, 2007

Let $A$, $B$, $C$ be the vertices, $I$ the in-center, $H$ the ortho-centre, $O$ the circum-center and $N$ the nine-point center of the triangle.

Claim 1: (b) $\Leftrightarrow IH = IO$ (for a non-equilateral triangle).

The line $NI$ joins the centre of nine-point circle and the centre of in-circle, so $NI$ is perpendicular to the common tangent to in-circle and nine-point circle (we use $N \neq I$, as triangle is not equilateral). Thus (b) is equivalent to the statement that $NI$ is perpendicular to the Euler-line.

Since $O$, $H$, $N$ are on the Euler-line and $N$ is the midpoint of $OH$, we have:
(b) $\Leftrightarrow NI$ is the perpendicular bisector of $OH$.

But the perpendicular bisector of $O$, $H$ is simply the locus of all points $P$ such that $OP = HP$. Thus (b) $\Leftrightarrow IH = IO$.

Claim 2: $IH = IO \Leftrightarrow AH = AO$ or $A$, $H$, $O$, $I$ are conicyclic.

Because $H$ and $O$ are isogonal conjugates $\angle HAI = \angle OAH$. Also $AI$ is common. Thus the condition $AH = AO$ gives that $\triangle AHI$ and $\triangle OHI$ are congruent. This gives $IH = IO$.

If on the other hand $A$, $H$, $O$, $I$ are con-cyclic, then $IH$ and $IO$ subtend the same angle at $A$. This implies that $IH = IO$.

Thus $AH = AO$ or $A$, $H$, $O$, $I$ con-cyclic implies that $IH = IO$.

Conversely, suppose $IH = IO$. Then $\frac{IH}{AI} = \frac{IO}{AI}$. This implies that
$$\frac{\sin \angle HAI}{\sin \angle AHI} = \frac{\sin \angle OAI}{\sin \angle AOI}$$
It follows that $\sin \angle AHI = \sin \angle AOI$ since $AI$ bisects $\angle HAO$. This tells that $\angle AHI + \angle AOI = \pi$ or $\angle AHI = \angle AOI$. In the first case $A$, $H$, $O$, $I$ are conicyclic. In the second case $AH = AO$.

Thus we have shown that $IH = IO$ if and only if $A$, $H$, $O$, $I$ are con-cyclic or $AH = AO$.

Claim 3: $IH = IO$ if and only if $AH = AO$, or $BH = BO$, or $CH = CO$

Suppose $IH = IO$. Clearly the circum-circle of $\triangle HOI$ does not pass through all of $A$, $B$, $C$, because then $O$ would lie on the circum-circle of $\triangle A$, $B$, $C$. Without loss of generality, assume that it does not pass through $A$. Then $AH = AO$, because $IH = IO$ and $A$, $H$, $O$, $I$ are not conicyclic. (In the other two cases, we have $BH = BO$ or $CH = CO$ depending on whether the circum-circle does not pass through $B$ or $C$ respectively.) The other part follows directly from claim 2.

Claim 4: $AH = AO$, $BH = BO$ or $CH = CO$ if and only if angles of $ABC$ are in arithmetic progression.

Let $R$ be the circum-radius so that $AO = BO = CO = R$. Then $AH = 2R\cos A$, $BH = 2R\cos B$, and $CH = 2R\cos C$. Thus $AH = AO \Leftrightarrow A = \frac{\pi}{3}$; $BH = BO \Leftrightarrow B = \frac{\pi}{3}$; $CH = CO \Leftrightarrow C = \frac{\pi}{3}$.

Now $AH = AO$, $BH = BO$ or $CH = CO$ implies that one of the angles of triangle $ABC$ is $\pi/3$. But clearly, one angle is $\pi/3$ if and only if angles are in arithmetic progression.

Thus (b) $\Leftrightarrow IH = IO$ (by claim 1) $\Leftrightarrow AH = AO$, $BH = BO$, or $CH = CO$ (by claim 3) $\Leftrightarrow$ angles are in arithmetic progression which is (a).