Shyam's house, his office and his gym are all equidistant from each other. The distance between any 2 of them is 4 km. Shyam starts walking from his gym in a direction parallel to the road connecting his office and his house and stops when he reaches a point directly east of his office. He then reverses direction and walks till he reaches a point directly south of his office. The total distance walked by Shyam is

Question

A, B, C, D, E, and F are the six police stations in an area, which are connected by streets as shown below. Four teams - Team 1, Team 2, Team 3 and
Team 4 patrol these streets continuously between 09:00 hrs. and 12:00 hrs. each day.

The teams need 30 minutes to cross a street connecting one police station to another. All four teams start from Station A at 09:00 hrs. and must return to Station A by 12:00 hrs. They can also pass via Station A at any point on their journeys.
The following facts are known.
1. None of the streets has more than one team traveling along it in any direction at any point in time.
2. Teams 2 and 3 are the only ones in stations E and D respectively at 10:00 hrs.
3. Teams 1 and 3 are the only ones in station E at 10:30 hrs.
4. Teams 1 and 4 are the only ones in stations B and E respectively at 11:30 hrs.
5. Team 1 and Team 4 are the only teams that patrol the street connecting stations A and E.
6. Team 4 never passes through Stations B, D or F.

Answer 1

We can confirm this result using coordinates. Place the Office at the origin $O = (0, 0)$.

Placing the vertices: House and Gym must be positioned symmetrically about the east-west line through the Office, which is what makes a point directly east and a point directly south of the Office both reachable along the two straight-line legs of the walk. Take the Office-House road at $30^\circ$ above east and the Office-Gym road at $30^\circ$ below east. With a side of 4 km, this gives House $H = (4\cos30^\circ, 4\sin30^\circ) = (2\sqrt3, 2)$ and Gym $G = (4\cos(-30^\circ), 4\sin(-30^\circ)) = (2\sqrt3, -2)$. All three sides $OH$, $OG$ and $HG$ work out to exactly 4 km, confirming the triangle is equilateral.
Walking to the east point: The direction of the Office-House road is the unit vector $(\cos30^\circ, \sin30^\circ)$. Starting at $G$ and moving along this direction by the vector $H - O$ gives the point $P_1 = G + (H - O) = (4\sqrt3, 0)$, which has $y = 0$, so it lies directly east of the Office. The distance covered, $|H - O|$, is exactly 4 km.
Walking to the south point: Reversing direction, Shyam retraces this same line. Moving back 4 km returns him to $G$. Continuing another 4 km past $G$ along the same line (applying the vector $O - H$ to $G$ once more) lands him at $P_2 = G - (H - O) = (0, -4)$, which has $x = 0$ and $y < 0$, so it lies directly south of the Office, exactly 4 km away. This second leg is $4 + 4 = 8$ km.

Adding both legs: $4 + 8 = 12$ km, confirming option (D). $\boxed{12 \text{ km}}$

Answer 2

We can confirm this result using coordinates. Place the Office at the origin $O = (0, 0)$.

Placing the vertices: House and Gym must be positioned symmetrically about the east-west line through the Office, which is what makes a point directly east and a point directly south of the Office both reachable along the two straight-line legs of the walk. Take the Office-House road at $30^\circ$ above east and the Office-Gym road at $30^\circ$ below east. With a side of 4 km, this gives House $H = (4\cos30^\circ, 4\sin30^\circ) = (2\sqrt3, 2)$ and Gym $G = (4\cos(-30^\circ), 4\sin(-30^\circ)) = (2\sqrt3, -2)$. All three sides $OH$, $OG$ and $HG$ work out to exactly 4 km, confirming the triangle is equilateral.
Walking to the east point: The direction of the Office-House road is the unit vector $(\cos30^\circ, \sin30^\circ)$. Starting at $G$ and moving along this direction by the vector $H - O$ gives the point $P_1 = G + (H - O) = (4\sqrt3, 0)$, which has $y = 0$, so it lies directly east of the Office. The distance covered, $|H - O|$, is exactly 4 km.
Walking to the south point: Reversing direction, Shyam retraces this same line. Moving back 4 km returns him to $G$. Continuing another 4 km past $G$ along the same line (applying the vector $O - H$ to $G$ once more) lands him at $P_2 = G - (H - O) = (0, -4)$, which has $x = 0$ and $y < 0$, so it lies directly south of the Office, exactly 4 km away. This second leg is $4 + 4 = 8$ km.

Adding both legs: $4 + 8 = 12$ km, confirming option (D). $\boxed{12 \text{ km}}$

Show Hint

The Correct Option is D

Solution and Explanation

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