Difference between revisions of "2009 USAMO Problems/Problem 5"

Revision as of 01:23, 2 October 2018

Problem

Trapezoid $ABCD$ , with $\overline{AB}||\overline{CD}$ , is inscribed in circle $\omega$ and point $G$ lies inside triangle $BCD$ . Rays $AG$ and $BG$ meet $\omega$ again at points $P$ and $Q$ , respectively. Let the line through $G$ parallel to $\overline{AB}$ intersect $\overline{BD}$ and $\overline{BC}$ at points $R$ and $S$ , respectively. Prove that quadrilateral $PQRS$ is cyclic if and only if $\overline{BG}$ bisects $\angle CBD$ .

Solution 1

We will use directed angles in this solution. Extend $QR$ to $T$ as follows:

$[asy] import cse5; import graph; import olympiad; dotfactor = 3; unitsize(1.5inch); path circle = Circle(origin, 1); draw(circle); pair A = (-.6, .8), B = (.6, .8), C = (.9, -sqrt(.19)), D = (-.9, -sqrt(.19)), G = bisectorpoint(C, B, D); draw(A--B--C--D--cycle); draw(B--D); dot("$A$", A, NW); dot("$B$", B, NE); dot("$C$", C, SE); dot("$D$", D, SW); dot("$G$", G, dir(40)); pair P = IP(L(A, G, 10, 10), circle, 1), Q = IP(L(B, G, 10, 10), circle, 1); draw(A--P--C); draw(B--Q); dot("$P$", P, SE); dot("$Q$", Q, S); pair R = IP((-1, G.y)--(1, G.y), B--D), S = IP((-1, G.y)--(1, G.y), B--C); draw(P--Q--R--S--cycle); dot("$R$", R, N); dot("$S$", S, E); pair T = IP(L(Q, R, 10, 10), circle, 1); draw(R--T--C, dashed); draw(T--B, dashed); dot("$T$", T, NW); [/asy]$

If:

Note that $\begin{align*}\measuredangle GBT+\measuredangle TRG&=\frac{m\widehat{TQ}}{2}+\measuredangle TRB+\measuredangle BRG\\ &=\frac{m\widehat{TQ}+m\widehat{DQ}+m\widehat{CB}+m\widehat{BT}}{2}.\\ \end{align*}$ Thus, $BTRG$ is cyclic.

Also, note that $GSCP$ is cyclic because $\begin{align*}\measuredangle CSG+\measuredangle GPC&=\measuredangle CBA+\measuredangle APC\\ &=180^\circ\text{ or }0^\circ, \end{align*}$ depending on the configuration.

Next, we have $T, G, C$ are collinear since $\measuredangle GTR=\measuredangle GBR=\frac{m\widehat{DQ}}{2}=\frac{m\widehat{QC}}{2}=\measuredangle CTQ.$

Therefore, $\begin{align*}\measuredangle RQP+\measuredangle PSR&=\frac{m\widehat{PBT}}{2}+\measuredangle PCG\\ &=\frac{m\widehat{PBT}+m\widehat{TDP}}{2}\\ &=180^\circ \end{align*},$ so $PQRS$ is cyclic.

Only If: These steps can be reversed.

Solution 2 (Projective)

Extend $QR$ to $K$ , and let line $l \parallel AB$ intersect $\omega$ at $K$ and another point $H$ , as shown:

$[asy] import cse5; import graph; import olympiad; dotfactor = 3; unitsize(1.5inch); path circle = Circle(origin, 1); draw(circle); pair A = (-.6, .8), B = (.6, .8), C = (.9, -sqrt(.19)), D = (-.9, -sqrt(.19)), G = bisectorpoint(C, B, D); draw(A--B--C--D--cycle); draw(B--D); dot("$A$", A, NW); dot("$B$", B, NE); dot("$C$", C, SE); dot("$D$", D, SW); dot("$G$", G, dir(40)); pair P = IP(L(A, G, 10, 10), circle, 1), Q = IP(L(B, G, 10, 10), circle, 1); draw(A--P); draw(B--Q); dot("$P$", P, SE); dot("$Q$", Q, S); pair R = IP((-1, G.y)--(1, G.y), B--D), S = IP((-1, G.y)--(1, G.y), B--C); draw(P--Q--R--S); dot("$R$", R, N); dot("$S$", S, E); pair T = IP(L(Q, R, 10, 10), circle, 1); draw(Q--T); pair V = IP(L(P, S, 10, 10), circle, 1); draw(T--V); draw(P--V, dotted); dot("$T$", T, NW); dot("$V$", V, NE); [/asy]$

If:

Suppose that $VP \cap CB = S'$ , and $AC \cap QV = R'$ . Pascal's theorem on the tuple $(V, P, A, C, B, Q)$ implies that the points $S'$ , $R'$ , and $G = PA \cap BQ$ are collinear. However, $AC$ and $BD$ are symmetrical with respect to the axis of symmetry of trapezoid $ABCD$ , and $TQ$ and $VQ$ are also symmetrical with respect to the axis of symmetry of $ABCD$ (as $Q$ is the midpoint of $\overset{\frown}{DC}$ , and $TV \parallel DC$ ). Since $R = BD \cap TQ$ , $R$ and $R'$ are symmetric with respect to the axis of symmetry of trapezoid $ABCD$ . This implies that line $R'G$ is equivalent to line $RG$ . Thus, $S'$ lies on line $RG$ . However, $S = BC \cap RG$ , so this implies that $S' = S$ .

Now note that $TVPQ$ is cyclic. Since $TV \parallel RS$ , $\measuredangle VTQ = \measuredangle SRQ$ . However, $\measuredangle VTQ + \measuredangle VPQ = 180^{\circ} = \measuredangle SRQ + \measuredangle SPQ$ . Therefore, $PQRS$ is cyclic.

Only If:

Consider the same setup, except $Q$ is no longer the midpoint of $\overset{\frown}{DC}$ . Note that $TV$ must be parallel to $RG$ in order for $PQRS$ to be cyclic. We claim that $S' = S$ and hope to reach a contradiction. Pascal's theorem on the tuple $(V, P, A, C, B, Q)$ implies that $S'$ , $R'$ , and $G = PA \cap BQ$ are collinear. However, there exists a unique point $Q$ such that $HQ$ , $AC$ , and $RG$ are concurrent. By If, $Q$ must be the midpoint of $\overset{\frown}{DC}$ in order for the concurrency to occur; hence, $R' \notin RS$ . Then $R'G \cap BC = S' \neq S$ , since $RG \cap BC = S$ . However, this is a contradiction, so therefore $TV$ cannot be parallel to $RG$ and $PQRS$ is not cyclic.

Solution by TheBoomBox77

2009 USAMO (Problems • Resources)
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Difference between revisions of "2009 USAMO Problems/Problem 5"

Revision as of 01:23, 2 October 2018

Contents

Problem

Solution 1

Solution 2 (Projective)

See Also

@@ Line 2: / Line 2: @@
 Trapezoid <math>ABCD</math>, with <math>\overline{AB}||\overline{CD}</math>, is inscribed in circle <math>\omega</math> and point <math>G</math> lies inside triangle <math>BCD</math>.  Rays <math>AG</math> and <math>BG</math> meet <math>\omega</math> again at points <math>P</math> and <math>Q</math>, respectively.  Let the line through <math>G</math> parallel to <math>\overline{AB}</math> intersect <math>\overline{BD}</math> and <math>\overline{BC}</math> at points <math>R</math> and <math>S</math>, respectively.  Prove that quadrilateral <math>PQRS</math> is cyclic if and only if <math>\overline{BG}</math> bisects <math>\angle CBD</math>.
-== Solution ==
+== Solution 1==
 We will use directed angles in this solution. Extend <math>QR</math> to <math>T</math> as follows:
 <center><asy>
@@ Line 51: / Line 51: @@
 '''Only If''':
 These steps can be reversed.
+== Solution 2 (Projective)==
+Extend <math>QR</math> to <math>K</math>, and let line <math>l \parallel AB</math> intersect <math>\omega</math> at <math>K</math> and another point <math>H</math>, as shown:
+<center><asy>
+import cse5;
+import graph;
+import olympiad;
+dotfactor = 3;
+unitsize(1.5inch);
+path circle = Circle(origin, 1);
+draw(circle);
+pair A = (-.6, .8), B = (.6, .8), C = (.9, -sqrt(.19)), D = (-.9, -sqrt(.19)), G = bisectorpoint(C, B, D);
+draw(A--B--C--D--cycle); draw(B--D);
+dot("$A$", A, NW); dot("$B$", B, NE); dot("$C$", C, SE); dot("$D$", D, SW); dot("$G$", G, dir(40));
+pair P = IP(L(A, G, 10, 10), circle, 1), Q = IP(L(B, G, 10, 10), circle, 1);
+draw(A--P); draw(B--Q);
+dot("$P$", P, SE); dot("$Q$", Q, S);
+pair R = IP((-1, G.y)--(1, G.y), B--D), S = IP((-1, G.y)--(1, G.y), B--C);
+draw(P--Q--R--S);
+dot("$R$", R, N); dot("$S$", S, E);
+pair T = IP(L(Q, R, 10, 10), circle, 1);
+draw(Q--T);
+pair V = IP(L(P, S, 10, 10), circle, 1);
+draw(T--V);
+draw(P--V, dotted);
+dot("$T$", T, NW);
+dot("$V$", V, NE);
+</asy></center>
+'''If''':
+Suppose that <math>VP \cap CB = S'</math>, and <math>AC \cap QV = R'</math>. Pascal's theorem on the tuple <math>(V, P, A, C, B, Q)</math> implies that the points <math>S'</math>, <math>R'</math>, and <math>G = PA \cap BQ</math> are collinear. However, <math>AC</math> and <math>BD</math> are symmetrical with respect to the axis of symmetry of trapezoid <math>ABCD</math>, and <math>TQ</math> and <math>VQ</math> are also symmetrical with respect to the axis of symmetry of <math>ABCD</math> (as <math>Q</math> is the midpoint of <math>\overset{\frown}{DC}</math>, and <math>TV \parallel DC</math>). Since <math>R = BD \cap TQ</math>, <math>R</math> and <math>R'</math> are symmetric with respect to the axis of symmetry of trapezoid <math>ABCD</math>. This implies that line <math>R'G</math> is equivalent to line <math>RG</math>. Thus, <math>S'</math> lies on line <math>RG</math>. However, <math>S = BC \cap RG</math>, so this implies that <math>S' = S</math>.
+Now note that <math>TVPQ</math> is cyclic. Since <math>TV \parallel RS</math>, <math>\measuredangle VTQ = \measuredangle SRQ</math>. However, <math>\measuredangle VTQ + \measuredangle VPQ = 180^{\circ} = \measuredangle SRQ + \measuredangle SPQ</math>. Therefore, <math>PQRS</math> is cyclic.
+'''Only If''':
+Consider the same setup, except <math>Q</math> is no longer the midpoint of <math>\overset{\frown}{DC}</math>. Note that <math>TV</math> must be parallel to <math>RG</math> in order for <math>PQRS</math>  to be cyclic. We claim that <math>S' = S</math> and hope to reach a contradiction. Pascal's theorem on the tuple <math>(V, P, A, C, B, Q)</math> implies that <math>S'</math>, <math>R'</math>, and <math>G = PA \cap BQ</math> are collinear. However, there exists a unique point <math>Q</math> such that <math>HQ</math>, <math>AC</math>, and <math>RG</math> are concurrent. By '''If''', <math>Q</math> must be the midpoint of <math>\overset{\frown}{DC}</math> in order for the concurrency to occur; hence, <math>R' \notin RS</math>. Then <math>R'G \cap BC = S' \neq S</math>, since <math>RG \cap BC = S</math>. However, this is a contradiction, so therefore <math>TV</math> cannot be parallel to <math>RG</math> and <math>PQRS</math> is not cyclic.
+Solution by '''TheBoomBox77'''
 == See Also ==