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# looking at this problem

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A right triangle has its legs parallel to the x and y axes as shown in the figure. If the hypotenuse has a slope of  -4/3, and the diameter of the bigger circle is 2020, what is the diameter of the smaller circle? Dec 10, 2020

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Let  the center of the larger circle  be (10,10)....and this is the incenter of the large triangle

Let A  be the angle at the  bottom right  vertex

The tangent of this angle is  4/3

But  the bisector of angle A   creates  an angle with 1/2 the measure of A

And the tangent of this angle is   sqrt   [  1 -cos A] / sqrt [1 + cos A]

And cos A = 3/5

So  tan (A/2)  =   sqrt  [ 1-3/5 ] / sqrt [1 + 3/5]   =  1/2

And

tan (A/2)   =10/20

cos (A/2)  =  2/sqrt (5)

Therefore, the distance  from the bottom right  vertex   of the large triangle to the  center of the larger circle can be found as

cos (A/2)  = 20 / D

D =  20 / (2 /sqrt (5) )  =  10sqrt (5)

And using reflexive symmetry of similar polygons, the angle  bisector will go through the center of both  circles....

So  using similar triangles we  have that

10/ (10sqrt (5)  )  =  r / [ 10sqrt (5)  - 10  - r ]

1/sqrt (5)  =  r / [10sqrt (5)  -10  - r ] =

r sqrt (5)  =  10sqrt (5)  - 10  - r

r (  sqrt (5) + 1  )  =  10 ( sqrt (5) - 1)

r  =  10 ( sqrt (5) -1) ( sqrt (5) -1) / 4

r = 10 ( 5 - 2sqrt (5) + 1) /4

r = 15 - 5sqrt (5)

Here's a pic  : Dec 10, 2020
edited by CPhill  Dec 10, 2020
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