Let's Start With An Easy One!

Algebra Level 2

a a and b b are real numbers such that

a 4 + a 2 b 2 + b 4 = 900 a^4+a^2b^2+b^4=900

a 2 + a b + b 2 = 45 a^2+ab+b^2=45

What is the value of 2 a b 2ab ?


This problem is from the OMO-2012.


This problem is from the set "Olympiads and Contests Around the World -1". You can see the rest of the problems here .


The answer is 25.

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7 solutions

a 2 + a b + b 2 = 45 a^2 + ab + b^2 = 45

( a 2 + a b + b 2 ) 2 = 4 5 2 \Rightarrow (a^2 + ab + b^2)^2 = 45^2

a 4 + a 2 b 2 + b 4 + 2 a 3 b + 2 a 2 b 2 + 2 a b 3 = 2025 \Rightarrow a^4 + a^2 b^2 + b^4 + 2a^3b + 2a^2b^2 + 2ab^3 = 2025

( 900 ) + 2 a b ( a 2 + a b + b 2 ) = 2025 \Rightarrow (900)+ 2ab(a^2 + ab + b^2) = 2025

2 a b × 45 = 1125 \Rightarrow 2ab\times 45 = 1125

2 a b = 25 \Rightarrow 2ab = 25

I got it!!

Oliver Daniel - 7 years ago

damn freaking hard. I almost forgot how to factorize this tricky polynomials.

Arvin Panagan - 6 years, 7 months ago
Jubayer Nirjhor
May 23, 2014

a 4 + a 2 b 2 + b 4 = ( a 2 + a b + b 2 ) ( a 2 a b + b 2 ) = 45 ( a 2 a b + b 2 ) = 900 a 2 a b + b 2 = 20 a^4+a^2b^2+b^4=(a^2+ab+b^2)(a^2-ab+b^2)=45(a^2-ab+b^2)=900~\implies~a^2-ab+b^2=20

2 a b = ( a 2 + a b + b 2 ) ( a 2 a b + b 2 ) = 45 20 = 25 \therefore ~~~ 2ab=(a^2+ab+b^2)-(a^2-ab+b^2)=45-20=\fbox{25}

Even I Did it this way

HariShankar PV - 7 years ago

Problem:

a a and b b are real numbers such that

a 4 + a 2 b 2 + b 4 = 900 a^4+a^2b^2+b^4=900

a 2 + a b + b 2 = 45 a^2+ab+b^2=45

What is the value of 2 a b 2ab ?

Solution:

Okay, seeing all square terms in the first equauon, we are motivated to find some factorization using difference of squares. Notice that we can complete a square by adding an extra a 2 b 2 a^2b^2 to the LHS: a 4 + a 2 b 2 + b 4 + a 2 b 2 = ( a 2 + b 2 ) 2 . a^4+a^2b^2+b^4+a^2b^2=(a^2+b^2)^2. So to return back to our original equation, by subtracting the extra term that we added: 900 = ( a 2 + b 2 ) 2 ( a b ) 2 = ( a 2 + a b + b 2 ) ( a 2 a b + b 2 ) . 900=(a^2+b^2)^2-(ab)^2=(a^2+ab+b^2)(a^2-ab+b^2). Amazing! We factorized our first equation, and that too contains a factor a 2 + a b + b 2 a^2+ab+b^2 , the value of which we know from the second equation! So, simple substitution yields: 900 = 45 ( a 2 a b + b 2 ) a 2 a b + b 2 = 20 . 900=45(a^2-ab+b^2)\Rightarrow \boxed{a^2-ab+b^2=20}. Now we are ready to fire the problem: 2 a b = a 2 + a b + b 2 ( a 2 a b + b 2 ) = 45 20 = 25 . 2ab=a^2+ab+b^2-(a^2-ab+b^2)=45-20=\boxed{25}. Thus the problem is solved. :)

A very well-written solution! Voted up!

Mursalin Habib - 6 years, 1 month ago

Hello,

given, a^4 + (ab)^2 + b^4 = 900 --------> a^4 + b^4 = 900 - (ab)^2 (1st)

a^2 + b^2 + ab = 45 ------> a^2 + b^2 = 45 - ab (2nd)

So i will take the below equation,

(a^2 + ab + b^2 )^2= (45)^2 -----> by squaring both sides,

a^4 + b^4 + 3(ab)^2 + 2ab(a^2+b^2) = 2025 (3rd),

by substituting (1st) & (2nd) into (3rd),

900 - (ab)^2 + 3(ab)^2 + 2ab(45-ab) = 2025

90ab = 2025 - 900

ab = 1125 / 90 , therefore for 2ab = 2 x (1125 / 90 ) = 25

Thanks...

Aliki Patsalidou
May 24, 2014

a^2+b^2=45-a b so a^4+b^4+a^2b^2=900 (a^2+b^2)^2 -a^2 b^2=900 (45-a b)^2 - a^2 b^2=900 2025 - 90a b + a^2 b^2 - a^2 b^2 =900 1125 = 90a b 2a*b=25

(a^2 + b^2 )^2- a^2b^2=900

(45-ab)^2- a^2b^2=900

Solving u get ab=25/2 2ab= 25

William Isoroku
Aug 6, 2014

Square the second equation. Luckily it includes the first equation inside of it. So basically just substitute and simplify and found the second equation in it as well!!! Then just divide.

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