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Rectangle ABCD has center O and AB/AD=k. A point is randomly chosen from the interior of rectangle ABCD. What is the probability that it is closer to O than to any of the four vertices?

MathCuber  Jul 16, 2018
 #1
avatar+27043 
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An interesting question!

 

Because of the symmetry of rectangle ABCD, we need consider only one quarter of it; the rectangle for which vertex A, say, is one vertex and for which centre O is the diagonally opposite vertex. Because of the symmetry between A and O in this reduced rectangle the probability of randomly choosing a point nearer one than the other is the same. That is the probability is 1/2. This is independent of the magnitude of k.

Alan  Jul 16, 2018
 #2
avatar+93654 
+3

Thanks Alan,

I just fancied trying to show this graphically. :)

I used a variable k and it makes no difference :)

 

Here is a diagram that i draw.

It might be a bit hard to understand but i have drawn a quarter of the rectangle and the shaded bit is where the point will be closeset to the vertice than to the middle. It is pretty clear that exactly half the time it will be closer to a vertice and half the time it will be closer to the centre.

 

https://www.desmos.com/calculator/od32ntpmfx

Melody  Jul 16, 2018
edited by Melody  Jul 16, 2018

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