0

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 $R,r$?
atzlt
  • 574
  • Have you drawn a diagram? Can you please share? Also can you show how you got to $IN$ from $OH$ and $IH$? – Math Lover Jun 09 '21 at 13:34
  • I think no diagrams are required for this problem - it's only talking about distances between points. I'll consider adding how I get $IN$. – atzlt Jun 09 '21 at 13:39
  • The nine-point circle has a radius equal to half the circumradius, so why denote it with a different symbol? Besides, if we substitute $\rho=R/2$, the $RHS$ of the first equation becomes negative... – krazy-8 Jun 09 '21 at 14:39
  • @dodoturkoz I realized that just after I read it. Then I'll ask for how to express $OH,IH$ using only $R,r$. – atzlt Jun 09 '21 at 15:11
  • I’m going to close this question because it's impossible to answer the question. The question has no answer. – atzlt Jun 11 '21 at 11:14

1 Answers1

1

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.

krazy-8
  • 2,294
  • Those formulas are great, but not what I'm looking for, sorry. – atzlt Jun 10 '21 at 09:45
  • @SketchySketch What was it that you were looking for? You somehow need to have the sides or angles involved. It doesn't seem possible to write these equations only in terms of the two radii. See this page for more, though it gives nearly identical formulas compared to mine for the ones you asked: https://mathworld.wolfram.com/Orthocenter.html – krazy-8 Jun 10 '21 at 10:51