# How can prove this equation.

if $a+b=c+d=e+f=\dfrac{\pi}{3}$,

$\dfrac{\sin{a}}{\sin{b}}\cdot\dfrac{\sin{c}}{\sin{d}}\cdot\dfrac{\sin{e}}{\sin{f}}=1$,

Prove that:

$\dfrac{\sin{(2a+f)}}{\sin{(2f+a)}}\cdot\dfrac{\sin{(2e+d)}}{\sin{(2d+e)}}\cdot\dfrac{\sin{(2c+b)}}{\sin{(2b+c)}}=1$

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What have you tried? –  ᴊ ᴀ s ᴏ ɴ Mar 13 '13 at 12:04
Challenging question !! –  lsp Mar 13 '13 at 12:12
Question is making him rich anyway. What have you tried? –  Inceptio Mar 13 '13 at 12:19
Yeah ! Hope the question is correct without any error ! @math110 Can you share what you have tried ? –  lsp Mar 13 '13 at 12:22
Also, where you came across the question? –  Gerry Myerson Mar 13 '13 at 12:25

Consider an equilateral triangle $ABC$, and let $D$ be on $BC$ so that $\angle{BAD}=a$, so $\angle{DAC}=\frac{\pi}{3}-a=b$. Let $E$ be on $AC$ so that $\angle{CBE}=c$, so $\angle{EBA}=\frac{\pi}{3}-c=d$. Let $F$ be on $AB$ so that $\angle{ACF}=e$, so $\angle{FCB}=\frac{\pi}{3}-e=f$.By the sine version of Ceva's theorem and the given condition $\frac{\sin{a}}{\sin{b}}\cdot \frac{\sin{c}}{\sin{d}}\cdot \frac{\sin{e}}{\sin{f}}=1$, $AD, BE, CF$ are concurrent at a point, which we shall call $P$.

Extend $AD$ to points $A_1, A_2$ s.t. $\angle{A_1CB}=a+f, \angle{A_2BC}=b+c$. We have $\angle{CA_1A}=\pi-\angle{A_1CA}-\angle{A_1AC}=\pi-b-(e+f+a+f)=e$. Similarly $\angle{BA_2A}=d$. Thus $APC$ is similar to $ACA_1$ and $APB$ is similar to triangle $ABA_2$. Therefore $\frac{AA_1}{AC}=\frac{AC}{AP}=\frac{AB}{AP}=\frac{AA_2}{AB}$, so $AA_1=AA_2$, so $A_1=A_2$. Now

$$\frac{\sin{(b+2c)}}{\sin{(a+2f)}}=\frac{\frac{A_1P}{\sin{(a+2f)}}}{\frac{A_2P}{\sin{(b+2c)}}}=\frac{\frac{CP}{\sin{e}}}{\frac{BP}{\sin{d}}}=\frac{CP\sin{d}}{BP\sin{e}}$$

Similarly, we get $$\frac{\sin{(d+2e)}}{\sin{(c+2b)}}=\frac{AP\sin{f}}{CP\sin{a}}$$ $$\frac{\sin{(f+2a)}}{\sin{(e+2d)}}=\frac{BP\sin{b}}{AP\sin{c}}$$ so multiplying gives the desired equality $$\frac{\sin{(2a+f)}}{\sin{(2f+a)}}\cdot\frac{\sin{(2e+d)}}{\sin{(2d+e)}}\cdot\frac{\sin{(2c+b)}}{\sin{(2b+c)}}=1$$

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Very nice. I had the trigonometric form of Ceva's theorem, but didn't find where to go then. –  Jean-Claude Arbaut Mar 13 '13 at 15:27
Very nice construction. How would you motivate creating the point $A_1$? –  Calvin Lin Mar 13 '13 at 16:03
Well the first thing that comes to mind is the sine version of Ceva, so I drew the triangle and put in the angles. The main idea was to have an angle of $2a+f$ somewhere, and the easiest way I could envision that happening was to add an angle of $a+f$ to $a$ and $f$. This led to construction of point $A_1$ such that $\angle{A_1CB}=a+f, \angle{A_1BC}=b+c$, so that $\angle{PCA_1}=a+2f, \angle{PBA_1}=b+2c$, and somehow use sin rule for $\frac{A_1P}{\sin{(a+2f)}}=\frac{A_1P}{\sin{(b+2c)}}$. I then got $\angle{CA_1B}=d+e$, so ideally I want $\angle{PA_1C}, \angle{PA_1B}$ to split nicely to $d, e$. –  Ivan Loh Mar 13 '13 at 16:19
I then realised that if $A, P, A_1$ were collinear, then the angles would indeed split up nicely. It turns out that it is then easier to define $A_1, A_2$ as intersection with extension of $AD$ and show that $A_1, A_2$ are actually the same point. –  Ivan Loh Mar 13 '13 at 16:23
very very nice! thank you , I think this problem have other theoms. –  math110 Mar 13 '13 at 16:25