<p>Ok. I got this MATH problem from school that seems totally insane and insolvable. Has anyone ever done it before? Any ideas about anything? Help!</p>
<p>The Lonely Hunter:</p>
<p>A hunter leaves his house and walks 10 miles due south. Not finding anything, he turns and walks 10 miles due east where he spots a bear. He shoots the bear, and carries it 10 miles due north, where he arrives back at his house. What color was the bear?</p>
<p>I would bet on tanman’s answer. in real life, you wouldn’t ever get back to your house no matter the conditions. But, if you look at a globe, you WOULD end up back at our house.</p>
<p>One problem that people might have with this is that it involves curved space. Most places on earth don’t have to worry about this, but at the north pole, if you go south a distance x, make a ninty degree turn and go a distance x, then turn north ninty degrees and walk a distance x, you’re back where you started. In flat two dimensional (Euclidian) space, that process would have resulted in an open square and you’d still be a distance x away from your start position. In spherical space, an equilateral triangle has three right angles.</p>
Suppose he starts at a spot 5 miles north of the equator. If he goes 10 miles south, then 10 east and then 10 north, he will wind up exactly 10 miles east of his starting point, which doesn’t satisfy the conditions of the problem.</p>
<ol>
<li><p>Now suppose he starts at a spot 10 miles north of the equator, and does the same 10 south/ 10 east/ 10 north trip. (Try to visualize it on a 3-D globe). He now winds up slightly <em>less</em> than 10 miles east of his starting point, because of the earth’s curvature. You can calculate this distance exactly if you want to, if you know the earth’s radius.</p></li>
<li><p>If you follow this general trend i.e. starting further & further north of the equator, you will find that the distance between the starting & ending points keeps shrinking steadily. It shrinks to 0 when the starting point is the north pole.</p></li>
</ol>