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I found this problem from an old math olympiad problem set and got stuck. Can anyone help?

Find the length of the shortest path on the plane from P(0,0) to Q(2,1) so that any point on this path is at least 1 unit away from (1,0), (1,1), (1,2), and (2,0).

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  • $\begingroup$ Have you tried drawing a picture? $\endgroup$ – Michael Biro Aug 27 '16 at 14:37
  • $\begingroup$ Yes, I have. Actually, I already know the answer. I peeked at the back for the answer hoping that I would understand how to solve the problem. But still, I do not know how to attack the problem. $\endgroup$ – Carl Terence Valdellon Aug 27 '16 at 14:41
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    $\begingroup$ I completely agree with @Michael Biro . Draw not only the points but circles with radii 1 with these points as centers, and wrap a rope around all this... Do you see why ? $\endgroup$ – Jean Marie Aug 27 '16 at 14:42
  • $\begingroup$ Ahhhh... thank you very much $\endgroup$ – Carl Terence Valdellon Aug 27 '16 at 14:43
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Start by drawing a picture. Include the start ($S$), the finish ($F$), the four points and the forbidden circles of radius $1$ around each of them:

Figure

You may not enter any of the red areas. Apart from that, you should follow the shortest possible path, the one a tight rubber band would use. There are essentially two options to consider: the upper path and the lower one. The upper is two units up, then a semicircle, then one unit down:

$$2 + \pi + 1 \approx 6.1416$$

The lower path has a quarter circle, then one unit right, then a semicircle:

$$\tfrac12\pi + 1 + \pi \approx 5.7123$$

so the lower path is the shorter one.

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  • $\begingroup$ We don't even need the numerical calculation. Surely the circumference of a circle is less than the perimeter of irs circumscribed square, thus $\pi<4$, and given that we find by simple algebra that the bottom path is shorter. $\endgroup$ – Oscar Lanzi Aug 28 '16 at 18:11

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