A point within an equilateral triangle

**redpilot1234** · December 6th 2010, 05:28 AM

Does someone know how to solve this 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.

Got far as generating 3 equations with 4 unknowns (3 angles, 1 side) using Cosine Law and I know it's not going to solve this one.

**Educated** · December 6th 2010, 11:48 PM

A point within an equilateral triangle-iscoceles-triangle.png

Using the cosine rule and the fact that angles at a point add to 360 degrees, you will have 4 equations with 4 unknowns.

$x^2 = 5^2 + 6^2 - 2(5)(6) \cos A$

$x^2 = 4^2 + 5^2 - 2(4)(5) \cos B$

$x^2 = 6^2 + 4^2 - 2(6)(4) \cos C$

$A + B + C = 360^{\circ}$

Substituting the first 3 equations into the forth gives us:

$\cos^{-1}\left(\dfrac{25+36-x^2}{60}\right)+\cos^{-1}\left(\dfrac{16+36-x^2}{48}\right)+\cos^{-1}\left(\dfrac{16+25-x^2}{40}\right)= 360^{\circ}$

Solve for x and you will know the length of the equaliteral triangle.

**Opalg** · December 7th 2010, 01:48 AM

As far as I can tell, this problem does not have a simple solution. I have a horrible method for tackling it, which leads to a horrible answer. The idea is to use coordinate geometry on an equilateral triangle ABC with sides of unit length, to find a point P for which the distances PA, PB, PC are in the ratio PA:PB:PC = 4:5:6, and then to scale the result up so that these distances become 4, 5 and 6.

So let $A = \bigl(\frac12,\frac{\sqrt3}2\bigl)$ , $B = (0,0)$ and $C = (1,0)$ . The set of points for which the ratio of distances PB:PC is 5:6 is a circle with equation $11(x^2+y^2) + 50x - 25 = 0$ . The set of points for which the ratio of distances PB:PA is 5:4 is a circle with equation $9(x^2+y^2) - 25x - 25\sqrt3y + 25 = 0$ . Those circles intersect at the point $P = (x,y)$ given by

$x = \dfrac{-245 + 165\sqrt{21}}{1204} \approx 0.4245,\quad y = \dfrac{2835 - 435\sqrt{21}}{1204\sqrt3} \approx 0.4036.$

The distance PB is then $d = \sqrt{x^2+y^2}\approx0.5857$ . If we scale up the diagram to make PB = 5 then the side of the triangle becomes $5/d\approx8.53635.$

If you do the calculation exactly, then the value for the side of the triangle comes out as $\boxed{\sqrt{\dfrac{602}{77 - 15\sqrt{21}}}}.$ Unless I have made arithmetical mistakes (something that unfortunately happens all too often), that expression cannot be simplified any further, which makes me think that this problem does not have a neat solution at all.

Thread: A point within an equilateral triangle

LinkBack

Thread Tools

Display

A point within an equilateral triangle

Similar Math Help Forum Discussions

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

Equilateral triangle

Area of equilateral triangle given known distances of a point from the sides.

Equilateral Triangle

Equilateral Triangle

Search Tags