More Heron’s Formula Extensions

Continuing on the theme from the last post: going-one-step-beyond-herons-formula  here’s another problem showing the power of the full conceptual framework.  Heron’s formula itself even makes an appearance again at the end.

Problem

Given a $\triangle{ABC} $ show that  $ \tan\left(\dfrac{a}{2}\right)\tan\left(\dfrac{b}{2}\right) + \tan\left(\dfrac{a}{2}\right)\tan\left(\dfrac{c}{2}\right) + \tan\left(\dfrac{b}{2}\right)\tan\left(\dfrac{c}{2}\right) = 1 $

Traditional ApproachThere are many ways to attack this problem using trig identities I’m going to demonstrate one that is very linear in the progression of ideas and that focus on a direct simplification of the left hand side.   To do that I’m going to take advantage of the triangle angles.- We know that a + b + c = 180 so   $ \dfrac{a}{2} + \dfrac{b}{2} + \dfrac{c}{2} = 90 \rightarrow  \dfrac{c}{2}  = 90 - \left( \dfrac{a}{2} + \dfrac{b}{2} \right)  $ So first isolate  $ \dfrac{c}{2} $   $ \tan\left(\dfrac{a}{2}\right)\tan\left(\dfrac{b}{2}\right) + \tan\left(\dfrac{a}{2}\right)\tan\left(\dfrac{c}{2}\right) + \tan\left(\dfrac{b}{2}\right)\tan\left(\dfrac{c}{2}\right) = $  $ \tan\left(\dfrac{a}{2}\right)\tan\left(\dfrac{b}{2}\right) + \tan\left(\dfrac{c}{2}\right) \cdot  \left( \tan\left(\dfrac{a}{2}\right) + \tan\left(\dfrac{b}{2}\right) \right)  $ Now looking at $ \tan\left(\dfrac{c}{2}\right) $  we already know that’s $ \tan\left(\ 90 - \left( \dfrac{a}{2} + \dfrac{b}{2} \right ) \right)  $ But  $\tan(90-x) =  \dfrac{\sin(90 - x)}{\cos(90 -x)}  =  \dfrac{\cos x}{\sin{x}} = \cot{x}  $ We can also now use tangent addition: $ \tan\left(\dfrac{a}{2} + \dfrac{b}{2}\right) = \dfrac{\tan\left(\dfrac{a}{2}\right) + \tan\left(\dfrac{b}{2}\right)} {1 - \tan\left(\dfrac{a}{2}\right) \cdot \tan\left(\dfrac{b}{2}\right)} $ So together we get $ \tan\left(\dfrac{c}{2}\right)  = \dfrac{1 - \tan\left(\dfrac{a}{2}\right) \cdot \tan\left(\dfrac{b}{2}\right)} {\tan\left(\dfrac{a}{2}\right) + \tan\left(\dfrac{b}{2}\right)}  $ Substituting into our equation we get:   $  \tan\left(\dfrac{a}{2}\right)\tan\left(\dfrac{b}{2}\right) +   \dfrac{1 - \tan\left(\dfrac{a}{2}\right) \cdot \tan\left(\dfrac{b}{2}\right)} {\tan\left(\dfrac{a}{2}\right) + \tan\left(\dfrac{b}{2}\right)}   \cdot  \left( \tan\left(\dfrac{a}{2}\right) + \tan\left(\dfrac{b}{2}\right) \right)  $ That has a very convenient cancellation:   $  \tan\left(\dfrac{a}{2}\right)\tan\left(\dfrac{b}{2}\right) + 1 - \tan\left(\dfrac{a}{2}\right) \cdot \tan\left(\dfrac{b}{2}\right) =  1  $

Overall this is fairly neat but it needs a few identities and doesn't give
any geometric notion about what's going on.

Using Ratios Instead

This time we're going to go back to the diagram and express the tangents as
a ratio. In general $ \tan\left(\dfrac{a}{2}\right) = \dfrac{r}{s-a}
$  where r is the inradius and s is the semiperimeter. 

 So substituting in we get: 

$ \tan\left(\dfrac{a}{2}\right) \tan\left(\dfrac{b}{2}\right) +
\tan\left(\dfrac{a}{2}\right)\tan\left(\dfrac{c}{2}\right) +
\tan\left(\dfrac{b}{2}\right)\tan\left(\dfrac{c}{2}\right)$
  
  

  
$  = \dfrac{r^2}{(s-a)(s-b)} + \dfrac{r^2}{(s-a)(s-c)} +
\dfrac{r^2}{(s-b)(s-c)} $
  
  

  
 $ = \dfrac{r^2 \cdot ((s - a) + (s -b ) + (s -c))}{(s-a)(s-b)(s-c)}
$
  
  

  
 $ = \dfrac{r^2 \cdot s }{(s-a)(s-b)(s-c)} = \dfrac{r^2 \cdot s^2
}{s(s-a)(s-b)(s-c)} $
  
  

  
 But both the numerator and denominator are expressions for the area of
the triangle: (semiperimeter $\cdot$ inradius and heron's formula)

So we get: $ \dfrac{[ABC]^2}{[ABC]^2} = 1 $

This way, for me at least, shows the symmetry of what’s going on  is very direct and the intermediate expressions are all recognizable in a geometric sense.