, 2025

Solution 1:

Since two of the numbers are $20$
and $30$, and the third number is
between 1 and 40, inclusive, and the three numbers are different, then
there are 38 possible values for the third number. We call the value
chosen $n$.

Since $\text{b lt; a}$ and $\text{b lt; c}$, then $b$ is the smallest, so we set $b$ equal to the smallest of 20, 30 and
$n$.

There are then 2 choices for ordering the assignment of the two
remaining numbers to $a$ and $c$.

Therefore, there are $38 \cdot 2 = 76$ possible combinations.

Solution 2:

From the given information, two of $a$, $b$
and $c$ are equal to $20$ and $30$.

Suppose that $a$ and $b$ are $20$ and $30$ in some order.

Since $\text{b lt; a}$, then $b = 20$ and $a = 30$.

Additionally, we know that $\text{c gt; b = 20}$, that $c \leq 40$, and
that $c \neq a = 30$.

This means that there are $19$
possible values for $c$ in this case
(the integers from $21$ to $40$, inclusive, excluding $30$).

Thus, in this case, there are $19$
possible combinations.

Suppose that $b$ and $c$ are $20$ and $30$ in some order.

Since $\text{b lt; c}$, then $b = 20$ and $c = 30$.

Additionally, we know that $\text{a gt; b = 20}$, that $a \leq 40$, and
that $a \neq c = 30$.

This means that there are $19$
possible values for $a$ in this
case.

Thus, in this case, there are $19$
possible combinations.

Suppose that $a$ and $c$ are $20$ and $30$ in some order.

If $a = 20$ and $c = 30$, then we know that $\text{b lt; a = 20}$ and $\text{b lt; c = 30}$ (which means that $\text{b lt;20}$) and $b \geq 1$. Here, the fact that the three
integers are different does not create additional restrictions.

There are $19$ possible values for
$b$ in this case (the integers from
$1$ to $19$, inclusive).

If $a = 30$ and $c = 20$, there will again be $19$ possible values for $b$.

Thus, in this case, there are $19 + 19 = 38$ possible combinations.

In total, there are $19 + 19 + 38 = 76$ possible combinations.
Using the fact that the values of the three given expressions
form a geometric sequence with no term equal to $0$, the following equations are
equivalent: $$\begin{align*} \dfrac{1 + \sin\theta}{2 - 2\cos\theta} & = \dfrac{2 + 2\cos\theta}{1 + \sin\theta} \\ (2 - 2 \cos\theta)(2 + 2\cos\theta) & = (1 + \sin\theta)^2 \\ 4 - 4\cos^2\theta & = 1 + 2\sin\theta + \sin^2\theta \\ 4(1 - \cos^2\theta) & = 1 + 2\sin\theta + \sin^2\theta \\ 4\sin^2\theta & = 1 + 2\sin\theta + \sin^2\theta \\ 3\sin^2\theta - 2\sin\theta - 1 & = 0 \\ (3\sin\theta + 1)(\sin\theta - 1) & = 0\end{align*}$$ Thus,
$\sin \theta = -\frac{1}{3}$ or $\sin\theta = 1$.

Since $\cos^2\theta = 1 - \sin^2\theta$, then $\cos^2 \theta = 1- (-\frac{1}{3})^2 = \frac{8}{9}$ or $\cos^2\theta = 0$.

Therefore, the possible values of $\cos\theta$ are $\sqrt{\frac{8}{9}}$, $-\sqrt{\frac{8}{9}}$, $0$.

These can be re-written as $\frac{2\sqrt{2}}{3}$, $-\frac{2\sqrt{2}}{3}$, $0$.

Note that $\cos \theta = 0$ gives the
constant geometric sequence $2$,
$2$, $2$.