The Rover’s Journey: ‍A Geometric Puzzle

A‌ rover exploring an exoplanet faces a navigational challenge: ⁤it must follow a specific path of increasing distances with 90-degree turns, and return to its starting point.This problem, posed by Ben Orlin,​ author of Math with Bad Drawings, highlights a fascinating interplay between geometry and simple instructions.

The Problem Explained

The rover is programmed to ​move in a series‍ of straight lines. On day one, it travels 1 kilometer, then turns 90 degrees. On day two, it travels ‍2 kilometers and⁢ turns 90 degrees,⁣ and so on, for a total of eight days. The rover‌ can choose to‌ turn either left⁣ or right at each turn. the question is: can the rover, after eight days,⁣ end up back‍ at its original landing site?

Solving ⁢the Puzzle

the key⁣ to solving this puzzle ​lies⁣ in understanding the rover’s​ net displacement in the horizontal and vertical directions. Each day’s movement⁢ contributes to either the eastward/westward or northward/southward displacement. ⁤To return to the starting point, the ⁤total eastward/westward ‍displacement must be zero, and the total⁢ northward/southward displacement must also be zero.

Let’s represent a right turn as +1 ​and a left turn as -1. The rover makes eight turns. For the rover to return to its starting point,⁢ the sum of these turns must be even (0, 2, 4, 6, or ​8). An even sum⁣ means the rover will have completed⁣ a whole number ‌of 360-degree rotations, effectively facing⁣ the same direction it started. however, this condition alone isn’t enough.

Consider the first four days. The rover travels ​1 km, 2 km, 3⁢ km, and 4 km. ⁣If⁤ it alternates right and left turns (e.g., ⁣R, L, R, L), its ⁤net displacement will​ be 0 in both directions.Though,for​ an ​eight-day mission,the‍ rover must⁢ complete a closed loop. This requires⁢ careful consideration of the turn sequence.

The‍ solution is that the‌ rover ⁢ can return ⁢to its starting point. one possible solution is to alternate right and left turns‌ consistently throughout the eight days. For example,Right,Left,Right,Left,Right,Left,Right,Left. This creates a spiral-like path that eventually​ brings the rover back to its ‌origin.

The Bonus Challenge: Rover Missions of Varying⁢ Lengths

The bonus question ​asks how many ⁣rovers, with mission lengths‍ ranging from ⁤1 to‌ 100 days, can return to their starting points. The rover can return to its starting point if and only if the sum of the turns is even. As each rover makes a number of turns equal to its mission length, the rover can return ⁣to its‌ starting point if and only if the mission length is even.

Therefore, out of 100 missions (1​ to 100 days), 50 rovers⁢ (those with mission lengths of 2, 4, 6… 100) can⁢ successfully return to their landing sites.

Key Takeaways

• This puzzle demonstrates how simple geometric principles can lead to surprisingly complex‌ problems.
• The concept of net displacement is crucial for understanding the rover’s path.
• The parity (evenness or oddness) ⁣of ​the number of turns determines​ whether⁤ the rover ⁤can return⁣ to its starting point.
• The problem highlights the importance of careful planning and consideration‍ of all possible outcomes.

This puzzle is ⁤a great example of how mathematical thinking can be applied to​ real-world scenarios, even those involving ⁢interplanetary exploration. It encourages a systematic approach to problem-solving and a deeper understanding of geometric concepts.