Difference between revisions of "Sharygin Olympiads, the best"
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== 2024 tur 2 klass 8 Problem 2== | == 2024 tur 2 klass 8 Problem 2== | ||
[[File:2024 final 8 2.png|450px|right]] | [[File:2024 final 8 2.png|450px|right]] | ||
− | Let <math>M</math> be the midpoint of side <math>AB</math> of an acute-angled triangle <math>ABC,</math> and <math>P</math> be the projection of the orthocenter <math>H</math> to the bisector of angle <math>C.</math> Prove that <math>MP</math> bisects the segment <math>CH.</math> (L.Emelyanov) | + | Let <math>M</math> be the midpoint of side <math>AB</math> of an acute-angled triangle <math>ABC,</math> and <math>P</math> be the projection of the orthocenter <math>H</math> to the bisector of angle <math>C.</math> |
+ | Prove that <math>MP</math> bisects the segment <math>CH.</math> (L.Emelyanov) | ||
<i><b>Solution</b></i> | <i><b>Solution</b></i> | ||
− | Denote <math>D</math> - the midpoint of <math>CH, A',B',</math> and <math>C'</math> the foots of the heights, <math>\angle A = 2 \alpha, \angle B = 2 \beta, \angle C = 2 \gamma, \omega</math> the Euler circle <math>A'DB'C'M | + | Denote <math>D</math> - the midpoint of <math>CH, A',B',</math> and <math>C'</math> the foots of the heights, <math>\angle A = 2 \alpha, \angle B = 2 \beta, \angle C = 2 \gamma, \omega</math> be the Euler circle <math>A'DB'C'M.</math> |
− | < | + | |
+ | <math>\Omega</math> is the circle <math>\odot CA'B'</math> with the diameter <math>CH.</math> | ||
+ | <cmath>HP \perp CP \implies P \in \Omega.</cmath> | ||
<cmath>AA' \perp BA' \implies A'M = BM \implies \angle BA'M = 2 \beta.</cmath> | <cmath>AA' \perp BA' \implies A'M = BM \implies \angle BA'M = 2 \beta.</cmath> | ||
<cmath>\triangle ABC \sim \triangle A'BC' \implies \angle BA'C' = 2 \alpha \implies</cmath> | <cmath>\triangle ABC \sim \triangle A'BC' \implies \angle BA'C' = 2 \alpha \implies</cmath> | ||
<cmath>\angle MA'C' = 2|\alpha - \beta| = \angle MDC'.</cmath> | <cmath>\angle MA'C' = 2|\alpha - \beta| = \angle MDC'.</cmath> | ||
− | <cmath>\angle ACC' = 90^\circ - 2 \alpha, \angle ACP = \gamma, \angle PCC' = |\angle ACC' - \angle ACP| = | + | <cmath>\angle ACC' = 90^\circ - 2 \alpha, \angle ACP = \gamma,</cmath> <cmath>\angle PCC' = |\angle ACC' - \angle ACP| = |
− | + | |90^\circ - 2 \alpha - \gamma| = | \alpha + \beta + \gamma - 2 \alpha - \gamma| = | \alpha - \beta|.</cmath> | |
<cmath>PD = CD \implies \angle PDH = 2 \angle PCD = 2|\alpha - \beta|.</cmath> | <cmath>PD = CD \implies \angle PDH = 2 \angle PCD = 2|\alpha - \beta|.</cmath> | ||
<math>\angle PDH = \angle MDH \implies</math> points <math>M,P,</math> and <math>D</math> are collinear. | <math>\angle PDH = \angle MDH \implies</math> points <math>M,P,</math> and <math>D</math> are collinear. |
Revision as of 12:54, 10 October 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]- 1 2024 tur 2 klass 8 Problem 2
- 2 2024, Problem 23
- 3 One-to-one mapping of the circle
- 4 2024, Problem 22
- 5 2024, Problem 21
- 6 2024, Problem 20
- 7 2024, Problem 19
- 8 2024, Problem 18
- 9 2024, Problem 17
- 10 2024, Problem 16
- 11 2024, Problem 15
- 12 2024, Problem 14
- 13 2024, Problem 12
- 14 2024, Problem 9
- 15 2024, Problem 8
- 16 2024, Problem 2
2024 tur 2 klass 8 Problem 2
Let be the midpoint of side of an acute-angled triangle and be the projection of the orthocenter to the bisector of angle Prove that bisects the segment (L.Emelyanov)
Solution
Denote - the midpoint of and the foots of the heights, be the Euler circle
is the circle with the diameter points and are collinear.
vladimir.shelomovskii@gmail.com, vvsss
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
Claim
Points and are collinear.
Proof
is the midpoint of arc Denote Therefore points and are collinear.
vladimir.shelomovskii@gmail.com, vvsss
One-to-one mapping of the circle
Let a circle two fixed points and on it and a point inside it be given. Then there is a one-to-one mapping of the circle onto itself, based on the following two theorems.
1. Let a circle two fixed points and on and a point inside be given.
Let an arbitrary point be given.
Let is the midpoint of the arc
Denote Prove that
2. Let a circle two fixed points and on and a point inside be given.
Let an arbitrary point be given.
Let is the midpoint of the arc
Denote
Denote Prove that
Proof
Points are collinear.
2. Points and are collinear (see Claim in 2024, Problem 23).
We use Pascal's theorem for points and crosspoints and get
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 20
Let a triangle points and be given, Points and are the isogonal conjugate of the points and respectively, with respect to
Denote and the circumradii of triangles and respectively.
Prove that where is the area of
Proof
Denote It is easy to prove that is equivalent to By applying the law of sines, we get
We need to prove that We make the transformations:
The last statement is obvious.
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 17
Let be not isosceles triangle, be its incircle.
Let and be the points at which the incircle of touches the sides and respectively.
Let be the point on ray such that
Let be the point on ray such that
The circumcircles of and intersect again at and respectively.
Prove that and are concurrent.
Proof
so points and are collinear (see Symmetry and incircle for details).
Therefore lines and are concurrent (see Symmetry and incircle A for details.)
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 15
The difference of two angles of a triangle is greater than Prove that the ratio of its circumradius and inradius is greater than
Proof
Suppose,
Let be the point on opposite be the midpoint of arc Then Incenter triangle lies on therefore
We use the Euler law
If then
If increases so decreases.
vladimir.shelomovskii@gmail.com, vvsss
2024, Problem 14
The incircle of a right-angled triangle touches the circumcircle of its medial triangle at point Let be the tangent to from the midpoint of the hypothenuse distinct from Prove that
Proof
Let and be the circumcircle and the incenter of
Let be nine-point center of be the point at such that
Denote
is the right-angled triangle, so is the midpoint
Let be the result of the homothety of the point centered in with the coefficient Then WLOG,
Let be the foot from to . Therefore points and are collinear. 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