Difference between revisions of "Sharygin Olympiads, the best"
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<math>O</math> is the circumcenter of <math>\triangle APC, O'</math> is the circumcenter of <math>\triangle BPC.</math> | <math>O</math> is the circumcenter of <math>\triangle APC, O'</math> is the circumcenter of <math>\triangle BPC.</math> | ||
− | Let <math>K</math> and <math>L</math> be the midpoints of the arcs <math>\overset{\Large\frown}{CB'}</math> of <math>\theta.</math> | + | Let <math>K</math> and <math>L</math> be the midpoints of the arcs <math>\overset{\Large\frown}{CB'}</math> of <math>\theta,D = AL \cap \omega.</math> |
− | Let <math>K'</math> and <math>L'</math> be the midpoints of the arcs <math>\overset{\Large\frown}{CA'}</math> of <math>\theta'.</math> | + | Let <math>K'</math> and <math>L'</math> be the midpoints of the arcs <math>\overset{\Large\frown}{CA'}</math> of <math>\theta', D' = BL' \cap \omega'.</math> |
These points not depends from position of point <math>P.</math> | These points not depends from position of point <math>P.</math> | ||
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'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
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==2024, Problem 22== | ==2024, Problem 22== | ||
[[File:2023 22 2.png|350px|right]] | [[File:2023 22 2.png|350px|right]] |
Revision as of 10:18, 31 March 2024
Igor Fedorovich Sharygin (13/02/1937 - 12/03/2004, Moscow) - Soviet and Russian mathematician and teacher, specialist in elementary geometry, popularizer of science. He wrote many textbooks on geometry and created a number of beautiful problems. He headed the mathematics section of the Russian Soros Olympiads. After his death, Russia annually hosts the Geometry Olympiad for high school students. It consists of two rounds – correspondence and final. The correspondence round lasts 3 months.
The best problems of these Olympiads will be published. The numbering contains the year of the Olympiad and the serial number of the problem. Solutions are often different from the original ones.
Contents
[hide]2024, Problem 23
A point moves along a circle Let and be fixed points of and be an arbitrary point inside
The common external tangents to the circumcircles of triangles and meet at point
Prove that all points lie on two fixed lines.
Solution
Denote
is the circumcenter of is the circumcenter of
Let and be the midpoints of the arcs of
Let and be the midpoints of the arcs of
These points not depends from position of point
Suppose, see diagram). Let Similarly,
Let
Therefore Similarly, if then
vladimir.shelomovskii@gmail.com, vvsss
2024, Problem 22
A segment is given. Let be an arbitrary point of the perpendicular bisector to be the point on the circumcircle of opposite to and an ellipse centered at touche
Find the locus of touching points of the ellipse with the line
Solution
Denote the midpoint the point on the line
In order to find the ordinate of point we perform an affine transformation (compression along axis which will transform the ellipse into a circle with diameter The tangent of the maps into the tangent of the Denote
So point is the fixed point ( not depends from angle
Therefore point lies on the circle with diameter (except points and
vladimir.shelomovskii@gmail.com, vvsss
2024, Problem 21
A chord of the circumcircle of a triangle meets the sides at points respectively. The tangents to the circumcircle at and meet at point and the tangents at points and meets at point The line meets at point
Prove that the lines and concur.
Proof
WLOG, Denote
Point is inside
We use Pascal’s theorem for quadrilateral and get
We use projective transformation which maps to a circle and that maps the point to its center.
From this point we use the same letters for the results of mapping. Therefore the segments and are the diameters of is the midpoint
preimage lies on preimage
vladimir.shelomovskii@gmail.com, vvsss
2024, Problem 19
A triangle its circumcircle , and its incenter are drawn on the plane.
Construct the circumcenter of using only a ruler.
Solution
We successively construct:
- the midpoint of the arc
- the midpoint of the arc
- the polar of point
- the polar of point
- the polar of the line
- the tangent to
- the tangent to
- the trapezium
- the point
- the point
- the midpoint of the segment
- the midpoint of the segment
- the diameter of
- the diameter of
- the circumcenter
vladimir.shelomovskii@gmail.com, vvsss
2024, Problem 18
Let be the altitudes of an acute-angled triangle be its excenter corresponding to be the reflection of about the line Points are defined similarly. Prove that the lines concur.
Proof
Denote the incenter of Points are collinear. We will prove that Denote - semiperimeter. The area Points are collinear, so the lines concur at the point
vladimir.shelomovskii@gmail.com, vvsss
2024, Problem 16
Let and be the bisectors of a triangle
The segments and meet at point Let be the projection of to
Points and on the sides and respectively, are such that
Prove that
Proof
is the common side)
is the midpoint
is the midpoint of (see Division of bisector for details.)
So Denote
Another solution see 2024_Sharygin_olimpiad_Problem_16
vladimir.shelomovskii@gmail.com, vvsss
2024, Problem 12
The bisectors of a with meet at point
The circumcircles of triangles meet at point
Prove that the line bisects the side
Proof
Denote the midpoint In triangles and , by applying the law of sines, we get
We use the formulas for circle and get
In triangles and , by applying the law of sines, we get
Therefore The function increases monotonically on the interval
This means and points and are collinear.
vladimir.shelomovskii@gmail.com, vvsss
2024, Problem 9
Let be a trapezoid circumscribed around a circle centered at which touches the sides and at points respectively.
The line passing trough and parallel to the bases of trapezoid meets at point
Prove that and concur.
Solution
Solution 1.
is the center of similarity of triangles and
Solution 2.
Denote
vladimir.shelomovskii@gmail.com, vvsss
2024, Problem 8
Let be a quadrilateral with and
The incircle of touches the sides and at points and respectively.
The midpoints of segments and are points
Prove that points are concyclic.
Solution
is the rotation of around a point through an angle
is the rotation of around a point through an angle
So is the rotation of around a point through an angle
vladimir.shelomovskii@gmail.com, vvsss
2024, Problem 2
Three distinct collinear points are given. Construct the isosceles triangles such that these points are their circumcenter, incenter and excenter (in some order).
Solution
Let be the midpoint of the segment connecting the incenter and excenter. It is known that point belong the circumcircle. Construction is possible if a circle with diameter IE (incenter – excenter) intersects a circle with radius OM (circumcenter – M). Situation when between and is impossible.
Denote points such that and
Suppose point is circumcenter, so is incenter. is midpoint BC. The vertices of the desired triangle are located at the intersection of a circle with center and radius with and a line
Suppose point is circumcenter, so is incenter. is midpoint The vertices of the desired triangle are located at the intersection of a circle with center and radius with and a line
Suppose point is circumcenter, so is incenter. is midpoint Suppose The vertices of the desired triangle are located at the intersection of a circle with center and radius with and a line
If there is not desired triangle.
vladimir.shelomovskii@gmail.com, vvsss