determine equilateral triangle's length using a point from the inside

**bibbonacci** · December 6th 2010, 12:54 AM

Problem:
A point P within an equilateral triangle has a distance of 4, 5, and 6 units respectively from the vertices. Determine the lengths of the sides of the equilateral triangle.

I was able to break it down into 3 triangles (coming from point P). So what I did was to use Cosine Law on those triangles and was able to get

2 sin 30 degrees x sin (60 - 2 theta) = -2/a

Am I on the right track. It seems that if I continue, I'd be getting 3 unknowns, 2 equations which is not feasible, unless I'm good at guessing. Do you have any other ways to solve this ?

Thank you in advance.

Regards,
Bibbo

**Grandad** · December 7th 2010, 01:37 AM

Hello bibbonacci

Originally Posted by bibbonacci

Problem:
A point P within an equilateral triangle has a distance of 4, 5, and 6 units respectively from the vertices. Determine the lengths of the sides of the equilateral triangle.

I was able to break it down into 3 triangles (coming from point P). So what I did was to use Cosine Law on those triangles and was able to get

2 sin 30 degrees x sin (60 - 2 theta) = -2/a

Am I on the right track. It seems that if I continue, I'd be getting 3 unknowns, 2 equations which is not feasible, unless I'm good at guessing. Do you have any other ways to solve this ?

Thank you in advance.

Regards,
Bibbo

If the angles at the point P are $\theta, \phi, \psi$ and the length of the side of the triangle is $\displaystyle a$ , then, using the Cosine Rule on all three triangles gives:

$\displaystyle \cos\theta=\frac{41-a^2}{40}$

$\displaystyle \cos\phi = \frac{61-a^2}{60}$

$\displaystyle \cos\psi = \frac{52-a^2}{48}=\cos(\theta+\phi)$ , since $\psi = 2\pi-(\theta+\phi)$

Eliminating $\displaystyle \theta$ and $\displaystyle \phi$ gives a pretty messy equation for $\displaystyle a$ , which I have solved using a spreadsheet (but perhaps that's cheating!), giving the answer $a \approx 8.536$ , to 3 d.p.

Perhaps someone can find an elegant algebraic solution?

Grandad

**Opalg** · December 7th 2010, 01:57 AM

Originally Posted by Grandad

If the angles at the point P are $\theta, \phi, \psi$ and the length of the side of the triangle is $\displaystyle a$ , then, using the Cosine Rule on all three triangles gives:

$\displaystyle \cos\theta=\frac{41-a^2}{40}$

$\displaystyle \cos\phi = \frac{61-a^2}{60}$

$\displaystyle \cos\psi = \frac{52-a^2}{48}=\cos(\theta+\phi)$ , since $\psi = 2\pi-(\theta+\phi)$

Eliminating $\displaystyle \theta$ and $\displaystyle \phi$ gives a pretty messy equation for $\displaystyle a$ , which I have solved using a spreadsheet (but perhaps that's cheating!), giving the answer $a \approx 8.536$ , to 3 d.p.

Perhaps someone can find an elegant algebraic solution?

Grandad

I agree with that answer. See this thread. But I don't think that an "elegant algebraic solution" is possible.

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