Let ABC be a triangle and let be its circumcircle. The internal bisectors of angles A, B and C intersect at A_1, ~B_1 and C_1 respectively and the internal bisectors of angles A_1, ~B_1 and C_1 of the triangle A_1 ~B_1

Correct answer: 55 aligned & A_1 ~B_1 C_1=(π)/(2)- ABC 2 & ~A_1 C_1 ~B_1=(π)/(2)- ACB 2 & ~B_1 ~A_1 C_1=(π)/(2)- BAC 2 & ~A_2 ~B_2 C_2=(π)/(2)- ((π)/(2)- ABC 2 ) 2 & =(π)/(4)+ ABC 4 aligned similarly A_2 C_2 ~B_2=(π)/(4)+ ACB 4 and B_2 ~A_2 C_2=(π)/(4)+ BAC 4 smaller angle of Δ A_2 B_2 C_2 is 45^ + ( 40^ 4 )=55^

Olympiad workbook

Exam: Olympiad workbook
Subject: IOQM
Chapter: Properties of Triangles

Olympiad workbook · IOQM · Properties of Triangles

Let ABC be a triangle and let be its circumcircle. The internal...

Let ABC be a triangle and let

\Omega

be its circumcircle. The internal bisectors of angles A, B and C intersect

\Omega

\mathrm{A}_1, \mathrm{~B}_1

and

\mathrm{C}_1

respectively and the internal bisectors of angles

\mathrm{A}_1, \mathrm{~B}_1

and

\mathrm{C}_1

of the triangle

\mathrm{A}_1 \mathrm{~B}_1 \mathrm{C}_1

intersect

\Omega

\mathrm{A}_2, \mathrm{~B}_2

and

\mathrm{C}_2

, respectively. If the smallest angle of triangle ABC is

40^{\circ}

, what is the magnitude of the smallest angle of triangle

\mathrm{A}_2 \mathrm{~B}_2 \mathrm{C}_2

in degrees?

Correct answer

Step-by-step solution

\begin{aligned} & \angle \mathrm{A}_1 \mathrm{~B}_1 \mathrm{C}_1=\frac{\pi}{2}-\frac{\angle \mathrm{ABC}}{2} \\ & \angle \mathrm{~A}_1 \mathrm{C}_1 \mathrm{~B}_1=\frac{\pi}{2}-\frac{\angle \mathrm{ACB}}{2} \\ & \angle \mathrm{~B}_1 \mathrm{~A}_1 \mathrm{C}_1=\frac{\pi}{2}-\frac{\angle \mathrm{BAC}}{2} \\ & \angle \mathrm{~A}_2 \mathrm{~B}_2 \mathrm{C}_2=\frac{\pi}{2}-\frac{\left(\frac{\pi}{2}-\frac{\angle \mathrm{ABC}}{2}\right)}{2} \\ & =\frac{\pi}{4}+\frac{\angle \mathrm{ABC}}{4}\end{aligned}

similarly

\angle \mathrm{A}_2 \mathrm{C}_2 \mathrm{~B}_2=\frac{\pi}{4}+\frac{\angle \mathrm{ACB}}{4}

and

\angle \mathrm{B}_2 \mathrm{~A}_2 \mathrm{C}_2=\frac{\pi}{4}+\frac{\angle \mathrm{BAC}}{4} \Rightarrow

smaller angle of

\Delta A_2 B_2 C_2

is

45^{\circ}+

\left(\frac{40^{\circ}}{4}\right)=55^{\circ}

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