I found a trigonometric solution too... But I decided to do it synthetically as I liked the problem. P22 Sharygin 2019 wrote: Let $AA_0$ be the altitude of the isosceles triangle $ABC (AB = AC).$ A circle $\gamma$ centered at the midpoint of $AA_0$ touches $AB$ and $AC.$ Let $X$ be an arbitrary point of line $BC.$ Prove that the tangents from $X$ to $\gamma$ cut congruent segments on lines $AB$ and $AC.$ Solution . Define the Newton line (also called Gauss line) of a quadrilateral as the line connecting the midpoints of its diagonals. Denote the Newton line of quadrilateral $WXYZ$ as $\ell_{WXYZ}.$ The result is trivial when $X\in\{B,C\}$, so let us assume $X\notin\{B,C\}.$ Claim 1. The incenter of a circumscribable quadrilateral lies on its Newton line. Proof. Well known. $\square$ Claim 2. Let $ABCD$ be a quadrilateral. Then the line connecting the midpoints of $AB$ and $CD$, line connecting the midpoints of $AD$ and $BC$ and the Newton line are concurrent. The concurrency point is called the Center of gravity of $ABCD.$ Moreover, if the concurrency point is $I,$ then $I$ is the midpoints of the line connecting the midpoints of $AB$ and $CD$ and the line connecting the midpoints of $AD$ and $BC$. Proof. We use complex numbers. Assign the complex numbers $a,b,c,d$ to $A,B,C,D$ respectively. Then the midpoint of $AB$ is $\tfrac{a+b}2,$ midpoint of $CD$ is $\tfrac{c+d}2,$ so the midpoint of the line connecting the midpoints of $AB$ and $CD$ is $\tfrac{\tfrac{a+b}2+\tfrac{c+d}2}{2}=\tfrac{a+b+c+d}4.$ Similarly we can do this for the second pair of opposite sides and for both the diagonals, so we conclude that the concurrency point is $\tfrac{a+b+c+d}4.$ Hence the claim. $\square$ Claim 3. Let $ABC$ be a triangle with incenter $I.$ Let its incircle be $\Gamma.$ Choose points $X$ and $Y$ on $\overline{AB}$ and $\overline{AC}$ respectively such that $\angle AIX=\angle AIY=90^{\circ}.$ Let $B'$ and $C'$ be the reflections of $B$ and $C$ over $X$ and $Y$ respectively. Then $B'C'$ is tangent to the $\Gamma.$ Proof. Assume the contrary that $B'C'$ is not tangent to $\Gamma.$ Let $B''\ne A$ and $C''\ne A$ be points on $AB$ and $AC$ such that $B'C''$ and $B''C'$ are tangent to $\Gamma.$ It is obvious that $I$ is the midpoint of $XY.$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(7cm); 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 = -6.810483357644759, xmax = 0.12944837703120488, ymin = -0.9860645749176019, ymax = 3.8406699533518407; /* image dimensions */ pen qqffff = rgb(0,1,1); /* draw figures */ draw((-4.206610164672261,3.4233675356969444)--(-6.009131136788336,-0.45866639901543504), linewidth(0.4)); draw((-6.009131136788336,-0.45866639901543504)--(-1.8334236644650364,-0.3961043400574436), linewidth(0.4)); draw((-1.8334236644650364,-0.3961043400574436)--(-4.206610164672261,3.4233675356969444), linewidth(0.4)); draw(circle((-4.048185727261299,0.8136140026331854), 1.2427612869874087), linewidth(0.8) + qqffff); draw((-5.458122404026126,0.7280241482765858)--(-2.638249050496473,0.899203856989785), linewidth(0.4)); draw((-4.954509025418634,1.8126407801897786)--(-3.3264285163414167,2.0067788895617813), linewidth(0.4) + linetype("4 4")); draw((-4.81371809081757,2.115857882390577)--(-3.3264285163414167,2.0067788895617813), linewidth(0.4) + linetype("4 4")); draw((-4.954509025418634,1.8126407801897786)--(-3.518817031589914,2.316414277739498), linewidth(0.4) + linetype("4 4")); /* dots and labels */ dot((-4.206610164672261,3.4233675356969444),dotstyle); label("$A$", (-4.180993651945735,3.4804658840780016), NE * labelscalefactor); dot((-6.009131136788336,-0.45866639901543504),dotstyle); label("$B$", (-5.988017399469493,-0.3977312617703316), NE * labelscalefactor); dot((-1.8334236644650364,-0.3961043400574436),dotstyle); label("$C$", (-1.8096501958929614,-0.33769725022469177), NE * labelscalefactor); dot((-4.048185727261299,0.8136140026331854),linewidth(4pt) + dotstyle); label("$I$", (-4.024905221927071,0.8629829806881049), NE * labelscalefactor); dot((-5.458122404026126,0.7280241482765858),linewidth(4pt) + dotstyle); label("$X$", (-5.435704493249607,0.7789353645242092), NE * labelscalefactor); dot((-2.638249050496473,0.899203856989785),linewidth(4pt) + dotstyle); label("$Y$", (-2.614105950604535,0.9470305968520006), NE * labelscalefactor); dot((-4.954509025418634,1.8126407801897786),dotstyle); label("$B'$", (-4.931418796266232,1.8715543746548542), NE * labelscalefactor); dot((-3.3264285163414167,2.0067788895617813),dotstyle); label("$C'$", (-3.304497083379393,2.0696666127554657), NE * labelscalefactor); dot((-3.518817031589914,2.316414277739498),linewidth(4pt) + dotstyle); label("$C''$", (-3.4966059203254405,2.363833269329101), NE * labelscalefactor); dot((-4.81371809081757,2.115857882390577),linewidth(4pt) + dotstyle); label("$B''$", (-4.7873371685566966,2.165721031228489), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Consider quadrilateral $BB'C'C,$ since $X$ is the midpoint of $BB'$, $Y$ is the midpoint of $CC'$ and $I$ is the midpoint of $XY,$ by Claim 2, $I$ is the center of gravity of quadrilateral $BB'C'C.$ So, $I$ lies on $\ell_{BB'C'C}.$ Consider quadrilateral $BB'C''C,$ note that it is a circumscribable quadrilateral, therefore, by Claim 1, $I$ lies on $\ell_{BB'C''C}.$ Note that $\ell_{BB'C'C}$ and $\ell_{BB'C''C}$ both go through the midpoint of $CB'.$ Hence we conclude that $\ell_{BB'C'C}$ and $\ell_{BB'C''C}$ are the same. Now, $\ell_{BB'C'C}$ passes through the midpoint of $BC'$ and $\ell_{BB'C''C}$ passes through the midpoint of $BC''.$ Hence, it follows that $\ell_{BB'C'C}$ is actually the $B-$midline of $\triangle BC'C'',$ in particular $\ell_{BB'C'C}\parallel AC.$ Analogously we obtain $\ell_{BB'C'C}\parallel AB,$ which would mean $AB\parallel AC.$ Contradiction! Thus, our assumption was wrong. $\square$ Coming back to the original problem, Let the two tangents intersect the sides $AB$ and $AC$ at points $P,S$ and $Q,R$ as shown in the diagram below. Let $K$ and $L$ be the midpoints of $AB$ and $AC$ respectively. Let $Q'$ and $P'$ be the reflections of $S$ and $R$ over $K$ and $L$ respectively. [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 = -9.772502611277053, xmax = 8.868900331677994, ymin = -5.422596250755891, ymax = 7.542531747631521; /* image dimensions */ pen qqffff = rgb(0,1,1); /* draw figures */ draw((-3.98,5.42)--(-8.24,-3.4), linewidth(0.4)); draw((-3.98,5.42)--(0.5535526989176085,-3.26256298140917), linewidth(0.4)); draw(circle((-3.9116118252705987,1.0443592546477072), 1.964636688110759), linewidth(0.8) + qqffff); draw((-3.98,5.42)--(-3.8432236505411974,-3.3312814907045856), linewidth(0.4)); draw((7.288710162084458,-3.157297226694199)--(-4.760627369085779,3.803771503442121), linewidth(0.8) + red); draw((-6.78764528316683,-0.39301206514822334)--(7.288710162084458,-3.157297226694199), linewidth(0.8) + red); draw((-5.43235471683317,2.4130120651482234)--(-3.149245230993478,3.8289562682866176), linewidth(0.4)); draw((-8.24,-3.4)--(7.288710162084458,-3.157297226694199), linewidth(0.4)); /* dots and labels */ dot((-3.98,5.42),dotstyle); label("$A$", (-3.9188440746916884,5.575186454343878), NE * labelscalefactor); dot((-8.24,-3.4),dotstyle); label("$B$", (-8.176050283117409,-3.2456158032654683), NE * labelscalefactor); dot((0.5535526989176085,-3.26256298140917),dotstyle); label("$C$", (0.6125004123068996,-3.1004837734327735), NE * labelscalefactor); dot((-3.8432236505411974,-3.3312814907045856),linewidth(4pt) + dotstyle); label("$A_0$", (-3.7737120448589936,-3.1972384599879033), NE * labelscalefactor); dot((-3.9116118252705987,1.0443592546477072),linewidth(4pt) + dotstyle); label("$M$", (-3.854340950321602,1.1728482160854654), NE * labelscalefactor); dot((7.288710162084458,-3.157297226694199),dotstyle); label("$X$", (7.353076908980956,-3.003729086877643), NE * labelscalefactor); dot((-6.78764528316683,-0.39301206514822334),linewidth(4pt) + dotstyle); label("$S$", (-6.724729984790459,-0.26234630114896185), NE * labelscalefactor); dot((-0.27720207008891307,-1.671519249695787),linewidth(4pt) + dotstyle); label("$R$", (-0.20991442341170538,-1.5362830074581726), NE * labelscalefactor); dot((-2.4343834163650837,2.459868239377773),linewidth(4pt) + dotstyle); label("$Q$", (-2.3707690898096088,2.591916952227371), NE * labelscalefactor); dot((-4.760627369085779,3.803771503442121),linewidth(4pt) + dotstyle); label("$P$", (-4.692881567132728,3.9303567829066686), NE * labelscalefactor); dot((-6.11,1.01),linewidth(4pt) + dotstyle); label("$K$", (-6.047447178904549,1.1405966539004222), NE * labelscalefactor); dot((-1.7132236505411957,1.078718509295415),linewidth(4pt) + dotstyle); label("$L$", (-1.6451089406461337,1.2050997782705088), NE * labelscalefactor); dot((-5.43235471683317,2.4130120651482234),linewidth(4pt) + dotstyle); label("$Q'$", (-5.370164373018638,2.543539608949806), NE * labelscalefactor); dot((-3.149245230993478,3.8289562682866176),linewidth(4pt) + dotstyle); label("$P'$", (-3.080303457880562,3.962608345091712), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Using Claim 3 on $\triangle ASR,$ we obtain that $P'Q'$ is tangent to $\gamma.$ Since $K$ and $L$ are midpoints and because of reflections, it follows that $AP'=CR$ and $AQ'=BS.$ Observe that if intersection of $PQ$ with $BC$ is in the exterior(interior) of segment $\overline{BC},$ then it must be true that intersection of $P'Q'$ with $BC$ must also be in the exterior(interior) of segment $\overline{BC}$ and vice-versa. Now, let $P''$ and $Q''$ denote the reflection of $P'$ and $Q'$ over $AA_0.$ Note that $AP''=CR,~AQ''=BS,~ AS=CQ''$ and $AR=BP''.$ Let $P''Q''\cap BC=X'.$ Then, by Menelaus theorem, $\tfrac{AS}{BS}\cdot\tfrac{BX}{XC}\cdot\tfrac{CR}{AR}=1$ and $\tfrac{AP''}{BP''}\cdot\tfrac{BX'}{X'C}\cdot\tfrac{CQ''}{AQ''}=1,$ comparing both we conclude that $\tfrac{BX}{XC}=\tfrac{BX'}{X'C},$ which means either $X=X'$ or $(B,C;X,X')$ is a harmonic bundle, for now, suppose $(B,C;X,X')$ is a harmonic bundle, then one of $X$ and $X'$ must be in the interior of segment $\overline{BC}$ and the other one must be in the exterior of segment $\overline{BC},$ but as $P'Q'$ and $P''Q''$ are just reflections over $AA_0,$ so one of $P'Q'\cap BC$ and $X$ must be in the interior and the other one in the exterior of segment $\overline{BC},$ which is clearly impossible. Hence, it follows that $X=X'.$ That means $X,P'',Q''$ are collinear, but as $P''Q''$ is tangent to $\gamma,$ it is evident that $P=P''$ and $Q=Q''.$ Because of definition of $P''$ and $Q''$, we have that $\overline{XPQ}$ and $\overline{XSR}$ cut equal segments on $AB$ and $AC.$ $~~\blacksquare$

Trig was pretty easy And I had tried this only after Vrangr told me this was very easy.

Attachments:
Problem 22.pdf (116kb)

Trig, trig for the win win. I'd be interested in seeing a proof of this using projective maps. Since, if $Y$ and $Z$ are two points on $AB$ and $AC$ such that $AY = CZ$ then $Y \mapsto Z$ is a projective map. However, I don't think the maps from using the tangents definition is projective.

Attachments:
22-trig.pdf (61kb)

Suppose that the tangents from $X$ to $\gamma$ intersect $AB$ and $AC$ at $\{V_1, V_2\}$ and $\{U_1, U_2\}$. Let the parallel from $X$ to $AB$ intersect $AC$ at $S$ and let $P$ be the point at infinity of line $AC$. [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 = -5.7622631664563615, xmax = 9.07668953519118, ymin = -6.147353275886057, ymax = 9.713838389652816; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen qqffff = rgb(0,1,1); pen ffqqtt = rgb(1,0,0.2); /* draw figures */ draw((-1.8345846898952407,6.684142416210509)--(-1.7113262728804806,0.9650150170289811), linewidth(2) + wrwrwr); draw(circle((-1.7729554813878607,3.824578716619745), 0.9811924706628136), linewidth(2) + wrwrwr); draw((-1.7113262728804806,0.9650150170289811)--(-0.7286186178281011,3.8470862251342357), linewidth(2) + wrwrwr); draw((-1.7113262728804806,0.9650150170289811)--(-2.8172923449476204,3.8020712081052546), linewidth(2) + wrwrwr); draw((-3.027305757008398,3.1861468229593477)--(3.7579230690758028,1.0828880716220681), linewidth(2) + qqffff); draw((3.7579230690758028,1.0828880716220681)--(-2.27615334693519,5.3891160609017055), linewidth(2) + qqffff); draw((-3.8,0.92)--(3.7579230690758028,1.0828880716220681), linewidth(2) + wrwrwr); draw((2.167383823022753,-3.5818230892639207)--(3.7579230690758028,1.0828880716220681), linewidth(2) + wrwrwr); draw((0.3773474542390385,1.0100300340579622)--(-0.11960598930405232,2.284829396614728), linewidth(2) + ffqqtt); draw((-2.27615334693519,5.3891160609017055)--(-1.8345846898952407,6.684142416210509), linewidth(2) + ffqqtt); draw((-3.8,0.92)--(-2.27615334693519,5.3891160609017055), linewidth(2) + wrwrwr); draw((-1.8345846898952407,6.684142416210509)--(-0.11960598930405232,2.284829396614728), linewidth(2) + wrwrwr); draw((2.167383823022753,-3.5818230892639207)--(0.3773474542390385,1.0100300340579622), linewidth(2) + wrwrwr); /* dots and labels */ dot((-1.8345846898952407,6.684142416210509),dotstyle); label("$A$", (-1.772233662235578,6.8449742006676155), NE * labelscalefactor); dot((-3.8,0.92),dotstyle); label("$B$", (-3.734272963897864,1.09075809747316), NE * labelscalefactor); dot((0.3773474542390385,1.0100300340579622),dotstyle); label("$C$", (0.4371215177875001,1.1731967235934244), NE * labelscalefactor); dot((-1.7113262728804806,0.9650150170289811),linewidth(4pt) + dotstyle); label("$A_0$", (-1.6403318604431556,1.09075809747316), NE * labelscalefactor); dot((-0.7286186178281011,3.8470862251342357),linewidth(4pt) + dotstyle); label("$N$", (-0.667556072224039,3.9761100116824144), NE * labelscalefactor); dot((-2.8172923449476204,3.8020712081052546),linewidth(4pt) + dotstyle); label("$M$", (-3.0912516801598042,3.9431345612343085), NE * labelscalefactor); dot((3.7579230690758028,1.0828880716220681),dotstyle); label("$X$", (3.8171051887183287,1.255635349713689), NE * labelscalefactor); dot((-0.11960598930405232,2.284829396614728),linewidth(4pt) + dotstyle); label("$U_2$", (-0.0575102389340846,2.4097761153973907), NE * labelscalefactor); dot((-2.27615334693519,5.3891160609017055),linewidth(4pt) + dotstyle); label("$V_1$", (-2.2174022432850045,5.525956182743386), NE * labelscalefactor); dot((-3.027305757008398,3.1861468229593477),linewidth(4pt) + dotstyle); label("$V_2$", (-3.4045184594168076,3.2341623766000347), NE * labelscalefactor); dot((-0.9649734556317169,4.453390498671544),linewidth(4pt) + dotstyle); label("$U_1$", (-0.8983842253607786,4.586155844972371), NE * labelscalefactor); dot((2.167383823022753,-3.5818230892639207),linewidth(4pt) + dotstyle); label("$S$", (2.234283567209258,-3.4433663391413827), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] By the dDIT we have that $\{XU_1, XU_2\}, \{XA, XA_0\} , \{XS, XP\}$ are reciprocal pairs of an involution, thus $\{U_1, U_2\}, \{A, C\} , \{S, P\}$ are reciprocal pairs of an involution on $AC \implies$ $$(U_1C, U_2S)=(U_2A,U_1P) \implies \frac{U_1U_2}{U_1S} \div \frac{CU_2}{CS}=\frac{U_1U_2}{U_1A} \implies $$$$\frac{CS}{CU_2}=\frac{U_1S}{U_1A}=\frac{SX}{AV_1}=\frac{CS}{AV_1} \implies AV_1=CU_2$$ Analogously, $BV_2=AU_1$, hence $U_1V_1=U_2V_2$ trivially follows.

I have a non trig solution for this problem

Attachments:
P22.pdf (403kb)

Vrangr wrote: Trig, trig for the win win. I'd be interested in seeing a proof of this using projective maps. Since, if $Y$ and $Z$ are two points on $AB$ and $AC$ such that $AY = CZ$ then $Y \mapsto Z$ is a projective map. However, I don't think the maps from using the tangents definition is projective. Can anybody tell me how can I make pdfs for solution and diagrams since I wrote all solutions and probably that's going to mess....

Plz anyone...

Let $D,E$ be the intersections of the tangents from $X$ and $AB,AC$ respectively. Let $X,Y$ be the points of tangency from $A$, $K,L$ from $X$, and $X',Y'$ from $A_0$. If $m$ is the line through $M$ parallel to $BC$, then clearly $X,X'$ are reflections about $m$, as are $Y,Y'$. Let $t$ be the line through $A,A_0$. Since $X$ lies on $BC$, then $X'Y' \cap KL$ lies on $t$, as does the point at infinity $KK' \cap YY'$. So by Pascal applied to $K'X'Y'YLK$, we have that $X'K' \cap LY$ also lies on $t$. But the pole of this point is $D'E$, where $D'$ is the reflection of $D$ in $m$. So $D'E$ is parallel to $BC$, which means that $D,E$ are equidistant from $m$ and we're done.

A solution for this problem.

Define points $B_0,B_1\in AC$ such that $A_0B_0$ tangents to $\gamma,$ and $XB_1\parallel AB,$ and points $C_0,C_1$ similarly. There exist unique involution which swaps pairs $(\overline{XA},\overline{XA_0}),$ $(\overline{XB_1},\overline{XC_1}),$ so under projection $AB\stackrel{X}{\mapsto} AC$ it gives a reflection over bisector of angle $BAC,$ followed by symmetry wrt midpoint of $AC.$ By DDIT on rhombus $AB_0A_0C_0$ involution also swaps tangents to $\gamma$ through $X,$ and conclusion follows.