Germany - Math Olympiad Problem | Be Careful! 

Theres no way this is an olympiad problem. This is like algebra ii at most.

Question: Solve for the value of a. Answer: x = ... Wait... what?!!! 😱🤔

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

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.

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.

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

Be Careful not to change the "a" to an "x" in your answer!

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.

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.

The oddity of this solution is a 4th-order polynomial with only 3 roots.

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.

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

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

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

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.

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

Feels like a problem from the International Olympiad in Algebra 2

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

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

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

So we say in the hypothesis (a belongs to C - complex one).

Dilly dalliy approach lengthening the presentation!

|a| = |a-1| -> a=a-1; a=-a+1 -> a=1/2

What happened to the 4th answer?

Almost correct explanation… √-4 = √(-1×4) = √(i² × 4) = √i² × √4 = i × 2 = 2i. Stating that √-1 = i is a sloppy shortcut ;)

What is the 4th solution?

The final answers are basically wrong because you are solving for a & not x-That's what my teacher would have said💔😭

Why does a fourth degree equation have only three solutions? The basic theorem of algebra is violated, however.

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.

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 ?

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. !

a = +/- (a-1)

I mean immediately looking at this question I knew one answer was 1/2

This task is solved by 2 steps with geometric considerations

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?

|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?

Cadê a quarta solução?

Force power?!?!?

"i" don't work as a solution.

what does this bracket sign mean? 8:23

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.

i just did this in 3 methods ._.

a=.5

It's obviously 0.5 . Why would you want to spend 10 minutes solving it?

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.

I don't believe this is a olympiad question.

1/2

This answer is incomplete, this equation should have 4 answers not just three

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.

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!

[ 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

Easy way is to fourth root both sides... So basically: a = a - 1, subtract a both sides... 0 = -1. So basically no solution.