# High school problem from IMO selection round

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 $$\triangle A_2B_2C_2$$.

I took $$A$$ to be smallest angle and from my geogebra sketch I found out that the smallest angle of $$\triangle A_2B_2C_2$$ is $$C_2$$ which is equal to 55 degrees. Please help me with this problem. But don't give me the solution as I can find that elsewhere but rather please give me some hints such as which theorems I might need to use. Sequential hints that lead to the answer would be even more helpful.

Here is a hint. For any triangle $$XYZ$$ inscribed in $$\Omega$$, let $$x$$, $$y$$, and $$z$$ denote the measures of the angles at $$X$$, $$Y$$, and $$Z$$, resp, of $$\triangle XYZ$$. Let $$X'$$ be the point on $$\Omega$$ s.t. $$XX'$$ internally bisects the angle $$X$$. The points $$Y'$$ and $$Z'$$ are defined similarly. If $$x'$$, $$y'$$, and $$z'$$ are the angles at $$X'$$, $$Y'$$, and $$Z'$$ of $$\triangle X'Y'Z'$$, then $$x'=\frac{y+z}{2}\wedge y'=\frac{z+x}{2}\wedge z'=\frac{x+y}{2}.$$ Apply the above result with $$\triangle ABC$$, and then $$\triangle A_1B_1C_1$$. (Note that $$55=\frac{180+40}{4}$$.)

• What is the meaning of ∧? Commented Apr 29, 2020 at 17:56
• It means "and". Commented Apr 29, 2020 at 17:57
• Oh I get it now. Commented Apr 29, 2020 at 18:04
• Could you explain how you got $x'=\frac{y+z}{2}$. I noticed that $\frac{y+z}{2}=90-\frac{x}{2}$ Commented Apr 29, 2020 at 18:08
• $x+y+z=180^\circ$, isn't it? Commented Apr 29, 2020 at 18:09

Observe that the full circumference is the sum of the four arcs $$CB, BC_1,\> C_1B_1,\>B_1C$$, which respectively correspond to the angles $$A, \>\frac C2,\> A_1, \> \frac B2$$, i.e.

$$180= A_1+A+\frac{B+C}2\implies A_1=90-\frac A2=70$$

assuming $$A=40$$, $$B,\> C \in(40,100)$$ without loss of generality. Then,

$$B_1=90- \frac B2\in (40,70) \>\>\>\>\> C_1= 90- \frac C2 \in (40,70)$$

and

$$A_2=90-\frac {A_1}2=55,\>\>\>\>\> B_2=90-\frac {B_1}2\in(55,70),\>\>\>\>\> C_2=90-\frac {C_1}2\in(55,70)$$

Thus, the smallest angle of △$$A_2B_2C_2$$ is 55 degrees.

• The exercise is to find the smallest angle of $\triangle A_2 B_2 C_2$, not to find the smallest that angle could be, over all possible triangles $ABC$ with $\angle A=40^\circ$. Commented Apr 29, 2020 at 19:37
• @RosieF - it is over all possible angles with A, B, and C $\ge$ 40. Commented Apr 29, 2020 at 20:05
• With your answer as now edited, the answers to the two questions are the same, so my earlier comment is moot. Commented Apr 30, 2020 at 5:48