## [Soln] 2015 Romania Junior Balkan TST Day 1 Problem 1

2015 Romania Junior Balkan TST 1.1. Let $ABC$ be an acute triangle with $AB \neq AC$. Also let $M$ be the midpoint of the side $BC$, $H$ the orthocenter of the triangle $ABC$, $O_1$ the midpoint of the segment $AH$ and $O_2$ the center of the circumscribed circle of the triangle $BCH$. Prove that $O_1AMO_2$ is a parallelogram.

The solution presented below is not elegant and arguably not insightful. (For an elegant solution, see this.) However, it reveals how I might have gone about solving this problem in a competition. I encourage you to look at the more elegant solutions to expand your range of tools and tricks, but don’t forget that there are also ugly methods that can help you solve problems under competition conditions.

(I must emphasise that even if you are able to solve a problem using an “ugly” brute-force method, make it a point to read up on the “elegant” solution. Elegant solutions are usually quicker and more illuminating; they will help you grasp math at a deeper conceptual level.)

First, let’s draw a diagram:

As I draw a geometric diagram I always ask myself, what “variables” are there in this diagram? Which quantities, if fixed, determine the rest of the diagram? In this problem fixing the original triangle $ABC$ determines the rest of the diagram: Once $ABC$ is fixed, its orthocenter $H$ and the midpoint $M$ are fixed. In turn, the points $O_1$ and $O_2$ by their definitions are fixed. Hence, in some sense it should be possible to express everything in terms of the properties of the original triangle $ABC$. All we need to express $ABC$ are the side lengths $a = BC$, $b = CA$, $c = AB$ and the 3 angles $A$, $B$ and $C$.

(Technically you don’t even need all 3 sides and all 3 angles. However, it’s usually easier to work with these 6 “variables” for a start to maintain some sort of symmetry.)

Now that I’ve established that in theory, I should be able to label all sides and angles in terms of $a$, $b$, $c$ and $A$, $B$ and $C$, I switch gears and ask myself, what could possibly help me to prove that $O_1AMO_2$ is a parallelogram? Looking at the diagram, it is obvious that $AO_1$ and $MO_2$ are parallel since they are both perpendicular to the common line $BC$. To complete the proof, I need to show that either (i) $AO_1 = MO_2$, or (ii) $O_1O_2$ is parallel to $AM$. Option (ii) seems more difficult as I may have a difficult time making sense of the line $O_1O_2$. There seems to be some hope in expressing $AO_1$ an $MO_2$ in my 6 “variables”, so I will try to prove (i).

I draw more lines, label more things and do some “angle-chasing”:

I am now ready to do some “side length-whacking”! Let’s start with $MO_2$:

\begin{aligned} \displaystyle\frac{BM}{MO_2} &= \tan A, \\ MO_2 &= \frac{BM}{\tan A} \\ &= \frac{a \cos A}{2 \sin A}. \end{aligned}

Next I tackle $AO_1$:

\begin{aligned} \displaystyle\frac{AE}{AH} &= \cos (90 - C), \\ AH &= \frac{AE}{\sin C}, \\ AO_1 &= \frac{AE}{2 \sin C}. \end{aligned}

From $\triangle AEB$, I can express $AE$ in terms of the 6 “variables”:

$\displaystyle\frac{AE}{AB} = \cos A, \hspace{1em} AE = c \cos A.$

Hence, we have

$AO_1 = \displaystyle\frac{c \cos A}{2 \sin C}.$

Comparing the expressions we have for $AO_1$ and $MO_2$, it’s easy to see that an application of the Sine Rule will complete the proof for us:

$MO_2 = \displaystyle\frac{a \cos A}{2 \sin A} = \frac{\frac{c \sin A}{\sin C} \cos A}{2 \sin A} = \frac{c \cos A}{2 \sin C} = AO_1.$

Hence, $AO_1 = MO_2$. Coupled with the fact that $AO_1$ is parallel to $MO_2$, we have shown that $O_1AMO_2$ is a parallelogram. Done!