Difference between revisions of "2022 IMO Problems/Problem 4"
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<cmath>\angle BTQ = 180^\circ - \angle BTC = 180^\circ - \angle DTE = \angle STE</cmath> | <cmath>\angle BTQ = 180^\circ - \angle BTC = 180^\circ - \angle DTE = \angle STE</cmath> | ||
<cmath>\angle ABT = \angle AET \implies \triangle TQB \sim \triangle TSE \implies</cmath> | <cmath>\angle ABT = \angle AET \implies \triangle TQB \sim \triangle TSE \implies</cmath> | ||
− | <cmath>\frac {QT}{ST}= \frac {TB}{TE} \implies QT \cdot TE =QT \cdot TC = ST \cdot TB= ST \cdot TD \implies</cmath> | + | <cmath>\angle PQC = \angle EST, \hspace{18mm}\frac {QT}{ST}= \frac {TB}{TE} \implies</cmath> |
+ | <cmath>QT \cdot TE =QT \cdot TC = ST \cdot TB= ST \cdot TD \implies</cmath> | ||
<math>\hspace{28mm}CDQS</math> is cyclic <math>\implies \angle QCD = \angle QSD.</math> | <math>\hspace{28mm}CDQS</math> is cyclic <math>\implies \angle QCD = \angle QSD.</math> | ||
<cmath>\angle QPR =\angle QPC = \angle QCD - \angle PQC =</cmath> | <cmath>\angle QPR =\angle QPC = \angle QCD - \angle PQC =</cmath> | ||
− | < | + | <cmath>\angle QSD - \angle EST = \angle QSR \implies</cmath> |
− | <math>PRQS</math> is cyclic. | + | <math>\hspace{43mm}PRQS</math> is cyclic. |
'''vladimir.shelomovskii@gmail.com, vvsss, www.deoma–cmd.ru''' | '''vladimir.shelomovskii@gmail.com, vvsss, www.deoma–cmd.ru''' |
Revision as of 18:54, 23 July 2022
Problem
Let be a convex pentagon such that
. Assume that there is a
point
inside
with
,
and
. Let line
intersect
lines
and
at points
and
, respectively. Assume that the points
occur on their
line in that order. Let line
intersect lines
and
at points
and
, respectively. Assume
that the points
occur on their line in that order. Prove that the points
lie on
a circle.
Solution
is cyclic
is cyclic.
vladimir.shelomovskii@gmail.com, vvsss, www.deoma–cmd.ru
Solution
https://www.youtube.com/watch?v=-AII0ldyDww [Video contains solutions to all day 2 problems]