Hi, I am Emon and I it's been long since I posted last. Today I will try to give you some ideas on how to work with a special type of triangle, known as "**Orthic Triangles**". In this post, I will mainly focus on problem-solving, but still, let me first give you some ideas on what exactly it is and what properties does it have...

**Definition (Orthic Triangle). **Let $ABC$ be a triangle and let $D, E, F$ be the foot of the perpendiculars from $A, B, C$ to $BC, CA$ and $AB$, respectively. Then, $\triangle DEF$ is known as the *orthic triangle *of $\triangle ABC$.

**Lemma $1$ (Orthic Triangle).**If $\triangle DEF$ is the orthic triangle of $\triangle ABC$ with orthocenter $H$, then the following conditions are satisfied :

$(i)$ $AEHF$ is a cyclic quadrilateral with circumdiameter $AH$.

$(ii)$ $BCEF$ is a cyclic quadrilateral with circumdiameter $BC$.

**Lemma $2$.**$\angle ABE = \angle ADE$ and $\angle ACF=\angle ADF$.

(We can prove this with the help of some inscribed angles)

Okay, so as I had mentioned earlier, we shall now move into some problem-solving without further delay.

**Problem $1$.**In the acute triangle $ABC$, $A',B',C'$ are the feet of the perpendicular from vertices $A,B,C$ onto the lines $BC,CA,AB$ respectively. Find all possibilities of the angles of the triangle $ABC$ for which the triangles $ABC$ and $A'B'C'$ are similar.

**Solution.**

We first have that $ABA'B'$ is cyclic. Hence, $\angle B'A'C=\angle BAC$. Similarly, since $ACA'C'$ is cyclic, we have, $\angle C'A'B=\angle BAC$.

Thus, we have, $\angle BAC=\angle B'A'C'=180^\circ-2\angle BAC$, since $\triangle ABC\sim \triangle A'B'C'$, which implies that $\angle BAC=60^\circ$.

Similarly, we get, $\angle ABC=\angle BCA=60^\circ$, so that $\triangle ABC$ is equilateral.

Hence, we can conclude that all equilateral triangles $ABC$ satisfy the given conditions.

**Problem $2$ (IberoAmerican SL $1995/1$).**In an acute triangle $ABC$, the distance from the orthocenter to the centroid is half the radius of the circumscribed circle. $D, E$, and $F$ are the feet of the altitudes of triangle $ABC$. If $r_1$ and $r_2$ are, respectively, the radii of the inscribed and circumscribed circles of the triangle $DEF$, determine $\frac{r_1}{r_2}$.

**Solution.**Let $H$ be the orthocentre, $O$ the circumcentre, $N$ be the nine point centre and $G$ be the centroid of $\triangle ABC$.

Now it is given that $HG= \frac{R}{2}=2GO$. So $HO =\frac{3R}{4}$. Hence, $NH= \frac{3R}{8}$ because $N$ is the midpoint of $OH$.

Now, since $(DEF)$ is the nine point circle of $\triangle ABC$, $r_2 =\frac{R}{2}$ (the nine point radius is half the circumradius by homothety.)

But, $H$ is the incentre of $\triangle DEF$ by

**Lemma $2(iii)$**. So, by Euler's formula $NH^2= (r_2)^2 -2r_1r_2$, as $N$ is the circumcentre of $DEF$ and $H$ is the incentre. Now putting the value of $NH$ and $r_2$, we get that $\displaystyle{r_1 =\frac{7R}{64}}$. Hence the ratio is $\boxed{\frac{7}{32}}$.**Problem $3$.**Show that the perimeter of the orthic triangle of a triangle is less than twice of any altitude.

**Solution.**Let $D, E, F$ be the foot of the altitudes on $BC, CA, AB$ from $A, B, C$, respectively.

We first observe that the perimeter of the orthic triangle $DEF$ of $\triangle ABC$ is $4R\sin A\sin B\sin C$, where $R$ denotes the circumradius of $\triangle ABC$.

Now, $2AD=2(2R\sin B\sin C)=4R\sin B\sin C$.

But, since $\sin A < 1$, we have, $DE+EF+FD=4R\sin A\sin B\sin C<4R\sin B\sin C=2AD$. Similarly, we have, $DE+EF+FD<2BE$ and $DE+EF+FD<2CF$, and hence we may conclude that, the perimeter of the orthic triangle $DEF$ is less than twice of any altitude $AD, BE, CF$, as desired.

That's all from my part, I guess. I would leave a few problems for you to try...

**Problem $4$ (Ukraine National MO $2000\ 11.3$).**Let $AA_1, BB_1, CC_1$ be the altitudes of the acute triangle $ABC$. Denote by $A_2, B_2, C_2$ the points of tangency of the circle inscribed in triangle $A_1B_1C_1$, with sides $B_1C_1$, $C_1A_1$, $A_1B_1$ respectively. Prove that the lines $AA_2$, $BB_2$, $CC_2$ intersect at one point.

**Problem $5$.**Let $\triangle XYZ$ be the orthic triangle of the orthic triangle of $\triangle ABC$. Show that $AX,BY,CZ$ meet on the Euler line of $\triangle ABC$.

**Problem $6$.**Let $\triangle{H_A H_B H_C}$ be the orthic triangle of $\triangle{ABC}$. Prove that the line perpendicular to $\overline{H_A H_B}$ intersecting $C$, the line perpendicular to $\overline{H_B H_C}$ intersecting $A$, and the line perpendicular to $\overline{H_C H_A}$ intersecting $B$ concur at the circumcenter of $\triangle{ABC}$.

**Problem $7$.**Let $\triangle{M_A M_B M_C}$ and $\triangle{H_A H_B H_C}$ be the medial and orthic triangles, respectively, of acute scalene triangle $\triangle{ABC}$ with shortest side $\overline{AB}$. The medial and orthic triangles intersect at four points, two of which lie on $\overline{H_A H_B}$. Prove that $M_A$, $M_B$, and those two points lie on a circle.

**Problem $8$.**Let $\triangle ABC$ be an acute angled triangle with $AB<AC$. Let $D,E,F$ be the intersections of angle bisector of $\angle A$, $\angle B$ and $\angle C$ with opposite sides of $\triangle ABC$. Let $\triangle H_AH_BH_C$ be the orthic triangle WRT $\triangle ABC$. Let $M$ be the midpoint of $BC$. Let $H_BH_C \cap EF$ $=$ $P$. Assume, $P$ to be inside $\triangle ABC$. Prove, $MP$ passes through one of the intersections of $\odot (H_AH_BH_C)$ and $\odot (DEF)$.

Okay, so that's it. There's nothing in the theory of orthic triangles as such. It can be better understood by solving more and more problems. I have tried to give you some ideas on how to solve questions related to it, and you should try some more on your own to get a good grasp on it. Hope you have enjoyed.

Bye bye until next time...

**Emon.**

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