Homothety...

Attachments:
Problem 8.pdf (144kb)

Homothety at the In-similicenter

Inversion also works reducing to an easy incircle configuration. Problem wrote: The circle $\omega_1$ passes through the vertex $A$ of the parallelogram $ABCD$ and touches the rays $CB, CD$. The circle $\omega_2$ touches the rays $AB, AD$ and touches $\omega_1$ externally at point $T$. Prove that $T$ lies on the diagonal $AC$ We will invert about $A$ with arbitary radius.Lines not through the point of inversion invert to circle so $CB \to (AC^*B^*)$ and $CD \to (AD^*C^*)$.Also observe that $\angle{AC^*D^*}=\angle{ADC}=\angle{ABC}=\angle{AC^*B^*}$,$\angle{C^*AB^*}=\angle{CAB}=\angle{ACD}=\angle{AD^*C^*}$ thus $C^*AB^* \sim C^*D^*A \implies (AD^*C^*)$ is tangent to $AB^*$ and $(AC^*B^*)$ is tangent to $AD^*$.The circle through $A$ tangent to $CD,CB$ must become a line tangent to $(AC^*D^*),(AB^*C^*)$ respectively.Thus it is the direct common tangent of $(AC^*D^*),(AB^*C^*)$.Let this tangent touch $(AC^*D^*),(AB^*C^*),AD^*,AB^*$ at $R,S,E,F$,So the image of the circle through $A$ is $EF$.Now the circle in the problem tangent to $AB,AD,\text{The circle through A}$ becomes circle tangent to $AB,AD,EF$ i.e. $AF,AE,EF$ so it must be the incircle(not the excircle because the problem stated it touched externally) of $AEF$.If $AC\cap{EF}=P$ then it suffices to show that $P$ is tangency point of the incircle with $EF$.It suffices to prove that $PE=\dfrac{AE+EF-AF}{2}$.But easy to see that $PR=PS$.Now since $FA,FR$ are tangents to $(AC^*D^*) \implies FA=FR$ and similarly $EA=ES$,therefore $PE=\dfrac{ES+EF-FR}{2}=\dfrac{ES+ER}{2}=\dfrac{EP-PR+EP+PS}{2}=\dfrac{2EP}{2}=EP$ so we are done.$\blacksquare$

My solve

Attachments:
Sharygin_P8_Pranjal.pdf (366kb)

Homothety forever.

Attachments:
H.pdf (156kb)

Inversion.

Attachments:
Sharygin- 8.pdf (33kb)

Its well known that $N-G-M$ are collinear as well as $O-G-K$ are collinear. Now let $DG \cap \omega_1=Q$ so we $(G,Q;O,N)=-1$ now $(G,Q;O,N) \overset{G}{=} (G,D;K,M)=-1 \implies G-D-B$ are collinear $\blacksquare$. Im quite n00b at projective so expecting this to be wrong.Pls check someone.

Attachments:
sharygin parallalegoram.pdf (3kb)

[asy][asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions (otherwise throws error due to the missing xmin) are done using bubu-asy.py. This adds back the dps, real xmin coordinates, changes the linewidth, the fontsize and adds the directions to the labellings. */ pair A = (-48.37454,183.79034); pair B = (-108.13975,-30.26500); pair C = (91.32239,-29.44420); pair D = (151.08760,184.61114); pair M = (21.47392,77.17306); pair T = (75.70674,-5.60834); import graph; size(15cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); draw(A--B, linewidth(0.5)); draw(B--C, linewidth(0.5)); draw(C--D, linewidth(0.5)); draw(D--A, linewidth(0.5)); draw(B--D, linewidth(0.5)); draw(A--C, linewidth(0.5) + red); draw(circle((5.14966,83.60037), 113.39823), linewidth(0.5) + blue); draw(circle((627.63842,-703.02859), 889.59323), linewidth(0.5) + blue); draw((-229.18460,-463.79976)--B, linewidth(0.5)); draw(D--(623.97772,186.55711), linewidth(0.5)); dot("$A$", A, NW); dot("$B$", B, dir(180)); dot("$C$", C, dir(0)); dot("$D$", D, NE); dot("$M$", M, dir(180)); dot("$T$", T, 1.5*N); label("$\omega_1$", circle((5.14966,83.60037), 113.39823), NW, fontsize(14) + blue); label("$\omega_2$", circle((627.63842,-703.02859), 889.59323), 90*dir(55), fontsize(14) + blue); clip((-300,-550)--(-300,250)--(750,250)--(750,-550)--cycle); [/asy][/asy] Let $M$ denote the intersection of the diagonals $AC$ and $BD$. Clearly $T$ is the insimilicenter of $\left\{\odot\omega_1,\odot\omega_2\right\}$. Now consider the homothety $\Psi_1$ centered at $A$ that maps $\odot\omega_2\mapsto\odot\omega_2'$ such that $\odot\omega_2'$ passes through $C$. Now consider the homothety $\Psi_2$ centered at $M$ with scale $=-1$. This maps $AD\mapsto CB$ and $AB\mapsto CD$. Now also let $\odot\omega_2'\mapsto\odot\omega_2^*$. So this gives that the circle $\odot\omega_2^*$ passes through $A$ and is tangent to $AD$ and $AB$. Finally let $\Psi_3$ be the homothety centered at $A$ that maps $\odot\omega_2^*\mapsto\odot\omega_1$. (This might be the identity map itself. We are doing this final homothety as there as not 1, but 2 circles possible for $\odot\omega_1$.) Now consider the composition of the three homotheties, $\Psi_3\circ(\Psi_2\circ\Psi_1)$. Note that the center of $\Psi_2\circ\Psi_1$ lies on $AM$ and that the scale is -ve in value. (We get that the scale of $\Psi_1$ is +ve and that the scale of $\Psi_2=-1$ which is -ve, and so, the product of their scales is going to be negative.) Now this furthermore gives that the center of $\Psi_3\circ(\Psi_2\circ\Psi_1)$ lies on $AM$ (as $\overline{A-M-C}$ are collinear) and that its scale is -ve due to a similar logic. So this homothety $\Psi_3\circ(\Psi_2\circ\Psi_1)$ must be the homothety centered at the insimilicenter as the scale is negative. Now as we already know that $T$ is the insimilicenter, we get that $\overline{A-M-T-C}$ are collinear and we are done.