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Team selection test for 23. BMO

Geometry Difficulty 8.4 Shortlist Prove it Bulgaria

Problem:

the incircle $k$ of $\triangle ABC$ is tangent to the sides $AB$ , $BC$ and $CA$ at points $C_1$ , $A_1$ and $B_1$ , respectively. The points $C_2$ , $A_2$ and $B_2$ are diametrically opposite to $C_1$ , $A_1$ and $B_1$ in $k$ .

a) Prove that the lines $AA_2$ , $BB_2$ and $CC_2$ are concurrent.

b) If the line $AA_2$ meets $k$ at $A_3$ , find the ratio in which the tangent line to $k$ at $A_3$ divides $BC$ .

Solution

Solution:

a) Let $A_4 = AA_2 \cap BC$ and let the tangent line to $k$ at $A_2$ meet $AB$ and $AC$ at points $X$ and $Y$ , respectively. Since $A_2A_1$ is a diameter, we have $XY \parallel BC$ , i.e. $\triangle AXY \sim \triangle ABC$ .

Since $k$ is an excircle of $\triangle AXY$ , it follows from above that $BA_4 = p - c$ and $CA_4 = p - b$ (we use the standard notation for $\triangle ABC$ ) and therefore $\frac{BA_4}{A_4C} = \frac{p-c}{p-b}$ . The corresponding equalities for the other two vertices and Ceva's theorem imply that the lines $AA_2$ , $BB_2$ and $CC_2$ are concurrent.

b) Denote by $Z$ the intersection point of the tangent line to $k$ at $A_3$ and $BC$ . We have $\angle A_1A_3A_4 = 90^\circ$ and $ZA_3 = ZA_1$ . It follows easily that $ZA_4 = ZA_1$ and therefore
$CZ = CA_1 + A_1Z = p-c + ZA_1 = BA_4 + ZA_4 = BZ$
Hence the desired ratio equals $1:1$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.