Let $\Gamma_1$ be a circle with centre $A$ and $\Gamma_2$ be a circle with centre $B$, with $A$ lying on $\Gamma_2$. On $\Gamma_2$ there is a (variable) point $P$ not lying on $AB$. A line through $P$ is a tangent of $\Gamma_1$ at $S$, and it intersects $\Gamma_2$ again in $Q$, with $P$ and $Q$ lying on the same side of $AB$. A different line through $Q$ is tangent to $\Gamma_1$ at $T$. Moreover, let $M$ be the foot of the perpendicular to $AB$ through $P$. Let $N$ be the intersection of $AQ$ and $MT$. Show that $N$ lies on a line independent of the position of $P$ on $\Gamma_2$.
Problem
Source:
Tags: geometry, circles, fixed
05.08.2019 02:31
@ admins everytime you move my IMO TST problems from High School Olympiads to High School Math, we lose the source as I write it in the source line I suggest adding a source line in this folder because I am editing all my problems to add the source at the end if i hadn't kept a backup of the sources, anyone might post the same problems again and again I though so far that writing the source helps others, that it is better than posting random source-less problems if it is to be deleted with the transfering of problems in another folder, then I shouldn't waste time rewriting the sources with edits, and keep the source to myself I do not mind having these IMO TST problems transfered in another folder (although they are considered IMO TST level) but I do care about losing the sources in every transfer edit: in order to avoid losing source -line during transfers, I shall place the source twice, and in the source line and after each problem, so problem fixed
17.03.2020 21:45
Bump////
18.03.2020 05:07
wait this problem is kind of hard for hsm imo. even though the solution is fairly simple, it took a really really long time for me to get because the configuration is so ripe; specifically, it's pretty difficult to prove that $A$ is the Miquel Point of $TQPM$ with the given conditions, which was a rabbit hole I went down for a while unfortunately. Let $F$ be the intersection of $QT$ with $\Gamma_2,$ and let $G$ be the antipode of $B$ with respect to $\Gamma_2.$ First, observe that since $\measuredangle ASQ=\measuredangle AMP=90^\circ,$ we know that $AMPS$ is cyclic, meaning that $\measuredangle AMS=\measuredangle APQ=\measuredangle AGQ,$ so $MS||GQ.$ However, we also know that $TS||GQ$ since $GQ\perp AQ$ and $TS$ is $Q$'s polar with respect to $\Gamma_1.$ Therefore, $M,T,N,S$ are collinear, so $N$ is the radical center of $(ASQT),\Gamma_1,\Gamma_2,$ implying that $N$ lies on the radical axis of $\Gamma_1$ and $\Gamma_2,$ which is obviously independent of $P.$ $\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.390543878616837, xmax = 10.343673929376408, ymin = -7.071364277729919, ymax = 7.060683879716323; /* image dimensions */ /* draw figures */ draw(circle((5,0), 5), linewidth(1)); draw(circle((0,0), 3.5874214915563782), linewidth(1)); draw((1.286959295808059,ymin)--(1.286959295808059,ymax), linewidth(1)); /* line */ draw((0,0)--(2.7261506972974385,-2.331886646838838), linewidth(1)); draw((0,0)--(8.085924042726855,-3.9340911024683134), linewidth(1)); draw(circle((0.7958022416532934,-1.829128207766219), 1.994745900676693), linewidth(1)); draw(circle((4.042962021363427,-1.9670455512341565), 4.496088311723551), linewidth(1)); draw(circle((4.04296202136343,1.967045551234153), 4.496088311723549), linewidth(1)); draw((8.085924042726855,-3.9340911024683134)--(8.085924042726852,3.934091102468313), linewidth(1)); draw((8.085924042726855,-3.9340911024683134)--(-0.1522321056813205,-3.5841900541238636), linewidth(1)); draw((1.5916044833065879,-3.658256415532439)--(8.085924042726852,3.934091102468313), linewidth(1)); draw((1.5916044833065879,-3.658256415532439)--(0,0), linewidth(1)); draw((-0.1522321056813205,-3.5841900541238636)--(8.085924042726855,0), linewidth(1)); draw((1.5916044833065879,-3.658256415532439)--(10,0), linewidth(1)); draw((10,0)--(0,0), linewidth(1)); draw((0,0)--(-0.1522321056813205,-3.5841900541238636), linewidth(1)); draw((0,0)--(8.085924042726852,3.934091102468313), linewidth(1)); draw((8.085924042726852,3.934091102468313)--(10,0), linewidth(1)); draw((2.726150697297438,2.3318866468388384)--(8.085924042726855,-3.9340911024683134), linewidth(1)); /* dots and labels */ dot((0,0),dotstyle); label("$A$", (0.06189422953583984,0.14125781092521703), NW * labelscalefactor); dot((5,0),dotstyle); label("$B$", (5.060886368217554,0.14125781092521703), NE * labelscalefactor); dot((8.085924042726855,-3.9340911024683134),dotstyle); label("$P$", (8.139444576789872,-3.787568855252784), SE * labelscalefactor); dot((8.085924042726855,0),linewidth(4pt) + dotstyle); label("$M$", (8.139444576789872,0.11193820893881402), NE * labelscalefactor); dot((-0.1522321056813205,-3.5841900541238636),linewidth(4pt) + dotstyle); label("$S$", (-0.09936358138937677,-3.465053233402351), NE * labelscalefactor); dot((1.5916044833065879,-3.658256415532439),linewidth(4pt) + dotstyle); label("$Q$", (1.6451527368016028,-3.5383522383683585), NE * labelscalefactor); dot((2.7261506972974385,-2.331886646838838),linewidth(4pt) + dotstyle); label("$T$", (2.7886172142713206,-2.218970148980224), SE * labelscalefactor); dot((1.286959295808059,-2.9580383504813503),linewidth(4pt) + dotstyle); label("$N$", (1.3519567169375726,-2.8346817906946864), NE * labelscalefactor); dot((8.085924042726852,3.934091102468313),linewidth(4pt) + dotstyle); label("$F$", (8.139444576789872,4.055424676110016), NE * labelscalefactor); dot((10,0),linewidth(4pt) + dotstyle); label("$G$", (10.059878506899269,0.11193820893881402), NE * labelscalefactor); dot((2.726150697297438,2.3318866468388384),linewidth(4pt) + dotstyle); dot((4.7207929373966095,0),linewidth(4pt) + dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]
18.03.2020 06:17
We‘ll prove that $N$ lies on the radical axis of the two circles. Let $AQ$ intersects the radical axis at $N'$, then from $R_{\Gamma_1}^2=AN'\times AQ$ we have $S, N', T$ is collinear. To prove that $N=N'$, it suffices to prove that $M$ lies on the polar reciprocal line of $Q$ wrt $\Gamma_1$, which is equivalent to prove that $L$ lies on the polar reciprocal line of $S$ wrt $\Gamma_1$, where $L$ is the projection of $Q$ on $AB$. Note that $A, L, Q, S$ is concyclic, so $$\angle LSQ+\angle SPA=\angle LAQ+\angle SPA=\angle A'AQ+\angle AA'Q=\pi/2,$$where $A'$ is the antipodal point of $A$ wrt $\Gamma_2$. This implies that $AP\perp SL$, then as $PS$ is tangent to $\Gamma_1$, $SL$ is the corresponding polar reciprocal line. $\square$