Let $a, b$ and $c$ be the lengths of the sides of an arbitrary triangle. Define $$x =\frac{ab + bc + ca}{a^2 + b^2 + c^2}.$$
Pick out the true statements.
(a) $1/2 ≤ x ≤ 2$.
(b) $1/2 ≤ x ≤ 1$.
(c) $1/2 < x ≤ 1$.

How can I able to solve this problem

  • 1
    $\begingroup$ What will the value of $x$ be if (i) $a=b=c$, or (ii) $a=b\gg c$? $\endgroup$
    – pharmine
    Jan 17, 2013 at 6:45

4 Answers 4

  1. We first rule out (a) by writing $(a-b)^2\geq 0$, $(b-c)^2\geq 0$, $(c-a)^2\geq 0 $ and adding side by side to get $2a^2+2b^2+2c^2\geq 2ab + 2bc+2ac$. This tells us $\frac{ab+bc+ac}{a^2+b^2+c^2}\leq 1 $. Therefore $x$ cannot be greater than 1. Therefore (a) is not your answer.
  2. We now rule out (b) by writing: from the Cosine Law $a^2+b^2-2ab\cos\gamma=c^2$, $b^2+c^2-2bc\cos\alpha=a^2$ and $c^2+a^2-2ca\cos\beta=b^2.$ Adding side by side gives $a^2+b^2+c^2=2ab\cos\gamma+2bc\cos\alpha+2ac\cos\beta<2ab+2bc+2ac.$ This implies $x=\frac{ab+bc+ac}{a^2+b^2+c^2}>\frac{1}{2}$. Notice the last inequality follows since the angles in a triangle add up to $180^o$.
  • $\begingroup$ Nice one! +1 ;-) $\endgroup$
    – Mikasa
    Jan 17, 2013 at 7:15
  • $\begingroup$ @mathemagician I am confused. Wouldn't c being true also mean that a and b are true? Kindly see my solution. $\endgroup$
    – Akash Gaur
    Jan 4, 2014 at 15:34
  • $\begingroup$ No, I think it's more an issue of where does $x$ live, i.e. give the best characterization for $x$ $\endgroup$ Jan 4, 2014 at 17:52

We know in every triangle $ABC$ there are some useful relations called Law of cosines:

$$a^2=b^2+c^2-2bc\cos(A)\\b^2=a^2+c^2-2ac\cos(B)\\c^2=a^2+b^2-2ab\cos(C)$$ By adding them we have: $$a^2+b^2+c^2=2(bc\cos(A)+ac\cos(B)+ab\cos(C))$$ and if we take $A=B=C=60^{~\text{o}}$ then $x=1$($ABC$ is a Equilateral).

Now take $A=90^{~\text{o}},B=45^{~\text{o}},C=45^{~\text{o}}$ and using $b=a\sin(B), c=a\cos(B)$ we have: $$x=\frac{a^2\sin(B)+a^2\sin(B)\cos(B)+a^2\cos(B)}{a^2}\sim 0.9$$ It seems that $x\leq1$.

  • $\begingroup$ Nice relations/identities to know! +1 $\endgroup$
    – amWhy
    Feb 16, 2013 at 0:12

because the expression is homogeneous of degree $0$ we may assume $a+b+c=1$

using $\sum$ to indicate symmetric sums we have, (with $\sum a=1)$: $$\frac{\sum 2ab}{\sum a^2} = \frac {(\sum a)^2-\sum a^2}{\sum a^2} $$ thus $$\frac{ab + bc + ca}{a^2 + b^2 + c^2} = \frac12 \left(\frac1{a^2 + b^2 + c^2}-1\right) $$ this expression, if $a,b,c$ are the sides of a triangle, has a maximum value of $1$ when the triangle is equilateral, and an unattained minimum of $\frac12$ in the degenerate case where one of the sides is zero

added: apologies, as I see Rhaldryn has already posted a more thorough answer using a similar line of argument. I will leave my answer here because the expression is slightly different, and may be a useful complement.

  • $\begingroup$ I do not understand the statement on your profile: "is to solve a quartic equation using only a hammer and chisel...". But this already exists(en.wikipedia.org/wiki/…) $\endgroup$
    – John Smith
    Jan 4, 2014 at 17:27
  • $\begingroup$ thank you Leslie - i had never seen that before! everyone should see it at least once. my profile statement was merely intended as an ironic reference to my incompetence with complicated computations. i shall change the profile when i decide that i am no longer a complete n00b. $\endgroup$ Jan 4, 2014 at 17:43

Triangle inequality $$\Rightarrow\left | a-b \right |<c \Rightarrow \sum_{a,b,c}(a-b)^{2}<\sum_{a,b,c}c^{2}\Rightarrow 2\sum_{a,b,c}a^{2}-2\sum_{a,b,c}ab<\sum_{a,b,c}a^{2}\Rightarrow \frac{1}{2}<\frac{\sum\limits_{a,b,c}ab}{\sum\limits_{a,b,c}a^{2}}=x$$ $x\in \mathbb{R}\Rightarrow x^{2}\geq 0$ $$\Rightarrow(a-b)^2\geq 0\Rightarrow \sum_{a,b,c}(a-b)^2\geq 0\Rightarrow 2\sum_{a,b,c}a^2-2\sum_{a,b,c}ab\geq 0\Rightarrow x=\frac{\sum\limits_{a,b,c}ab}{\sum\limits_{a,b,c}a^{2}}\leq 1$$ Answer:

The true statements are:

(a) $1/2 ≤ x ≤ 2$
(b) $1/2 ≤ x ≤ 1$
(c) $1/2 < x ≤ 1$

(c) $1/2 < x ≤ 1$ gives the tightest bounds, but since the other two intervals are supersets of this interval, (a) and (b) are also satisfied. To further clarify:

$\forall\Delta ABC$, with sides a,b,c

$x =\frac{ab + bc + ca}{a^2 + b^2 + c^2}\Rightarrow 1/2 < x ≤ 1\Rightarrow 1/2 ≤ x ≤ 1\Rightarrow 1/2 ≤ x ≤ 2$

Also, $\forall x\in (\frac{1}{2},1],\exists \Delta ABC$ with sides $ a,b,c \ni x =\frac{ab + bc + ca}{a^2 + b^2 + c^2}$ since condition (c) gives the tightest bound.

However, $\exists x\in[\frac{1}{2},1],\forall\Delta ABC$ with sides $a,b,c ;\frac{ab + bc + ca}{a^2 + b^2 + c^2}\not=x$ and hence $\exists x\in[\frac{1}{2},2],\forall\Delta ABC$ with sides $a,b,c ; \frac{ab + bc + ca}{a^2 + b^2 + c^2}\not=x.$ Thus the implication is one sided for conditions (a) and (b).

$\therefore x=\frac{ab + bc + ca}{a^2 + b^2 + c^2}\Leftrightarrow1/2 < x ≤ 1 $but $1/2 ≤ x ≤ 1\vee1/2 ≤ x ≤ 2\not\Rightarrow \frac{ab + bc + ca}{a^2 + b^2 + c^2}=x $


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