We know the bisector length of angle $C$ is $\dfrac{ 2ab \cos C/2 }{a+b};$

In triangle ACB the $\angle C$ is trisected making segments $(c_1,c_2,c_3) $ on the side opposite. Please help to express these three segments in terms of $ (a,b,c, A,B,C).$


Can I now ask also now to find lengths of $(c_1,c_2... c_n)?$ when $\angle C$ is poly-sectioned $\angle C/n?$

  • 1
    $\begingroup$ have you tried anything yourself? $\endgroup$ – David Quinn Apr 8 '18 at 17:30
  • $\begingroup$ Hoping that more powerful vector approaches might be in existence I avoided a laborious ( inelegant ?) conventional trig/algebraic calculation. $\endgroup$ – Narasimham Apr 8 '18 at 17:42
  • $\begingroup$ Do you consider it too laborious to coordinatize, with $C = (0,0)$, $A = (b,0)$, $B=(a\cos 3\theta, a\sin 3\theta)$, and then to intersect $\overleftrightarrow{AB}$ with lines $y = x\,\tan k \theta$? $\endgroup$ – Blue Apr 8 '18 at 17:49
  • $\begingroup$ Not reeally. can I ask for generalization,? Angle $C$ poly-sected ( equal $\angle C/n $ with segments $( c_1,c_2,... c_n )$ in a while. $\endgroup$ – Narasimham Apr 8 '18 at 19:26
  • $\begingroup$ I was just wondering for $n$ whether segment lengths make a nice set with trig/algebraic quantities. $\endgroup$ – Narasimham Apr 8 '18 at 19:35

If the lines of trisection meet $AB$ in points $P$ and $Q$, then we can apply the sine rule in triangle $CPB$ and obtain $$c_1=\frac{a\sin\frac 13C}{\sin(\frac 13C+B)}$$

For $c_3$ you can exchange $B$ for $A$ and $a$ for $b$.

For $c_2$ you can use $c=c_1+c_2+c_3$

To generalise further, suppose the angle $C$ is divided into $n$ equal parts, creating segments $c_1,c_2,...c_n$ on the side $AB$

Let $u_r=c_1+c_2+...+c_r$

Then by exactly the same reasoning as before, $$u_r=\frac{a\sin\frac rnC}{\sin(\frac rnC+B)}$$

From this, one could deduce the $c_i$

  • $\begingroup$ Can I now ask to generalize when angle $C$ is divided into $n$ equal parts? $\endgroup$ – Narasimham Apr 8 '18 at 19:29
  • $\begingroup$ @Narasimham I have added a bit more... $\endgroup$ – David Quinn Apr 8 '18 at 19:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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