Srsly grade 11 problem 6 and 7 are both geo?

GorgonMathDota wrote: Srsly grade 11 problem 6 and 7 are both geo? That is a typing error. Henry_2001 wrote: Let $ABCD$ be a given convex quadrilateral in a plane. Prove that there exist a line with four different points $P,Q,R,S$ on it and a square $A’B’C’D’$ such that $P$ lies on both line $AB$ and $A’B’,$ $Q$ lies on both line $BC$ and $B’C’,$ $R$ lies on both line $CD$ and $C’D’,$ $S$ lies on both line $DA$ and $D’A’.$ Wait it looks trivially trivial. It indeed is possible for any four collinear points. \(\overline{P,Q,R,S}\) draw a line from \(Q\) making \(\angle \theta\) the only condition should be \(PS \cos\theta = QR \sin \theta\). And complete the square. [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 = -8.933393727086195, xmax = 10.719550498276064, ymin = -4.743185416351953, ymax = 7.076485907826339; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); /* draw figures */ draw((-5.06,3.03)--(0.48,3.37), linewidth(2) + wrwrwr); draw((0.48,3.37)--(0.84,-1.71), linewidth(2) + wrwrwr); draw((0.84,-1.71)--(-5.22,-0.83), linewidth(2) + wrwrwr); draw((-5.22,-0.83)--(-5.06,3.03), linewidth(2) + wrwrwr); draw((xmin, 24.125*xmin + 125.10249999999988)--(xmax, 24.125*xmax + 125.10249999999988), linewidth(2) + wrwrwr); /* line */ draw((xmin, -0.14521452145214522*xmin-1.588019801980198)--(xmax, -0.14521452145214522*xmax-1.588019801980198), linewidth(2) + wrwrwr); /* line */ draw((xmin, -0.785596746736963*xmin + 0.7182775889005671)--(xmax, -0.785596746736963*xmax + 0.7182775889005671), linewidth(2) + wrwrwr); /* line */ draw((xmin, 1.2729176949288263*xmin + 10.996904020527792)--(xmax, 1.2729176949288263*xmax + 10.996904020527792), linewidth(2) + wrwrwr); /* line */ draw(circle((0.7072939540428336,0.16262975961779422), 10.92963619383511), linewidth(2) + wrwrwr); draw((xmin, 1.2729176949288263*xmin-0.7376972299995048)--(xmax, 1.2729176949288263*xmax-0.7376972299995048), linewidth(2) + wrwrwr); /* line */ draw((xmin, 0.5311802879311687*xmin + 4.795095746891031)--(xmax, 0.5311802879311687*xmax + 4.795095746891031), linewidth(2) + rvwvcq); /* line */ draw((xmin, -1.882599981062516*xmin-4.759306736810882)--(xmax, -1.882599981062516*xmax-4.759306736810882), linewidth(2) + rvwvcq); /* line */ draw((xmin, -1.882599981062516*xmin + 4.669067597145297)--(xmax, -1.882599981062516*xmax + 4.669067597145297), linewidth(2) + rvwvcq); /* line */ draw((xmin, 0.5311802879311687*xmin-0.2130708465426529)--(xmax, 0.5311802879311687*xmax-0.2130708465426529), linewidth(2) + rvwvcq); /* line */ /* dots and labels */ dot((-5.06,3.03),dotstyle); label("$A$", (-4.986859288321765,3.229611429636777), NE * labelscalefactor); dot((0.48,3.37),dotstyle); label("$B$", (0.5542345196404137,3.5684552955912983), NE * labelscalefactor); dot((0.84,-1.71),dotstyle); label("$C$", (0.9130103777099072,-1.5142026937265174), NE * labelscalefactor); dot((-5.22,-0.83),dotstyle); label("$D$", (-5.14631522524154,-0.6371950406677571), NE * labelscalefactor); dot((-4.993225319959166,4.640939155985119),dotstyle); label("$P$", (-4.907131319861877,4.844102790949495), NE * labelscalefactor); dot((3.6014387967358266,-2.111001013387381),dotstyle); label("$S$", (3.6835572816909963,-1.912842536025954), NE * labelscalefactor); dot((-3.0960580649995517,3.1505307324729523),linewidth(4pt) + dotstyle); label("$Q$", (-3.01359206893955,3.3093393980966646), NE * labelscalefactor); dot((0.7072939540428336,0.16262975961779422),linewidth(4pt) + dotstyle); label("$R$", (0.7934184250200761,0.31954058085089054), NE * labelscalefactor); dot((7.4592341234153325,8.757293876312787),linewidth(4pt) + dotstyle); label("$E$", (5.9358723906828175,6.75757403398679), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

@above It's G10 P7 I have corrected it.

Either I have some mistake in my computation or this problem is really trivial for a CSMO P7. Same as above, this is indeed possible for any 4 collinear points : $P,Q,R,S$ where $P \in AB$, $Q \in BC$, $R \in CD$, $S \in DA$. Main Claim. Given 4 collinear points $P,Q,R,S$ (in that order) such that $|PQ| : |QR| : |RS| = a : b : c$ where $|AB|$ denotes the length of segment $AB$. Then, there exists at least 1 square $ABCD$ such that $P,Q,R,S$ is in each of $AB, BC, CD, DA$ exactly once. Proof. Let $\alpha$ be the acute angle formed by the side of the square with the line (not the extension of the side of the square). Take a square of side $(a+b+c) \cos \alpha$ where $\alpha = \tan^{-1} \left( \frac{a+b+c}{b} \right)$ and this indeed works.

This one is the same as for any convex quadrilateral $ABCD$ in a plane,there's a square $A'B'C'D'$ that is perspective to $ABCD$(by Desargues).(quite interesting!)