## 25 Oct Simple concept Tough problem – 1

Many tough mathematical problems require just basic mathematics for their solutions. This series of articles will discuss such problems on a regular basis. The problem for the first edition is taken from the Regional Mathematics Olympiad (RMO) 2019. It’s question number 2 in the RMO paper. While there are surely several ways to solve, I found a solution that requires some basic results from school geometry.

Question : Let $ABC$ be a triangle with circumcircle $\Omega$ and let $G$ be the centroid of triangle $ABC$. Extend $AG, BG$ and $CG$ to meet the circle $\Omega$ again in $A_1, B_1, C_1$ respectively. Suppose $\angle BAC = \angle A_1B_1C_1, \angle ABC = \angle A_1 C_1B_1$ and $\angle ACB = \angle B_1 A_1 C_1$. Prove that $ABC$ and $A_1B_1C_1$ are equilateral triangles.

Well, the first step for any geometry problem is to draw a neatly labelled diagram. Do remember to keep the diagram as general as possible. In this problem, a very frequent mistake is to draw $ABC$ and $A_1B_1C_1$ as equilateral triangles. Never do that. If you do, the medians will look like altitudes and there will be a tendency to assume wrong things. So, let us have a look below at the diagram drawn from the given information :

Let $AD,BE,CF$ be the medians of $\Delta ABC$. Next we mark some angles. Let $\angle BAA_1=\alpha$ and $\angle A_1AC=\beta$. Immediately, from the well known result that angles on the same segment of a circle are equal, we get a few other angles. $\angle BB_1A_1=\alpha$ and $\angle A_1C_1C=\beta$. But $\angle BAC=\angle A_1B_1C_1$, so that $\alpha +\beta=\alpha +\angle BB_1C_1$. This gives $\angle BB_1C_1=\beta$ and so $\angle BCC_1=\beta$. Also, let $\angle ACC_1=\gamma$ so that $\angle AA_1C_1=\gamma$. Then using $\angle ACB=\angle \angle B_1A_1C_1$ we get $\angle AA_1B_1=\beta$ so that $\angle ABB_1=\beta$. Next, let $\angle B_1BC=\delta$ so that $\angle B_1C_1C=\delta$. Using $\angle ABC=\angle A_1C_1B_1$, we get $\angle A_1C_1C=\beta$. We mark the angles in the following diagram.

Ah ! The diagram looks a bit decorated now ! Let’s do some construction. Mark the line segments $A_1B$ and $A_1C$. Also mark $AC_1$ and $BC_1$. Mark the obvious angles that are formed. Most angles will have familiar names, thanks to the theorem stated earlier on angles on the same segment.

Thus we get $\angle A_1BC=\beta$. Hey wait, that means $\angle A_1BC=\angle BCC_1$ but they are alternate angles formed by the transversal $BC$ intersecting the line segments $A_1B$ and $CC_1$. Doesn’t that mean $A_1B || CC_1$? Yes indeed !

Next, consider $\Delta A_1BD$ and $\Delta GDC$. We have $\angle A_1BD=\angle GCD=\beta$, $BD=DC$ (as AD is the median) and $\angle A_1DB=\angle GDC$ (vertically opposite angles). So, by $ASA$ congruency axiom, $\Delta A_1BD\cong \Delta GDC$. Thus, $A_1B=CC_1$. Here we are ! $A_1BGC$ is actually a parallelogram. So its opposite angles will be equal. We have, $\angle A_1BG=\angle A_1CG$ so that $\delta+\beta=\beta+\alpha$ i.e. $\delta=\alpha$.

Similarly, we can show $BC_1AG$ is a parallelogram. So, $\alpha=\gamma$. Thus, all angles of $\Delta ABC$ and $\Delta A_1B_1C_1$ are equal to $\alpha+\beta$. So, $\Delta ABC$ and $\Delta A_1B_1C_1$ are equilateral.

So, we see that almost half of the problem was solved just by drawing the properly labelled diagram. The concepts required were of parallel lines, congruence of triangles and a single theorem of circles ! Watch out for the next problem of SCTP series soon ! Stay tuned to GonitSora !!

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