A Fun One from the Vietnamese IMO

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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Postfach 11 15
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[baoinhquoc]

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

[esbenthomas]

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.

[abbashashimov]

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.

[picipi]

When the trees start speaking algebruh

[gautruc]

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.

[aviratnakumar]

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

[thanglongnguyenvu]

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.

[NguyenTai-qwnv]

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

[aahaanchawla]

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

[kittisakchooklin]

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.

[Noname-]

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

=>(ab+bc+ac)(a+b+c)=abc

Raise the first equation to the 2023 then multiply with third equation give


Substitute the equality above and done

[Olise]

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.

[KhangNguyen-gdzw]

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!

[jessiechua]

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.

[parthibhayat]

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

[withertodream]

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

[fordchang]

(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.

[Sams_V.]

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

[jonxyrussims]

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

[jucom]

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