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 = ?

$$k = \dfrac{u+v+w}{2}$$, $$k-u = \dfrac{-u+v+w}{2}$$, $$k-v = \dfrac{u-v+w}{2}$$, $$k-w = \dfrac{u+v-w}{2}$$.

Substituting in we get

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

Simplifying yields

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

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

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

This is a quadratic equation in $$w^2$$. Substitute in the values for $$a,u,v$$ 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 $$u=4$$, $$v=15$$ and $$a=24$$, you should get the equation

$$w^4-482w^2+52897=0$$ which factors into

$$(w-13)(w+13)(w^2-313)=0$$

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

 

[Wilmer]

Gotcha! Thanks again...