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Floating Tangent (Posted on 2009-10-02) Difficulty: 3 of 5

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Solution Solution Comment 2 of 2 |

In addition to the figure -
let I be the enter of the circle
and P the point of tangency of RT
with the circle.
If t = tan((angle PIC)/2), then it 
is easy to show that the area of
triangle RST is
   sqrt(3)*(3 - t^2)
  -------------------
       1 - 3*t^2
Note: Angle PIC can be measured CW 
      or CCW. Thus, we can have one
      or two positions for point P.
1a)   sqrt(3)*(3 - t^2)
     ------------------- = 3*sqrt(3).
          1 - 3*t^2
     or t = 0.
     Thus, P = C.
1b)   sqrt(3)*(3 - t^2)
     ------------------- = 8.
          1 - 3*t^2
                   8 - 3*sqrt(3)
     or t = sqrt[ --------------- ].
                    24 - sqrt(3)
     Thus, angle PIC =
                      8 - 3*sqrt(3)
     2*arctan{ sqrt[ --------------- ] }
                       24 - sqrt(3)
     ~= 39.073817 degrees.
1c)   sqrt(3)*(3 - t^2)
     ------------------- = 6*sqrt(3)
          1 - 3*t^2
     or t = sqrt[ 3/17 ].
     Thus, angle PIC = 
     2*arctan{ sqrt[ 3/17 ] }
     ~= 45.572996 degrees.   
2)   For angle RTS to be a right angle,
     angle PIC must equal 30 degrees.
     Thus, area [RST] 
        sqrt(3)*(3 - tan(15)^2)
     = -------------------------
            1 - 3*tan(15)^2
     = 2*sqrt(3) + 3
     ~= 6.464102 
 

  Posted by Bractals on 2009-10-03 14:51:58
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