[SOLVED]Triangle sides challenge

anemone

MHB POTW Director
Staff member
It is given that the ratio of angles $A,\,B$ and $C$ is $1:2:4$ in a $\triangle ABC$, prove that $(a^2-b^2)(b^2-c^2)(c^2-a^2)=(abc)^2$.

Well-known member
We see that angles $A,\,B$ and $C$ are $\frac{\pi}{7}$, $\frac{2\pi}{7}$ and $\frac{4\pi}{7}$

and using laws of sines we need to prove

$(\sin ^2 A -\sin ^2 B)(\sin ^2 B - \sin ^2 C)(\sin ^2 C - \sin ^2 A)= (\sin\,A \sin\,B \sin\, C)^2$.

where $A= \frac{\pi}{7}$ $B= \frac{2\pi}{7}$ and $C= \frac{4\pi}{7}$

to avoid fraction let $x =\frac{\pi}{7}$ so $A= x$ $B= 2x$ and $C= 4x$

so $\sin\,x = \sin 6x$ and $\sin 3x = \sin 4x$
we shall be using the following

$\sin\, X + \sin \,Y = 2 \sin \frac{X+Y}{2} \cos \frac{X-Y}{2}\cdots(1)$
$\sin\, X - \sin \,Y = 2 \sin \frac{X-Y}{2} \cos \frac{X+Y}{2}\cdots(2)$

So

$\sin ^2 X - \sin ^2 Y$
= $(\sin\, X + \sin \,Y)(\sin\, X - \sin \,Y)$
= $(2 \sin \frac{X+Y}{2} \cos \frac{X-Y}{2})(2 \sin \frac{X-Y}{2} \cos \frac{X+Y}{2})$
= $(2 \sin \frac{X+Y}{2} \cos \frac{X+Y}{2})(2 \sin \frac{X-Y}{2} \cos \frac{X-Y}{2})$
= $\sin (X+Y)\sin(X-Y)$

Now
LHS
= $(\sin ^2 A -\sin ^2 B)(\sin ^2 B - \sin ^2 C)(\sin ^2 C - \sin ^2 A)$
= $(\sin ^2 x -\sin ^2 2x)(\sin ^2 2x - \sin ^2 4x)(\sin ^2 4x - \sin ^2 x)$
= $(\sin ^2 2x -\sin ^2 x)(\sin ^2 4x - \sin ^2 2x)(\sin ^2 4x - \sin ^2 x)$ multiplying 1st and 2nd terms by -1
= $\sin 3x \sin \,x \sin 6x \sin 2x \sin 3x \sin \,x$
= $\sin 4x \sin \,x \sin x \sin 2x \sin 3x \sin \,x$ as $\sin 3x = \sin 4x$ and $\sin 6x =\sin \,x$
= $(\sin\,x \sin 2x \sin 4x)^2$
= $(\sin\,A \sin\,B \sin\, C)^2$
= RHS
Proved

Opalg

MHB Oldtimer
Staff member
\begin{tikzpicture}
\draw circle (4cm) ;
\coordinate [label=right:{$1=\omega^7$}] (A) at (4,0) ;
\coordinate [label=above right:$\omega$] (B) at(51.4:4cm) ;
\coordinate [label=above:$\omega^2$] (C) at(102.8:4cm) ;
\coordinate [label=left:$\omega^3$] (D) at(154.3:4cm) ;
\coordinate [label=left:$\omega^4$] (E) at(205.7:4cm) ;
\coordinate [label=below:$\omega^5$] (F) at(257.1:4cm) ;
\coordinate [label=below right:$\omega^6$] (G) at(308.5:4cm) ;
\draw (A) -- node [above right]{$a$} (B) -- node[above]{$b$} (D) -- node[below]{$c$} (A) ;
\foreach \point in {A,B,C,D,E,F,G} \fill (\point) circle (2pt) ;
\end{tikzpicture}
Let $\omega = e^{2\pi i/7}$. The triangle with vertices at $1$, $\omega$ and $\omega^3$ has the correct angles. Its sides have lengths $a = |\omega-1|$, $b = |\omega^3 - \omega|$, $c = |\omega^3-1|$. Then (using the fact that $|z|^2 = z\overline{z}$) $$a^2 = (\omega-1)(\omega^6-1) = 2 - \omega - \omega^6, \\b^2 = (\omega^3-\omega)(\omega^4-\omega^6) = 2 - \omega^2 - \omega^5, \\c^2 = (\omega^3-1)(\omega^4-1) = 2 - \omega^3 - \omega^4,$$ $$a^2-b^2 = \omega^2 + \omega^5 - \omega - \omega^6 = -\omega(\omega-1)(\omega^4-1), \\b^2-c^2 = \omega^3 + \omega^4 - \omega^2 - \omega^5 = -\omega^2(\omega-1)(\omega^2-1), \\c^2-a^2 = \omega + \omega^6 - \omega^3 - \omega^4 = \omega^8 + \omega^6 - \omega^{10} - \omega^4= -\omega^4(\omega^2-1)(\omega^4-1).$$ Therefore $(a^2-b^2)(b^2-c^2)(c^2-a^2) = - (\omega-1)^2(\omega^2-1)^2(\omega^4-1)^2.$

Also, $$a^2 = (\omega-1)(\omega^6-1) = (\omega-1)(\omega^6-\omega^7) = -\omega^6(\omega-1)^2, \\b^2 = (\omega^3-\omega)(\omega^4-\omega^6) = -\omega^5(\omega^2-1)^2, \\c^2 = (\omega^3-1)(\omega^4-1) = (\omega^3-\omega^7)(\omega^4-1) = -\omega^3(\omega^4-1)^2,$$ from which $(abc)^2 = - (\omega-1)^2(\omega^2-1)^2(\omega^4-1)^2.$

Comparing those two outcomes, $(a^2-b^2)(b^2-c^2)(c^2-a^2) = (abc)^2.$

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