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

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

Lines

Algebra

Angles

But try the problem first...

Answer: is 55.

Source

Suggested Reading

PRMO, 2019, Question 10

Geometry Vol I to IV by Hall and Stevens

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.

- https://www.cheenta.com/smallest-perimeter-of-triangle-aime-2015-question-11/
- https://www.youtube.com/watch?v=ST58GTF95t4&t=140s

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