Difference between revisions of "2009 USAMO Problems/Problem 5"
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dot("$T$", T, NW); | dot("$T$", T, NW); | ||
</asy></center> | </asy></center> | ||
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+ | '''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\ | ||
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\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 | + | 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> |
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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. | ||
== See Also == | == See Also == |
Revision as of 16:18, 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 intersect and at points and , respectively. Prove that quadrilateral is cyclic if and only if bisects .
Solution
We will use directed angles in this solution. Extend to as follows:
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.
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.