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A point P is chosen at random inside equilateral triangle ABC Find the probability that P is closer to the center of triangle than to any of the vertices of the triangle.

(In other words, let O be the center of the triangle. Find the probability that OP is shorter than all of AP, BP and CP)

 Oct 4, 2019
 #1
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See the image here :

 

 

Let ABC  be an equilateral triangle  of side 6

 

The area of this triangle  is   (√3/4) *side^2  = (√3/4)*6^2  = √3 (36/4) = 9√3

 

Any point  falling into  [ ILMNKJ]  will be nearer to the center O  than to any vertex

 

Looking at triangle  ICL....

 

IL  = 2     and this triangle  is similar to triangle ABC

 

So.....the scale factor  of ICL to ABC  = (2/6) =   (1/3)

 

So....the area of triangle  ICL  =  area of triangle ABC * (scale factor)^2  = 9√3 * (1/3)^2  =  √3

 

And triangles AJK and BNM  are congruent to ICL  so they have the same areas

 

So  [ ILMNKJ] has an area  of [ABC]  - [ ICL] - [ AJK] - [ BNM]  =

 

9√3 - √3 - √3 - √3   =  

 

9√3 - 3√3  =

 

6√3

 

So.....the probability that P is closer to the center O than to  any vertex  =

 

[ ILMNKJ]         6√3          6            2

________  =   _____  =  ___ =    ___

[ ABC ]             9√3          9            3

 

 

cool cool cool

 Oct 4, 2019
edited by CPhill  Oct 4, 2019

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