Difference between revisions of "2024 USAJMO Problems/Problem 1"

Latest revision as of 09:38, 24 April 2024

Problem

Let $ABCD$ be a cyclic quadrilateral with $AB = 7$ and $CD = 8$ . Points $P$ and $Q$ are selected on segment $AB$ such that $AP = BQ = 3$ . Points $R$ and $S$ are selected on segment $CD$ such that $CR = DS = 2$ . Prove that $PQRS$ is a cyclic quadrilateral.

Solution 1

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 $ABCD$ .

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -12.19, xmax = 24.94, ymin = -15.45, ymax = 6.11; /* image dimensions */ pen wrwrwr = rgb(0.38,0.38,0.38); /* draw figures */ draw(circle((2.92,-3.28), 5.90), linewidth(2) + wrwrwr); draw((-2.52,-1.01)--(3.46,2.59), linewidth(2) + wrwrwr); draw((7.59,-6.88)--(-0.29,-8.22), linewidth(2) + wrwrwr); draw((3.46,2.59)--(7.59,-6.88), linewidth(2) + wrwrwr); draw((-0.29,-8.22)--(-2.52,-1.01), linewidth(2) + wrwrwr); draw((0.03,0.52)--(1.67,-7.89), linewidth(2) + wrwrwr); draw((5.61,-7.22)--(0.89,1.04), linewidth(2) + wrwrwr); /* dots and labels */ dot((2.92,-3.28),dotstyle); label("$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$", (3.73,-7.39), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$

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 $R$ .

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -12.19, xmax = 24.94, ymin = -15.45, ymax = 6.11; /* image dimensions */ pen wrwrwr = rgb(0.38,0.38,0.38); /* draw figures */ draw(circle((2.92,-3.28), 5.90), linewidth(2) + wrwrwr); draw((-2.52,-1.01)--(3.46,2.59), linewidth(2) + wrwrwr); draw((7.59,-6.88)--(-0.29,-8.22), linewidth(2) + wrwrwr); draw((3.46,2.59)--(7.59,-6.88), linewidth(2) + wrwrwr); draw((-0.29,-8.22)--(-2.52,-1.01), linewidth(2) + wrwrwr); draw((0.03,0.52)--(1.67,-7.89), linewidth(2) + wrwrwr); draw((5.61,-7.22)--(0.89,1.04), linewidth(2) + wrwrwr); draw((0.46,0.78)--(2.92,-3.28), linewidth(2) + wrwrwr); draw((2.92,-3.28)--(3.64,-7.55), linewidth(2) + wrwrwr); draw((2.92,-3.28)--(7.59,-6.88), linewidth(2) + wrwrwr); draw((5.61,-7.22)--(2.92,-3.28), linewidth(2) + wrwrwr); draw((2.92,-3.28)--(3.46,2.59), linewidth(2) + wrwrwr); draw((2.92,-3.28)--(0.89,1.04), linewidth(2) + wrwrwr); /* dots and labels */ dot((2.92,-3.28),dotstyle); label("$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$", (3.73,-7.39), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$

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

~Technodoggo

Solution 2

We can consider two cases: $AB \parallel CD$ or $AB \nparallel CD.$ The first case is trivial, as $PQ \parallel RS$ and we are done due to symmetry. For the second case, WLOG, assume that $A$ and $C$ are located on $XB$ and $XD$ respectively. Extend $AB$ and $CD$ to a point $X,$ and by Power of a Point, we have $XA\cdot XB = XC \cdot XD,$ which may be written as $XA \cdot (XA+7) = XC \cdot (XC+8),$ or $XA^2 + 7XA = XC^2 + 8XC.$ We can translate this to $XA^2 + 7XA +12 = XC^2 + 8XC +12,$ so $XP\cdot XQ = (XA+3)(XA+4)=(XC+2)(XC+6)= XR\cdot XS,$ and therefore by the Converse of Power of a Point $PQRS$ is cyclic, and we are done.

- spectraldragon8

Solution 3

All 4 corners of $PQRS$ have equal power of a point ( $12$ ) with respect to the circle $(ABCD)$ , with center $O$ .

Draw diameters (of length $AQ$ ) of circle $(ABCD)$ through $Q$ and $S$ , with length $A$ . Let $q$ be the distance from $Q$ to the circle along a diameter, and likewise $s$ be distance from $S$ to the circle.

Then $q(AQ-q) = s(AQ-s) = 12$ and $q,s < AQ/2$ (radius). Therefore, $q=s$ and $AQ/2 -q = AQ/2 -s$ . But $AQ-q=OQ$ , $AQ-s=OS$ , $OQ = OP$ and $OS = OR$ by symmetry around the perpendicular bisectors of $PQ$ and $RS$ , so $P,Q,R,S$ are all equidistant from $O$ , forming a circumcircle around $PQRS$ .

-BraveCobra22aops and oinava

Solution 4 (Coord Bash)

[Will add when have time] ~KevinChen_Yay

2024 USAJMO (Problems • Resources)
Preceded by First Question	Followed by Problem 2
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All USAJMO Problems and Solutions

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