Tricky equation a^2 = k(k-u)(k-v)(k-w) where 2k = u+v+w

[Wilmer]

a^2 = k(k-u)(k-v)(k-w) where 2k = u+v+w

Given a, u and v, w = ?

 

[mrtwhs]

a^2 = k(k-u)(k-v)(k-w) where 2k = u+v+w

Given a, u and v, w = ?

\(\displaystyle k = \dfrac{u+v+w}{2}\), \(\displaystyle k-u = \dfrac{-u+v+w}{2}\), \(\displaystyle k-v = \dfrac{u-v+w}{2}\), \(\displaystyle k-w = \dfrac{u+v-w}{2}\).

Substituting in we get

\(\displaystyle a^2=\dfrac{1}{16}(u+v+w)(u+v-w)(w+(u-v))(w-(u-v))\)

Simplifying yields

\(\displaystyle 16a^2=((u+v)^2-w^2)(w^2-(u-v)^2)\)

\(\displaystyle -16a^2=(w^2-(u+v)^2)(w^2-(u-v)^2)\)

\(\displaystyle w^4-2(u^2+v^2)w^2+(u^2-v^2)^2+16a^2=0\)

This is a quadratic equation in \(\displaystyle w^2\). Substitute in the values for \(\displaystyle a,u,v\) and use the quadratic formula.

 

[Wilmer]

I was able to get to:
w = SQRT[u^2 + v^2 + 2SQRT((uv)^2 - 4a^2)]

As you probably surmised, this is Heron's triangle area in disguise!

Example: triangle sides u,v,w: u=4, v=13, w=15 : a = area = 24
My solution will give correctly w = 15 using u=4:v=13 or u=13:v=4
But not if u=4, v=15: does not yield 13

Can you tell me why...thanks in advance...

 

[mrtwhs]

I was able to get to:
w = SQRT[u^2 + v^2 + 2SQRT((uv)^2 - 4a^2)]

As you probably surmised, this is Heron's triangle area in disguise!

Example: triangle sides u,v,w: u=4, v=13, w=15 : a = area = 24
My solution will give correctly w = 15 using u=4:v=13 or u=13:v=4
But not if u=4, v=15: does not yield 13

Can you tell me why...thanks in advance...

Double check your arithmetic.
With \(\displaystyle u=4\), \(\displaystyle v=15\) and \(\displaystyle a=24\), you should get the equation

\(\displaystyle w^4-482w^2+52897=0\) which factors into

\(\displaystyle (w-13)(w+13)(w^2-313)=0\)

This appears to yield two possible answers \(\displaystyle w=13\) and \(\displaystyle w=\sqrt{313}\).

 

[Wilmer]

Gotcha! Thanks again...