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Can anyone help with this?

 

Let PQR be an equilateral triangle, centered at O.  A point X is chosen at random inside the triangle.  Find the probability that X is closer to O than to any of the sides.  (In other words, find the probability that XO is shorter than XA, XB, and XC.)

 
 Jan 25, 2023
 #1
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This should work I think

the side length of the dark little triangle is half that of the large triange.

All the points inside the little triangle will be closer to the centre (which in my pic is J) than they will to any side ot the big triangle.

 

The sides of the little to big are in the ratio 1:2

So the are of the little to big will be 1:4

The area of the little to the area in the big, not including the little wil be  1:4-1   =    1:3

So probability that the distance to the centre is less than the distance to the edge is 1:3   i.e.    \(33.\dot3\;\%\)

 

 

 

 
 Jan 28, 2023
edited by Melody  Jan 28, 2023

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