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 $BD$?
+128408 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)
