Let $T_1, T_2, T_3, T_4$ be pairwise distinct collinear points such that $T_2$ lies between $T_1$ and $T_3$, and $T_3$ lies between $T_2$ and $T_4$. Let $\omega_1$ be a circle through $T_1$ and $T_4$; let $\omega_2$ be the circle through $T_2$ and internally tangent to $\omega_1$ at $T_1$; let $\omega_3$ be the circle through $T_3$ and externally tangent to $\omega_2$ at $T_2$; and let $\omega_4$ be the circle through $T_4$ and externally tangent to $\omega_3$ at $T_3$. A line crosses $\omega_1$ at $P$ and $W$, $\omega_2$ at $Q$ and $R$, $\omega_3$ at $S$ and $T$, and $\omega_4$ at $U$ and $V$, the order of these points along the line being $P,Q,R,S,T,U,V,W$. Prove that $PQ + TU = RS + VW$ Geza Kos, Hungary
Problem
Source: RMM 2021/1
Tags: geometry, RMM
13.10.2021 16:37
Didn't like this. Here's my soln sketch Let $O_i$ be center of the corresponding circle. The main idea is to construct the midpoints of $QR$, $ST$, $UV$ and $PW$ - i.e. these are projections of the centers. Let them be $A_2$, $A_3$, $A_4$, $A_1$, respectively. Note that $T_1T_2O_2$, $T_2T_3O_3$, $T_3T_4O_4$, $T_1T_4O_1$ are isosceles triangles, using the fact that the circles are tangent and the collinearity of centers. Thus $O_1O_2O_3O_4$ is parallelogram due to angle chasing from the isosceles triangles. Now we calculate: $PQ=PA_1-QA_2-A_2A_1$ and the others in similar manner. Now we see that the desired equality is equivalent to $A_2A_3+A_1A_2=A_3A_4+A_1A_4$, equivalent to $A_2A_3=A_1A_4$. Let $O_3O_2$ meet $PW$ at $N$, and $O_1O_4$ at $M$. Due to the parallel sides, triangles $NO_2A_2$ and $MO_4A_4$ are similar, and finish with Thales theorem in triangles $NO_3A_3$ and $MO_1A_1$.
13.10.2021 18:34
Suppose $O_{i}$ is the center of $\omega_{i}$ for $i=1,2,3,4$.Then $\triangle O_{i}T_{i}T_{i+1} (i=1,2,3,4)$ are similiar triangles.Then $\omega_{1}$ and $\omega_{4}$ is tangent at $T_{4}$ and $O_{1}O_{2}O_{3}O_{4}$ is a parallelogram. Suppose $A,B,C,D$ are the projections of $O_{1},O_{2},O_{3},O_{4}$ on the line $PW$.Then $ABCD$ is a degenerate parallelogram.That is $AB=CD$. Since $A,B,C,D$ are also the mid points of $PW,QR,ST,UV$,we have $PQ-RS=BP-BS=(AP-AB)-(CB-CS)=(AW-AD)-(CD-CT)=DW-DT=VW-TU$,done.
Attachments:

13.10.2021 18:44
I had a terrible calculation using power of a point at first, because I did not notice the midpoints... now that I've been told, I have no motivation to write that mess up All indices are taken modulo $4$. To begin: Claim: Circle $\omega_4$ is tangent at $T_4$ to $\omega_1$. Proof. Follows by Monge d'Alembert theorem on the composed homothety $\omega_4 \to \omega_3 \to \omega_2 \to \omega_1$. $\blacksquare$ Let $O_i$ denote the center of $\omega_i$. Claim: We have a parallelogram $O_1O_2O_3O_4$. Proof. Let $t_i$ denote the common tangent at $T_i$. Then $t_i$ and $t_{i+1}$ are reflections about the perpendicular bisector of $T_i T_{i+1}$. Hence the $t_1 \parallel t_3$, $t_2 \parallel t_4$ and from $\overline{O_i O_{i-1}} \perp t_i$ we're done. $\blacksquare$ [asy][asy] real r = 0.65; real s = 0.70; size(12cm); pair T_1 = dir(155); pair T_4 = dir(25); pair O_2 = r*T_1; pair O_1 = origin; pair O_4 = s*T_4; pair O_3 = O_2+O_4; filldraw(unitcircle, invisible, black); filldraw(CR(O_2, 1-r), invisible, black); filldraw(CR(O_4, 1-s), invisible, black); filldraw(CR(O_3, r+s-1), invisible, black); pair T_2 = extension(O_2, O_3, T_1, T_4); pair T_3 = extension(O_4, O_3, T_1, T_4); draw(T_1--T_4, red); filldraw(O_1--O_2--O_3--O_4--cycle, invisible, red); pair P = dir(137); pair W = dir(5); draw(P--W, blue+1); pair Q = IP(P--W, CP(O_2, T_2)); pair R = OP(P--W, CP(O_2, T_2)); pair S = IP(P--W, CP(O_3, T_3)); pair T = OP(P--W, CP(O_3, T_3)); pair U = IP(P--W, CP(O_4, T_4)); pair V = OP(P--W, CP(O_4, T_4)); pair M_2 = midpoint(Q--R); pair M_3 = midpoint(S--T); pair M_4 = midpoint(U--V); pair M_1 = midpoint(P--W); draw(O_1--M_1, blue); draw(O_2--M_2, blue); draw(O_3--M_3, blue); draw(O_4--M_4, blue); clip(box((-1.1,-0.1),(1.1,1.1))); dot("$T_1$", T_1, dir(T_1)); dot("$T_4$", T_4, dir(T_4)); dot("$O_2$", O_2, dir(O_2)); dot("$O_1$", O_1, dir(270)); dot("$O_4$", O_4, dir(O_4)); dot("$O_3$", O_3, dir(O_3)); dot("$T_2$", T_2, dir(250)); dot("$T_3$", T_3, dir(290)); dot(P); dot(W); dot(Q); dot(R); dot(S); dot(T); dot(U); dot(V); dot("$M_2$", M_2, dir(70)); dot("$M_3$", M_3, dir(250)); dot("$M_4$", M_4, dir(270)); dot("$M_1$", M_1, dir(290)); /* TSQ Source: !real r = 0.65; !real s = 0.70; !size(12cm); T_1 = dir 155 T_4 = dir 25 O_2 = r*T_1 O_1 = origin R270 O_4 = s*T_4 O_3 = O_2+O_4 unitcircle 0.1 yellow / black CR O_2 1-r 0.2 palecyan / black CR O_4 1-s 0.2 palecyan / black CR O_3 r+s-1 0.2 green / black T_2 = extension O_2 O_3 T_1 T_4 R250 T_3 = extension O_4 O_3 T_1 T_4 R290 T_1--T_4 red O_1--O_2--O_3--O_4--cycle 0.1 lightred / red P .= dir 137 W .= dir 5 P--W blue+1 Q .= IP P--W CP O_2 T_2 R .= OP P--W CP O_2 T_2 S .= IP P--W CP O_3 T_3 T .= OP P--W CP O_3 T_3 U .= IP P--W CP O_4 T_4 V .= OP P--W CP O_4 T_4 M_2 = midpoint Q--R R70 M_3 = midpoint S--T R250 M_4 = midpoint U--V R270 M_1 = midpoint P--W R290 O_1--M_1 blue O_2--M_2 blue O_3--M_3 blue O_4--M_4 blue !clip(box((-1.1,-0.1),(1.1,1.1))); */ [/asy][/asy] To finish, let $M_i$ be the projection of $O_i$ on $PQRSTUVW$ (it's a midpoint) and compute the vector sum \begin{align*} \vec P - \vec Q - \vec R + \vec S + \vec T - \vec U - \vec V + \vec W &= (\vec P + \vec W) - (\vec Q + \vec R) + (\vec S + \vec T) - (\vec U + \vec V) \\ &= 2 \vec{M_1} - 2 \vec{M_2} + 2\vec{M_3} - 2 \vec{M_4} = \vec 0 \end{align*}
14.10.2021 00:00
RMM 2021/1 wrote: Let $T_1, T_2, T_3, T_4$ be pairwise distinct collinear points such that $T_2$ lies between $T_1$ and $T_3$, and $T_3$ lies between $T_2$ and $T_4$. Let $\omega_1$ be a circle through $T_1$ and $T_4$; let $\omega_2$ be the circle through $T_2$ and internally tangent to $\omega_1$ at $T_1$; let $\omega_3$ be the circle through $T_3$ and externally tangent to $\omega_2$ at $T_2$; and let $\omega_4$ be the circle through $T_4$ and externally tangent to $\omega_3$ at $T_3$. A line crosses $\omega_1$ at $P$ and $W$, $\omega_2$ at $Q$ and $R$, $\omega_3$ at $S$ and $T$, and $\omega_4$ at $U$ and $V$, the order of these points along the line being $P,Q,R,S,T,U,V,W$. Prove that $PQ + TU = RS + VW$ Geza Kos, Hungary Kinda annoying not gonna lie. Also, is it just me that think this is harder than IMO 2021/4 [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(14cm); 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 = -10.305443580156439, xmax = 3.2912639947835385, ymin = -3.482149717840559, ymax = 4.429523330992136; /* image dimensions */ pen wwwwww = rgb(0.4,0.4,0.4); /* draw figures */ draw((-9.06,1.14)--(0.6027164168119578,0.603968847799081)); draw(circle((-4.070065641727407,3.730542849728922), 5.622308881247185)); draw(circle((-7.943396453268674,1.7196888548586333), 1.25811074592831), wwwwww); draw(circle((-4.222478994087539,-0.7699893808173304), 3.2189110828366103), wwwwww); draw(circle((-0.3491481825462704,1.2408646140529587), 1.1452870524822651), wwwwww); draw((-9.411382749871688,1.9753537697730914)--(1.263201629080057,1.951043657741222)); draw((-9.06,1.14)--(-4.070065641727407,3.730542849728922), linetype("2 2")); draw((-4.070065641727407,3.730542849728922)--(0.6027164168119578,0.603968847799081), linetype("2 2")); draw((-7.943396453268674,1.7196888548586333)--(-4.222478994087539,-0.7699893808173304), linetype("2 2")); draw((-4.222478994087539,-0.7699893808173304)--(-0.3491481825462704,1.2408646140529587), linetype("2 2")); /* dots and labels */ dot((-9.06,1.14),dotstyle); label("$T_1$", (-9.014466989852904,1.2553790539155454), NE * labelscalefactor); dot((0.6027164168119578,0.603968847799081),dotstyle); label("$T_4$", (0.6974669922470795,0.4973744504345686), NE * labelscalefactor); dot((-6.897762453892018,1.0200516714786554),dotstyle); label("$T_2$", (-6.847047576774493,1.1369408346216427), NE * labelscalefactor); dot((-1.3656179877410137,0.7131605609030186),dotstyle); label("$T_3$", (-1.3041389138198682,0.8882205741044473), NE * labelscalefactor); dot((-7.943396453268674,1.7196888548586333), dotstyle); label("$O_2$", (-8.101147728490223,1.4120386845968877), NE * labelscalefactor); dot((-4.222478994087539,-0.7699893808173304),dotstyle); label("$O_3$", (-4.1703438207323025,-1.0751639205750672), NE * labelscalefactor); dot((-0.3491481825462704,1.2408646140529587), dotstyle); label("$O_4$", (-0.2974140498216993,1.3382858074212773), NE * labelscalefactor); dot((-9.411382749871688,1.9753537697730914),dotstyle); label("$P$", (-9.36978164773461,2.096290410902254), NE * labelscalefactor); dot((1.263201629080057,1.951043657741222),dotstyle); label("$W$", (1.3133457325753712,2.0726027670434735), NE * labelscalefactor); dot((-9.17536749580886,1.97481627275552), dotstyle); label("$Q$", (-9.132905209146806,2.048915123184693), NE * labelscalefactor); dot((-6.7102761504024,1.9692023167785748), dotstyle); label("$R$", (-6.65754642590425,2.0370713012553026), NE * labelscalefactor); dot((-5.916032480954411,1.9673935201173245), dotstyle); label("$S$", (-5.864010356635104,2.0370713012553026), NE * labelscalefactor); dot((-2.516475002907848,1.9596514272904555), dotstyle); label("$T$", (-2.4648334629001103,2.0252274793259124), NE * labelscalefactor); dot((-1.243121548220608,1.9567515143720047), dotstyle); label("$U$", (-1.1975445164553562,2.0252274793259124), NE * labelscalefactor); dot((0.5480765897779837,1.9526722710159934), dotstyle); label("$V$", (0.5908725948825675,2.0252274793259124), NE * labelscalefactor); dot((-4.070065641727407,3.730542849728922), dotstyle); label("$O_1$", (-4.02821795757962,3.825488412593232), NE * labelscalefactor); dot((-7.94282182310563,1.9720092947670476), dotstyle); label("$A$", (-7.901147728490223,2.0726027670434735), NE * labelscalefactor); dot((-4.21625374193113,1.96352247370389), dotstyle); label("$B$", (-4.3703438207323025,2.0607589451140833), NE * labelscalefactor); dot((-0.34752247922131213,1.954711892693999), dotstyle); label("$C$", (-0.2974140498216993,2.048915123184693), NE * labelscalefactor); dot((-4.074090560395815,1.9631987137571567), dotstyle); label("$D$", (-4.02821795757962,2.0607589451140833), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let us denote $O_1, O_2, O_3, O_4$ as centers of $\omega_1, \omega_2, \omega_3$ and $\omega_4$ respectively. Claim 01. $\omega_4$ is tangent to $\omega_1$. Proof. Indeed, since the set of circles $(\omega_1, \omega_2), (\omega_2, \omega_3), (\omega_3, \omega_4)$ are tangents, then we have $(O_1, O_2, T_1), (O_2, O_3, T_2), (O_3, O_4, T_3)$ being collinear. We aim to show that $O_1, O_4, T_4$ being collinear, which would imply our claim. Indeed, we have \begin{align*} \measuredangle O_1 T_4 T_1 &= \measuredangle T_4 T_1 O_1 \equiv \measuredangle T_2 T_1 O_2 = \measuredangle O_2 T_2 T_1 = \measuredangle O_3T_2T_3 = \measuredangle T_2 T_3 O_3 = \measuredangle T_4T_3 O_4 = \measuredangle O_4 T_4 T_3 \equiv \measuredangle O_4 T_4 T_1, \end{align*}as desired. Claim 02. $O_1O_2O_3O_4$ is a parallelogram. Proof. We have $\measuredangle T_4T_1 O_1 = \measuredangle T_2T_1 O_2 = \measuredangle T_4T_3O_4$ and $(T_1, T_3, T_4), (O_1, O_4, T_4)$ being collinear which implies $O_3 O_4 \equiv T_3 O_4 \parallel T_1O_1 \equiv O_2O_1$. Similarly, we could get $O_1O_4 \parallel O_2O_3$. Now, let $A,B,C,D$ be the midpoints of $QR, ST, UV, PW$. Note that it suffices to prove that $AB = CD$, because this implies \begin{align*} AB &= CD \\ 2 \cdot AB &= 2 \cdot CD \\ AB + AD &= BC + CD \\ (PA + AD) + (AB + CW) &= (BC + CW) + (CD + AP) \\ AB + CW &= CD + AP \\ (AR + RS + SB) + (CV + VW) &= (AQ + QP) + (BT + TU + UC) \\ RS + VW &= PQ + TU \end{align*}which is what we wanted. Claim 03. $AB = CD$. Proof. Let $A,B,C,D$ lie on line $\ell$, then $A,B,C,D$ are just projections of vertices of parallelogram $O_2,O_3,O_1,O_4$ on line $\ell$, in that order. There are a lot of cases: we'll just do one of the cases, and do the others analogously. Let $E$ and $F$ be such that $ABO_3E$ and $O_1DCF$ are rectangles. This implies $O_4, C, F$ and $O_2, A, E$ collinear. Furthermore, $\angle (O_3 E, O_2O_3) = \angle(O_2 O_3, \ell) = \angle(O_1 O_4, \ell) = \angle(O_1 O_4, O_1 F)$, and $O_2 O_3 = O_1 O_4$ implies $O_4 F = O_2 E$, and therefore $AB = EO_3 = O_1 F = CD$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -12.347299876313178, xmax = 10.249136556825986, ymin = -5.507180782097362, ymax = 7.641268292238014; /* image dimensions */ pen wwwwww = rgb(0.4,0.4,0.4); pen ccqqqq = rgb(0.8,0,0); /* draw figures */ draw((-8.595511200047737,0.39427252872227925)--(1.4541014611494365,2.6846293850272627), wwwwww); draw((-7.943396453268674,1.7196888548586333)--(-4.222478994087539,-0.7699893808173304)); draw((-4.222478994087539,-0.7699893808173304)--(-0.3491481825462704,1.2408646140529587)); draw((-7.943396453268674,1.7196888548586333)--(-4.070065641727407,3.730542849728922)); draw((-4.070065641727407,3.730542849728922)--(-0.3491481825462704,1.2408646140529587)); draw((-4.222478994087539,-0.7699893808173304)--(-4.690641683476388,1.284211770339787), ccqqqq); draw((-7.688441292112145,0.600998285507887)--(-7.943396453268674,1.7196888548586333), ccqqqq); draw((-3.57070486944915,1.539450956874771)--(-4.070065641727407,3.730542849728922), ccqqqq); draw((-0.3491481825462704,1.2408646140529587)--(-0.5729052608133901,2.222664441706671),ccqqqq); draw((-7.2202786027232975,-1.4532028656492304)--(-7.688441292112145,0.600998285507887), linetype("2 2") + ccqqqq); draw((-7.2202786027232975,-1.4532028656492304)--(-4.222478994087539,-0.7699893808173304), linetype("2 2") + ccqqqq); draw((-4.070065641727407,3.730542849728922)--(-1.072266033091647,4.413756334560823), linetype("2 2") + ccqqqq); draw((-1.072266033091647,4.413756334560823)--(-0.5729052608133901,2.222664441706671), linetype("2 2") + ccqqqq); /* dots and labels */ dot((-7.943396453268674,1.7196888548586333), dotstyle); label("$O_2$", (-7.8595058808813585,1.8740593419681562), NE * labelscalefactor); dot((-4.222478994087539,-0.7699893808173304), dotstyle); label("$O_3$", (-4.139360858352349,-0.606037339717858), NE * labelscalefactor); dot((-0.3491481825462704,1.2408646140529587), dotstyle); label("$O_4$", (-0.26174937984327645,1.401659974027963), NE * labelscalefactor); dot((-4.070065641727407,3.730542849728922), dotstyle); label("$O_1$", (-3.9818944023722853,3.881756655713977), NE * labelscalefactor); dot((-7.688441292112145,0.600998285507887), dotstyle); label("$A$", (-7.603622889913755,0.7521108431101974), NE * labelscalefactor); dot((-4.690641683476388,1.284211770339787), dotstyle); label("$B$", (-4.611760226292541,1.4410265880229791), NE * labelscalefactor); dot((-0.5729052608133901,2.222664441706671), dotstyle); label("$C$", (-0.4979490638133723,2.3858253239033655), NE * labelscalefactor); dot((-3.57070486944915,1.539450956874771), dotstyle); label("$D$", (-3.489811727434586,1.6969095789905837), NE * labelscalefactor); dot((-7.2202786027232975,-1.4532028656492304), dotstyle); label("$E$", (-7.15090682897107,-1.2949530846306399), NE * labelscalefactor); dot((-1.072266033091647,4.413756334560823), dotstyle); label("$F$", (-0.9900317387510719,4.570672400626759), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Motivational Remark. The first two claims are immediate, just like usual, when dealing with so many points, it's good to divide the problems into several part: the part where points are fixed at first to see what property the figure has, and the part where the other points get involved. The "magical" leap of faith is introducing $A,B,C,D$ to the diagram. This is definitely not what I tried at first; I tried considering $PW \cap T_1 T_4$ and thought of using power of point, but the bash demotivated me. The best way to think about this is to get rid of a lot of points at once, and same as IMO 2021/4, by adding some segment of the same "length" to simplify it. The first thing I tried was extending $O_2T_2 \cap QR = L$, and try to work on there, but this didn't work because no equal length came up, but why bother with finding equal length, when you could just add midpoints to the diagram? After simplification, $8$ points reduced to $4$ points, which is much cleaner. However, we know something more about this four points: the line joining the center of the circle midpoint of a chord in the respective circle, must be perpendicular to the chord. Now, this problem reduces to something much familiar for me, which is something resembling INAMO 2017 P1, from which we could just finish immediately.
14.10.2021 03:59
For $i=1,2,3,4$, let $O_i$ be the center of $\omega_i$, and let $t_i$ be the tangent of $\omega_i$ at $T_i$. Also, let $\ell = \overline{T_1T_2T_3T_4}$. We first prove that: Claim: $\omega_4$ and $\omega_1$ are tangent. Proof: From $\measuredangle(t_1,\ell) \stackrel{\omega_2}{=} -\measuredangle(t_2,\ell) \stackrel{\omega_3}{=} \measuredangle(t_3,\ell) \stackrel{\omega_4}{=} -\measuredangle(t_4,\ell)$, we get that $t_4$ is indeed tangent to $\omega_1$. $\blacksquare$ Claim: $O_1O_2O_3O_4$ is a parallelogram. Proof: $O_1O_2$ and $O_3O_4$ are perpendicular to $t_1$ and $t_3$, respectively; they are parallel due to the homothety at $T_2$. Analogously, we get that $O_2O_3\parallel O_4O_1$. $\blacksquare$ Finally, notice that the projections of $O_1$, $O_2$, $O_3$, $O_4$ on to the line $\overline{PQRSTUVW}$ are midpoints of $PW$, $QR$, $ST$, $UV$, respectively. This implies that \begin{align*} &\left(\frac{P+W}{2}\right) + \left(\frac{S+T}{2}\right) = \left(\frac{Q+R}{2}\right) + \left(\frac{U+V}{2}\right) \\ \implies & (Q-P) + (U-T) = (S-R) + (W-V), \end{align*}which gives the result.
24.10.2021 03:54
Solved with Jeffrey Chen and Nathan Cho. Let \(O_i\) be the center of \(\omega_i\). Evidently \(\omega_1\) and \(\omega_4\) are also tangent at \(T_4\) since \(\measuredangle T_1O_1T_4=\measuredangle T_1O_2T_2=\measuredangle T_3O_3T_2=\measuredangle T_3O_4T_4\) by homothety. Moreover, by homothety \(\overline{O_2T_2}\parallel\overline{O_1T_4}\) and \(\overline{O_4T_3}\parallel\overline{O_1T_1}\), so \(O_1O_2O_3O_4\) is a parallelogram. [asy][asy] size(7cm); defaultpen(fontsize(10pt)); pen pri=blue; pen pri2=heavyblue; pen sec=lightblue; pen tri=purple; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair O1,T1,O2,T2,O3,T3,T4,O4; pair P,Q,R,SS,T,U,V,WW; O1=origin; T1=dir(160); O2=.6*T1; T2=reflect(O2,O2+(0,1))*T1; O3=extension(O2,T2,O1,dir(75)); T3=reflect(O3,O3+(0,1))*T2; T4=reflect( (0,0),(0,1))*T1; O4=extension(O3,T3,O1,T4); P=dir(175); WW=dir(15); Q=intersectionpoint(P--foot(O2,P,WW),circle(O2,abs(T1-O2))); R=intersectionpoint(WW--foot(O2,P,WW),circle(O2,abs(T1-O2))); SS=intersectionpoint(P--foot(O3,P,WW),circle(O3,abs(T2-O3))); T=intersectionpoint(WW--foot(O3,P,WW),circle(O3,abs(T2-O3))); U=intersectionpoint(P--foot(O4,P,WW),circle(O4,abs(T3-O4))); V=intersectionpoint(WW--foot(O4,P,WW),circle(O4,abs(T3-O4))); draw(T1--O1--T4,tri); draw(O2--O3--O4,tri); fill(O1--O2--O3--O4--cycle,tfil); filldraw(unitcircle,fil,pri); filldraw(circle(O2,abs(T1-O2)),fil,pri); filldraw(circle(O3,abs(T2-O3)),fil,pri); filldraw(circle(O4,abs(T3-O4)),fil,pri); draw(P--WW,sec); draw(T1--T4,pri2); dot("\(O_1\)",O1,S); dot("\(O_2\)",O2,N); dot("\(O_3\)",O3,N); dot("\(O_4\)",O4,S); dot("\(T_1\)",T1,T1); dot("\(T_2\)",T2,NE); dot("\(T_3\)",T3,NW); dot("\(T_4\)",T4,T4); dot("\(P\)",P,P); dot("\(Q\)",Q,NE); dot("\(R\)",R,NW); dot("\(S\)",SS,NE); dot("\(T\)",T,NW); dot("\(U\)",U,NE); dot("\(W\)",WW,SW); dot("\(V\)",V,E); real t=-.2; clip( (-100,t)--(100,t)--(100,100)--(-100,100)--cycle); [/asy][/asy] Now let \(M_i\) be the projection of \(O_i\) onto \(\overline{PV}\). Since \(O_1+O_3=O_2+O_4\), we have \(M_1+M_3=M_2+M_4\), thus \[\frac{P+V}2+\frac{S+T}2=\frac{Q+R}2+\frac{U+W}2,\]hence \[(P-Q)+(T-U)=(R-S)+(V-W),\]as desired.
06.11.2021 06:11
A comedy. Let the center of $\omega_i$ be $O_i$. Let the line through $P$ and $W$ be $\ell$. Observe that $O_1O_2\parallel O_3O_4$ because \[\measuredangle T_3T_4O_1=\measuredangle O_1T_1T_4 = \measuredangle O_2T_1T_2 = \measuredangle T_1T_2O_2 = \measuredangle T_3T_2O_3 = \measuredangle O_3T_3T_2 = \measuredangle O_4T_3T_2 = \measuredangle O_4T_3T_4 = \measuredangle T_3T_4O_4.\]Moreover, $O_1O_4\parallel O_2O_3$ so $O_1O_2O_3O_4$ is a parallelogram. Let the perpendicular line from $O_i$ to $\ell$ intersect $\ell$ at $A_i$. In vectors, $\vec A_1 = (\vec P+\vec W)/2$, $\vec A_2 = (\vec Q+\vec R)/2$, $\vec A_3 = (\vec S+\vec T)/2$, and $\vec A_4 = (\vec U+\vec V)/2$. By the parallelogram condition, $\vec A_2+\vec A_4 = \vec A_1 + \vec A_3$. Equivalently, $\vec P+\vec W + \vec S+\vec T = \vec Q+\vec R+\vec U+\vec V$. This rearranges to \[\vec P - \vec Q + \vec T - \vec U = \vec R - \vec S + \vec V - \vec W,\]which is the desired.
03.01.2022 02:46
First non-in-test geo in a while
Let $O_i$ be the center of $\omega_i$ for each $i$. Let $\ell$ be the line through $P,Q,R$,etc., and let $t$ be the line through the $T_i$'s. The main idea is to get rid of the points $P,Q,R,S$, etc. and instead make the problem about the $O_i$'s and $\ell$. Let $X=\ell\cap t$. We will use directed lengths to avoid the need for cases. Note $PQ=PX-QX$, and $RS=RX-SX$, and $TU=XU-XT$, and $VW=XW-XV$. So, defining $M_{\bullet *}$ to be the midpoint of $\bullet$ and $*$, we have \begin{align*} PQ+TU=RS+VW&\iff (PX-QX) + (XU-XT) = (RX-SX)+(XW-XV) \\ &\iff (PX-XW) + (-QX-RX) + (SX-XT) + (XU+XV) = 0 \\ &\iff 2M_{PW} X + 2XM_{QR} + 2M_{ST}X + 2XM_{UV} = 0 \\ &\iff XM_{QR} + XM_{UV} = XM_{PW} + XM_{ST} \\ &\iff XM_{M_{QR}M_{UV}} = XM_{M_{PW}M_{ST}}. \qquad (\heartsuit) \end{align*}Notice now that $M_{QR}$ is the foot from $O_2$ to $\ell$, since $\overline{QR}$ is a chord of $\omega_2$. Similarly, each of the midpoints in $(\heartsuit)$ is a foot from $O_i$ to $\ell$. Rename $M_{QR}\to M_2$, $M_{PW}\to M_1$, $M_{ST}\to O_3$, and $M_{UV}\to M_4$. We will in fact show that the points in $(\heartsuit)$ are the same, i.e. \[ M_{M_2M_4}=M_{M_1M_3}. \qquad (\spadesuit)\]Claim: $O_1O_2O_3O_4$ is a parallelogram. Proof: Easy coordinates. Let $t$ be the $x$-axis, and let $T_1=(0,0)$, $T_2=(a,0)$, $T_3=(b,0)$, and $T_4=(1,0)$. Then $O_2=(a/2, ar/2)$ for some constant $r$, and $O_1=(1/2,r/2)$. Then $O_3=((a+b)/2,-(b-a)r/2)$ and $O_4=((b+1)/2,(1-b)r/2)$. Now $O_2+O_4=(1/2+a/2+b/2, r/2-br/2+ar/2)=O_1+O_3$, proving the claim. $\blacksquare$
So the problem is reduced to this: we have a parallelogram $O_1O_2O_3O_4$, and a line $\ell$, and $M_i$ is the foot from $O_i$ to $\ell$. We want to show $M_{M_2M_4}=M_{M_1M_3}$ $(\spadesuit)$. This is equivalent to $M_2+M_4=M_1+M_3$ as vectors. Since $O_2+O_4=O_1+O_3$ the wanted statement follows by the linearity of projections.
21.03.2022 19:37
03.02.2024 19:31
Let $O_i$ denote the center of $\omega_i$. Setting $\overline{T_1T_2T_3T_4}$ as the $x$-axis, it is clear that $\overline{O_1T_4}$ is the negative of the slope of $\overline{T_1O_1O_2}$ is the negative of the slope of $\overline{O_2O_3}$ is the negative of the slope of $\overline{O_3O_4}$ is the negative of the slope of $\overline{O_4T_4}$, hence $\overline{O_1T_4}$ and $\overline{O_4T_4}$ coincide $O_1O_2O_3O_4$ is a parallelogram. Now let $M_i$ be the foot from $O_i$ to $\overline{T_1T_2T_3T_4}$. We have $$(P+W)+(S+T)-(Q+R)-(U+V)=2M_1+2M_3-2M_2-2M_4=0,$$since $\tfrac{M_1+M_3}{2}=\tfrac{M_2+M_4}{2}$ are both the projection of $\overline{O_1O_3} \cap \overline{O_2O_4}$ onto $\overline{T_1T_2T_3T_4}$. $\blacksquare$ Remark: This problem becomes significantly more tractable if one thinks in coordbash terms