## Sunday, May 3, 2015

### Evaluate ab+bc+ca

Given real numbers $a,\,b,\,c$ such that

$a^2+ab+b^2=2$, $b^2+bc+c^2=3$, $c^2+ca+a^2=5$

Evaluate $ab+bc+ca$.

If nothing that we could do to algebraically manipulate this problem that comes to mind, as this is a hard problem, then what is left for us to be done, is, we can sum the three given equations, to get:

$2(a^2+b^2+c^2)+ab+bc+ac=2+3+5$

$\therefore ab+bc+ac=10-2(a^2+b^2+c^2)$

We might consider the identity below at this point in time:

$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+a)$

$\implies ab+bc+a=\dfrac{(a+b+c)^2-(a^2+b^2+c^2)}{2}$

Now, we need to pause for a moment, as it looks as though we will hit a bump if we continue with what we have worked out and what we are given simply because we can never obtain the value for $a^2+b^2+c^2$ without knowing explicitly the values for each variables of $a,\,b$ and $c$. However, our intuition tells us this is not the problem asking us to do, we are of course not at all ask to find the values for $a,\,b$ and $c$.

That means we need to cast our net a little wider, we need to borrow help from other topics in mathematics rather than limiting ourselves with algebraic method to solve for the problem.

The thing is, what other related topic that can be of help for this problem? Matrices? Trigonometry? Or Geometry?

Speaking of trigonometry, aha! If we have a triangle with sides $a,\,b$ and $c$ and angles $A,\,B$ and $C$ the cosine rule can be written as:

$a^2=b^2+c^2-2bc\cos A$

How do we relate $b^2+bc+c^2=3$ with the above, so to make full use of that?

If we let $\cos A=\cos 150^{\circ}=-\dfrac{1}{2}$, we will have:

$b^2+c^2-2bc\left(-\dfrac{1}{2}\right)=3$

$b^2+c^2+bc=(\sqrt{3})^2$

We can then build a triangle below:
Similarly, we can let $\cos B=\cos C=\cos 150^{\circ}=-\dfrac{1}{2}$ to get:

$a^2+b^2+ab=(\sqrt{2})^2$

$a^2+c^2+ac=(\sqrt{5})^2$ with the diagrams illustrated the situation as below:

Putting the diagram together, we discover that we are actually dealing with a right-angled triangle:

Notice that $(\sqrt{2})^2+(\sqrt{3})^2=(\sqrt{5})^2$, therefore the biggest triangle that we see in the last diagram shows that it is actually a right-angled triangle with the side that has $\sqrt{5}$ as its length be the hypotenuse side.

We are not done yet, our target is to evaluate the expression $ab+bc+ca$, looking at the diagram, we can use the area of the big triangle is equivalent to the sum of the three smaller triangles inside it that observation to get what we aim for.

$\text{Area of triangle$PQR$}=\text{Area of triangle$PSR$}+\text{Area of triangle$PSQ$}+\text{Area of triangle$QSR$}$

$\dfrac{\sqrt{2}\sqrt{3}}{2}=\dfrac{bc\sin 120^{\circ}}{2}+\dfrac{ab\sin 120^{\circ}}{2}+\dfrac{ac\sin 120^{\circ}}{2}$

$\sqrt{2}\sqrt{3}=\sin 120^{\circ}(bc+ab+ac)$

$\sqrt{2}\sqrt{3}=\dfrac{\sqrt{3}}{2}(bc+ab+ac)$

Therefore

$bc+ab+ac=2\sqrt{2}$