Difference between revisions of "Sharygin Olympiads, the best"
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- the circumcenter <math>O = DM \cap EM'.</math> | - the circumcenter <math>O = DM \cap EM'.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==2024, Problem 18== | ||
+ | [[File:2024 18 1.png|300px|right]] | ||
+ | Let <math>AH, BH', CH''</math> be the altitudes of an acute-angled triangle <math>ABC, I_A</math> be its excenter corresponding to <math>A, I'_A</math> be the reflection of <math>I_A</math> about the line <math>AH.</math> Points <math>I'_B, I'_C</math> are defined similarly. Prove that the lines <math>HI'_A, H'I'_B, H''I'_C</math> concur. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Denote <math>I</math> the incenter of <math>\triangle ABC.</math> Points <math>A, I, I_A</math> are collinear. | ||
+ | We will prove that <math>I \in HI'_A.</math> | ||
+ | Denote <math>D \in BC, ID \perp BC, D' \in I'_AI_A, ID' \perp BC, E = BC \cap AI_A,</math> | ||
+ | <math>F \in BC, I_AF \perp BC, AH = h_A, ID = r, I_AF = r_A, BC = a, s</math> - semiperimeter. | ||
+ | <cmath>\frac {HD}{HF} = \frac {AI}{AI_A} = \frac {h_A - r}{h_A + r}.</cmath> | ||
+ | The area <math>[ABC] = r \cdot s = r_A (s - a) = \frac {a h_A}{2} \implies</math> | ||
+ | <cmath>\frac {1}{r} = \frac {1}{r_A} + \frac {2}{h_A} \implies \frac {h_A - r}{h_A+ r} = \frac {r_A}{r}.</cmath> | ||
+ | <cmath>\frac {I'_AD'}{HD} = \frac {HD + HF}{HD} = 1 + \frac {HF}{HD} = 1 + \frac {r_A}{r}= \frac {r_A + r}{r} \implies</cmath> | ||
+ | Points <math>I, H, I'_a</math> are collinear, so the lines <math>HI'_A, H'I'_B, H''I'_C</math> concur at the point <math>I.</math> | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Revision as of 12:02, 25 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
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