If $a,b,c\in \mathbb{R}$ satisfy the system

$a^2+ab+b^2=9$;

$b^2+bc+c^2=16$;.

$c^2+ac+a^2=25$.

Find $ab+ac+bc$

• Are you sure it isn't $c^2 + ac + a^2$? Dec 29, 2013 at 0:51
• Are these supposed to be integers? Real numbers? What? Dec 29, 2013 at 1:14
• If you define $a,b,c\in\mathbb R$, then Raghav's answer is not sufficient. There are the other solutions. see the first one and the fourth one in the following page. wolframalpha.com/input/… Dec 29, 2013 at 3:15

The following is a geometric solution. Let $P$ be a point, and consider three line segments $PA,PB$ and $PC$ making an angle of $120$ degrees with each other. Thus $\angle APB = \angle BPC = \angle CPA = 120^o.$ Here $|PA| = a, |PB| = b$ and $|PC| = c.$

Then by the law of cosines, triangle $ABC$ has sides $3,4$ and $5.$

Further the area of the triangle $ABC$ is $\frac{1}{2}\cdot 3 \cdot 4 = 6.$ This can be calculated another way, namely $\frac{1}{2}\left( ab + bc + ca\right) \sin 120,$ since area of $ABC =$ areas of $PAB + PBC + PCA.$

Comparing the two, we are done.

• so, you are assuming $a,b,c\gt0$? But there are the other solutions. wolframalpha.com/input/… Dec 29, 2013 at 3:02
• But we aren't interested in finding $a,b,c.$ We are only interested in $ab+ bc+ca.$ Note this is preserved by the other solutions you've mentioned. Dec 29, 2013 at 3:04
• I'm talking about the value of $ab+bc+ca$. See the first one and the fourth one in the page I wrote. Dec 29, 2013 at 3:08
• If P is exterior at triangle ABC , $ab+bc+ac=-8\sqrt{3}$ Dec 29, 2013 at 19:16

If you subtract the first equation from the second, you get $c^2-a^2 + bc - ab = 7,$ so $(c-a)(c+a +b) = 7.$ You get similar equations (all with an $a+b+c$ factor) when you subtract the second from the third, etc, which gives you a linear system in $a, b, c$ (you know $a+b+c \neq 0,$ from the first sentence).

• Another identity can demand usage $$\sum_{\text{cyc }}(a-b)(a^2+ab+b^2)=0$$ :) Dec 29, 2013 at 15:54
• $$a^2+ab+b^2=9$$ $$(c-a)(a+b+c)=7$$ $$(c-b)(a+b+c)=16$$ Dec 29, 2013 at 19:13