Geometry problem

[HappyMutant]

Some point P is on the interior of equilateral triangle ABC, such that PA = 3, PB = 4, and PC = 5. Find the length of the sides of the triangle.

(We had ARML practice today, and this was the hardest (and coolest) problem.)

[Da5id]

Well, I have an aproximate value. I'll try to calculate it another way to get a precise value.
L=~6.7664325675223

[HappyMutant]

Yes, one can find an exact answer.

[Da5id]

I've been out of school for over a decade, I'm pretty rusty. Have to dust off the cobwebs to find some of the formulas....

[Griffin]

I get: sqrt[25 + 12*sqrt(3)], which works out to be the same as what Da5id got.

[HappyMutant]

Yes.

Feel free to explain. (The solution makes me happy.)

[Griffin]

I'm not sure if this is the most elegant solution, but it still works out pretty nice (SPOILER SPACE):

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If you rotate the figure around point A so that C' lines up with B, you get what's in the picture above. Clearly angle P'AP is a 60 degree angle. This makes triangle P'AP an equilateral triangle, so this means the line segment P'P is equal to PA and P'A, or 3. Now you have a nice 3-4-5 right triangle in triangle BP'P. Since BPP'is 90 degrees and PP'A is 60 degrees, BPA must be 150 degrees.

Now we just have a simple law of cosines problem

a^2 = b^2 + c^2 - 2bc(cos A)
x^2 = 3^2 + 4^2 - 2*3*4(cos 150)
x^2 = 25 - 24 * (-sqrt(3)/2)
x^2 = 25 + 12sqrt(3)

This was a tough problem. I must have worked on it acouple hours till I finally figured out the 3-4-5 right triangle trick.

[Da5id]

Nice.

[Alter Ego]

I must say I spent a little while on this, went the co-od geometry way and ended up with a cubic equation with the same solution - but this is SO much more elegant. Griffin, VERY NICELY done.

And excellent little puzzle there Happy.