At some time between 9:30 and 10 o'clock the triangle determined by the minute hand and the hour hand is an isosceles triangle (see diagram). If the two equal angles in this triangle are each twice as large as the third angle, what is the time?
(A) $9: 35$
(B) 9:36
(C) $9: 37$
(D) $9: 38$
(E) 9:39

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-149.jpg?height=298&width=290&top_left_y=2022&top_left_x=1400)
At some time between 9:30 and 10 o'clock the triangle determined by the minute hand and the hour hand is an isosceles triangle (see diagram). If the two equal angles in this triangle are each twice as large as the third angle, what is the time?
(A) $9: 35$
(B) 9:36
(C) 9:37
(D) $9: 38$
(E) 9:39

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-308.jpg?height=320&width=330&top_left_y=1927&top_left_x=1296)

## Solution

Let $x$ represent the angle, in degrees, between the hour and the minute hands. We are told that the triangle in the diagram is isosceles, with the two equal angles each twice as large as the third angle.

Thus, $x+x+\frac{1}{2} x=180$

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-309.jpg?height=328&width=336&top_left_y=275&top_left_x=1312)

$$
\begin{aligned}
\frac{5}{2} x & =180 \\
x & =72
\end{aligned}
$$

For each minute that passes, the minute hand moves through an angle of $360^{\circ} \div 60=6^{\circ}$, and the hour hand moves through an angle of $\left(360^{\circ} \div 12\right) \div 60=\frac{1}{2}^{\circ}$.

At 9:00 there is an angle of $270^{\circ}$ between the hour and the minute hands. At the time shown in the diagram there is an angle of $72^{\circ}$ between the hour and the minute hands. Since the minute hand gains $5 \frac{1}{2}^{\circ}$ on the hour hand every minute, it takes $\frac{270-72}{5 \frac{1}{2}}=36$ minutes from 9:00 for the hour and minute hands to reach the given position. Thus, the time is 9:36.

ANSWER: (B)