Euler's formula

There is a remarkable formula due to Euler expressing the distance between the incenter and circumcenter. Before the theorem lets recall another theorem (due to Euler too).

Theorem 1. If two lines through a point P meet a circle with center O and radius R at points A and A' (possibly coincident) and B and B' (possibly coincident), respectively, then PA.PA'=PB.PB'=|R²-d²|, where d=OP.

Sketch of the proof. No matter where is the point P - inside or outside of the circle we have two similar triangles: PAB' and PA'B, which yield the first identity. For the case when B=B' (if P is outside the circle this is the case when PB is tangent to the circle) use the similarity of the triangles PAB and PA'B. For the second if P is inside the circle, taking BB' to be the diameter through P, using the already proven identity we have that PA.PA'=PB.PB'=(R+d)(R-d). Similarly for the case when P is outside the circle.

For any circle of radius R and any point P, distant d from the center, we call |R²-d²| the power of P with respect to the circle. The power of a point has numerous properties, which are very useful for proving various identities and theorems

Theorem (Euler). Let O and I be the circumcenter and incenter, respectively, of a triangle with circumradius R and inradius r; let d be the distance OI. Then

d²=R²-2rR.

In other words the power of the incenter with respect to the circumcircle is 2Rr.

Proof. Lets extend the internal bisector of $\angle A$ to meet the circumcircle at L, the midpoint of the arc BC not containing A (see Figure 2).

  

Figure 2: Euler's Theorem

If LM is the diameter perpendicular to BC. Writing for convenience $\alpha=\frac{1}{2} \angle A$ and $\beta=\frac{1}{2} \angle B$, we notice that $\angle BML=\angle BAL=\alpha$ and $\angle LBC=\angle LAC=\alpha$. Since the exterior angle of $\Delta ABI$ at I is

$$\angle BIL =\alpha+\beta =\angle LBI,$$

$\Delta LBI$ is isosceles: LI=LB. Thus using that the triangles LMB and AYI are right we obtain:

$$R^2-d^2=LI.IA=LB.IA= \frac{LM}{\sin \alpha}. IY \sin \alpha=2Rr,$$

as we wished to prove.

Note. The above theorem for the case when $\Delta ABC$ is isoscles is given on the IV IMO.