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Difference between revisions of "2009 USAMO Problems/Problem 5"
m (→‎Solution 2 (Projective))
 
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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>.
 
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:
 
We will use directed angles in this solution. Extend <math>QR</math> to <math>T</math> as follows:
 
<center><asy>
 
<center><asy>
Line 15: Line 15:
  
 
pair A = (-.6, .8), B = (.6, .8), C = (.9, -sqrt(.19)), D = (-.9, -sqrt(.19)), G = bisectorpoint(C, B, D);
 
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);
+
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));
 
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);
 
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);
+
draw(A--P--C); draw(B--Q);
 
dot("$P$", P, SE); dot("$Q$", Q, S);
 
dot("$P$", P, SE); dot("$Q$", Q, S);
  
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dot("$R$", R, N); dot("$S$", S, E);
 
dot("$R$", R, N); dot("$S$", S, E);
  
−
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);
 
</asy></center>
 
</asy></center>
 +
 +
'''If''':
  
 
Note that <cmath>\begin{align*}\measuredangle GBT+\measuredangle TRG&=\frac{m\widehat{TQ}}{2}+\measuredangle TRB+\measuredangle BRG\\
 
Note that <cmath>\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}.\\
 
&=\frac{m\widehat{TQ}+m\widehat{DQ}+m\widehat{CB}+m\widehat{BT}}{2}.\\
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
Thus, <math>BTRG</math> is cyclic iff <math>\overline{BG}</math> bisects <math>\angle CBD</math> since that would imply <math>m\widehat{DQ}=m\widehat{QC}</math>.
+
Thus, <math>BTRG</math> is cyclic.
  
 
Also, note that <math>GSCP</math> is cyclic because <cmath>\begin{align*}\measuredangle CSG+\measuredangle GPC&=\measuredangle CBA+\measuredangle APC\\
 
Also, note that <math>GSCP</math> is cyclic because <cmath>\begin{align*}\measuredangle CSG+\measuredangle GPC&=\measuredangle CBA+\measuredangle APC\\
Line 40: Line 42:
 
\end{align*}</cmath> depending on the configuration.
 
\end{align*}</cmath> depending on the configuration.
  
−
Next, we have <math>T, G, C</math> are collinear since <cmath>\measuredangle GTR=\measuredangle GBR=\frac{m\widehat{DQ}}{2}=\frac{m\widehat{QC}}{2}=\measuredangle CTQ,</cmath> iff <math>\overline{BG}</math> bisects <math>\angle CBD</math>.
+
Next, we have <math>T, G, C</math> are collinear since <cmath>\measuredangle GTR=\measuredangle GBR=\frac{m\widehat{DQ}}{2}=\frac{m\widehat{QC}}{2}=\measuredangle CTQ.</cmath>
−
 
 
  
 
Therefore, <cmath>\begin{align*}\measuredangle RQP+\measuredangle PSR&=\frac{m\widehat{PBT}}{2}+\measuredangle PCG\\
 
Therefore, <cmath>\begin{align*}\measuredangle RQP+\measuredangle PSR&=\frac{m\widehat{PBT}}{2}+\measuredangle PCG\\
 
&=\frac{m\widehat{PBT}+m\widehat{TDP}}{2}\\
 
&=\frac{m\widehat{PBT}+m\widehat{TDP}}{2}\\
 
&=180^\circ
 
&=180^\circ
−
\end{align*}</cmath>
+
\end{align*},</cmath> so <math>PQRS</math> is cyclic.
−
iff <math>\overline{BG}</math> bisects <math>\angle CBD</math>, as desired. <math>\blacksquare</math>
+
 
 +
'''Only If''':
 +
These steps can be reversed.
 +
 
 +
== Solution 2 (Projective)==
 +
 
 +
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>
 +
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 ==
 
== See Also ==

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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