Difference between revisions of "2009 USAMO Problems/Problem 5"
Line 3: | Line 3: | ||
== Solution == | == Solution == | ||
− | {{ | + | We will use directed angles in this solution. |
+ | <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(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></center> | ||
+ | |||
+ | 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}.\ | ||
+ | \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>. | ||
+ | |||
+ | Also, note that <math>GSCP</math> is cyclic because <cmath>\begin{align*}\measuredangle CSG+\measuredangle GPC&=\measuredangle CBA+\measuredangle APC\ | ||
+ | &=180^\circ\text{ or }0^\circ, | ||
+ | \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,</cmath> iff <math>\overline{BG}</math> bisects <math>\angle CBD</math> (this implies <math>T, G, C</math> are collinear). | ||
+ | |||
+ | |||
+ | Therefore, <cmath>\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*}</cmath> | ||
+ | iff <math>\overline{BG}</math> bisects <math>\angle CBD</math>, as desired. | ||
== See Also == | == See Also == |
Revision as of 10:58, 23 March 2014
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 intersects and at points and , respectively. Prove that quadrilateral is cyclic if and only if bisects .
Solution
We will use directed angles in this solution.
Note that Thus, is cyclic iff bisects since that would imply .
Also, note that is cyclic because depending on the configuration.
Next, we have iff bisects (this implies are collinear).
Therefore,
iff bisects , as desired.
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.