[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!

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