Comparing Incircles

Geometry Level 4

A triangle $ABC$ has incentre $I$ . Consider lines $l_a$ passing through $A$ , $l_b$ passing through $B$ and $l_c$ passing through $C$ and respectively parallel to sides $BC$ , $CA$ and $AB$ . By the intersections of these lines, we obtain a triangle $\Delta_1$ .

Now, reflect the line $l_a$ across line $AI$ to obtain line $L_a$ , reflect the line $l_b$ across line $BI$ to obtain line $L_b$ and reflect the line $l_c$ across line $CI$ to obtain line $L_c$ . By the intersections of these new lines, we obtain a triangle $\Delta_2$ .

Then the correct relationship is:

(* Definitions : The reflection of a point $P$ across a line $l$ is another point $P'$ such that $l$ is the perpendicular bisector of segment $PP'$ . The reflection of a line $l_x$ across a line $l$ is another line $L_x$ such that the reflection of each point of $l_x$ across line $l$ lies on $L_x$ .)

Incircle of

\Delta_1

is always congruent to the incircle of

\Delta_2

. Incircle of

\Delta_1

is smaller than or congruent to the incircle of

\Delta_2

. None. It depends on the triangle. Incircle of

\Delta_2

is smaller than or congruent to the incircle of

\Delta_1

1 solution

Aditya Kumar
Jun 24, 2015

Recall the famous Euler inequality for a triangle:

$R \geq 2 \times r$ where $R$ is its circumradius and $r$ the inradius.

The triangle $\Delta_1$ is actually similiar to the triangle $ABC$ and each side (and therefore each linear property) is twice as long as that of the triangle $ABC$ . (Prove it easily using the numerous parallelograms in the figure!) So, the inradius of this triangle is twice the inradius of triangle $ABC$ i.e. $2 \times r$

Inradius of $\Delta_1 = 2\times r$ ...(1)

Suppose in the first case $l_a$ and $l_b$ meet at $C'$ , $l_b$ and $l_c$ meet at $A'$ and $l_c$ and $l_a$ meet at $B'$ . And in the latter case, $L_a$ and $L_b$ meet at $C''$ , $L_b$ and $L_c$ meet at $A''$ and $L_c$ and $L_a$ meet at $B''$ . Note that before:

$\angle B'AC$ = $\angle ACB=x$ , suppose.

After reflection, $\angle B'AC$ now becomes $\angle C''AB$ .

So, $\angle ACB=x=\angle C''AB$

Clearly, this means that line $C''A$ is tangent to the circumcircle of $ABC$ at $A$ , i.e. the line $L_a$ is the tangent line of the circumcircle of $ABC$ at $A$ . Similiarly, the line $L_b$ is the tangent line of the circumcircle of $ABC$ at $B$ and the line $L_c$ is the tangent line of the circumcircle of $ABC$ at $C$ .

So, the circumcircle of $ABC$ is actually the incircle of the triangle formed by these lines, i.e. $\Delta_2$ . So, the inradius of the triangle $\Delta_2$ is actually the circumradius of $ABC$ i.e. $R$

Inradius of $\Delta_2 = R$ ...(2)

Using the Euler's inequality and the results in (1) and (2), the relation follows.

Comparing Incircles

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