Euler's Triangle Formula

trieuler.html

Exploration

If T is a triangle, and P and r are the center and radius of the circle inscribed in T, and Q and R are the center and radius of the circle circumscribing T, show that

where d is the distance from P to Q.

Approach

Construct the required geometry using a triangle in a special, but sufficiently general, position. Show the equation is true by showing that the difference of the left and right side of the equation is identically zero.

Initialize

To initialize Descarta2D, select the input cell bracket and press SHIFT-Enter.

This initialization assumes that the Descarta2D software has been copied into one of the standard directories for AddOns which are on the Mathematica search path, $Path.

<<Descarta2D`

Solution

Construct the required geometry and find symbolic expressions for d, r and R.

Clear[a,b];
P1=Point2D[0,0];
P2=Point2D[1,0];
P3=Point2D[a,b];
T=Triangle2D[P1,P2,P3];
Ci=Circle2D[T,Inscribed2D];
Cc=Circle2D[T,Circumscribed2D];
{P,r}={Point2D[Ci],Radius2D[Ci]};
{Q,R}={Point2D[Cc],Radius2D[Cc]};
d=Distance2D[P,Q];

In this step we take a slight diversion to show that . We will use this substitution in a subsequent step. Variables A and B are the lengths of the sides of the triangle, that is and

Clear[A,B];
Factor[b^2+(a-a^2-b^2)^2] //. {
   1-2*a+a^2->(a-1)^2,
   (-1+a)^2+b^2->B^2,
   a^2+b^2->A^2}

The expression should be zero if the equation is true, so we will apply a series of simplifications to show that the expression is identically zero. Notice throughout expressions of the form whenever Z is known to be positive.

e1=d^2-(R^2-2*r*R) //. {
   a^2+b^2->A^2,
   1-2a+a^2+b^2->(1-a)^2+b^2,
   (1-a)^2+b^2->B^2,
   Sqrt[A^2]->A,
   Sqrt[B^2]->B,
   b^2+(a-a^2-b^2)^2->A^2*B^2}

Make substitutions to remove some of the radicals. S is the semi-perimeter of the triangle, S=(1+A+B)/2.

Clear[S];
e2=e1 //. {
   1+A+B->2*S,
   Sqrt[A^2*B^2/b^2]->A*B/b}

This is the crucial substitution. Using Heron's formula for the area of a triangle,

Area =

(C=1 in this case) and the standard formula for area,

Area = base × height = 1×b/2=b/2,

we can eliminate the remaining radical.

e3=e2 //. {
   Sqrt[(-1+S)(-A+S)(-B+S)/S]->Area1/S,
   Area1->b/2}

If the expression is a fraction, we don't care what value the denominator is, so long as the numerator is zero.

e4=Numerator[Together[e3]]

Repeated expansions and substitutions confirm that the expression is zero and that the original equation is an identity.

FixedPoint[
   (Expand[(# //. {A^2->a^2+b^2,
                   B^2->(1-a)^2+b^2,
                   S->(1+A+B)/2,
                   A^4->(a^2+b^2)^2})])&,
   e4]