Argentina competition problems

The journalist is right for $p \ge 73$ and wrong for $p \le 72$.

Let $a, b, c$ denote the number of votes for $A, B, C$ in the first round, and let $N = a + b + c$ be the total number of voters in this round. By hypothesis $a = \frac{44}{100}(b + c) = \frac{11}{25}(N - a)$, hence $a = \frac{11}{36}N$; also $c < a$.

The number of voters in the second round is $N' = N - \frac{p}{100}c$. There are $\left(1 - \frac{p}{100}\right)c$ persons who voted for $C$ in the first round and participate in the second round. For brevity call them and their votes additional. Since $B$'s supporters voted for him in both rounds, the most $A$ can achieve is that his own supporters vote for him again in the second round, and also the additional voters. So the maximum number of votes $A$ can get is $a_{\text{max}} = a + \left(1 - \frac{p}{100}\right)c$.

We are interested in the difference $N' - 2a_{\text{max}}$ (whose sign determines the chances of $A$):
$$N' - 2a_{\text{max}} = \left(N - \frac{p}{100}c\right) - 2 \cdot \frac{11}{36}N - 2\left(1 - \frac{p}{100}\right)c = \frac{7}{18}N - \frac{200-p}{100}c.$$

Let us say already here that the different outcomes for $p \ge 73$ and $p \le 72$ are due to the inequalities
$$\frac{127}{100} + \frac{11}{36} < \frac{7}{18} + \frac{128}{100} < \frac{11}{36}$$
(which hold as $127 \cdot 11 = 1397 < 1400 = 100 \cdot 2 \cdot 7 < 128 \cdot 11 = 1408$).

Suppose that $p \ge 73$. Then $\frac{200-p}{100}c < \frac{127}{100}c < \frac{127}{100}a = \frac{127}{100}N$. Since $\frac{127}{100} < \frac{11}{36} < \frac{7}{18}$, it follows that $\frac{200-p}{100}c < \frac{7}{18}N$, i.e. $N' - 2a_{\text{max}} > 0$. So $A$ cannot win even if he gets the maximum possible number of votes. Therefore $B$ wins with certainty, and the journalist is right.

For $p \le 72$ there are examples showing that either candidate can win. In this case $\frac{200-p}{100}c \ge \frac{128}{100}c$.
Now the inequality $\frac{7}{18} < \frac{128}{100} \cdot \frac{11}{36}$ (see above) implies $\frac{100}{128} \cdot \frac{7}{18} < \frac{11}{36}N$. Take $N$ such that both sides of the inequality are integers differing by more than 1, for instance $N = 2\operatorname{lcm}(36,128)$. Then an integer $c$ can be chosen so that $\frac{100}{128} < \frac{7}{18}N < \frac{11}{36}N$. The condition $c < a$ for the first round is satisfied. For this choice of $c$ we have $\frac{7}{18}N < \frac{128}{100}c$, and $\frac{200-p}{100}c \ge \frac{128}{100}c$ was shown above for $p \le 72$, which implies $N' - 2a_{\text{max}} < 0$. So $A$ wins if he gets $a_{\text{max}}$ votes. This is possible if all of his supporters vote for him again, and also all additional voters.

On the other hand it is clear that $B$ is a possible winner for any $p$, for instance if $A$ gets no votes at all (which is not excluded by the conditions). It is of more substance to note that $B$ can also win for $p \le 72$ even if all of $A$'s supporters vote for him again in the second round. Indeed if $p \le 72$ then $c < a$ implies $N' = N - \frac{p}{100}c > N - \frac{72}{100}a = N - \frac{72}{100} \cdot \frac{11}{36}N = \frac{39}{50}N$. This is greater than $2a = \frac{11}{18}N$, so if all additional votes go to $B$ then $B$ wins.

Remark. There are values of $N$ for which the situation can describe actual elections, for instance, $N = 1800000$. Then $a = \frac{11}{36}N = 550000$ and, for $p \le 72$, the key number for the construction is $\frac{100}{128} \cdot \frac{7}{18}N = 546875 < 550000$. So there are plenty of (integer) choices for $c$ in $[546875, 550000]$.