, a = -ia + i, and a + ia = i. Multiplying by 1 - i, we get 2a = 1 + i, so a = (1 + i)/2
Solutions: a = 1/2, (1-i)/2 or (1+i)/2
I was fine until 4:22 then I couldn't understand the mumbling accent. After 30 years, I need a slightly slower clearer explanation. You were saying what you were doing but you were not explaining what you were doing. I will have to go to one of the other maths channels.
Looking at the thumbnail for 10 seconds, I get 1/2. I'll skip the 10-minute video. I'm 66 years old.
Am 61, watched the video and did not regret it.
I think it was about actually going step by step and getting a methodical solution instead of just having an intuitive solution. And the complex solution also of course
you didnt get the complex solutions
A polynomial to n degree should have n roots...
guy's a realist @@kjzdv5044
Jeeeeeeez you completely overcomplicated this poor unsignifiant thing. Here is how to destroy this:
We notice that a cannot be equal to 1 then we can rewrite the equation like this: [a/(a-1)]^4=1
Now let's call z=a/(a-1). We know that the solution of z^4=1 are 1, -1, i and -I (or we can see it very quickly by using polar writing of z and 1, really easy stuff there).
All we need now is to find a from these solutions.
So let's solve a/(a-1)=t for a: a=at-t then a=t/(1-t)
If t=1, we have no solution. If t=-1, the solution is 1/2. If t=i, a=i/(1-i)=i.(1+i)/[(1-i)(1+i)]=(i-1)/2. If t=-i, a=-i/(1+i)=-i.(1-i)/[(1+i)(1-i)]=(1-i)/2.
Problem destroyed.
Another method:
(a²)²=[(a-1)²]² then a²=(a-1)² or a²=-(a-1)²
If a²=(a-1)² then a=a-1 (impossible) or a=-(a-1) which leads to a=1/2.
If a²=-(a-1)² then a²=[I.(a-1)]² then a=i.(a-1) or a=-i.(a-1), which leads to the two complex solutions we already saw.
There must be 4 solutions !
The answer is impossible because no number can equal a. Tell me what number a equals. Answering with a variable is not a real answer.
Io con 2 passaggi la risolvo più veloce di te! Non mi piace il tuo approccio all’algebra!
what are your two steps?
Trivial. Looks like a quartic equation, but is really just a cubic ( expand the LHS and subtract a^4 from both sides). a =1/2 is an obvious solution. Take out the factor (a+1/2) and solve the remaining quadratic equation using the well known formula.
[ 1 year early ] I did advanced Maths ⇒ 1984 they put me in ⊆ forced Manchester labour camp + in addition to the college Physics prize - Germany 1,5% of wealth / 50% of population
what
@yousefabdelmonem3788 “Year”-One orbit of Earth’s star ~365.256 days “Maths”-Abbreviation for Mathematics in England UK
@@OghamTheBold I know what these words mean, but the OP isn’t coherent as a whole, care to explain what it meant?
@yousefabdelmonem3788 1st part is Plain English enow-Forced me in a labour camp contrary to UDHR in 1984 with advanced Maths and the college Physics prize-Luckily ChatGPT just output a structured argument using UDHR/ECHR against the latest abomination-DWP work coach said "I don't care who lives or doesn't"-2nd is economic inequality data-I even used the correct decimal format when I saw the word German in the video
Easy way is to fourth root both sides...
So basically:
a = a - 1, subtract a both sides...
0 = -1. So basically no solution.
Watch the video