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Three equations with square variables

Let

a^2+b^2+ab=9

b^2+c^2+bc=16

a^2+c^2+ac=25

What is the value of ab + bc +ac?

Note by Sinuhé Ancelmo
4 years, 1 month ago

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3 votes

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Take a point \(O\) from which three line segments \(OA, OB, OC\), lying on the same plane, emanate such that their respective lengths are \(a, b, c\) and the angle between any two of them is 120 degrees. Use Cosine rule and the mentioned equations to find \(AB, BC, CA\) which will have values 3,4 and 5 units. Now this triangle \(ABC\) is right angled( by the converse of Pythagoras' Theorem), so it's area will be 6 sq. units. Also, the area of the three triangles \(AOB, BOC, COA\) can be found out by the sine rule (individually), and now equate it with the earlier found area. The answer comes out to be \(8 \sqrt{3}\). [ the calculations are to be done by you].

Shourya Pandey - 4 years, 1 month ago

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Is it possible to solve this algebraically?

One Top - 4 years, 1 month ago

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Well, your idea is interesting, but are you sure about the value of ab + bc +ac? shouldn' it be 8*sqrt(3)? Could you chek it again, plz??

Sinuhé Ancelmo - 4 years, 1 month ago

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Could you show any algebraic, only algebraic solution, friend?

Sinuhé Ancelmo - 4 years, 1 month ago

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Edited.

Shourya Pandey - 4 years, 1 month ago

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@Shourya Pandey You can realize that - 8 * sqrt (3) is also a valid answer for ab + bc + ac, right?

Sinuhé Ancelmo - 4 years, 1 month ago

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@Shourya Pandey Thanks! Already read your edition, friend!

Sinuhé Ancelmo - 4 years, 1 month ago

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what will be the algebric solution Help me

Anit Singh - 2 years, 6 months ago

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The solution could be algebraical or geometrical, Thanks in advance for help!

Sinuhé Ancelmo - 4 years, 1 month ago

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