Solution to Problem 7

Subtracting the second equation from the first yields

$$y-x=\frac{a^2-b^2}{p},\qquad \mbox{ where }\ p=x+y+z;$$

hence

$$2(x^2+xy+y^2)-(y-x)^2=2c^2-\frac{(a^2-b^2)^2}{p^2}$$

and similarly

$$2(y^2+xy+z^2)-(z-y)^2=2a^2-\frac{(b^2-c^2)^2}{p^2},$$

and

$$2(z^2+zx+x^2)-(x-z)^2=2b^2-\frac{(c^2-a^2)^2}{p^2}.$$

The sum of the three equations reduces to

$$2p^2=2(a^2+b^2+c^2)-\frac{(a^2-b^2)^2+(b^2-c^2)^2+(c^2-a^2)^2}{p^2}$$

and the solutions of this quadratic equation in p² are

$$(\circledast)\qquad\ 2p^2=a^2+b^2+c^2\pm\sqrt{6(a^2b^2+b^2c^2+ c^2a^2)-3(a^4+b^4+c^4)}.$$

In a triangle ABC, with sides a,b,c, let P be any point inside and let x,y,z denote the distances from P to the vertices.

It is well known, but not easy to prove, that if all the angles of triangle ABC are less than 120^o, then the point P for which x+y+z is a minimum is located such that

$$\angle APB=\angle BPC=\angle CPA=120^o.$$

In this case the equations clearly hold - they are just statements of the cosine law for triangles, and $(\circledast)$ simplifies to

$$p=\sqrt{\frac{a^2+b^2+c^2+4\Delta\sqrt{3}}{2}},$$

where $\Delta$ denotes the area of triangle ABC.

Questions and/or comments should be directed to Judy Downey or Griff Elder

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Last modified:   Mon Oct 9 09:10:59 CDT 2000