concurrent wanted, tangents to circumcircle, touchpoint of incircle

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concurrent wanted, tangents to circumcircle, touchpoint of incircle

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Source: 2015 Iberoamerican Shortlist G9 https://artofproblemsolving.com/community/c2533295_iberoamerican_shortlist_2011_geometry

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#1 h Nov 4, 2021, 7:19 PM • 1 Y

Y by centslordm

Let $ABC$ be a triangle and $\Gamma$ its circumcircle. Let $A_1$ be the midpoint of arc $BC$ containing $A$ and $A_2$ be the touchpoint of the incircle of $ABC$ with $BC$ . Line $A_1A_2$ intersects $\Gamma$ at $A_3$ . Similarly, are defined $B_3$ and $C_3$ . Let us take the points $P, Q$ and $R$ so that $QR, PR$ and $PQ$ they are tangent to $\Gamma$ at $A_3$ , $B_3$ and $C_3$ respectively. Prove that $PA$ , $QB$ , and $RC$ concur.

This post has been edited 1 time. Last edited by parmenides51, Nov 4, 2021, 7:20 PM

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#2 h Nov 6, 2021, 2:51 AM • 2 Y

Y by centslordm, parmenides51

Let $XYZ$ be the tangential triangle of $A_1B_1C_1$ . Note that $ABC$ and $XYZ$ are homothetic, thus all $\overline{A_1A_2},\overline{B_1B_2},\overline{C_1C_2},\overline{AX},\overline{BY},\overline{CZ}$ are concurrent. By Pascal's theorem on $B_3B_3B_1C_3C_3C_1$ and $B_1B_1B_3C_1C_1C_3$ , we get that $P$ lies on $\overline{AX}$ . Similarly, $Q$ lies on $\overline{BY}$ and $R$ lies on $\overline{CZ}$ . We conclude desired.

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Eyed

#3 h Nov 6, 2021, 6:53 PM • 2 Y

Y by centslordm, parmenides51

Consider circle $\omega_{A}$ tangent to $(ABC)$ at $A_3$ and tangent to $BC$ . If $\omega_{A}\cap BC = A_2'$ , then by homothety, $A_3, A_2', A_1$ are collinear, so $A_2' = A_2$ . Define $\omega_{B}, \omega_{C}$ similarly.

Now, observe that $AC_2^2 = AB_2^2$ , so the power of $A$ wrt $\omega_B, \omega_C$ is the same. Furthermore, $PB_3^2 = PC_3^2$ , so $P$ also has the same power wrt $\omega_B, \omega_C$ . Thus, $AP$ is the radical axis of $\omega_B, \omega_C$ . We get similar results for $QB, CR$ , so we conclude that $AP, BQ, CR$ intersect on the radical center of the three circles.

Remark: It turns out that $A_1A_2, B_1B_2, C_1C_2, IO$ are concurrent (though I did not use this fact); if $\omega$ is the incircle and $X$ the the insimilecenter of $\omega, (ABC)$ , by monge of $\omega, \omega_{A}, (ABC)$ , we have $X, A_3, A_2$ are collinear. Replacing $A$ with $B,C$ gives all the lines goes through $X$ , and so does $OI$ .

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