Finding value of $\frac{49b^2-33bc+9c^2}{a^2}$

Question

If $a,b,c$ are positive real numbers such that

$a^2+ab+b^2=9, b^2+bc+c^2=52,$

$c^2+ac+a^2=49$. Then finding $\displaystyle \frac{49b^2-33bc+9c^2}{a^2}$ is

Try: let $O$ be a point inside the Triangle $ABC$ such that angle $OAB$ and angle $OBC$ and $OCA$ is $120^\circ$. So we have

$OA=a,OB=b,OC=c$ and using cosine formula we have $AB=3,BC=7,CA=\sqrt{52}$.

Now using Area of triangle is $\displaystyle ab+bc+ca=\frac{42}{\sqrt{3}}$.

Could some help me to solve it , Thanks

Answer 1

$a^2+ab+b^2=9 \implies a^2 = 9 - ab - b^2 $

$b^2+bc+c^2=52 \implies b^2 = 52 -bc - c^2$

$c^2+ac+a^2=49 \implies c^2 = 49 - ac - a^2$

Substituting $a^2$ in $(iii)$

$(c-b)(a+b+c) = 40 --(iv)$

Similarly,

$(b-a)(a+b+c) = 3 -- (v)$

$(c-a)(a+b+c)= 43 -- (vi)$

Now adding $(i),(ii),(iii)$

$2(a^2 + b^2 + c^2) + ab + bc + ca = 110$

$a^2 + b^2 + c^2 + 2ab + 2bc + 2ca + a^2 + b^2 + c^2- ab - bc - ca = 110$

$(a+b+c)^2 + \cfrac{(a-b)^2 + (b-c)^2 + (c-a)^2}{2} = 110$

Now, substitute $(a-b),(b-c),(c-a)$ from equation $(iv),(v),(vi)$,

$(a+b+c)^2 + \cfrac{3^2+40^2+ 43^2}{2(a+b+c)^2}= 110$

Take,$(a+b+c)^2 = t$ and solving quadratic,

$t = 91,19$

$(a+b+c) = \sqrt{91}$ or $\sqrt{19}$

clearly, $a+b+c$ can't be $\sqrt{19}$

Now, substitue $a+b+c$ in $(iv),(v),(vi)$

we get, the values of (c-b),(b-a),(c-a)

put the values of b,c in $a+b+c = \sqrt{91} $

We will get the value of a,b,c.

Answer 2

We can treat the equations as quadratic with variables $a,b,c$, and thus we solve for each of their values using the quadratic formula: $$\begin{align} a\to& \frac{1}{2} \left(\sqrt{36-3 b^2}-b\right)\tag{1}\\ b\to& \frac{1}{2} \left(\sqrt{208-3 c^2}-c\right)\tag{2}\\ c\to& \frac{1}{2} \left(\sqrt{196-3 a^2}-a\right)\tag{3}\\ \end{align}$$ Since $a,b,c \in \mathbb{R}^+$, then we can discard the negative values of $a,b,c$. We can rewrite $(3)$ in terms of $a$, and we get: $$a\to \frac{1}{2} \left(\sqrt{196-3 c^2}-c\right)\tag{4}$$ Then we substitute $(4)$ in $(1)$ and we get: $$\frac{1}{2} \left(\sqrt{196-3 c^2}-c\right)=\frac{1}{2} \left(\sqrt{36-3 b^2}-b\right)\tag{5}$$ Using $(2)$ and substituting it in $(5)$, we have: $$2 \sqrt{196-3 c^2}+\sqrt{208-3 c^2}=\sqrt{6} \sqrt{c^2+\sqrt{208-3 c^2} c-80}+3 c$$ For which we get $c=\frac{58}{\sqrt{91}}$, using that we get $a$ and $b$ from $(4)$ and $(2)$, respectively: $$\begin{align} a=&\frac{15}{\sqrt{91}}\\ b=&\frac{18}{\sqrt{91}}\\ \end{align}$$ And thus, we can now solve the problem: $$\begin{align}\displaystyle \frac{49b^2-33bc+9c^2}{a^2}\Rightarrow &\displaystyle \frac{49(\frac{18}{\sqrt{91}})^2-33(\frac{18}{\sqrt{91}})(\frac{58}{\sqrt{91}})+9(\frac{58}{\sqrt{91}})^2}{(\frac{15}{\sqrt{91}})^2}\\ &\Rightarrow 52 \end{align}$$

Answer 3

Applying cosine rule on $\triangle OBC$, we have

$$BC=\sqrt{b^2+bc+c^2}=\sqrt{52}$$

Similarly,

$$CA=7$$

By applying cosine rule on $\triangle ABC$

$$BC^2=AB^2+AC^2-2AB\cdot AC \cos \angle CAB$$

$$52=9+49-2(3)(7)\cos \angle CAB$$

$$\cos \angle CAB =\frac{1}{7} $$

Let's apply inversion at point $A$ with radius $1$, and let $O'$ be the image of $O$, $B'$ be the image of $B$ and $C'$ be the image of $C$.

Then $\triangle ABC \sim \triangle AC'B'$ , $\triangle AOB \sim \triangle AB'O'$, and $\triangle AOC \sim \triangle AC'O'$.

Hence $$O'B'=\frac{BO \cdot AB' }{AO}=\frac{b\cdot AB'\cdot AB}{AO\cdot AB}=\frac{b}{3a}$$

$$O'C'=\frac{CO \cdot AC' }{AO}=\frac{c\cdot AC'\cdot AC}{AO\cdot AC}=\frac{c}{7a}$$

$$B'C'=\frac{CB \cdot AC' }{AB}=\frac{CB\cdot AC'\cdot AC}{AC\cdot AB}=\frac{CB}{7\cdot 3}=\frac{\sqrt{52}}{21}$$

Let's compute $\cos \angle B'O'C'$, from quadrilateral $AO'B'C'$.

$$\angle AB'O'+ \angle B'O'C'+\angle O'C'A + \angle C'AB'= 2\pi $$

$$\angle AOB+ \angle B'O'C'+\angle COA + \angle CAB= 2\pi $$

$$\frac{2\pi}3+ \angle B'O'C'+ \frac{2\pi}3+ \angle CAB= 2\pi $$

$$ \angle B'O'C'= \frac{2\pi}3 - \angle CAB$$

\begin{align}\cos \angle B'O'C' &= -\frac12 \cos \angle CAB +\frac{\sqrt3}2 \sin \angle CAB\\ &=-\frac1{14} +\frac{\sqrt3}2 \frac{\sqrt{48}}{7}\\ &= \frac{11}{14}\end{align}

Applying cosine rule on $\triangle O'B'C'$,

$$B'C'^2=O'B'^2+O'C'^2-2\cdot O'B'\cdot O'C' \cos \angle B'O'C'$$

$$\frac{52}{21^2}=\left(\frac{b}{3a} \right)^2+\left(\frac{c}{7a} \right)^2-2\cdot \left(\frac{b}{3a} \right)\cdot \left(\frac{c}{7a} \right) \cdot \frac{11}{14}$$

$$\frac{52}{7^2}=\left(\frac{b}{a} \right)^2+\left(\frac{3c}{7a} \right)^2-2\cdot \left(\frac{b}{a} \right)\cdot \left(\frac{c}{7a} \right) \cdot \frac{33}{14}$$

$$\frac{52}{7^2}=\left(\frac{7b}{7a} \right)^2+\left(\frac{3c}{7a} \right)^2-\frac{33bc}{(7a)^2}$$

$$\frac{(7b)^2+(3c)^2-33bc}{a^2}=52$$

Remark:

In general, if $$a^2+ab+b^2=\gamma$$ $$b^2+bc+c^2=\alpha$$ $$c^2+ac+a^2=\beta$$

then

$$a^2\alpha =\beta b^2 + \gamma c^2 - Dbc $$

where $$D=\left[ - \frac{ \beta + \gamma - \alpha}{2} \right] + \frac{\sqrt3}{2} \sqrt{4 \beta \gamma - (\alpha+\beta-\gamma)^2}$$