Indian National Mathematical Olympiad

Solution:

The answer is 'yes'. To prove this, we will first prove two lemmas:

Lemma 1 Given any two points $A, B$, their midpoint $M$, and any point $C$, Vikram can draw a line parallel to $A B$ through $C$.

Proof. If $C$ is on line $A B$ we are already done. If not, extend $B C$ to $X$ as shown, draw $P = A C \cap X M$, and then draw $D = B P \cap A X$. We claim $C D$ is the desired line. Indeed, using Ceva's theorem on triangle $A B X$ and the fact $A M = M B$, we see that
$$\frac{A M}{M B} \cdot \frac{B C}{C X} \cdot \frac{X D}{D A} = 1 \Longrightarrow \frac{X C}{C B} = \frac{X D}{D A}$$
This means $C D \parallel A B$.

Lemma 2 Given two non-parallel segments $A B, B C$ and their midpoints $M, N$, Vikram can draw the midpoint of any other segment $X Y$.

Proof. Assume first $X Y$ is not parallel to $A B$ or $B C$. Using lemma 1, draw lines $\ell_1$ and $\ell_2$ through $X$ parallel to $A B$ and $B C$ respectively, and similarly draw $m_1$ and $m_2$ through $Y$ parallel to $A B$ and $B C$ respectively. If we draw $P = \ell_1 \cap m_2$ and $Q = \ell_2 \cap m_1$, then $X P Y Q$ is a parallelogram, so intersecting $P Q$ and $X Y$ gives the midpoint of $X Y$.

As for the remaining case, one can draw $A C$ and construct the midpoint $P$ of $A C$ by the construction described above. Since $X Y$ can be parallel to at most one of the sides $A B, B C$ and $A C$, we can pick the two non-parallel sides, and use the above construction to draw the midpoint of $X Y$.

Now for the main problem, note that if no two of the $1011$ chosen segments share an endpoint, then we have at least $2 \cdot 1011 = 2022$ distinct endpoints, a contradiction. Thus there must be two segments $A B$ and $B C$ which have their midpoints marked. Since no three of the chosen $2021$ points were collinear, $A B$ and $B C$ are not parallel, so using lemma 2, Vikram can construct the midpoint of any other segment, in particular, the segment chosen by Betal.

Solution:

As in the previous solution, note that there exist $A B$ and $A C$ whose midpoints $C^{\prime}$ and $B^{\prime}$ are marked. Using the straightedge, Vikram can draw the two medians $A C^{\prime}$ and $A B^{\prime}$ and obtain their intersection, the centroid $G$ of $\triangle A B C$. Now intersecting $A G$ with $B C$ gives $A^{\prime}$, the midpoint of $B C$.

Lemma Given a point $P$ not on $A B, A C$, Vikram can draw the midpoint of $P A$.

Proof. If $P B \parallel A C$ and $P C \parallel A B$, then $P B A C$ is a parallelogram, in which case $A^{\prime}$ constructed above is the midpoint of $P A$. Without loss of generality, we may assume $P B \nparallel A C$.

Using the straightedge, one can mark the points $D = P B \cap A C$ and $P B \cap A^{\prime} C^{\prime} = D^{\prime}$. Since $C A \parallel A^{\prime} C^{\prime}$, we have
$$\frac{B D^{\prime}}{D^{\prime} D} = \frac{B C^{\prime}}{C^{\prime} A} = 1$$
so $D^{\prime}$ is the midpoint of $B D$. Now in $\triangle A B D$, two midpoints $C^{\prime}$ and $D^{\prime}$ are known, so the midpoint $Q^{\prime}$ of $A D$ can be constructed using the centroid construction outlined before. Let $P^{\prime} = C^{\prime} Q^{\prime} \cap P A$; this exists as $C^{\prime} Q^{\prime} \parallel B P \nparallel A P$. As before, $C^{\prime} P^{\prime} \parallel B P$, so
$$\frac{A P^{\prime}}{P^{\prime} P} = \frac{A C^{\prime}}{C^{\prime} B} = 1$$
which means $P^{\prime}$ is the desired midpoint of $P A$.

Now suppose we need to find the midpoint of $P Q$. If $P, Q$ are different points from $A$, then one can draw the midpoints of $A P$ and $A Q$ using the lemma. Then by using the centroid of $\triangle A P Q$, one can find the midpoint of $P Q$ as we did for $B C$. If $P$ or $Q$ is $A$, the above lemma immediately yields the required midpoint.