Difference between revisions of "2022 IMO Problems/Problem 4"
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==Solution== | ==Solution== | ||
[[File:2022 IMO 4.png|400px|right]] | [[File:2022 IMO 4.png|400px|right]] | ||
− | <cmath>TB = TD, TC = TE, BC = DE \implies \triangle TBC = \triangle TDE</cmath> | + | <cmath>TB = TD, TC = TE, BC = DE \implies</cmath> |
− | <cmath> | + | <cmath>\triangle TBC = \triangle TDE \implies \angle BTC = \angle DTE.</cmath> |
− | <cmath>\angle ABT = \angle | + | <cmath>\angle BTQ = 180^\circ - \angle BTC = 180^\circ - \angle DTE = \angle STE</cmath> |
− | <cmath>\frac {QT}{ST}= \frac {TB}{TE} \implies QT \cdot TC = ST \cdot TD \implies</cmath> | + | <cmath>\angle ABT = \angle AET \implies \triangle TQB \sim \triangle TSE \implies</cmath> |
− | <math>CDQS</math> is cyclic <math>\implies \angle QCD = \angle QSD.</math> | + | <cmath>\frac {QT}{ST}= \frac {TB}{TE} \implies 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> | ||
<cmath>\angle QPR =\angle QPC = \angle QCD - \angle PQC =</cmath> | <cmath>\angle QPR =\angle QPC = \angle QCD - \angle PQC =</cmath> | ||
<math>\angle QSD - \angle EST = \angle QSR \implies</math> | <math>\angle QSD - \angle EST = \angle QSR \implies</math> |
Revision as of 18:46, 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]