Difference between revisions of "2009 USAMO Problems/Problem 5"
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== Problem == | == Problem == | ||
− | 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> | + | 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. | + | 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; | ||
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( | + | 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); | ||
Line 26: | Line 26: | ||
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, | + | 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 | + | 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 <cmath>\measuredangle GTR=\measuredangle GBR=\frac{m\widehat{DQ}}{2}=\frac{m\widehat{QC}}{2}=\measuredangle CTQ | + | 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. |
− | + | ||
+ | '''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 19:24, 18 October 2018
Problem
Trapezoid , with
, is inscribed in circle
and point
lies inside triangle
. Rays
and
meet
again at points
and
, respectively. Let the line through
parallel to
intersect
and
at points
and
, respectively. Prove that quadrilateral
is cyclic if and only if
bisects
.
Solution 1
We will use directed angles in this solution. Extend to
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]](http://latex.artofproblemsolving.com/1/1/7/117688c890349ac2b07c260b638aa6973c0f0c25.png)
If:
Note that
Thus,
is cyclic.
Also, note that is cyclic because
depending on the configuration.
Next, we have are collinear since
Therefore, so
is cyclic.
Only If: These steps can be reversed.
Solution 2 (Projective)
Extend to
, and let line
intersect
at
and another point
, 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]](http://latex.artofproblemsolving.com/f/b/9/fb97d4173f22b4e32b34e763023736dea8d8b663.png)
If:
Suppose that , and
. Pascal's theorem on the tuple
implies that the points
,
, and
are collinear. However,
and
are symmetrical with respect to the axis of symmetry of trapezoid
, and
and
are also symmetrical with respect to the axis of symmetry of
(as
is the midpoint of
, and
). Since
,
and
are symmetric with respect to the axis of symmetry of trapezoid
. This implies that line
is equivalent to line
. Thus,
lies on line
. However,
, so this implies that
.
Now note that is cyclic. Since
,
. However,
. Therefore,
is cyclic.
Only If:
Consider the same setup, except is no longer the midpoint of
. Note that
must be parallel to
in order for
to be cyclic. We claim that
and hope to reach a contradiction. Pascal's theorem on the tuple
implies that
,
, and
are collinear. However, there exists a unique point
such that
,
, and
are concurrent. By If,
must be the midpoint of
in order for the concurrency to occur; hence,
. Then
, since
. However, this is a contradiction, so therefore
cannot be parallel to
and
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.