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

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== Solution ==
 
== Solution ==
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We will use directed angles in this solution.
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We will use directed angles in this solution. Extend <math>QR</math> to <math>T</math> as follows:
 
<center><asy>
 
<center><asy>
 
import cse5;
 
import cse5;
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&=180^\circ
 
&=180^\circ
 
\end{align*}</cmath>
 
\end{align*}</cmath>
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iff <math>\overline{BG}</math> bisects <math>\angle CBD</math>, as desired.
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iff <math>\overline{BG}</math> bisects <math>\angle CBD</math>, as desired. <math>\blacksquare</math>
  
 
== See Also ==
 
== See Also ==

Revision as of 14:29, 23 March 2014

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(D--B--G); 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--cycle); dot("$R$", R, N); dot("$S$", S, E); pair T = IP(L(Q, R, 10, 10), circle, 0); draw(R--T--C, dashed); draw(T--B, dashed); dot("$T$", T, NW); [/asy]

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 iff $\overline{BG}$ bisects $\angle CBD$ since that would imply $m\widehat{DQ}=m\widehat{QC}$.

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 \[\measuredangle GTR=\measuredangle GBR=\frac{m\widehat{DQ}}{2}=\frac{m\widehat{QC}}{2}=\measuredangle CTQ,\] iff $\overline{BG}$ bisects $\angle CBD$.


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*} iff $\overline{BG}$ bisects $\angle CBD$, as desired. $\blacksquare$

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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