Skip to main content

Some Wandering Through Orthic Triangles

 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 $BC$.
$(iii)$ $H$ is the incenter of $\triangle DEF$.

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.

Comments

Popular posts from this blog

EGMO solutions, motivations and reviews ft. Atul, Pranjal and Abhay

The  European Girls' Mathematical Olympiad a.k.a EGMO 2022 just ended. Congrats to Jessica Wan from USA, Taisiia Korotchenko, and Galiia Sharafetdinova for the perfect scores! Moreover, the Indian girls brought home 4 bronze medals! By far, this is the best result the EGMO India Team has ever achieved! To celebrate the brilliant result, here's a compilation of EGMO 2022 solutions and motivations written by my and everyone's favorite IMOTCer Atul ! And along with that, we also have reviews of each problem written by everyone's favorite senior, Pranjal !  These solutions were actually found by Atul, Pranjal,  and Abhay  during the 3-hour live solve. In the live solve, they solved all the 6 problems in 3 hours 😍!!! Okie Dokie, I think we should get started with the problems! Enjoy! Problem 1:  Let $ABC$ be an acute-angled triangle in which $BC<AB$ and $BC<CA$. Let point $P$ lie on segment $AB$ and point $Q$ lie on segment $AC$ such that $P \neq B$, $Q \neq C$ and

Kőnig-Egerváry theorem

Graph theory has been my most favourite thing to learn in maths and and in this blog i hope to spread the knowledge about Kőnig's theorem. It is advised that the readers are aware about basic graph theory terminologies including bipartite graphs. Before going on to the theorem i would like to go on about matchings and vertex cover which we are going to use in the theorem  Matchings A matching $M$ of a graph $G$ is a subset of the edges of a graph in which no two edge share a common vertex (non adjacent edges). For example :- The Matching $M$ over here is edge $\{ 3 - 5 , 1-2 , \}$ or $\{ 1 - 2 , 4 - 3 \}$ etc .  Maximum Matching is a matching that contains the largest possible number of edges for instance in the above example the maximum matching is 2 edges as there cannot be a subset of non adjacent edges having greater than 2 edges (Readers are advised to try so they can convince themselves) A Perfect Matching  is a matching that matches all vertices of the graph or in other sen

Algorithms, or Mathematics?!

Hi everyone! In my last blog, I spoke about a couple of very interesting algorithmic techniques and how they can be used to solve a variety of very interesting problems. This blog however, is going to be completely different.   When we’re young we begin by learning the steps to add – we’re given the rules and we must learn to repeat them – no questions asked. Why does carry forward work the way it does? Well, no questions asked. To some extent, it is important to know how exactly to do a certain set of things while first learning maths. We may not be able to get anywhere in a subject if we’re unable to learn a few basic rules and know how to use them. However, after a certain point it is important to bring in the spirit of mathematical thinking within each student too – something missing in almost every form of school math education. Mathematical miseducation is so common, we wouldn’t even see it. We practically expect a math class to look like repetition and memorisation of disjointed