Difference between revisions of "2019 IMO Problems/Problem 2"
(Created page with "In triangle <math>ABC</math>, point <math>A_1</math> lies on side <math>BC</math> and point <math>B_1</math> lies on side <math>AC</math>. Let <math>P</math> and <math>Q</math...") |
|||
(6 intermediate revisions by 2 users not shown) | |||
Line 1: | Line 1: | ||
+ | ==Problem== | ||
+ | |||
In triangle <math>ABC</math>, point <math>A_1</math> lies on side <math>BC</math> and point <math>B_1</math> lies on side <math>AC</math>. Let <math>P</math> and <math>Q</math> be points on segments <math>AA_1</math> and <math>BB_1</math>, respectively, such that <math>PQ</math> is parallel to <math>AB</math>. Let <math>P_1</math> be a point on line <math>PB_1</math>, such that <math>B_1</math> lies strictly between <math>P</math> and <math>P_1</math>, and <math>\angle PP_1C=\angle BAC</math>. Similarly, let <math>Q_1</math> be the point on line <math>QA_1</math>, such that <math>A_1</math> lies strictly between <math>Q</math> and <math>Q_1</math>, and <math>\angle CQ_1Q=\angle CBA</math>. | In triangle <math>ABC</math>, point <math>A_1</math> lies on side <math>BC</math> and point <math>B_1</math> lies on side <math>AC</math>. Let <math>P</math> and <math>Q</math> be points on segments <math>AA_1</math> and <math>BB_1</math>, respectively, such that <math>PQ</math> is parallel to <math>AB</math>. Let <math>P_1</math> be a point on line <math>PB_1</math>, such that <math>B_1</math> lies strictly between <math>P</math> and <math>P_1</math>, and <math>\angle PP_1C=\angle BAC</math>. Similarly, let <math>Q_1</math> be the point on line <math>QA_1</math>, such that <math>A_1</math> lies strictly between <math>Q</math> and <math>Q_1</math>, and <math>\angle CQ_1Q=\angle CBA</math>. | ||
− | Prove that points <math>P,Q,P_1</math>, and <math>Q_1</math> are concyclic | + | Prove that points <math>P,Q,P_1</math>, and <math>Q_1</math> are concyclic. |
+ | |||
+ | ==Solution== | ||
+ | [[File:2019 IMO 2.png|420px|right]] | ||
+ | [[File:2019 IMO 2a.png|420px|right]] | ||
+ | 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. | ||
+ | |||
+ | We assume that the intersection point of <math>AP</math> and <math>BQ</math> lies on the segment <math>PA_1.</math> If it lies on segment <math>AP,</math> then the proof is the same, but some angles will be replaced with additional ones up to <math>180^\circ</math>. | ||
+ | |||
+ | Let the circumcircle of <math>\triangle ABC</math> be <math>\Omega</math>. Let <math>A_0</math> and <math>B_0</math> be the points of intersection of <math>AP</math> and <math>BQ</math> with <math>\Omega</math>. Let <math>\angle BAP = \delta.</math> | ||
+ | |||
+ | <cmath>PQ||AB \implies \angle QPA_0 = \delta.</cmath> | ||
+ | |||
+ | <math>\angle BAP = \angle BB_0A_0 = \delta</math> since they intersept the arc <math>BA_0</math> of the 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> (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''' | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{IMO box|year=2019|num-b=1|num-a=3}} |
Latest revision as of 01:49, 19 November 2023
Problem
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.
We assume that the intersection point of and
lies on the segment
If it lies on segment
then the proof is the same, but some angles will be replaced with additional ones up to
.
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.
(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
See Also
2019 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |