Israel 2018 Q6 - Equal circles inside a triangle

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Israel 2018 Q6 - Equal circles inside a triangle

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Source: Israel National Olympiad 2018 Q6

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Cuubic

#1 h Aug 7, 2019, 9:45 PM • 2 Y

Y by Adventure10, Mango247

In the corners of triangle $ABC$ there are three circles with the same radius. Each of them is tangent to two of the triangle's sides. The vertices of triangle $MNK$ lie on different sides of triangle $ABC$ , and each edge of $MNK$ is also tangent to one of the three circles. Likewise, the vertices of triangle $PQR$ lie on different sides of triangle $ABC$ , and each edge of $PQR$ is also tangent to one of the three circles (see picture below). Prove that triangles $MNK,PQR$ have the same inradius.

https://i.imgur.com/bYuBabS.png

This post has been edited 1 time. Last edited by Cuubic, Aug 7, 2019, 9:59 PM
Reason: Picture modified

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tk1

#2 h Aug 8, 2019, 12:15 AM • 2 Y

Y by Adventure10, Mango247

Let the radius of the three circles be $r$ , and the inradii of $\triangle PQR, \triangle MNK$ be $\rho_1, \rho_2$ , respectively. Taking into account that the area of a triangle is equal to its semi-perimeter times its inradius, we have: $(ABC)=(AQR)+(BRP)+(PCQ)+(QPR)=\frac{1}{2}\cdot r\cdot(a+b+c)+\frac{1}{2}\cdot(r+\rho_1)\cdot(QP+PR+RQ).$ Similarly we find that $(ABC)=(MAN)+(NBK)+(CMK)+(MNK)=\frac{1}{2}\cdot r\cdot(a+b+c)+\frac{1}{2}\cdot(r+\rho_2)\cdot(MN+NK+KM).$ However, $QR+RP+PQ=(AQ+AR-2r/\tan\frac{A}{2})+(BR+BP-2r/\tan\frac{B}{2})+(CP+CQ-2r/\tan\frac{C}{2})=a+b+c-2r(1/\tan\frac{A}{2}+1/\tan\frac{B}{2}+1/\tan\frac{C}{2})=MN+NK+KM.$

Therefore $\rho_1=\rho_2.$

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#3 h Aug 22, 2019, 3:28 AM • 1 Y

Y by Adventure10

First we'll define some more points. Let $I_a, I_b, I_c, I_1, I_2$ be the incenters of $\triangle ARQ, \triangle BRP, \triangle CPQ, \triangle MNK, \triangle PQR,$ respectively. Let $\omega_a, \omega_b, \omega_c$ be the incircles of $\triangle ARQ, \triangle BRP, \triangle CPQ,$ respectively. Let $A_1 = I_aI_1 \cap MN, B_1 = I_bI_1 \cap NK, C_1 = IcI_1 \cap KM.$ Define $A_2, B_2, C_2$ analogously. Let $r_1, r_2$ be the inradii of $\triangle MNK, \triangle PQR$ respectively. Let $r$ be the radius of the three congruent circles.

Now, observe that $\triangle I_aI_bI_c \sim \triangle ABC$ , and the two triangles are homothetic. By Monge's Theorem, it's easy to see that $A_1B_1 || I_aI_b$ , and so $\triangle A_1B_1C_1$ is homothetic to $\triangle I_aI_bI_c.$ Furthermore, it can be seen that $\triangle A_1B_1C_1$ has an area which is $\frac{r_1^2}{(r+r_1)^2}$ times the area of $\triangle I_aI_bI_c.$ Analogously, $\triangle A_2B_2C_2$ is also homothetic to $\triangle I_aI_bI_c$ and has an area which is $\frac{r_2^2}{(r+r_2)^2}$ times the area of $\triangle I_aI_bI_c.$ We hence have:

$\frac{[\triangle A_1B_1C_1]}{[\triangle A_2B_2C_2]} = \frac{\frac{r_1^2}{(r+r_1)^2}}{\frac{r_2^2}{(r+r_2)^2}}, \qquad (1)$
where $[\triangle]$ denotes the area of $\triangle.$

Now, we will prove the following important lemma.

Lemma. If $X, Y, Z$ lie on sides $BC, CA, AB$ respectively so that the area of $\triangle XYZ$ is $k < 1$ times that of $\triangle ABC$ , then the unique triangle $X'Y'Z'$ so that $X' \in YZ, Y' \in XZ, Z' \in XY,$ and $\triangle X'Y'Z'$ is homothetic to $\triangle ABC$ has an area which is $k^2$ times that of $\triangle ABC.$

Proof. Let $t < 1$ be a constant such that $\frac{Y'Z'}{BC} = \frac{X'Z'}{AC} = \frac{X'Y'}{AB}.$ Now, if we define constants $\alpha, \beta, \gamma$ such that $\frac{[\triangle Y'BC]}{[\triangle ABC]} = \alpha, \frac{[\triangle Z'CA]}{[\triangle ABC]} = \beta, \frac{[\triangle X'AB]}{[\triangle ABC]} = \gamma,$ then we have $\alpha + \beta + \gamma = 1-t.$ Then, it's clear that $[\triangle Y'Z'X] = t \alpha, [\triangle Z'X'Y] = t \beta, [\triangle X'Y'Z] = t \gamma \Rightarrow [\triangle XYZ] = t (\alpha + \beta + \gamma) + [\triangle X'Y'Z'].$ Since $[\triangle X'Y'Z'] = t^2,$ we've that $[\triangle XYZ] = t(1-t) + t^2 = t$ . In other words, we have $t = k$ , as desired.
$\blacksquare$

Now, let $c_1, c_2<1$ be constants such that $[\triangle MNK]$ has an area which is $c_1$ times that of $[\triangle ABC],$ and $[\triangle PQR]$ has an area which is $c_2$ times that of $[\triangle ABC].$ Since $\frac{CP + CQ - PQ}{2}$ is the length of the tangent from $C$ to $\omega_c$ , adding this with two similar equations gives us that $\frac{AB + BC + CA - PQ - QR - RP}{2}$ is the sum of the lengths from $A, B, C$ to their corresponding circles. Similarly, so is $\frac{AB + BC + CA - MN - NK - KM}{2}$ , and so hence $\triangle PQR, \triangle MNK$ have equal perimeters. Since the area of a triangle is radius times semiperimeter, we have that $\frac{c_1}{c_2} = \frac{r_1}{r_2}.$ From the lemma, it then follows that

$\frac{[\triangle A_1B_1C_1]}{[\triangle A_2B_2C_2]} = \frac{r_1^2}{r_2^2}. \qquad (2)$
Comparing with $(1)$ yields that $(r+r_1)^2 = (r+r_2)^2 \Rightarrow r_1 = r_2,$ as desired.

$\square$

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