Analyzing an uncertain triangle

Let the smallest side of the triangle be $a$ . Since the sides are in AP, then we can easily determine that the other sides are $a + k$ and $a+2k$ for some positive number $k$ .

We know that the perimeter is $1$ , so

$a + (a+ k) + (a +2k) = 1$

$3(a+k) = 1$

$a + k = \frac {1}{3}$

This makes statement $A$ correct.

Now, by virtue of triangle inequalities, we have to make sure that the sum of the smallest and the second side of the triangle is greater than the third side. This automatically makes statement $D$ false as the longest side cannot have a length of half the perimeter or more.

Setting the inequality, we get

$a + (a+k) > a + 2k$

$2a + k > a + 2k$

$a > k$

With this statement, we have established that statement $C$ is true, and statement $B$ is false.

Now, going back to the first statement that was found. that is,

$a + k = \frac {1}{3}$

Apart from the fact that both $a$ and $k$ have to be positive reals, no other restrictions exist. Thus, they can be rational or not for as long their sum is $\frac{1}{3}$ , which makes statement $E$ true.

Finally, since it is possible to assign values for $a$ and $k$ that will satisfy all conditions given above, statement $F$ is false.

And so the answer is $A, C, E$ .