Difference between revisions of "2019 IMO Problems/Problem 2"
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==Solution== | ==Solution== | ||
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The essence of the proof is to build a circle through the points <math>P, Q,</math> and two additional points <math>A_0</math> and <math>B_0,</math> then we prove that the points <math>P_1</math> and <math>Q_1</math> lie on the same circle. | The essence of the proof is to build a circle through the points <math>P, Q,</math> and two additional points <math>A_0</math> and <math>B_0,</math> then we prove that the points <math>P_1</math> and <math>Q_1</math> lie on the same circle. | ||
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<math>\angle QPA_0 = \angle QB_0A_0 \implies QPB_0A_0</math> is cyclic (in circle <math>\omega.</math>) | <math>\angle QPA_0 = \angle QB_0A_0 \implies QPB_0A_0</math> is cyclic (in circle <math>\omega.</math>) | ||
+ | |||
+ | Let <math>\angle BAC = \alpha, \angle AA_0B_0 = \varphi.</math> | ||
+ | |||
+ | <math>\angle PP_1C = \alpha, \angle BB_0C = \alpha</math> since they intersept the arc <math>BC</math> of the circle <math>\Omega.</math> | ||
+ | So <math>B_0P_1CB_1</math> is cyclic. | ||
+ | |||
+ | <math>\angle ACB_0 = \angle AA_0B_0 = \varphi</math> (since they intersept the arc <math>A_0B_0</math> of the circle <math>\Omega.</math> | ||
+ | |||
+ | <math>\angle B_1CB_0 = \varphi.</math> | ||
+ | <math>\angle B_1P_1B_0 = \angle B_1CB_0 = \varphi</math> (since they intersept the arc <math>B_1B_0</math> of the circle <math>B_0P_1CB_1).</math> | ||
+ | |||
+ | Hence <math>\angle PA_0B_0 = \angle PP_1B_0 = \varphi,</math> the point <math>P_1</math> lies on <math>\omega.</math> | ||
+ | |||
+ | Similarly, point <math>Q_1</math> lies on <math>\omega.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss, www.deoma–cmd.ru''' |
Revision as of 12:15, 13 August 2022
In triangle , point
lies on side
and point
lies on side
. Let
and
be points on segments
and
, respectively, such that
is parallel to
. Let
be a point on line
, such that
lies strictly between
and
, and
. Similarly, let
be the point on line
, such that
lies strictly between
and
, and
.
Prove that points , and
are concyclic.
Solution
The essence of the proof is to build a circle through the points and two additional points
and
then we prove that the points
and
lie on the same circle.
Let the circumcircle of be
. Let
and
be the points of intersection of
and
with
. Let
since they intersept the arc
of the circle
.
is cyclic (in circle
)
Let
since they intersept the arc
of the circle
So
is cyclic.
(since they intersept the arc
of the circle
(since they intersept the arc
of the circle
Hence the point
lies on
Similarly, point lies on
vladimir.shelomovskii@gmail.com, vvsss, www.deoma–cmd.ru