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In triangle $\triangle ABC$, $AB = 37, AC = 33$, and $BC = 39$.  A circle $\Gamma$ is drawn with center $A$ and radius $AC$.  Let $D$ be the point on $\Gamma$ that is furthest from $B$. What is the distance $BD$?

 Jan 7, 2021
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See the following :

 

 

Let A  = (0,0)   B = (37,0)  

 

And  the equation of the  circle  is x^2 + y^2  = 33^2

 

It's clear  that   the greatest distance  from B to D  will  be when T  =(-33,0)

 

And this distance  is  just   37 + 33   = 70

 

We can see  that if we construct a  circle with a radius of  70 centered at B....it will  only intersect  the circle with a radius of AC at   D  =(-33,0)

 

 

cool cool cool

 Jan 7, 2021

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