For triangle $ABC$, the following four statements were made, and we know that two of the statements are true and two are false. What could be the perimeter of triangle $ABC$ if $BC=1$?

I. Triangle $ABC$ is a right triangle. - II. The angle at $A$ is $30^{\circ}$. - III. $AB=2BC$. - IV. $AC=2BC$.
The communication regarding the truth or falsity of the statements can only be utilized if we consider the possible cases for the pairs of true statements one by one.

1. Let's try to assume that III. and IV. are true. In this case, \( AB = AC = 2BC = 2 \), the triangle is isosceles, so there is no right angle either at the base or between the legs, since \( BCC' \), thus \( \angle CAC' < \angle ACC' = \angle AC'C \), and therefore \( \angle CAC' < 60^\circ \), \( \angle CAB < 30^\circ \).) The perimeter of the found triangle is \( 3 + \sqrt{5} \) units.

If, however, (also in the case where III. and I. are true) we take \( C \) as the right angle vertex, similarly to the above, it can be seen that \( \angle BAC = 30^\circ \), II. is also true, so the triangle does not match. - Finally, we do not need to try to place the right angle at \( A \), since \( BC < AB \) implies \( \angle BAC < \angle ACB \), and thus \( \angle BAC < 90^\circ \).

5. Essentially, we arrive at the last case if we consider IV. and I. to be true, only \( B \) and \( C \) need to be swapped: if \( \angle ACB = 90^\circ \) and \( AC = 2BC \), then \( \angle A < 30^\circ \) and \( AB = \sqrt{5} \), II. and III. are false, the perimeter is \( 3 + \sqrt{5} \).
6. Finally, no \( ABC \) triangle can match in which II. and I. are true: \( \angle BAC = 30^\circ \) and either a right angle is at \( B \) or at \( C \), because then IV. or III. is also true. There is no other possibility for selecting two true statements.

According to all of this, based on the statements I.-IV. and the "2 true, 2 false" communication, the perimeter of a suitable triangle can be 5 and can be \( 3 + \sqrt{5} \) units.

Zsuzsanna Kósa (Budapest, Móricz Zs. Gymnasium, 1st year)