Heron's formula?!

How I did the problem (elementarily):

Let I, J be points on BC so that $\frac{IB}{IC}=\frac{JB}{JC}=11$ , which means that AI is the internal bisector of angle BAC, and AI is the external bisector of BAC, which means that AI is always perpendicular to AJ, or the locus of A is a circle with diameter IJ. From here, we can calculate the value of IJ, which is $\frac { 120 }{ 11+1 } +\frac { 120 }{ 11-1 } = 22$ .

Let AH be the altitude to BC, we can see that AH is at most half the diameter IJ or $AH=11$ , which means the maximum of the triangle is $660$ .

P/S: If $\frac{AB}{AC}=a$ and $BC=b$ , the maximum value of the triangle is $\frac{ab^2}{2(a^2-1)}$ .

$s =\frac{a + b + c}{2}\$ where $a = 120$ and $c = 11b$

Using Heron's formula, we get,

$\sqrt {(60 + 6b)(60 + 5b)(60 - 5b)(60 - 6b)}$ = $\sqrt {(36{b}^2 - 3600)(3600 - 25{b}^2)}$ = $30 * \sqrt {({b}^2 - 100)(144 - {b}^2)}$

Now, from AM-GM inequality,

$\frac{{b}^2 - 100 + 144 - {b}^2}{2}\ \geq \sqrt {({b}^2 - 100)(144 - {b}^2)}$

Therefore, the maximum value of $30 * \sqrt {({b}^2 - 100)(144 - {b}^2)} = 30 * \frac{144 - 100}{2}\ = \boxed {660}$

Let's consider $B=(0,0)$ , $C=(120,0)$ and $A=(x,y)$ . Since $AB=11AC$ we can write

$\displaystyle \sqrt{(x-120)^2+y^2}=11\sqrt{x^2+y^2}$ .

Considering $x \geq 0$ and $y \geq 0$ , we can square both sides. After some algebra the equation simplifies to

$\displaystyle (x+1)^2+y^2=121 \hspace{5pt} \Longrightarrow \hspace{5pt} y=\sqrt{121-(x+1)^2}$ .

The area $A$ is

$\displaystyle A(x)=\frac{1}{2} \cdot 120 \cdot \sqrt{121-(x+1)^2}$ .

Let's maximize it, finding its critical point

$\displaystyle \frac{dA}{dx}=-\frac{60(x+1)}{\sqrt{121-(x+1)^2}}=0 \hspace{5pt} \Longrightarrow \hspace{5pt} x=-1$

Eventually $\displaystyle A(-1)=\boxed{660}$

$Let~the~sides~be~~~~(a,~b,~c)=(120,~ x,~ 11x).\\ \therefore~S=\frac 1 2 (120+12x)=60+6x.\\ But~Area=\sqrt {S(S-a)(S-b)(S-c)}.\\ \therefore~Area=\sqrt{6*(x+10)*6*(x-10)*5*(x+12)*5*( - x+12)} \\ =30\sqrt{(x^2-100)*( - x^2+144 )} \\ = 30\sqrt{ - x^4+244x^2-14400} \\ = 30\sqrt{22^2 -( x^2-122)^2} \\ Maximum~area~when~~ ( x^2-122)^2=0.\\ \therefore~~Area_{max}=\Large~~\color{#D61F06}{660}.$

Here's the solution that uses Heron's area formula:

(This isn't necessary, but I set $AC=x$ , so $AB=11x$ .)

For Heron's formula, you need to first find the semi-perimeter (half of the perimeter), $s=\frac{AB+AC+BC}{2}=\frac{11x+x+120}{2}=\frac{12x+120}{2}=6x+60$ .

Heron's area formula for a triangle with sides $a,b,c$ and semi-perimeter $s$ is

$A=\sqrt{s(s-a)(s-b)(s-c)}$ .

Plugging in what we have so far, we get that

$A=\sqrt{(6x+60)(6x+60-120)(6x+60-x)(6x+60-11x)}=\sqrt{(6x+60)(6x-60)(60+5x)(60-5x)}=\sqrt{(36x^2-60^2)(60^2-25x^2)}$

Finding the maximum by finding the critical points (I'm going to skip the differentiation and simplification steps, they're not the important part here):

$\frac{dA}{dx}=\frac{3600x(122-x^2)}{60\sqrt{(x^2-100)(144-x^2)}}$

You might be tempted to plug in $x=\sqrt{122}$ . And you can. But, you can save a little time if you notice that we have the area formula in terms of $x^2$ , so you can also plug in $x^2=122$ .

Regardless, plugging in gives us $A=660$