I recently saw in a book which said: if $O,H,I,N$ are the circumcenter, the orthocenter, the incenter and the center of the nine point circle of a triangle, $R,r,\rho$ are the radii of the circumcircle, incircle and the nine point circle, one have $$ OH^2=R^2-4R\rho,\\ IH^2=2r^2-2R\rho,\\ IN=\frac{1}{2}R-r. $$ I can deduce the third one from the previous two using the length theorem of centerline. Since $ON=NH$, we have $$ \begin{align} IN^2&=\frac{1}{2}IO^2+\frac{1}{2}IH^2-\frac{1}{4}OH^2\\ &=\frac{1}{2}(R^2-2Rr)+\frac{1}{2}(2r^2-2R\rho)-\frac{1}{4}(R^2-4R\rho)\\ &=\frac{1}{4}R^2-Rr+r^2\\ &=(\frac{1}{2}R-r)^2, \end{align} $$ which is equivalent to $IN=\frac{1}{2}R-r$.
How can I prove the first two?
dodoturkoz's comment suggests that the book is wrong. Then my question is, how to express $OH,IH$ using only $R,r$?
There are two identities that I have came across before. It doesn't seem possible to write these equations only in terms of the inradius and the circumradius.
Distance between circumcenter and orthocenter:
$$OH^2=9R^2-(a^2+b^2+c^2) \tag{1}$$
Outline of proof:
Use the following well-known result (can be proven using Stewart's Theorem):
Let $G$ be the centroid of $\triangle ABC$. For any point $M$: $$MA^2 + MB^2 + MC^2 =3MG^2 + AG^2 + BG^2 + CG^2$$
Let $O$ be $M$:
$$3R^2 = AG^2 + BG^2 + CG^2 + 3 OG^2$$
Now, use Apollonius' Theorem and the fact that $OH = 3OG$ (recall the Euler Line), and we are done.
There is also a very neat proof using complex numbers here.
Distance between incenter and orthocenter:
$$ IH^2=2r^2-4R^2\cos A\cos B\cos C\ \tag{2}$$
Outline of proof:
Use the law of cosines and trigonometry. See here for a more detailed proof.