Let $O$ be the center of $(ABCD)$ with radius $R$. Clearly $OP = OQ$ and $OR = OS$. I claim that in fact $O$ is the center of $PQRS$. Let $X$ and $Y$ be the feet from $O$ to $\overline{AB}$ and $\overline{CD}$. Note that $X$, $Y$ are the midpoints of $\overline{PQ}$ and $\overline{RS}$. Now \begin{align*} OX^2 &= R^2 - (3.5)^2 \implies OP^2 = XP^2 + OX^2 = (0.5)^2 + R^2 - (3.5)^2 = R^2 - 12 \\ OY^2 &= R^2 - 4^2 \implies OR^2 = YR^2 + OY^2 = 2^2 + R^2 - 4^2 = R^2 - 12 \end{align*}so $OP = OR$. Similarly $OQ = OS$, done.

Let $O$ be the center of $(ABCD)$ which has radius $R$. Let $x$ and $y$ be the distance of the perpendicular line from $O$ to $AB$ and $CD$ respectively. We can see that \[4^2 + y^2 = R^2 = \left(\frac 72 \right)^2 + x^2 \Rightarrow y^2 + \frac{15}{4} = x^2.\]We can see that $O$ lies on the perpendicular bisector of both $PQ$ and $RS$. We just need to prove that $OP = OS$ or that $x^2 + \left(\frac 12 \right)^2 = y^2 + 2^2$ which is evident. $\blacksquare$

First, let $E$ and $F$ be the midpoints of $AB$ and $CD$, respectively. It is clear that $AE=BE=3.5$, $PE=QE=0.5$, $DF=CF=4$, and $SF=RF=2$. Also, let $O$ be the circumcenter of $ABCD$. [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 = -12.19, xmax = 24.94, ymin = -15.45, ymax = 6.11; /* image dimensions */ pen wrwrwr = rgb(0.38,0.38,0.38); /* draw figures */ draw(circle((2.92,-3.28), 5.90), linewidth(2) + wrwrwr); draw((-2.52,-1.01)--(3.46,2.59), linewidth(2) + wrwrwr); draw((7.59,-6.88)--(-0.29,-8.22), linewidth(2) + wrwrwr); draw((3.46,2.59)--(7.59,-6.88), linewidth(2) + wrwrwr); draw((-0.29,-8.22)--(-2.52,-1.01), linewidth(2) + wrwrwr); draw((0.03,0.52)--(1.67,-7.89), linewidth(2) + wrwrwr); draw((5.61,-7.22)--(0.89,1.04), linewidth(2) + wrwrwr); /* dots and labels */ dot((2.92,-3.28),dotstyle); label("$O$", (2.43,-3.56), NE * labelscalefactor); dot((-2.52,-1.01),dotstyle); label("$A$", (-2.91,-0.91), NE * labelscalefactor); dot((3.46,2.59),linewidth(4pt) + dotstyle); label("$B$", (3.49,2.78), NE * labelscalefactor); dot((7.59,-6.88),dotstyle); label("$C$", (7.82,-7.24), NE * labelscalefactor); dot((-0.29,-8.22),linewidth(4pt) + dotstyle); label("$D$", (-0.53,-8.62), NE * labelscalefactor); dot((0.03,0.52),linewidth(4pt) + dotstyle); label("$P$", (-0.13,0.67), NE * labelscalefactor); dot((0.89,1.04),linewidth(4pt) + dotstyle); label("$Q$", (0.62,1.16), NE * labelscalefactor); dot((5.61,-7.22),linewidth(4pt) + dotstyle); label("$R$", (5.70,-7.05), NE * labelscalefactor); dot((1.67,-7.89),linewidth(4pt) + dotstyle); label("$S$", (1.75,-7.73), NE * labelscalefactor); dot((0.46,0.78),linewidth(4pt) + dotstyle); label("$E$", (0.26,0.93), NE * labelscalefactor); dot((3.64,-7.55),linewidth(4pt) + dotstyle); label("$F$", (3.73,-7.39), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] By properties of cyclic quadrilaterals, we know that the circumcenter of a cyclic quadrilateral is the intersection of its sides' perpendicular bisectors. This implies that $OE\perp AB$ and $OF\perp CD$. Since $E$ and $F$ are also bisectors of $PQ$ and $RS$, respectively, if $PQRS$ is indeed a cyclic quadrilateral, then its circumcenter is also at $O$. Thus, it suffices to show that $OP=OQ=OR=OS$. Notice that $PE=QE$, $EO=EO$, and $\angle QEO=\angle PEO=90^\circ$. By SAS congruency, $\Delta QOE\cong\Delta POE\implies QO=PO$. Similarly, we find that $\Delta SOF\cong\Delta ROF$ and $OS=OR$. We now need only to show that these two pairs are equal to each other. Draw the segments connecting $O$ to $B$, $Q$, $C$, and $R$. [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 = -12.19, xmax = 24.94, ymin = -15.45, ymax = 6.11; /* image dimensions */ pen wrwrwr = rgb(0.38,0.38,0.38); /* draw figures */ draw(circle((2.92,-3.28), 5.90), linewidth(2) + wrwrwr); draw((-2.52,-1.01)--(3.46,2.59), linewidth(2) + wrwrwr); draw((7.59,-6.88)--(-0.29,-8.22), linewidth(2) + wrwrwr); draw((3.46,2.59)--(7.59,-6.88), linewidth(2) + wrwrwr); draw((-0.29,-8.22)--(-2.52,-1.01), linewidth(2) + wrwrwr); draw((0.03,0.52)--(1.67,-7.89), linewidth(2) + wrwrwr); draw((5.61,-7.22)--(0.89,1.04), linewidth(2) + wrwrwr); draw((0.46,0.78)--(2.92,-3.28), linewidth(2) + wrwrwr); draw((2.92,-3.28)--(3.64,-7.55), linewidth(2) + wrwrwr); draw((2.92,-3.28)--(7.59,-6.88), linewidth(2) + wrwrwr); draw((5.61,-7.22)--(2.92,-3.28), linewidth(2) + wrwrwr); draw((2.92,-3.28)--(3.46,2.59), linewidth(2) + wrwrwr); draw((2.92,-3.28)--(0.89,1.04), linewidth(2) + wrwrwr); /* dots and labels */ dot((2.92,-3.28),dotstyle); label("$O$", (2.43,-3.56), NE * labelscalefactor); dot((-2.52,-1.01),dotstyle); label("$A$", (-2.91,-0.91), NE * labelscalefactor); dot((3.46,2.59),linewidth(1pt) + dotstyle); label("$B$", (3.49,2.78), NE * labelscalefactor); dot((7.59,-6.88),dotstyle); label("$C$", (7.82,-7.24), NE * labelscalefactor); dot((-0.29,-8.22),linewidth(1pt) + dotstyle); label("$D$", (-0.53,-8.62), NE * labelscalefactor); dot((0.03,0.52),linewidth(1pt) + dotstyle); label("$P$", (-0.13,0.67), NE * labelscalefactor); dot((0.89,1.04),linewidth(1pt) + dotstyle); label("$Q$", (0.62,1.16), NE * labelscalefactor); dot((5.61,-7.22),linewidth(1pt) + dotstyle); label("$R$", (5.70,-7.05), NE * labelscalefactor); dot((1.67,-7.89),linewidth(1pt) + dotstyle); label("$S$", (1.75,-7.73), NE * labelscalefactor); dot((0.46,0.78),linewidth(1pt) + dotstyle); label("$E$", (0.26,0.93), NE * labelscalefactor); dot((3.64,-7.55),linewidth(1pt) + dotstyle); label("$F$", (3.73,-7.39), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Also, let $r$ be the circumradius of $ABCD$. This means that $AO=BO=CO=DO=r$. Recall that $\angle BEO=90^\circ$ and $\angle CFO=90^\circ$. Notice the several right triangles in our figure. Let us apply Pythagorean Theorem on $\Delta BEO$. We can see that $EO^2+EB^2=BO^2\implies EO^2+3.5^2=r^2\implies EO=\sqrt{r^2-12.25}.$ Let us again apply Pythagorean Theorem on $\Delta QEO$. We can see that $QE^2+EO^2=QO^2\implies0.5^2+r^2-12.25=QO^2\implies QO=\sqrt{r^2-12}.$ Let us apply Pythagorean Theorem on $\Delta CFO$. We get $CF^2+OF^2=OC^2\implies4^2+OF^2=r^2\implies OF=\sqrt{r^2-16}$. We finally apply Pythagorean Theorem on $\Delta RFO$. This becomes $OF^2+FR^2=OR^2\implies r^2-16+2^2=OR^2\implies OR=\sqrt{r^2-12}$. This is the same expression as we got for $QO$. Thus, $OQ=OR$, and recalling that $OQ=OP$ and $OR=OS$, we have shown that $OP=OQ=OR=OS$. We are done. QED

of day 1 oops

Denote the center of $(ABCD)$ as $O$, the radius of $(ABCD)$ as $R$. As the problem states, we have $OP^2=R^2-(\frac{7}{2})^2+(\frac{1}{2})^2=R^2-12=OQ^2; OS^2=R^2-4^2+2^2=R^2-12=OR^2=OP^2=OQ^2$. Thus, $P,Q,R,S$ are concyclic with center $O$. Aime problem 1 more likely

why maa, why.

Two approaches, which probably constitute upwards of 95% of submitted solutions. Perpendicular Bisectors: Let $O$, $R$ denote the center and radius of $(ABCD)$. Drawing the perpendicular bisectors of $AB$ and $CD$, we find that $O$ is simply the center of $(PQRS)$, as \[OP = OQ = R^2 - \left(\frac 72\right)^2 + \left(\frac 12\right)^2 = R^2 - 4^2 + 2^2 = OR = OS. \quad \blacksquare\] Power of a Point: If $AB \parallel CD$, notice $PQRS$ is an isosceles trapezoid. Otherwise, consider the power of $AB \cap CD$, and conclude by noting \[x(x+7) = y(y+8) \iff (x+3)(x+4) = (y+2)(y+6). \quad \blacksquare\]

Headsolved in 1 minute. This must have came from the AMC 8 Shortlist Solution: Note that the perpendicular bisectors of the chords are the same as the perpendicular bisectors of $PQ$ and $RS$, so if the claim is true, then the original circle's center must be the circumcenter. We only need to show that $OP = OQ = OR = OS$, which is equivalent to $(R^2-3.5^2)+0.5^2 = (R^2-4^2)+2^2$, which is true.

extend + intersect + pop kills only solve of day 1

for PoP solution people: don't forget AB parallel to CD configuration there's no intersection when extended instead it's just isosceles trapezoid ---> cyclic

This problem is completely ridiculous. Drawing the circumcircle of $ABCD$ with center $O$ and the perpendicular bisectors to each side reveals that the perpendicular bisector from $O$ to $AB$ is also the perpendicular bisector from $O$ to $PQ$, and the perpendicular bisector from $O$ to $BC$ is also the perpendicular bisector from $O$ to $RS$. It remains to show these two lengths are equal, which can be done with simple length chasing. Thus $PQRS$ is cyclic, done.

AlexWin0806 wrote: for PoP solution people: don't forget AB parallel to CD configuration there's no intersection when extended instead it's just isosceles trapezoid ---> cyclic how many points docked if you didnt include it?

AlexWin0806 wrote: for PoP solution people: don't forget AB parallel to CD configuration there's no intersection when extended instead it's just isosceles trapezoid ---> cyclic Oh boy.

I literally just started my journey into oly and head-solve-sketch-ed it... I missed JMO slightly but it's good to know I could've kept my "nonzero score" promise pretty easy

wait. which thread is the actual usajmo p1 thread? prob not this one cuz title and forum rules? at this rate, everyone who qualed jmo should be getting Honorable Mentions at least.

They wanted to give cheaters something even they could solve :skull: Let AB, CD have midpoints M, N, respectively. $AM^2+MO^2 = DN^2+NO^2$. But then $PM^2+MO^2 = AM^2+MO^2 - 12 = DN^2+NO^2 - 12 = SN^2 + NO^2$ and we’re done.

no joke, MAA is more considerate of cheaters than honest hardworking students. Disqualifying honest hardworking students? Totally ok. Exposing and removing cheaters? No! violation of their privacy. jellybeanzzz wrote: They wanted to give cheaters something even they could solve :skull:

Solution

i was so confused when i saw this yesterday...

Observe that $P,Q,R,S$ all have equal power $-12$ with respect to the circle through $A,B,C,D$, so they lie on a circle concentric with the previously mentioned circle. $\square$

If $\overline{AB} \parallel \overline{CD}$ then the conclusion follows as $PQRS$ is an isosceles trapezoid. Otherwise let $X = \overline{AB} \cap \overline{CD}$ and note that, \begin{align*} XA \cdot XB &= XC \cdot XD\\ \iff (XB + 7) \cdot XB &= XC \cdot (XC + 8) \end{align*}Now it suffices to show that, \begin{align*} XP \cdot XQ &= XR \cdot XS\\ \iff (XB + 4) \cdot (XB + 3) &= (XC + 2) \cdot (XC + 6)\\ \iff (XB + 7) \cdot XB &= XC \cdot (XC + 8) \end{align*}so the conclusion follows.

Whatttt ! Solved within few minutes ! Solution 2 is shorter than Solution 1 Solution 1 : Let $O$ be the center of circle $(ABCD)$ . Join $OA,OB,OP,OQ$ . We know $OA=OB$. Now Using Stewart's theorem observe that $$ AB(OP^2+AP.BP)=OA^2.BP+OB^2.AP \Longrightarrow 7(OP^2+12)=OA^2(BP+AP)=7OA^2 \Longrightarrow OP^2=OA^2-12$$Similarly Using Stewart's theorem we also get $$AB(OQ^2+AQ.BQ)=OA^2.BQ+OB^2.AQ \Longrightarrow 7(OQ^2+12)=OA^2(BQ+AQ)=7OA^2 \Longrightarrow OQ^2=OA^2-12$$Therefore $OP=OQ$ and in same way we get $OR=OS$ Now drop $OY\perp CD$ so now we know $CY=DY=4$. Using Pythagorus Theorem we get $$OS^2=R^2-(4)^2+(2)^2=R^2-12 \Longrightarrow OP=OQ=OR=OS$$Therefore, $PQRS$ is a cyclic quadrilateral. Solution 2 : Let $AB \cap CD = Z$ . By PoP, we get $$ZA.ZB=ZC.ZD \Longrightarrow ZA(ZA+7)=ZD(ZD+8)$$$$\Longrightarrow ZA^2+7ZA+12=ZD^2+8ZD+12 \Longrightarrow (ZA+3)(ZA+4)=(ZD+2)(ZD+6) $$$$\Longrightarrow ZP.ZQ=ZS.ZR$$Hence $PQRS$ is a cyclic quadrilateral.

Solution from Twitch Solves ISL: We distinguish between two cases. Case where $AB$ and $CD$ are not parallel. We let lines $AB$ and $CD$ meet at $T$. Without loss of generality, $A$ lies between $B$ and $T$ and $D$ lies between $C$ and $T$. Let $x=TA$ and $y=TD$, as shown below. [asy][asy] size(8cm); pair A = dir(170); pair B = dir(80); pair C = dir(323.912853); pair D = dir(216.087147); pair T = extension(A, B, C, D); filldraw(unitcircle, invisible, blue); draw(B--T--C--B, blue); draw(A--D, blue); pair P = 4*A/7+3*B/7; pair Q = 3*A/7+4*B/7; pair R = 6*C/8+2*D/8; pair S = 2*C/8+6*D/8; filldraw(circumcircle(P, Q, R), invisible, dashed+red); label("$x$", midpoint(T--A), dir(A+B), deepgreen); label("$3$", midpoint(A--P), dir(A+B), deepgreen); label("$1$", midpoint(P--Q), -dir(A+B), deepgreen); label("$3$", midpoint(Q--B), dir(A+B), deepgreen); label("$y$", midpoint(T--D), dir(C+D), deepgreen); label("$2$", midpoint(D--S), dir(C+D), deepgreen); label("$4$", midpoint(S--R), -dir(C+D), deepgreen); label("$2$", midpoint(R--C), dir(C+D), deepgreen); dot("$A$", A, dir(120)); dot("$B$", B, dir(B)); dot("$C$", C, dir(300)); dot("$D$", D, dir(240)); dot("$T$", T, dir(T)); dot("$P$", P, dir(160)); dot("$Q$", Q, dir(80)); dot("$R$", R, dir(270)); dot("$S$", S, dir(270)); /* -----------------------------------------------------------------+ | TSQX: by CJ Quines and Evan Chen | | https://github.com/vEnhance/dotfiles/blob/main/py-scripts/tsqx.py | +-------------------------------------------------------------------+ A 120 = dir 170 B = dir 80 C 300 = dir 323.912853 D 240 = dir 216.087147 T = extension A B C D unitcircle / 0.1 lightcyan / blue B--T--C--B / blue A--D / blue P 160 = 4*A/7+3*B/7 Q 80 = 3*A/7+4*B/7 R 270 = 6*C/8+2*D/8 S 270 = 2*C/8+6*D/8 circumcircle P Q R / 0.1 yellow / dashed red !label("$x$", midpoint(T--A), dir(A+B), deepgreen); !label("$3$", midpoint(A--P), dir(A+B), deepgreen); !label("$1$", midpoint(P--Q), -dir(A+B), deepgreen); !label("$3$", midpoint(Q--B), dir(A+B), deepgreen); !label("$y$", midpoint(T--D), dir(C+D), deepgreen); !label("$2$", midpoint(D--S), dir(C+D), deepgreen); !label("$4$", midpoint(S--R), -dir(C+D), deepgreen); !label("$2$", midpoint(R--C), dir(C+D), deepgreen); */ [/asy][/asy] By power of a point, \begin{align*} \text{$ABCD$ cyclic} \iff x(x+7) &= y(y+8) \\ \text{$PQRS$ cyclic} \iff (x+3)(x+4) &= (y+2)(y+6). \end{align*}However, the latter equation is just the former with $12$ added to both sides. (That is, $(x+3)(x+4) = x(x+7)+12$ while $(y+2)(y+6)=y(y+8)+12$.) So the conclusion is immediate. Case where $AB$ and $CD$ are parallel. In that case $ABCD$ is an isosceles trapezoid. Then the entire picture is symmetric around the common perpendicular bisector of the lines $AB$ and $CD$. Now $PQRS$ is also an isosceles trapezoid, so it's cyclic too. [asy][asy] size(4cm); pair A = dir(135); pair B = dir(45); pair C = dir(323.912853); pair D = dir(216.087147); filldraw(unitcircle, invisible, blue); draw(A--B--C--D--cycle, blue); pair P = 4*A/7+3*B/7; pair Q = 3*A/7+4*B/7; pair R = 6*C/8+2*D/8; pair S = 2*C/8+6*D/8; filldraw(circumcircle(P, Q, R), invisible, dashed+red); dot("$A$", A, dir(120)); dot("$B$", B, dir(B)); dot("$C$", C, dir(300)); dot("$D$", D, dir(240)); dot("$P$", P, dir(90)); dot("$Q$", Q, dir(80)); dot("$R$", R, dir(270)); dot("$S$", S, dir(270)); [/asy][/asy]

feel like it's too easy

thought process during problem: 1. its not barybashable 2. tried to complex bash. i forgot all the formulas and stuff so i rederived cylic quad condition. ended up with no conjugates stuff. i dont know how to complex bash facepalm 3. found easy pythag sol

KevinYang2.71 wrote: thought process during problem: 1. its not barybashable 2. tried to complex bash. i forgot all the formulas and stuff so i rederived cylic quad condition. ended up with no conjugates stuff. i dont know how to complex bash facepalm 3. found easy pythag sol relatable

Notice that we can use Pythagorean Theorem to show that $OP^2=OQ^2=OR^2=OS^2=r^2-12$, where $r$ is the circumradius of $(ABCD)$.

Glad to see that Evan also found the PoP solution to be more natural than the unintuitive and inferior Pythagorean Theorem solution, which requires testers to observe the difficult equality $4^2-2^2 = 3.5^2-0.5^2$.

Naive power of a point for $P$ and $R$ along diameters works too (to get that the distance to circle and therefore distance to center is the same for both sides' endpoints) without even knowing (essentially, we're proving) the theorem that having a common power implies they form a concentric circle.

For God's sake is this AMC 8?

Let $\omega$ be the circumcircle of quadrilateral $ABCD$. Then the powers of $P,Q,R,S$ with respect to $\omega$ are all $-12$, so they are concyclic on a circle concentric with $\omega$.

Trivial by PoP. $a(a+7)=b(b+8)$ implies $(a+3)(a+4)=(b+2)(b+6)$.

So can you guys help me out with the Nigerian math olympiad 2023 it's missing

USAJMO used to have some more general problems, not computational right? $-------------------------------------------------------$ Let $AB\cap CD=T$. Suppose that $TB=a$ and $TC=b$. Then $a(a+7)=b(b+8)$ which implies $(a+3)(a+4)=(b+2)(b+6)$. The rest follows easily by Reverse PoP as desired.