# 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
6 years, 1 month ago

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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].

- 6 years, 1 month ago

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??

- 6 years, 1 month ago

Edited.

- 6 years, 1 month ago

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

- 6 years, 1 month ago

- 6 years, 1 month ago

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

- 6 years, 1 month ago

Is it possible to solve this algebraically?

- 6 years, 1 month ago

what will be the algebric solution Help me

- 4 years, 5 months ago

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

- 6 years, 1 month ago

The point "O" is known as "Fermat" or Torricelli Point of the triangle.

- 1 year, 2 months ago