How to solve this Math Olympiad Question? Here comes the best way! The method can deal this kind of math questions. Watch the video and learn how to solve this problem!
I did it this way. We have a+b+c=3.multiplied by 2, we get 2a+2b+2c=6..(A) we have also a^2+b^2+c^2=3.. (B). Now (B)--(A) is (a^2--2a) +(b^2--2b) +(c^2--2c) =--3.add 1 to each of these terms to get perfect squares I. e (a--1) ^2+(b--1) ^2+(c--1) ^2=--3+3=0(3 added to rhs to equalise). Now, a=1, b=1, c=1.therefore, a^2022+b^2022+c^2022=1^2022+1^2022+1^2022=3.ans.
When i saw the question i got the answer by using common sense. Adding a, b ,c gives 3 and ,adding squares of a ,b ,c gives 3 ,so 1 is the only number with square that equals to the number itself..so a=b=c=1...but for different value in the question, your method is really usefull
The 1/1/1 answer is obvious, proving that it is the only (real) solution is less trivial. Now 1 is not the only number equals to its own square, 0 is too. But even that doesn't prove that a=b=c. Indeed if you allow complex solutions, there are solutions where abc. To prove a=b=c when working with real numbers, you need something like he did in the video.
a, b, and c values are 1. Since (a+b+c)²=a²+b²+c²+2ab+2ac+2bc, we'd have this: 1²+1²+1²+2×1×1+2×1×1+2×1×1. Also since 2 is being added 3 times could be easily written as 2×3, and 1 added 3 times can be written as 1×3. So 1×3+2×3 is equal to 3+6=9. So the answer for a²⁰²²+b²⁰²²+c²⁰²² would be 3
Thanks for this puzzle of a problem. I'm still struggling to find another way of proving that the obvious solution is the only one. It really looks like it should be easier to prove, but I can't seem to get there.
The first equation represents a 2D plane in a 3D Euclidean space, which intersects the coordinate axes at (3,0,0), (0,3,0), and (0,0,3). The point on this plane which is closest to the origin is the point (1,1,1), and is a distance of sqrt(3) from the origin. The second equation has solutions consisting of points that lie on the 2D surface of a sphere of radius sqrt(3) centered about the origin. If the sphere and plane intersect, then the given expression will have one or more values. The (1,1,1) point is the ONLY point where the surfaces intersect. Consequently, the value of the given expression must be equal to 3.
There is very a creative way to solve this. (a+b+c)²=3² a²+b²+c²+2ab+2bc+2ca=9 3+2ab+2bc+2ca=9 2ab+2bc+2ca = 6 ab+ bc + ca= 3 a+b+c= ab + bc + ca for ab to be same as a, b = 1 for bc to be same as b , c = 1 For ca to be same as c , a =1 Now just do the problem 1²⁰²²+1²⁰²²+1²⁰²²=3 1+1+1=3 3=3 Which is true
Why you are flashing the writting over the sum/statement on the screen, when you are doing the math, your flashing should be suitably place so that one can follow your writtings
There actually are at least two solutions: a=b=c=1 and a=b=c=0! The “1” solution seems too straightforward. Personally, I prefer a little subtlety, i.e., “0!”.
Much easier way is to find just 1 solution that works, e.g. (a,b,c)=(1,1,1) So if this problem is valid, then a^2022 + b^2022 + c^2022 = 1^2022 + 1^2022 + 1^2022 = 3
no, because you have to find all possible solutions in the Real numbers, and you have not proven that that is the only solution. Notice that I. say in the real numbers because the. implication 4:07 is valid only if x^2 can't be negative. That is the point of maths you have to proof stuff.
@@jaimeduncan6167 All I am saying is that in math olympiad, if you get a question like this, and the final answer is expected to be a value, then choosing the degenerate case always gets you to the answer. Since (a,b,c)=(1,1,1) is an easily observable solution, if there is an answer it can only be 3.
@@graemedurie9094 I'm not sure we're understanding each other. I'm asking whether you found an easier way to prove your solution, which I would like to know how to do, or you simply had an intuition for the solution, saw that it checked out, and left it at that without actually proving anything.
@@requemao My "method" started with noting that the sum of a,b and c was equal to the sum of each of them squared. That meant that a had to equal a squared and so forth. The only number for which that is possible is 1. It could not be even -1. I'd not call it intuition but basic knowledge.
You need to define question better. Could be complex numbers etc., but you did not include 0 either and the question does not seem to state positive integers only and you did not state your assumptions. Sorry but 4 out 10 on this question. Better luck on the rest of the exam... lol
