Let a circle ω (not labelled in the graph) centered at P tangent to AB, and T is point of tangency. ∠ A P B = 90 ∘ . Let K (not labelled in the graph) be some point on the circle ω , the semicircle with diameter BK intersects PB at Q. Let R be the radius of that semi-circle. If 4 R 2 − A T ⋅ T B = 10 and P Q = 2 , calculate BQ.

Question

Let a circle   ω (not labelled in the graph) centered at P tangent to AB, and T is point of tangency.   ∠  A  P  B  =      90    ∘   . Let K (not labelled in the graph) be some point on the circle   ω , the semicircle with diameter BK intersects PB at Q. Let R be the radius of that semi-circle. If   4      R    2    −  A  T  ⋅  T  B  =  10 and   P  Q  =      2   , calculate BQ.

candelo6a · Accepted Answer

Step 1Since Q be the intersection of semi-circle and BP, we have   K  Q  ⊥  B  P . So, we have   K      P    2    −  B      K    2    =  (  K      P    2    −  Q      K    2    )  −  (  B      K    2    −  Q      K    2    )  =  P      Q    2    −  B      Q    2   . Notice that   ∠  A  P  B  =      90    ∘   and   T  P  ⊥  A  B , we have   K      P    2    =  K      T    2    =  A  T  ×  T  B , and   B      K    2    =  4      R    2   . So we have   B      Q    2    =  P      Q    2    +  (  B      K    2    −  K      P    2    )  =  P      Q    2    +  (  4      R    2    −  A  T  ×  T  B  )  =  2  +  10  =  12 , thus   B  Q  =  2      3   .

fabios3 · Accepted Answer

Step 1Another solution:  4      R    2    =  10  +  A  T  .  T  B  (  I  )    △  A  P  B  :  P      T    2    =  A  T  .  T  B  (  I  I  )    △  P  D  Q  :  P      D    2    =  P      Q    2    +  Q      D    2    =  (      2        )    2    +  Q      D    2    =  2  +  Q      D    2    (  I  I  I  )    △  Q  B  D  :  B      D    2    =  B      Q    2    +  Q      D    2      ⟹    4      R    2    −  B      Q    2    =  Q      D    2    (  I  V  )    (  I  I  )  e  m  (  I  )  :  4      R    2    =  10  +                              P          T                ⏟                    =      P      D        2    =  10  +  P      D    2    =  10  +  2  +  Q      D    2    (  V  )    (  I  I  I  )  e  m  (  V  )  :  4      R    2    =  12  +  4      R    2    −  B      Q    2      ⟹                                B          Q          =                      1                    2          =          2                      3

Let a circle ω (not labelled in the graph) centered at P tangent to AB, and T

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