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A rectangle measures 6 centimeters by 25 centimeters. The probability that a point randomly chosen inside the rectangle is closer to a side of length 25 than a side of length 6 is \(\frac p q\), where \(p\) and \(q\) are relatively prime positive whole numbers. What is the sum of \(p\) and \(q\)?

 Mar 13, 2021
 #1
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I AM NOT  totally  sure  about this one, TGO,  but it's an interesting problem......here's  my best attempt

 

Look at  the image  below  :

 

 

Note  that  any point lying  within  triangles  ADE  and  BCF   will be CLOSER  to  a side of  6  than to  a side of 25

 

Each of  these triangles has  a base of 6  and a height of  3

 

So....their total  area is  6 * 3   =18

 

And the total area of  the rectangle  = 25 * 6  =150

 

So....the  region  that a point  could fall into and  be  closer to a  side of 25  than a side of 6 has an area of :  150   -18 =   132

 

So....  p  / q   =  132   /150   =  22 /  25

 

So..... p + q  =   22 + 25  = 47

 

cool cool cool

 Mar 13, 2021

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