Let ABC be a triangle and Ω be its circumcircle, the internal bisectors of angles A, B, C intersect Ω at A 1 , B 1 , C 1 . The internal bisectors of A 1 , B 1 , C 1 intersect Omega A 2 , B 2 , C 2 . If the smallest angle of △ A B C is 40 degrees, find the smallest angle of △ A 2 B 2 C 2 .

Question

Let ABC be a triangle and   Ω be its circumcircle, the internal bisectors of angles A, B, C intersect   Ω at       A    1    ,      B    1    ,      C    1  . The internal bisectors of       A    1    ,      B    1    ,      C    1   intersect Omega       A    2    ,      B    2    ,      C    2  . If the smallest angle of   △  A  B  C is 40 degrees, find the smallest angle of   △      A    2        B    2        C    2  .

Katelyn Chapman · Accepted Answer

Step 1Observe that the full circumference is the sum of the four arcs   C  B  ,  B      C    1    ,        C    1        B    1    ,        B    1    C, which respectively correspond to the angles   A  ,        C    2    ,        A    1    ,        B    2  , i.e.  180  =      A    1    +  A  +            B      +      C        2      ⟹        A    1    =  90  −      A    2    =  70assuming   A  =  40,   B  ,    C  ∈  (  40  ,  100  ) without loss of generality.Step 2Then,       B    1    =  90  −      B    2    ∈  (  40  ,  70  )                C    1    =  90  −      C    2    ∈  (  40  ,  70  )and      A    2    =  90  −            A      1        2    =  55  ,                B    2    =  90  −            B      1        2    ∈  (  55  ,  70  )  ,                C    2    =  90  −            C      1        2    ∈  (  55  ,  70  )Thus, the smallest angle of   △       A    2        B    2        C    2   is 55 degrees.

pulpenoe · Accepted Answer

Step 1For any triangle XYZ inscribed in   Ω, let x, y, and z denote the measures of the angles at X, Y, and Z, resp, of   △  X  Y  Z. Let X′ be the point on   Ω s.t. XX′ internally bisects the angle X.Step 2The points Y′ and Z′ are defined similarly. If x′, y′, and z′ are the angles at X′, Y′, and Z′ of   △      X    ′        Y    ′        Z    ′  , then       x    ′    =            y      +      z        2    ∧      y    ′    =            z      +      x        2    ∧      z    ′    =            x      +      y        2    ..Apply the above result with   △  A  B  C, and then   △      A    1        B    1        C    1  . (Note that   55  =            180      +      40        4  .)

Let ABC be a triangle and Omega be its circumcircle, the internal bisectors of angles A, B, C intersect Omega at A_1, B_1, C_1. The internal bisectors of A_1, B_1, C_1 intersect Omega A_2, B_2, C_2. If the smallest angle of triangle ABC is 40 degrees, find the smallest angle of triangle A_2 B_2 C_2.

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