in $\Delta ABC$,$AB=c,AC=b,BC=a$and such


show that $\Delta ABC$ is an equilateral triangle

this problem I have solution,But not nice, and I think this problem have more nice methods,Thank you everyone.

my solution: $$ab^2\cdot\dfrac{b^2+c^2-a^2}{2bc}=bc^2\cdot\dfrac{a^2+c^2-b^2}{2ac}=ca^2\cdot\dfrac{a^2+b^2-c^2}{2ab}$$ $$\Longrightarrow a=b=c$$

My other idea:

$$\Longleftrightarrow\dfrac{\sin{B}}{\sin{C}}\cos{A}=\dfrac{\sin{C}}{\sin{A}}\cos{B}=\dfrac{\sin{A}}{\sin{B}}\cos{C}$$ $$\Longleftrightarrow \sin{(2A)}\sin{B}=\sin{(2B)}\sin{C}=\sin{(2C)}\sin{A}$$ then How prove $$A=B=C$$ and have other nice methods? Thank you

  • $\begingroup$ Can you show more details of your solution, so that I can understand what you consider 'not nice'? $\endgroup$ – Calvin Lin Sep 12 '13 at 4:27

Dividing throughout by $abc$, we get

$$ \frac{b^2+c^2 - a^2}{c^2} = \frac{a^2 + c^2 - b^2}{a^2} = \frac{a^2+b^2-c^2}{b^2}.$$

Subtracting 1 from each term,

$$ \frac{b^2- a^2}{c^2} = \frac{ c^2 - b^2}{a^2} = \frac{a^2-c^2}{b^2}.$$

Applying Componendo et dividendo, these fractions are equal to

$$ \frac{ b^2-a^2+c^2-b^2+a^2-c^2} { c^2 + a^2 + b^2}$$

Hence, all of them are 0, which means $a=b=c$.

  • 1
    $\begingroup$ @math110 Judging from your comment on quarkine, is this the solution that you are thinking of? This is pretty 'nice', unless you want a more synthetic approach. $\endgroup$ – Calvin Lin Sep 12 '13 at 4:23
  • $\begingroup$ Yes,It's very nice,and Thank you +1 $\endgroup$ – math110 Sep 12 '13 at 4:29
  • $\begingroup$ and I think it maybe can use $$\dfrac{\sin{B}}{\sin{C}}\cos{A}=\dfrac{\sin{C}}{\sin{A}}\cos{B}=\dfrac{\sin{A}}{\sin{B}}\cos{C}$$ $\endgroup$ – math110 Sep 12 '13 at 4:31
  • $\begingroup$ @math110 Thanks. The motivation behind this is that since the equations are not symmetric, looking at just 2 of them is not sufficient to conclude that $a=b$. Com et div is the simplest way to combine 3 fractions, which is why I thought about using it. It could be used initially, to show that each fraction is 1. $\endgroup$ – Calvin Lin Sep 12 '13 at 4:35
  • $\begingroup$ Oh,Thank you very much,I know,+1 $\endgroup$ – math110 Sep 12 '13 at 4:36

Hint:divide all the three by abc and then proceed.

  • $\begingroup$ this idea is same as me.But Thank you $\endgroup$ – math110 Sep 12 '13 at 4:17

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