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

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== Solution 2 (Projective)==
 
== Solution 2 (Projective)==
  
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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:
+
Extend <math>QR</math> to <math>T</math>, and let line <math>l \parallel AB</math> intersect <math>\omega</math> at <math>K</math> and another point <math>V</math>, as shown:
  
 
<center><asy>
 
<center><asy>

Latest revision as of 18:24, 18 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 $T$, and let line $l \parallel AB$ intersect $\omega$ at $K$ and another point $V$, 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

See Also

2009 USAMO (Problems • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

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