Archive for February, 2016

Problem:

Suppose that a regular tetrahedron with edge length of s is inscribed in a sphere, then find the radius of the sphere.

Solution:

To start with, let’s draw a cube with a tetrahedron inside it as shown in the diagram.

Screenshot - 02162016 - 04:13:34 AM

This setup will simplify our work to find the radius because the cube ABCDEFGH is also inscribed in the sphere. And obviously the green line segment AH is the diameter of the sphere. Finding AH is extremely easy.

AE=s

AF=EF=EH=\dfrac{s}{\sqrt2}

AH = \sqrt{3\cdot AF^2} = \sqrt{\dfrac{3s^2}{2}} = \dfrac{\sqrt6 s}{2}

\therefore \text{radius} = \dfrac{1}{2}AH = \dfrac{\sqrt6 s}{4}

Problem:

x^2 + xy + y^2 = 3

y^2 + yz + z^2 = 4

z^2 + zx + x^2 = 5

x + y + z = \; ?

Solution:

Although there is a 3-4-5 ratio, it has nothing to do with Pythagoras Theorem. First let’s try to play around with the variables.

(z-x)(x+y+z)=z^2+yz-x^2-xy=4-3=1

(x-y)(x+y+z)=x^2+zx-y^2-yz=5-4=1

Obviously x+y+z \neq 0, hence

z-x=\dfrac{1}{x+y+z}=x-y

\therefore y+z=2x

\therefore x+y+z=3x

Then we have the following,

z-x = \dfrac{1}{3x} = x-y

\therefore y=x-\dfrac{1}{3x} \quad \text{and} \quad z=x+\dfrac{1}{3x}

Substitute these equations into y^2+yz+z^2=4,

\left(x-\dfrac{1}{3x}\right)^2 + \left(x-\dfrac{1}{3x}\right)\left(x+\dfrac{1}{3x}\right) + \left(x+\dfrac{1}{3x}\right)^2 = 4

2x^2 + \dfrac{2}{9x^2} + x^2 - \dfrac{1}{9x^2} = 4

3x^2 + \dfrac{1}{9x^2} - 4 = 0

27x^4-36x^2+1=0

By quadratic equation,

x^2=\dfrac{36\pm\sqrt{36^2-4\cdot27}}{2\cdot27}

x^2 = \dfrac{36\pm\sqrt{36^2-4\cdot9\cdot3}}{2\cdot3\cdot9}

x^2=\dfrac{6\pm\sqrt{33}}{9}

\therefore x=\dfrac{\pm\sqrt{6\pm\sqrt{33}}}{3}

Therefore,

x+y+z=3x=\pm\sqrt{6\pm\sqrt{33}}

Four solution for x+y+z, they are

\sqrt{6+\sqrt{33}}\; ,\; \sqrt{6-\sqrt{33}}\; , \;-\sqrt{6+\sqrt{33}}\; ,\; -\sqrt{6-\sqrt{33}}