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

(→‎Solution 4 (Coord Bash))
(→‎Solution 4 (Coord Bash))
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Let's list what we know from the givens:
 
Let's list what we know from the givens:
−
\begin{itemize}
+
Since the radii of a circle are equal in length, we can let <cmath>k=a_1^2+a_2^2=b_1^2+b_2^2=c_1^2+c_2^2=d_1^2+d_2^2.</cmath>
−
    \item Since the radii of a circle are equal in length, we can let
 
−
    \begin{equation}
 
−
        k=a_1^2+a_2^2=b_1^2+b_2^2=c_1^2+c_2^2=d_1^2+d_2^2.
 
−
    \end{equation}
 
−
   
 
−
    \item From the distance formula on <math>AB</math> and <math>CD</math>, we can simplify and get
 
−
    \begin{equation}
 
−
        a_1^2+a_2^2+b_1^2+b_2^2-2(a_1b_1+a_2b_2)=\frac{49}{4} \\
 
−
        \Rightarrow 2(a_1b_1+a_2b_2)=2k-\frac{49}{4}
 
−
    \end{equation}
 
−
    and
 
−
    \begin{equation}
 
−
        c_1^2+c_2^2+d_1^2+d_2^2-2(c_1d_1+c_2d_2)=16 \\
 
−
        \Rightarrow 2(c_1d_1+c_2d_2)=2k-16
 
−
    \end{equation}
 
−
    using the above substitution with <math>k</math>.
 
  
−
    \item Since <math>P</math>, <math>Q</math>, <math>R</math>, and <math>S</math> are weighted points on <math>AB</math> and <math>CD</math>, we can get <cmath>P\left(\frac{8a_1+6b_1}{7},\frac{8a_2+6b_2}{7}\right) \ \text{and} \ Q\left(\frac{6a_1+8b_1}{7},\frac{6a_2+8b_2}{7}\right)</cmath> along with <cmath>R\left(\frac{3c_1+d_1}{2},\frac{3c_2+d_2}{2}\right) \ \text{and} \ S\left(\frac{c_1+3d_1}{2},\frac{c_1+3d_1}{2}\right).</cmath>
+
From the distance formula on <math>AB</math> and <math>CD</math>, we can simplify and get <cmath>a_1^2+a_2^2+b_1^2+b_2^2-2(a_1b_1+a_2b_2)=\frac{49}{4} \Rightarrow 2(a_1b_1+a_2b_2)=2k-\frac{49}{4}</cmath> and <cmath>c_1^2+c_2^2+d_1^2+d_2^2-2(c_1d_1+c_2d_2)=16 \Rightarrow 2(c_1d_1+c_2d_2)=2k-16</cmath> using the above substitution with <math>k</math>.
−
\end{itemize}
+
 
 +
Since <math>P</math>, <math>Q</math>, <math>R</math>, and <math>S</math> are weighted points on <math>AB</math> and <math>CD</math>, we can get <cmath>P\left(\frac{8a_1+6b_1}{7},\frac{8a_2+6b_2}{7}\right) \ \text{and} \ Q\left(\frac{6a_1+8b_1}{7},\frac{6a_2+8b_2}{7}\right)</cmath> along with <cmath>R\left(\frac{3c_1+d_1}{2},\frac{3c_2+d_2}{2}\right) \ \text{and} \ S\left(\frac{c_1+3d_1}{2},\frac{c_1+3d_1}{2}\right).</cmath>
  
 
Now, we want to show that the perpendicular bisectors of the sides of quadrilateral <math>PQRS</math> are concurrent to prove that it is cyclic. Moreover, we suspect that the circumcenter of quadrilateral <math>PQRS</math> is the circumcenter of quadrilateral <math>ABCD</math>, both of them being <math>(0,0)</math>.
 
Now, we want to show that the perpendicular bisectors of the sides of quadrilateral <math>PQRS</math> are concurrent to prove that it is cyclic. Moreover, we suspect that the circumcenter of quadrilateral <math>PQRS</math> is the circumcenter of quadrilateral <math>ABCD</math>, both of them being <math>(0,0)</math>.
Line 170: Line 155:
 
     \Rightarrow k-3&=k-3,
 
     \Rightarrow k-3&=k-3,
 
\end{align*}
 
\end{align*}
−
which is true, implying that (0,0) does satisfy the equation.
+
which is true, implying that <math>(0,0)</math> does indeed satisfy the equation.
  
 
Analogously, we can show that the perpendicular bisector of line <math>QR</math> also passes through the origin.
 
Analogously, we can show that the perpendicular bisector of line <math>QR</math> also passes through the origin.

Revision as of 17:20, 29 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)

Let $A(2a_1,2a_2)$, $B(2b_1,2b_2)$, $C(2c_1,2c_2)$, $D(2d_1,2d_2)$, and the circumcircle center be $O(0,0)$.

Let's list what we know from the givens: Since the radii of a circle are equal in length, we can let \[k=a_1^2+a_2^2=b_1^2+b_2^2=c_1^2+c_2^2=d_1^2+d_2^2.\]

From the distance formula on $AB$ and $CD$, we can simplify and get \[a_1^2+a_2^2+b_1^2+b_2^2-2(a_1b_1+a_2b_2)=\frac{49}{4} \Rightarrow 2(a_1b_1+a_2b_2)=2k-\frac{49}{4}\] and \[c_1^2+c_2^2+d_1^2+d_2^2-2(c_1d_1+c_2d_2)=16 \Rightarrow 2(c_1d_1+c_2d_2)=2k-16\] using the above substitution with $k$.

Since $P$, $Q$, $R$, and $S$ are weighted points on $AB$ and $CD$, we can get \[P\left(\frac{8a_1+6b_1}{7},\frac{8a_2+6b_2}{7}\right) \ \text{and} \ Q\left(\frac{6a_1+8b_1}{7},\frac{6a_2+8b_2}{7}\right)\] along with \[R\left(\frac{3c_1+d_1}{2},\frac{3c_2+d_2}{2}\right) \ \text{and} \ S\left(\frac{c_1+3d_1}{2},\frac{c_1+3d_1}{2}\right).\]

Now, we want to show that the perpendicular bisectors of the sides of quadrilateral $PQRS$ are concurrent to prove that it is cyclic. Moreover, we suspect that the circumcenter of quadrilateral $PQRS$ is the circumcenter of quadrilateral $ABCD$, both of them being $(0,0)$.

We only list the linear equations of the perpendicular bisectors of lines $PS$ and $QR$, as the perpendicular bisectors of lines $PQ$ and $RS$ are the same as the perpendicular bisectors of lines $AB$ and $CD$, respectively.

First, we analyze the perpendicular bisector of line $PS$. Notice that it has slope $-\frac{16a_1+12b_1-7c_1-21d_1}{16a_2+12b_2-7c_2-21d_2}$ and point $(\frac{16a_1+12b_1+7c_1+21d_1}{28}, \frac{16a_2+12b_2+7c_2+21d_2}{28})$. The point-slope form equation would thus be \[y-\frac{16a_2+12b_2+7c_2+21d_2}{28}=-\frac{16a_1+12b_1-7c_1-21d_1}{16a_2+12b_2-7c_2-21d_2}(x-\frac{16a_1+12b_1+7c_1+21d_1}{28}).\] Since we claim that $(0,0)$ is on the line, we substitute to get \[(16a_1+12b_1+7c_1+21d_1)(16a_1+12b_1-7c_1-21d_1)=-(16a_2+12b_2+7c_2+21d_2)(16a_2+12b_2-7c_2-21d_2),\] and this simplifies to \[16\left(16(a_1^2+a_2^2)+24(a_1b_1+a_2b_2)+9(b_1^2+b_2^2)\right)=49\left((c_1^2+c_2^2)+6(c_1d_1+c_2d_2)+9(d_1^2+d_2^2)\right).\] Using substitutions from equations (1), (2), and (3), this becomes \begin{align*} 

   16(16k+12(2k-\frac{49}{4})+9k)&=49(k+3(2k-16)+9k) \\
   \Rightarrow 16(49k-3\cdot49)&=49(16k-3\cdot16) \\
   \Rightarrow k-3&=k-3,

\end{align*} which is true, implying that $(0,0)$ does indeed satisfy the equation.

Analogously, we can show that the perpendicular bisector of line $QR$ also passes through the origin.

Since the perpendicular bisectors of the sides of quadrilateral $PQRS$ all intersect at the same point, namely (0,0), which is also the circumcenter of quadrilateral $ABCD$, we can conclude that $PQRS$ is a cyclic quadrilateral.

~KevinChen_Yay

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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