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.

Averi Fields

Averi Fields

Answered question

2022-09-26

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 .

Answer & Explanation

Katelyn Chapman

Katelyn Chapman

Beginner2022-09-27Added 13 answers

Step 1
Observe 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 = 70
assuming A = 40, B , C ( 40 , 100 ) without loss of generality.
Step 2
Then, 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

pulpenoe

Beginner2022-09-28Added 1 answers

Step 1
For 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 2
The 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 .)

Do you have a similar question?

Recalculate according to your conditions!

New Questions in High school geometry

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?