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2024 USAJMO Problems/Problem 1

Revision as of 19:45, 22 March 2024 by Virjoy2001 (talk | contribs) (→‎Solution 1)

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. 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

See Also

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

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