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

m
(→‎Solution)
Line 26: Line 26:
 
dot("$R$", R, N); dot("$S$", S, E);
 
dot("$R$", R, N); dot("$S$", S, E);
  
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pair T = IP(L(Q, R, 10, 10), circle, 0);
+
pair T = IP(L(Q, R, 10, 10), circle, 1);
 
draw(R--T--C, dashed); draw(T--B, dashed);
 
draw(R--T--C, dashed); draw(T--B, dashed);
 
dot("$T$", T, NW);
 
dot("$T$", T, NW);

Revision as of 18:18, 23 August 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

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.

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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