Triangles and Internal bisectors | PRMO 2019 | Question 10

Content

Try this beautiful problem from the PRMO, 2019 based on triangles and internal bisectors.

Triangles and internal bisectors - PRMO 2019

Let ABC be a triangle and let D be its circumcircle, The internal bisectors of angles A,B and C intersect D at \(A_1,B_1 and C_1\) the internal bisectors of \(A_1,B_1,C_1\) of the triangle \(A_1B_1C_1\) intersect D at \(A_2,B_2,C_2\). If the smallest angle of triangle ABC is 40 find the magnitude of the smallest angle of triangle \(A_2B_2C_2\) in degrees.

is 107
is 55
is 840
cannot be determined from the given information

Key Concepts

Lines

Algebra

Angles

Check the Answer

But try the problem first...

Answer: is 55.

Source

Try with Hints

First hint

angle \(A_1B_1C_1=90 - \frac{ABC}{2}\) angle \(A_1C_1B_1=90-\frac{ACB}{2}\)

Second Hint

angle \(B_1A_1C_1\)=90-\(\frac{BAC}{2}\)

Final Step

then angle \(A_2B_2C_2=90-\frac{90-\frac{ABC}{2}}{2}\)=45+\(\frac{ABC}{4}\)=55.

Triangles and internal bisectors - PRMO 2019

Let ABC be a triangle and let D be its circumcircle, The internal bisectors of angles A,B and C intersect D at \(A_1,B_1 and C_1\) the internal bisectors of \(A_1,B_1,C_1\) of the triangle \(A_1B_1C_1\) intersect D at \(A_2,B_2,C_2\). If the smallest angle of triangle ABC is 40 find the magnitude of the smallest angle of triangle \(A_2B_2C_2\) in degrees.