Brain Teaser No 1
Posted by: Don Atkinson on 16 November 2001
THE EXPLORER
An explorer set off on a journey. He walked a mile south, a mile east and a mile north. At this point he was back at his start. Where on earth was his starting point? OK, other than the North Pole, which is pretty obvious, where else could he have started this journey?
Cheers
Don
Hi Dan,
Since my previous answer disappeared in The Great Forum Mishap of 2004, here's another one. This one's a little neater (I was having a bad brain day last time):
Labels:
P, Q, R: the edges of the triangle.
p, q, r: the vertices of the triangle opposite edges P, Q and R respectively
As before,if any triangle has a vertex (p, say) that's not on the rectangle then a larger triangle can be made by translating vertex p away from edge P along a line perpendicular to P until it reaches the rectangle. [Then the distance moved by p is the increase in height of the triangle].
Therefore, all triangles of maximum size must have all three vertices on the rectangle.
Suppose that the a triangle has all three vertices on the rectangle, and none of the triangle's edges is parallel to an edge of the rectangle.
Then, we can translate p along a line parallel to P until p is within the rectangle. As seen above, this cannot be the largest triangle. But the distance between p and P remains unchanged by the translation, so the area remains the same. Hence the original triangle cannot have been the largest possible.
Therefore, a triangle of maximum size must have at least one edge parallel to an edge of the rectangle.
Suppose a triangle has all three vertices on the rectangle, and edge P of the triangle is parallel to edges S and T of the rectangle. If P is not on S or T then a larger triangle can be made by translating P away from p until it is on S or T. [Then the distance moved by P is the increase in height of the triangle].
Therefore, a triangle of maximum size must have one edge on an edge of the rectangle.
Suppose a triangle has all three vertices on the rectangle, and edge P of the triangle is on edge S of the rectangle. If the vertices q and r of P don't coincide with the edges of S then they can be translated away from each other until they do, to make a larger triangle. [Then the distance moved by q and r is the increase in base of the triangle].
Therefore a triangle of maximum size has all three vertices on the rectangle, with at least one edge of the triangle coinciding exactly with one edge of the rectangle.
So if q and r are the vertices on the coincident edge, P, then vertex p must be on the parallel edge of the rectangle. Wherever p is on that edge, its distance from P is the same. Hence the base and height of the maximum sized triangle must be x and y or y and x respectively, so its area is x*y/2.
There must be an easier way...
Since my previous answer disappeared in The Great Forum Mishap of 2004, here's another one. This one's a little neater (I was having a bad brain day last time):
Labels:
P, Q, R: the edges of the triangle.
p, q, r: the vertices of the triangle opposite edges P, Q and R respectively
As before,if any triangle has a vertex (p, say) that's not on the rectangle then a larger triangle can be made by translating vertex p away from edge P along a line perpendicular to P until it reaches the rectangle. [Then the distance moved by p is the increase in height of the triangle].
Therefore, all triangles of maximum size must have all three vertices on the rectangle.
Suppose that the a triangle has all three vertices on the rectangle, and none of the triangle's edges is parallel to an edge of the rectangle.
Then, we can translate p along a line parallel to P until p is within the rectangle. As seen above, this cannot be the largest triangle. But the distance between p and P remains unchanged by the translation, so the area remains the same. Hence the original triangle cannot have been the largest possible.
Therefore, a triangle of maximum size must have at least one edge parallel to an edge of the rectangle.
Suppose a triangle has all three vertices on the rectangle, and edge P of the triangle is parallel to edges S and T of the rectangle. If P is not on S or T then a larger triangle can be made by translating P away from p until it is on S or T. [Then the distance moved by P is the increase in height of the triangle].
Therefore, a triangle of maximum size must have one edge on an edge of the rectangle.
Suppose a triangle has all three vertices on the rectangle, and edge P of the triangle is on edge S of the rectangle. If the vertices q and r of P don't coincide with the edges of S then they can be translated away from each other until they do, to make a larger triangle. [Then the distance moved by q and r is the increase in base of the triangle].
Therefore a triangle of maximum size has all three vertices on the rectangle, with at least one edge of the triangle coinciding exactly with one edge of the rectangle.
So if q and r are the vertices on the coincident edge, P, then vertex p must be on the parallel edge of the rectangle. Wherever p is on that edge, its distance from P is the same. Hence the base and height of the maximum sized triangle must be x and y or y and x respectively, so its area is x*y/2.
There must be an easier way...
OFF TOPIC ALERT:
do you want to share the details of your new camera.....or are the details elsewhere on the Forum ?
....and how can you listen to music, play with a camera AND do brain teasers simultaneaously.....
And this "work" stuff keeps getting in the way.
New camera is an Voigtlander Bessa R combined with a 50mm/F2.0 Russian screw mount lens. First roll came out quite well, but I need to work on my framing of pictures using the framelines in the viewfinder. The new body was relatively inexpensive at the equivalent of around 140UKP. Light, compact, and no gimmicks. Only negative so far is that the russian lens is a bit stiff and so focusing is slower than I like. I'll play around some more before getting the Nokton (F1.5) however.
cheers,
Dan
do you want to share the details of your new camera.....or are the details elsewhere on the Forum ?
....and how can you listen to music, play with a camera AND do brain teasers simultaneaously.....
And this "work" stuff keeps getting in the way.
New camera is an Voigtlander Bessa R combined with a 50mm/F2.0 Russian screw mount lens. First roll came out quite well, but I need to work on my framing of pictures using the framelines in the viewfinder. The new body was relatively inexpensive at the equivalent of around 140UKP. Light, compact, and no gimmicks. Only negative so far is that the russian lens is a bit stiff and so focusing is slower than I like. I'll play around some more before getting the Nokton (F1.5) however.
cheers,
Dan