We'll prove the statement by contradiction. Name $a_i$ exactly one of the actors or actresses that got $i$ votes, for $1 \le i \le 100$. The total number of votes is clearly $6732$ and $5050$ of them went to $a_1,...,a_{100}$. We now separate critics into two groups: -the first group are the critics who gave at most one vote to $a_{34},...,a_{100}$. They are at most $6732-5050$ (number of actors and actresses out of the $a_i$-s) $+1+...+33$ (votes given to $a_1,...a_{33}$ at most). So they are at most $2243$; -the second group are the critics who gave both votes to $a_{34},...,a_{100}$. By our initial assumption, they are at most the number of couples actor-actress, so they are at most $33 \cdot 34=1122$. Our two groups must include all the critics, but we see that $1122+2243<3366$, so we are done.

let $a_1, a_2 ... a_{100}$ be the actors who are voted for 1, 2, ...100 times respectively. Then among them there are at exactly $ \frac{100*101}{2}=5050$ votes. Then $5050-3366=1684$ critics voted for both an actor and an actress within the $a$. Now, assume for the sake of contradiction that no two critics voted for the same actor/actress pair. Let the $a$ be the vertices of a graph, and let each pair of votes cast by each of the 1684 critics who voted for two people in the $a$ represent an edge, between the actor and actress that he voted for. Clearly, there are 1684 edges, since no two critics voted for the same pair. Clearly the graph is bipartite because we can divide it into the group of actors and of actresses, and no edge joins two actors or two actresses. Now, consider the actors or actresses $a_1, a_2, a_3, a_4... a_{33}$. Note that each $a_i$ has degree at most i. Then there are at most $\frac{33*34}{2}=561$ edges containing one of these people. Now, we count the edges not containing any of those 33 people. Since the graph is bipartite, we can divide $a_{34},a_{35},a_{36}...a_{100}$ into two groups such that there are no edges connecting two people in the same group. Moreover, these two groups have a total of 67 people. Thus, the maximum number of edges between these 67 people is equal to the product of the number of people in the groups. Thus, the number of edges between these 67 people is at most $33*34$ = 1122. Then, the total number of edges between all 100 people in this graph is at most $1122+561=1683$, since each edge either contains one of $a_1,a_2...a_{33}$ or does not. This is a contradiction, since the graph has 1684 edges if no two critics voted for the same pair of people.

Then $5050-3366=1684$ critics voted for both an actor and an actress within the $a$. And what if one critic gave both of his votes to people, that are not in $a_i$

The solution in post#2 is brilliant, can someone tell me the motivation of that separation?

Why must '33' and how did you get this number. Thankyou

AmirAlison wrote: Then $5050-3366=1684$ critics voted for both an actor and an actress within the $a$. And what if one critic gave both of his votes to people, that are not in $a_i$ Sorry to bring this up but like he said what if there are more than 100 actor and actress it doesnt say that there are only 100

Hiitsmemath wrote: AmirAlison wrote: Then $5050-3366=1684$ critics voted for both an actor and an actress within the $a$. And what if one critic gave both of his votes to people, that are not in $a_i$ Sorry to bring this up but like he said what if there are more than 100 actor and actress it doesnt say that there are only 100 There are at least 1684 critics who voted for an $a_i$.

We'll prove the statement by contradiction. Let's define each voter as a point in the plane with coordinates $(x, y)$, which represent the actor and actress they voted for. It is sufficient to prove that it's impossible to choose 3366 points such that there exist lines parallel to $x$-axis or $y$-axis containing $1, 2, \ldots, 100$ points. Claim: We need at least 3367 points to meet the conditions. We are going to place the lines with 100, 99, 98, $\ldots$ points on the plane in this order. We will also add 2 variables $z_x$ and $z_y$ - the longest "chain" of points not in use, parallel to the $x$-axis or $y$-axis respectively. Let the line with 100 points have points $(0, 0), (1, 0), (2, 0), \ldots, (100, 0)$. (It could also be parallel to the $y$-axis.) This means that now $z_y = 1$. Adding the 99-line, we have the choice to add 99 points parallel to the $x$-axis (and make $z_y = 2$), add 99 points parallel to the $y$-axis (and make $z_x = 1$), or add 98 points parallel to the $y$-axis (and make $z_x = 1$). We will have the same choice for the $n$-line: add $n - a$ points parallel to the $x$-axis and increase $z_y$ by 1 (optional) ($a \in [0, z_x]$), or add $n - b$ points parallel to the $y$-axis and increase $z_x$ by 1 (optional) ($b \in [0, z_y]$). Because we want to minimize the number of points, we should always choose to add either $n - z_x$ or $n - z_y$ points. Now all that is left is to see which lines should be parallel to the $x$-axis and which to the $y$-axis. An important note to add here is that we add new lines until line $\max(z_x, z_y)$. When we have finished counting the points, let $z_x = a$ and $z_y = b$. WLOG, assume $a \geq b$. That means the total number of points is at a minimum: \[ \frac{100 \times 101}{2} - \frac{a \times (a+1)}{2} - {b \times (b+1)} \] We also know that $100 - (a + b) = a$ or $a + 1$ (depending on the parity of $b$), and $a \in [33, 50]$. Substituting, we get that the number of points is \[ \frac{100 \times 101}{2} - \frac{a \times (a+1)}{2} - {(100 - 2a) \times (101 - 2a)} \] or \[ \frac{100 \times 101}{2} - \frac{a \times (a+1)}{2} - {(99 - 2a) \times (100 - 2a)} \] Calculating the minimum of both functions, we see that we need 3367 points ($a = 33, b = 33$) or 3399 points ($a = 34, b = 32$). Therefore, we're done because we can't do it with less than $3367$ points.