Olympiad Question. Can You Solve this? | Simple & In-Depth Explanation

#OlympiadMathematics #OlympiadPreparation #CollegeEntranceExam

[sandanadurair]

a-1/a = b : a/b-1/ab =1
b-1/b = c : b/c-1/bc
c-1/c = a : c/a-1/ca =
adding 1, 2 and 3 and rearranging we get
a/b+b/c+c/a = 3 + 1/ab +1/bc + 1/ca
1/a = a-b : 1/b= b-c : 1/c=c-a from given
sunstituting all these in 4 we get
a(b-c)+b(c-a)+c(a-b) = 3+1/ab+1/bc+1/ca
expanding we get
ab-ca+bc-ab+ca-bc = 3+1/ab+1/bc+1/ca
0 = 3 + 1/ab+1/bc+1/ca
rearranging we get
1/ab+1/bc+1/ca = -3

[math]

this multivariable rational equation took a lot of steps, very well done, thanks for sharing

[sberacatalin]

Frumos! Cu artificii cu tot tacamul! Super! Next?!

[seema]

Can you make more such Olympiad videos? Because I am preparing for Olympiads and your questions and their explanation is amazing.

[ssudhiravinesh]

Such amazing efforts by you sir! Keep it up! Our math family will grow soon!

[HUNTER-qnff]

Greetings from Thailand i always found you video on recommended and the problem look interesting and i like it!😊

[GiuseppeAriano]

1/a = a-b; 1/b=b-c; 1/c=c-a; ab=a^2-1; bc=b^2-1; ca=c^2—1.
1/(ab)+1(bc)+1(ac)= (a-b) (b-c) +(b-c) (c-a) +(a-b) (c-a) = ab+bc+ca-(a^2+b^2+c^2)=-3.
Thank you for your work, Sir. I amuse with your questions.

[ABDxLM]

How did you know squaring will help in solving the problem ?

[TheRaghu]

a-1/a=b can be written as a^2-1=ab => 1/ab=1/a^2-1 similarly if we do for other 2 equations fallowed by adding them and rearranging can get the same solution without using algebraic formulae.

[usman_mmalik]

math has been always the toughest for me
you have made it easy

[ramanivenkata]

Classic working. It will not occur to all the way it occurred to Professor. Many will take a long time to solve.

[laylahaddaad]

Thank u so much profecor please what is the program u are using to explain the problem??

[walts]

Would be helpful to talk about the symmetry that motivates your algebra choices. Otherwise it looks like a magic trick that students cannot apply to other problems. Very good channel content overall!

[pedroloures]

WOW!! This was a cool problem! Thanks!

[zplusacademy]

Very simple and easy calculation 🧮🧮🧮🧮🧮 great 👍👍👍👍👍 sir ❤️🙏🙏🙏

[marioperic]

Here is my way (without looking the video):
summing also equations we get 1/a+!/b+1/c=0!

if we do square on the initial set of equations :
a^2+1/a^2-2=b^2; b^2+1/b^2=c^2-2; c^2+1/c^2-2=a^2;
Summing the equations we get 1/a^2+1/b^2+1/c^2=6.

since 1/a+1/b+1/c=0; => [1/a+1/b+1/c]^2=0 too.
or

at the end we get 1/ab+1/bc'1/ac=-3

[wackojacko]

🤔hmmmm....a perfect-square trinomial...(a+b+c)^2...a perfectly square tri-anything, is an oxymoron. In this case though the oxymoron is a paradox because (a+b+c)^2 is absolutely congruent...🙂

[gol-r]

Idk why it is smooth but it is smooth. Good question.