Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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:

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

share|improve this question
    
What have you tried? –  ᴊ ᴀ s ᴏ ɴ Mar 13 '13 at 12:04
1  
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

1 Answer 1

up vote 6 down vote accepted

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

Diagram for illustration

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.