In quadrilateral $P Q R S$, diagonals $P R$ and $S Q$ intersect at $O$ inside $P Q R S, S P=S Q=S R=1$, and $\angle Q S R=2 \angle Q S P$. Marc determines the measure of the twelve angles that are the interior angles of $\triangle P O S$, $\triangle P O Q, \triangle R O S$, and $\triangle R O Q$. The measure of each of these angles, in degrees, is a positive integer, and exactly six of these integers are prime numbers. How many different quadrilaterals have these properties and are not rotations or translations of each other?
(A) 7
(B) 5
(C) 9
(D) 6
(E) 8

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-021.jpg?height=426&width=326&top_left_y=1988&top_left_x=1344)
We begin by labeling $\angle Q S P=2 \theta$. Since $\angle Q S R=2 \angle Q S P$, $\angle Q S R=2 \times 2 \theta=4 \theta$.

We will now find the measures of all 12 angles in terms of $\theta$.

Let $x=\angle S P R$. Since $\triangle S P R$ has $S P=S R$, it is isosceles which means $\angle S R P=\angle S P R=x$.

Also, $\angle P S R=2 \theta+4 \theta=6 \theta$, so $180^{\circ}=6 \theta+x+x$.

Rearranging this equation, we have $2 x=180^{\circ}-6 \theta$, so $x=90^{\circ}-3 \theta$. In $\triangle S P Q$, we know $\angle S P Q+\angle S Q P+\angle P S Q=180^{\circ}$, so rearranging and using $\angle P S Q=2 \theta$ gives $\angle S P Q+\angle S Q P=180^{\circ}-2 \theta$.

Since $S P=S R, \triangle S P Q$ is isosceles which means $\angle S P Q=\angle S Q P$.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-103.jpg?height=469&width=418&top_left_y=812&top_left_x=1493)

Substituting this into the above equation, we have $2 \angle S Q P=180^{\circ}-2 \theta$ or $\angle S Q P=90^{\circ}-\theta$.

Note that this means $\angle S P Q=90^{\circ}-\theta$ as well, so $\angle S P R+\angle R P Q=90^{\circ}-\theta$.

Substituting $\angle S P R=90^{\circ}-3 \theta$, we get that $90^{\circ}-3 \theta+\angle R P Q=90^{\circ}-\theta$, so $\angle R P Q=2 \theta$.

Since $\angle P O Q+\angle P Q O+\angle O P Q=180^{\circ}$, substitution gives $\angle P O Q+\left(90^{\circ}-\theta\right)+2 \theta=180^{\circ}$.

Solving this, we get $\angle P O Q=90^{\circ}-\theta$.

Angles $\angle P O Q$ and $\angle R O S$ are opposite, which means $\angle R O S=\angle P O Q=90^{\circ}-\theta$. Also, $\angle P O Q+\angle Q O R=180^{\circ}$, so $\angle Q O R=180^{\circ}-\angle P O Q=180^{\circ}-\left(90^{\circ}-\theta\right)=90^{\circ}+\theta$.

Since $\angle P O S$ and $\angle R O Q$ are opposite, they are equal, so $\angle P O S=90^{\circ}+\theta$.

We now set $\angle S Q R=y$.

Since $S Q=S R, \triangle S Q R$ is isosceles, which means

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-103.jpg?height=474&width=466&top_left_y=1484&top_left_x=1488)
$\angle S R Q=\angle S Q R=y$.

Substituting this and $\angle Q S R=4 \theta$ into $\angle Q S R+\angle S R Q+\angle S Q R=180^{\circ}$ gives $4 \theta+2 y=180^{\circ}$, so $2 y=180^{\circ}-4 \theta$ or $y=90^{\circ}-2 \theta$.

Finally, we have $\angle R O Q+\angle O Q R+\angle Q R O=180^{\circ}$, so $\angle Q R O=180^{\circ}-\left(90^{\circ}+\theta\right)-\left(90^{\circ}-2 \theta\right)=\theta$.

In summary, the 12 angles in terms of $\theta$ are

$$
\begin{aligned}
& \angle Q S P= \angle R P Q=2 \theta \\
& \angle Q S R=4 \theta \\
& \angle S R P= \angle R P S=90^{\circ}-3 \theta \\
& \angle Q R P== \\
& \angle R Q S=90^{\circ}-2 \theta \\
& \angle S Q P=\angle P O Q=\angle R O S=90^{\circ}-\theta \\
& \angle P O S=\angle Q O R=90^{\circ}+\theta
\end{aligned}
$$

We are given that the measure of each angle in degrees is an integer.

In particular, since $\angle Q R P=\theta$, we get that $\theta$ must be an integer number of degrees.

We also know that $P R$ and $Q S$ intersect inside $P Q R S$, so it must be that $6 \theta=\angle P S R<180^{\circ}$. This means $\theta<30^{\circ}$.

Since $4 \theta$ is even (and greater than 2), it cannot be a prime number for any integer $\theta$.

The integers $2 \theta$ and $3 \theta$ are only prime numbers when $\theta=1^{\circ}$.

In this case, the twelve angles are

$$
2^{\circ}, 2^{\circ}, 4^{\circ}, 87^{\circ}, 87^{\circ}, 1^{\circ}, 88^{\circ}, 89^{\circ}, 89^{\circ}, 89^{\circ}, 91^{\circ}, 91^{\circ}
$$

of which five are prime numbers: two copies of $2^{\circ}$ and three copies of $89^{\circ}$.

Therefore, $\theta \neq 1^{\circ}$.

By factoring, we have that $90^{\circ}-2 \theta=2\left(45^{\circ}-\theta\right)$ which is even.

Therefore, this number can only be prime when $45^{\circ}-\theta=1$ or $\theta=44^{\circ}$.

We know that $\theta<30^{\circ}$, so $90^{\circ}-2 \theta$ must be composite.

Similarly, $90^{\circ}-3 \theta=3\left(30^{\circ}-\theta\right)$ can only be a prime number when $30^{\circ}-\theta=1^{\circ}$ or $\theta=29^{\circ}$.

When $\theta=29^{\circ}$, the twelve angles are

$$
58^{\circ}, 58^{\circ}, 116^{\circ}, 3^{\circ}, 3^{\circ}, 29^{\circ}, 32^{\circ}, 61^{\circ}, 61^{\circ}, 61^{\circ}, 119^{\circ}, 119^{\circ}
$$

and exactly 6 of these are prime: $29^{\circ}$, the two copies of $3^{\circ}$, and the three copies of $61^{\circ}$.

When $\theta<30^{\circ}$ is different from $1^{\circ}$ and $29^{\circ}$, each of the six angles

$$
2 \theta, 4 \theta, 90^{\circ}-3 \theta, 90^{\circ}-2 \theta, 2 \theta, 90^{\circ}-3 \theta
$$

is composite.

In order to satisfy the condition that the measures of exactly 6 of the angles in degrees is a prime number, we therefore require that each of $\theta, 90^{\circ}-\theta$, and $90^{\circ}+\theta$ is prime.

The prime numbers less than 30 are $2,3,5,7,11,13,17,19,23$, and 29.

We have already investigated what happens when $\theta=29^{\circ}$, so the values for the angles $\theta, 90^{\circ}+\theta$, and $90^{\circ}-\theta$ in the other cases are calculated in the table:

| $\theta$ | $90^{\circ}-\theta$ | $90^{\circ}+\theta$ | All 3 angles prime? |
| :---: | :---: | :---: | :---: |
| $2^{\circ}$ | $88^{\circ}$ | $92^{\circ}$ | $\times$ |
| $3^{\circ}$ | $87^{\circ}$ | $93^{\circ}$ | $\times$ |
| $5^{\circ}$ | $85^{\circ}$ | $95^{\circ}$ | $\times$ |
| $7^{\circ}$ | $83^{\circ}$ | $97^{\circ}$ | $\checkmark$ |
| $11^{\circ}$ | $79^{\circ}$ | $101^{\circ}$ | $\checkmark$ |
| $13^{\circ}$ | $77^{\circ}$ | $103^{\circ}$ | $\times$ |
| $17^{\circ}$ | $73^{\circ}$ | $107^{\circ}$ | $\checkmark$ |
| $19^{\circ}$ | $71^{\circ}$ | $109^{\circ}$ | $\checkmark$ |
| $23^{\circ}$ | $67^{\circ}$ | $113^{\circ}$ | $\checkmark$ |

Of the angles that appear in the second two columns, the numbers which are prime are

$$
67^{\circ}, 71^{\circ}, 73^{\circ}, 79^{\circ}, 83^{\circ}, 97^{\circ}, 101^{\circ}, 103^{\circ}, 107^{\circ}, 109^{\circ}, 113^{\circ}
$$

Therefore, all three angles are prime when $\theta$ is one of $7^{\circ}, 11^{\circ}, 17^{\circ}, 19^{\circ}, 23^{\circ}$.

Combining these with $\theta=29^{\circ}$ from earlier, this gives a total of 6 quadrilaterals with the given properties.