2024 USAJMO Problems/Problem 1

Problem

sus

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

2024 USAJMO (Problems • Resources)
Preceded by First Question	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6
All USAJMO Problems and Solutions

Page

Toolbox

Search

2024 USAJMO Problems/Problem 1

Contents

Problem

Solution 1

See Also