### Brain Teasers

# Undiscovered Planet

You land on an unknown planet which has a radius of 10km and is completely round and smooth. You start from your ship and walk straight for 17.345 kilometers. You turn left and walk straight for 47.124 kilometers. How far are you from your ship?

### Answer

14.142135623730950488016887242097Or square root of 200!

If you imagine you are facing a globe head on, you have the north pole at the top of the globe and the south pole at the bottom. Imagine a "west pole" and "east pole" to the left and right of the globe.

If you are at the north pole and turn left, you will walk through the "west pole" (right and you go through the "east pole")

If you walk on an axis from the north pole towards the south pole and make a left or right turn you will always walk through these imaginary east and west poles. So in the question 17.345 kilometers is irrelevant, only the turn.

(It also doesn't matter where you start on the planet, the distance at the end will still be the same.)

47.124 is approximately 3/4 of the circumference of this planet, which will leave you directly on the imaginary east pole. The distance is now the hypotenuse of a right angle triangle of side 10km.

Answer: root 200 or approximately 14km!

(If you got the distance as 15.7km - the distance you would have to walk back, then that's ok too!)

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## Comments

This one's easy!

Just add them together!

Just add them together!

You obviously didn't read the answer.

how is 47.145km three-fourths of a 20km diameter?

circumference sorry. But if you can understand this puzzle, it should be obvious

This is the best teaser I've seen on this site so far. Really cool. I would only suggest that you shouldn't put so many significant figures in your answer, since you only gave a few in the question, and 47.175 km is only approximately 3/4 of the circumference.

Easy But Very Good. I Just Added It To My Favourites. Well Done cathalmccabe.

WO!!! I am SOOOOOOOOOOOOO cofuseded!

I think I can simplify thsi a little.

Imagine that you've landed on the north pole (it doesn't really matter, as we're not worried about the rotation of the "planet"). The first leg takes you from the pole down to the equator. Now you make a right-angle turn. From here, it doesn't matter how for you walk, as long as you keep going in a straight line. You're walking around the equator, staying the same distance from your ship.

If you were to walk back to your ship, you'd have to do it along the surface. This would be 17.45km, just like the first leg. However, the straight-line distance (through the "planet") is the hypotenuse of a right triangle. The right angle is at the center of the "planet", with the two legs running to you and the ship. Each of these legs is 10km.

Does that help?

Imagine that you've landed on the north pole (it doesn't really matter, as we're not worried about the rotation of the "planet"). The first leg takes you from the pole down to the equator. Now you make a right-angle turn. From here, it doesn't matter how for you walk, as long as you keep going in a straight line. You're walking around the equator, staying the same distance from your ship.

If you were to walk back to your ship, you'd have to do it along the surface. This would be 17.45km, just like the first leg. However, the straight-line distance (through the "planet") is the hypotenuse of a right triangle. The right angle is at the center of the "planet", with the two legs running to you and the ship. Each of these legs is 10km.

Does that help?

It says the planet is completely round and flat. This sounds like a two dimensional planet to me. In that case, the greatest distance that could be walked in one direction is 20 km, from one side of the circle to the other. Then you'd fall off the edge!

Something is WRONG!

As soon as I read, "round and flat," I anticipated a mistake. The suggestion that we "walk on an axis from the north pole" forebodes a mess. Let's assume "round and smooth," realize longitude is not axis, and thank norcekri for clarifying that the distance sought was "tunnel" distance.

Try the following on for truth.

You land at a pole. You travel along some line of longitude. At 15.708 km (10km * pi / 2), you reach the equator and the so-called answer works. By stretching this leg of your walk to the 17.345 km dictated by the question, you reach a point 1.637 km beyond the equator. A left turn of 90 degrees will set you walking on a path that is orbital. Because you walk straight, your path will be along a true, full circumference of the planet. 15.708 km later, you again reach the equator, and the so-called answer works. 15.708 km after that, you are 1.637 km closer to the pole. This, of course, puts you 14.071 surface km from your ship and still closer by tunnel. The so-called answer works, again, 15.708 km later as your oblique travel away from the ship again brings you back to the equator.

Now, we come to the biggest problem. The leftward part of your walk, so far, totals (just under) 47.124 km. The extra 51 meters in the question's 47.175 km plugs into the relation: maximum drop below equator / maximum distance beyond 'equinox' a.k.a. truth point = extra distance to pole / traveled distance from 'equinox.' 1637 / 15708 = x / 51. So x = 5.3... at the angles involved, an extra 5 meters of surface travel will add about 3.5 meters to the tunnel distance. This is a big stinky mistake from someone waving 27 extra decimal places in our faces.

As soon as I read, "round and flat," I anticipated a mistake. The suggestion that we "walk on an axis from the north pole" forebodes a mess. Let's assume "round and smooth," realize longitude is not axis, and thank norcekri for clarifying that the distance sought was "tunnel" distance.

Try the following on for truth.

You land at a pole. You travel along some line of longitude. At 15.708 km (10km * pi / 2), you reach the equator and the so-called answer works. By stretching this leg of your walk to the 17.345 km dictated by the question, you reach a point 1.637 km beyond the equator. A left turn of 90 degrees will set you walking on a path that is orbital. Because you walk straight, your path will be along a true, full circumference of the planet. 15.708 km later, you again reach the equator, and the so-called answer works. 15.708 km after that, you are 1.637 km closer to the pole. This, of course, puts you 14.071 surface km from your ship and still closer by tunnel. The so-called answer works, again, 15.708 km later as your oblique travel away from the ship again brings you back to the equator.

Now, we come to the biggest problem. The leftward part of your walk, so far, totals (just under) 47.124 km. The extra 51 meters in the question's 47.175 km plugs into the relation: maximum drop below equator / maximum distance beyond 'equinox' a.k.a. truth point = extra distance to pole / traveled distance from 'equinox.' 1637 / 15708 = x / 51. So x = 5.3... at the angles involved, an extra 5 meters of surface travel will add about 3.5 meters to the tunnel distance. This is a big stinky mistake from someone waving 27 extra decimal places in our faces.

The corrected numbers are better.

I don't know enough about spherical geometry to interpret this idea of turning 90 degrees on the surface of a sphere. Does this dictate that you will move in a great circle? The question assumes this is so. If you look at a map using Mercator's projection of the Earth, a right angle turn would place you on a small circle, traveling parallel to an equator. I understand the explanation given here but I am not sure this is consistent with definitions of global navigation. It's a thought provoking puzzle.

(I do know that if three points are joined on the surface of a sphere, the sum of the angles in the spherical triangle is greater than 180 degrees so you cannot flatten out the surface and expect plane geometry to work)

(I do know that if three points are joined on the surface of a sphere, the sum of the angles in the spherical triangle is greater than 180 degrees so you cannot flatten out the surface and expect plane geometry to work)

OK. I have learned something new and that is if you travel in a straight line on the surface of a sphere you will always traverse a great circle. But the answer given here is wrong. Stil is correct. If you had walked 15.708 km initially, this would be one quarter of a great circle. Then, if you turned 90 degrees and walked in a straight line the plane of the great circle you are traversing is at 90 degrees to the plane of you original traverse and no matter how far you went you would always be a constant distance from the starting point.

Imagine leaving the North Pole and travelling to the equator. Now turn and travel along the equator. You are always the same distance from the North Pole.

The problem here that Stil has identified is that if you initially travel 17.345 km you pass the equator. Now if you turn 90 degrees and traverse a great circle, the plane of this circle is inclined a 99 degrees 23 minutes to the orthogonal plane of your original journey.

To see this, imagine you had actually walked almost (but not quite) 31.41 km initially. You would almost be at the South Pole. Now if you turned and walked a great circle you would certainly not remain a constant distance from the North Pole in fact you would pass very close to it on your way back.

The concept in this teaser is very good, but like many puzzles it has been ruined by poor arithmetic. Just changing the 17.345 km to 15.708 km would fix the puzzle IMHO.

Imagine leaving the North Pole and travelling to the equator. Now turn and travel along the equator. You are always the same distance from the North Pole.

The problem here that Stil has identified is that if you initially travel 17.345 km you pass the equator. Now if you turn 90 degrees and traverse a great circle, the plane of this circle is inclined a 99 degrees 23 minutes to the orthogonal plane of your original journey.

To see this, imagine you had actually walked almost (but not quite) 31.41 km initially. You would almost be at the South Pole. Now if you turned and walked a great circle you would certainly not remain a constant distance from the North Pole in fact you would pass very close to it on your way back.

The concept in this teaser is very good, but like many puzzles it has been ruined by poor arithmetic. Just changing the 17.345 km to 15.708 km would fix the puzzle IMHO.

Sorry - forget the turn 90 degrees bit. The puzzle says turn left. The amount depends on which direction you must take in order to traverse a great circle.

I know that 4 consecutive posts is a bit over the top but....

I must apologise to cathalmccabe!

I have just modelled this problem with a texta and a golf ball and although I set out to explain why the answer was wrong, I discovered it is in fact correct! Unbelievably so.

I began by imagining you landed 1.637 km (17.345 - 15.70 from the N pole. You travel the 1.637 km to teh North Pole and keep going another 15.708 to arrive at the equator. Turn left and travel 3/4 of a great circle around the equator thereby keeping a constant distance from the North pole.

Now I had supposed that you could not be the same distance from the North Pole AND at the same distance from your starting point 1.637 km away from the North Pole.

It turns out that is exactly what happens because you finish at a point which is perpendicular to the plane of your original traverse. That means, from your finishing point, all points on your original traverse are like the equator is to the Noth Pole - they are all the same distance away.

Hence Cathal's assertion that the 17.345 km original journey is irrelevant is quite correct.

This teaser has astounded me and I now rate it one of the best on this site. Thanks Cathal!

I must apologise to cathalmccabe!

I have just modelled this problem with a texta and a golf ball and although I set out to explain why the answer was wrong, I discovered it is in fact correct! Unbelievably so.

I began by imagining you landed 1.637 km (17.345 - 15.70 from the N pole. You travel the 1.637 km to teh North Pole and keep going another 15.708 to arrive at the equator. Turn left and travel 3/4 of a great circle around the equator thereby keeping a constant distance from the North pole.

Now I had supposed that you could not be the same distance from the North Pole AND at the same distance from your starting point 1.637 km away from the North Pole.

It turns out that is exactly what happens because you finish at a point which is perpendicular to the plane of your original traverse. That means, from your finishing point, all points on your original traverse are like the equator is to the Noth Pole - they are all the same distance away.

Hence Cathal's assertion that the 17.345 km original journey is irrelevant is quite correct.

This teaser has astounded me and I now rate it one of the best on this site. Thanks Cathal!

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