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#1 2011-04-22 01:18:26

From: Bumpkinland
Registered: 2009-04-12
Posts: 108,540

Geogebra and jumping frogs.


Posted elsewhere is the problem of the jumping frog. Geogebra makes the problem easy and shows that it is more geometric than anything else.

A bullfrog leaps 2 meters in some direction. Does not like its location so it randomly leaps n meters again in some direction. If its odds of being within 1 meter from its original position are 1 / 6 then what is n?

Basically what you do is draw a unit circle around the origin. The origin represents the starting point of the frog. Draw another concentric circle with radius 2 (shaded circle). This represents all possible first jumps. WLOG pick a point on the shaded circle and call it B. Draw two tangents from B to the smaller circle. Call the tangent points F and G. Now it is just a geometry proof. Angle FBG is 60 degrees. The red circle represents all possible 2nd jumps. This angle of 60 degrees represents 1 / 6 of all possible second jumps.

Line segments AB and BC are

long. That is the length of n.

View Image: hello.gif

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.


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