MathDB

Let triangle

ABC

be such that its circumradius is

R = 1.

Let

r

be the inradius of

ABC

and let

p

be the inradius of the orthic triangle

A'B'C'

of triangle

ABC.

Prove that

p \leq 1 - \frac{1}{3 \cdot (1+r)^2}.

[hide="Similar Problem posted by Pascual2005"]
Let

ABC

be a triangle with circumradius

R

and inradius

r

. If

p

is the inradius of the orthic triangle of triangle

ABC

, show that

\frac{p}{R} \leq 1 - \frac{\left(1+\frac{r}{R}\right)^2}{3}

.
Note. The orthic triangle of triangle

ABC

is defined as the triangle whose vertices are the feet of the altitudes of triangle

ABC

.
SOLUTION 1 by mecrazywong:

p=2R\cos A\cos B\cos C,1+\frac{r}{R}=1+4\sin A/2\sin B/2\sin C/2=\cos A+\cos B+\cos C

. Thus, the ineqaulity is equivalent to

6\cos A\cos B\cos C+(\cos A+\cos B+\cos C)^2\le3

. But this is easy since

\cos A+\cos B+\cos C\le3/2,\cos A\cos B\cos C\le1/8

.
SOLUTION 2 by Virgil Nicula:
I note the inradius

r'

of a orthic triangle.
Must prove the inequality

\frac{r'}{R}\le 1-\frac 13\left( 1+\frac rR\right)^2.

From the wellknown relations

r'=2R\cos A\cos B\cos C

and

\cos A\cos B\cos C\le \frac 18

results

\frac{r'}{R}\le \frac 14.

But

\frac 14\le 1-\frac 13\left( 1+\frac rR\right)^2\Longleftrightarrow \frac 13\left( 1+\frac rR\right)^2\le \frac 34\Longleftrightarrow

\left(1+\frac rR\right)^2\le \left(\frac 32\right)^2\Longleftrightarrow 1+\frac rR\le \frac 32\Longleftrightarrow \frac rR\le \frac 12\Longleftrightarrow 2r\le R

(true).
Therefore,

\frac{r'}{R}\le \frac 14\le 1-\frac 13\left( 1+\frac rR\right)^2\Longrightarrow \frac{r'}{R}\le 1-\frac 13\left( 1+\frac rR\right)^2.

SOLUTION 3 by darij grinberg:
I know this is not quite an ML reference, but the problem was discussed in Hyacinthos messages #6951, #6978, #6981, #6982, #6985, #6986 (particularly the last message).

Problem Statement