www.8foxes.com
Fox 355
Henk Reuling
Solution
Add a coordinate system.
Take AD = BC = 2 and AB = CD = 4.
So A(0, 0), B(4, 0), C(4, 2) and D(0, 2).
2
2
Big semicircle: (x – 2) + y = 4
2
2
Small semicircle: x + (y – 1) = 1
 x 2  4 x  4  y 2  4
x 2  y 2  4 x  0



 4 x  2 y  0 
 y  2x
 2
 2
2
2
 x  y  2 y  1  1
x  y  2 y  0
(x – 2)2 + (2x)2 = 4 → x2 – 4x + 4 + 4x2 = 4 → 5x2 – 4x = 0 → x(5x – 4) = 0
4
4
8
40
So for point S it gives 5x – 4 = 0 → x = /5 and y = 2· /5 = /5 → S ( 45 , 85 )  ( 20
25 , 25 ) .
T lies on line AC, with ST perpendicular to line AC.
  4 
8
8
16
 2
AC    so   is a normal vector of line ST → equation line ST is 2x + y = /5 + /5 = /5.
2
1
 
 
Intersect lines ST and AC with equation y = ½x:
16
32
16
16
2x + ½x = /5 → 25x = 32 → x = /25; y = /25 → T ( 32
25 , 25 ) .
2
2
20 2
16 40 2
12 2
24 2
12 2
12
ST  ( 32
25  25 )  ( 25  25 )  ( 25 )  (  25 )  ( 25 )  (1  2 )  25 5
2
2
2
16 2
68 2
34 2
34 2
34
TC  (4  32
25 )  (2  25 )  ( 25 )  ( 25 )  ( 25 )  (2 1 )  25 5
ST
tan  

TC
12
25
34
25
5
5

12 6

34 17
So the correct answer is
A.