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6 discussions

Geometry

Separating 2019 arbitrary points with discs: what configuration is worst?

An information bound gives

2^k\ge2019

, but it ignores which membership patterns discs can realize. I'm looking for a stronger obstruction, perhaps with all points on a line, and a matching construction for arbitrary point sets. Problem: Several points are given in the plane. A child wants to draw

k

(closed) discs in such a manner, that for any two points

A, B

(

A \neq B

) there exists a disc that contains only one of these points. What is the minimum

k

, such that for any initial configuration of 2019 points it is possible to draw the

k

discs with the above property? Source: Baltic Way 2019 · MNET-01FF

aeriths_chair · Sep 18, 20260 replies

GeometryResolved

A sinusoid squeezed between two shifted sines — strictness matters

The lower and upper envelopes have corners. How can a single sinusoid lie between them for every x and touch only where the original two sines agree? Problem: Let

a

be a real number. Prove that there exist real numbers

b

and

c

such that the inequalities

\min \{\sin(x), \sin(a+x)\} \le b \sin(x+c) \le \max \{\sin(x), \sin(a+x)\}

hold for all real numbers

x

and the equalities hold only if

\sin(x) = \sin(a+x)

. Source: Baltic Way 2021 Shortlist · MNET-01HD

mangoaggro · Jul 13, 20261 reply

Geometry

Local concyclicity for every five points — how does it become global?

The hypothesis only gives four concyclic points at a time. I can prove useful things once five points share a circle, but getting that fifth point seems to be the main obstacle. Problem: There are

2021

points on a plane, no three of which are collinear. For every

5

points there exists at least

4

among them which are concyclic. Is it necessarily true that at least

2020

of the points are concyclic? Source: Baltic Way 2021 Shortlist · MNET-01HW

quicksave_q · Feb 12, 20263 replies

Geometry

Three circles through P: checking the converse of the radical-axis argument

I can see why a second common point should force the three opposite-vertex diagonals to concur. In the converse, how do I know the second intersection is real and distinct from P rather than a tangency? I'd like to settle that before calling the proof complete. Problem: Let

ABCDEF

be a cyclic and convex hexagon. A point

P

is called *admissible* if it does not lie on the circumcircle of

ABCDEF

or on any of the lines

AD

BE

and

CF

. An admissible point is called *fantastic* if

\odot(ADP)

\odot(BEP)

and

\odot(CFP)

intersect in exactly two points. Prove that if there exists a fantastic point, then all admissible points are fantastic. Source: Baltic Way 2023 Shortlist · MNET-01J8

midnaMain · Jan 8, 20260 replies

Geometry

Reflections of an interior point and a surprising collinearity

I have several equal angles from reflecting P in the sides, but I can't connect the lines MD and LE to the line AK. Would an inversion about A simplify the two second intersections, or is there a direct spiral-similarity configuration? I would appreciate a construction hint before a full solution. Problem: Let

\triangle ABC

be an acute triangle and

P

be an inner point of

\triangle ABC

. Let

K

L

and

M

be the reflections of

P

across

BC

AC

and

AB

, respectively. Let

D

and

E

be the second points of intersection of

\odot(PBC)

with lines

AB

and

AC

, respectively. Let lines

MD

and

LE

intersect at

F

. Prove that

F

A

and

K

are collinear. Source: Baltic Way 2023 Shortlist · MNET-01J6

empty_estus · May 9, 20250 replies

GeometryResolved

Two Apollonius circles whose radical axis goes through the circumcenter

The ratios 21:20:19 look numerical, but the conclusion feels structural. Could this be an Apollonius-circle argument rather than a long coordinate expansion? Problem: Let

A

B

C

P

and

Q

be five pairwise different points in the plane. Suppose that

A

B

and

C

are not collinear and that

\frac{AP}{BP} = \frac{AQ}{BQ} = \frac{21}{20},

\frac{BP}{CP} = \frac{BQ}{CQ} = \frac{20}{19}.

Prove that the line

PQ

contains the circumcentre of the triangle

ABC

. Source: Baltic Way 2020 · MNET-01H2

1hpClutch · Mar 2, 20254 replies

Loading discussions…

Separating 2019 arbitrary points with discs: what configuration is worst?

An information bound gives

2^k\ge2019

k

(closed) discs in such a manner, that for any two points

A, B

(

A \neq B

) there exists a disc that contains only one of these points. What is the minimum

k

, such that for any initial configuration of 2019 points it is possible to draw the

k

discs with the above property? Source: Baltic Way 2019 · MNET-01FF

aeriths_chair · Sep 18, 20260 replies

A sinusoid squeezed between two shifted sines — strictness matters

The lower and upper envelopes have corners. How can a single sinusoid lie between them for every x and touch only where the original two sines agree? Problem: Let

a

be a real number. Prove that there exist real numbers

b

and

c

such that the inequalities

\min \{\sin(x), \sin(a+x)\} \le b \sin(x+c) \le \max \{\sin(x), \sin(a+x)\}

hold for all real numbers

x

and the equalities hold only if

\sin(x) = \sin(a+x)

. Source: Baltic Way 2021 Shortlist · MNET-01HD

mangoaggro · Jul 13, 20261 reply

Local concyclicity for every five points — how does it become global?

The hypothesis only gives four concyclic points at a time. I can prove useful things once five points share a circle, but getting that fifth point seems to be the main obstacle. Problem: There are

2021

points on a plane, no three of which are collinear. For every

5

points there exists at least

4

among them which are concyclic. Is it necessarily true that at least

2020

of the points are concyclic? Source: Baltic Way 2021 Shortlist · MNET-01HW

quicksave_q · Feb 12, 20263 replies

Three circles through P: checking the converse of the radical-axis argument

ABCDEF

be a cyclic and convex hexagon. A point

P

is called *admissible* if it does not lie on the circumcircle of

ABCDEF

or on any of the lines

AD

BE

and

CF

. An admissible point is called *fantastic* if

\odot(ADP)

\odot(BEP)

and

\odot(CFP)

intersect in exactly two points. Prove that if there exists a fantastic point, then all admissible points are fantastic. Source: Baltic Way 2023 Shortlist · MNET-01J8

midnaMain · Jan 8, 20260 replies

Reflections of an interior point and a surprising collinearity

\triangle ABC

be an acute triangle and

P

be an inner point of

\triangle ABC

. Let

K

L

and

M

be the reflections of

P

across

BC

AC

and

AB

, respectively. Let

D

and

E

be the second points of intersection of

\odot(PBC)

with lines

AB

and

AC

, respectively. Let lines

MD

and

LE

intersect at

F

. Prove that

F

A

and

K

are collinear. Source: Baltic Way 2023 Shortlist · MNET-01J6

empty_estus · May 9, 20250 replies

Two Apollonius circles whose radical axis goes through the circumcenter

The ratios 21:20:19 look numerical, but the conclusion feels structural. Could this be an Apollonius-circle argument rather than a long coordinate expansion? Problem: Let

A

B

C

P

and

Q

be five pairwise different points in the plane. Suppose that

A

B

and

C

are not collinear and that

\frac{AP}{BP} = \frac{AQ}{BQ} = \frac{21}{20},

\frac{BP}{CP} = \frac{BQ}{CQ} = \frac{20}{19}.

Prove that the line

PQ

contains the circumcentre of the triangle

ABC

. Source: Baltic Way 2020 · MNET-01H2

1hpClutch · Mar 2, 20254 replies