, 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 $b < a$ and $b < c$, 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$b < a$, then$b = 20$and$a = 30$.

Additionally, we know that $c > b = 20$, 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 $b < c$, then $b = 20$ and $c = 30$.

Additionally, we know that $a > b = 20$, 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 $b < a = 20$ and $b < c = 30$ (which means that $b<20$) 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:

$\text{1 + 2 - 2 = 2 + 2 1 + (2 - 2 )(2 + 2 ) = (1 + ) 2 4 - 4 2 = 1 + 2 + 2 4(1 - 2 ) = 1 + 2 + 2 4 2 = 1 + 2 + 2 3 2 - 2 - 1 = 0 (3 + 1)( - 1) = 0}$$Thus,$$\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$,$2$,$2$.