Difference between revisions of "Sharygin Olympiads, the best"

Revision as of 10:07, 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.

2024, Problem 23

A point $P$ moves along a circle $\Omega.$ Let $A$ and $B$ be fixed points of $\Omega,$ and $C$ be an arbitrary point inside $\Omega.$

The common external tangents to the circumcircles of triangles $\triangle APC$ and $\triangle BCP$ meet at point $Q.$

Prove that all points $Q$ lie on two fixed lines.

Solution

Denote $A' = AC \cap \Omega, B' = BC \cap \Omega, \omega = \odot APC, \omega' = \odot BPC.$ $\theta = \odot ACB', \theta' = \odot BCA'.$

$O$ is the circumcenter of $\triangle APC, O'$ is the circumcenter of $\triangle BPC.$

Let $K$ and $L$ be the midpoints of the arcs $\overset{\Large\frown}{CB'}$ of $\theta.$

Let $K'$ and $L'$ be the midpoints of the arcs $\overset{\Large\frown}{CA'}$ of $\theta'.$

These points not depends from position of point $P.$

Suppose, $P \in \overset{\Large\frown} {B'ABA'} ($ see diagram). $\angle A'BC = 2 \alpha = \angle B'AC \implies \angle D'BC = \angle DAC = \alpha \implies \angle DOC = \angle D'O'C = 2 \alpha.$ $O'D' = O'C, OC = OD \implies \triangle OCD \sim \triangle O'D'C \implies OC||O'D'.$ Let $F= CD \cup OO' \implies \frac {FO}{FO'} = \frac {OC}{O'D'} \implies Q = F.$ $\angle LCB' = \alpha = \angle B'BL' \implies LC || L'B.$ Similarly, $AL || CL' \implies \triangle DLC \sim \triangle CL'D' \implies \frac {LC}{L'D'} = \frac {DC}{CD'} = \frac {OC}{O'D'}.$

Let $F' = LL' \cap DD' \implies \frac {F'C}{F'D'} = \frac {LC}{L'D'} = \frac {OC}{O'D'}= \frac {FC}{FD'} \implies F' = F.$

Therefore $Q \in LL'.$ Similarly, if $P \in \overset{\Large\frown} {B'A'}$ then $Q \in KK'.$

vladimir.shelomovskii@gmail.com, vvsss

2024, Problem 22

A segment $AB$ is given. Let $C$ be an arbitrary point of the perpendicular bisector to $AB,$ $O$ be the point on the circumcircle of $\triangle ABC$ opposite to $C,$ and an ellipse centered at $O$ touche $AB, BC, CA.$

Find the locus of touching points $P$ of the ellipse with the line $BC.$

Solution

Denote $M$ the midpoint $AB, D$ the point on the line $CO, DO = MO, \alpha = \angle CBM, b = OM.$

$\angle CBO = 90^\circ \implies \angle COB = \alpha, MB = b \tan \alpha,$ $CB = \frac {b \sin \alpha}{\cos^2 \alpha}, CO = \frac {b} {\cos^2 \alpha}, CD = b \left (1 + \frac {1} {\cos^2 \alpha} \right ).$ In order to find the ordinate of point $P,$ we perform an affine transformation (compression along axis $AB)$ which will transform the ellipse $MPD$ into a circle with diameter $MD.$ The tangent of the $CP$ maps into the tangent of the $CE, E = \odot CBO \cap \odot MD, PF \perp CO.$ $\angle OEF = \angle ECO \implies OF = OE \sin \angle OEF = OE \sin \angle ECO = b \cos^2 \alpha.$ $CP = \frac {CF}{\sin \alpha} = \frac {b}{\sin \alpha}\left ( \frac {1} {\cos^2 \alpha} - \cos^2 \alpha \right ) = b \sin \alpha \left ( \frac {1}{\cos^2 \alpha } + 1 \right).$ $\frac {CP}{CD} = \sin \alpha , \angle PCD = 90^\circ - \alpha \implies \angle CPD = 90^\circ.$ $BP = CP - CB = b \sin \alpha.$ Denote $Q = AB \cap DP \implies BQ = \frac {BP}{\cos \alpha} = b \tan \alpha = MB.$

So point $Q$ is the fixed point ( $P$ not depends from angle $\alpha, \angle BPQ = 90^\circ ).$

Therefore point $P$ lies on the circle with diameter $BQ$ (except points $B$ and $Q.)$

vladimir.shelomovskii@gmail.com, vvsss

2024, Problem 21

A chord $PQ$ of the circumcircle of a triangle $ABC$ meets the sides $BC, AC$ at points $A', B',$ respectively. The tangents to the circumcircle at $A$ and $B$ meet at point $X,$ and the tangents at points $P$ and $Q$ meets at point $Y.$ The line $XY$ meets $AB$ at point $C'.$

Prove that the lines $AA', BB',$ and $CC'$ concur.

Proof

WLOG, $P \in \overset{\Large\frown} {AC}.$ Denote $\Omega = \odot ABC, Z = BB' \cap AA', D = AQ \cap BP.$

Point $D$ is inside $\Omega.$

We use Pascal’s theorem for quadrilateral $APQB$ and get $D \in XY.$

We use projective transformation which maps $\Omega$ to a circle and that maps the point $D$ to its center.

From this point we use the same letters for the results of mapping. Therefore the segments $AQ$ and $BP$ are the diameters of $\Omega, C'D \in XY || AP \implies C'$ is the midpoint $AB.$

$AB|| PQ \implies AB || B'A' \implies C' \in CZ \implies$

preimage $Z$ lies on preimage $CC'.\blacksquare$

vladimir.shelomovskii@gmail.com, vvsss

2024, Problem 19

A triangle $ABC,$ its circumcircle $\Omega$ , and its incenter $I$ are drawn on the plane.

Construct the circumcenter $O$ of $\triangle ABC$ using only a ruler.

Solution

We successively construct:

- the midpoint $D = BI \cap \Omega$ of the arc $AB,$

- the midpoint $E = CI \cap \Omega$ of the arc $AC,$

- the polar $H'H''$ of point $H \in DE,$

- the polar $G'G''$ of point $G \in DE,$

- the polar $F = H'H'' \cap G'G''$ of the line $DE,$

- the tangent $FD || AC$ to $\Omega,$

- the tangent $FE || AB$ to $\Omega,$

- the trapezium $ACDF,$

- the point $K = AF \cap CD,$

- the point $L = AD \cap CF,$

- the midpoint $M = AC \cap KL$ of the segment $AB,$

- the midpoint $M'$ of the segment $AC,$

- the diameter $DM$ of $\Omega,$

- the diameter $EM'$ of $\Omega,$

- the circumcenter $O = DM \cap EM'.$

vladimir.shelomovskii@gmail.com, vvsss

2024, Problem 18

Let $AH, BH', CH''$ be the altitudes of an acute-angled triangle $ABC, I_A$ be its excenter corresponding to $A, I'_A$ be the reflection of $I_A$ about the line $AH.$ Points $I'_B, I'_C$ are defined similarly. Prove that the lines $HI'_A, H'I'_B, H''I'_C$ concur.

Proof

Denote $I$ the incenter of $\triangle ABC.$ Points $A, I, I_A$ are collinear. We will prove that $I \in HI'_A.$ Denote $D \in BC, ID \perp BC, D' \in I'_AI_A, ID' \perp BC, E = BC \cap AI_A,$ $F \in BC, I_AF \perp BC, AH = h_A, ID = r, I_AF = r_A, BC = a, s$ - semiperimeter. $\frac {HD}{HF} = \frac {AI}{AI_A} = \frac {h_A - r}{h_A + r}.$ The area $[ABC] = r \cdot s = r_A (s - a) = \frac {a h_A}{2} \implies$ $\frac {1}{r} = \frac {1}{r_A} + \frac {2}{h_A} \implies \frac {h_A - r}{h_A+ r} = \frac {r_A}{r}.$ $\frac {I'_AD'}{HD} = \frac {HD + HF}{HD} = 1 + \frac {HF}{HD} = 1 + \frac {r_A}{r}= \frac {r_A + r}{r} \implies$ Points $I, H, I'_a$ are collinear, so the lines $HI'_A, H'I'_B, H''I'_C$ concur at the point $I.$

vladimir.shelomovskii@gmail.com, vvsss

2024, Problem 16

Let $AA', BB',$ and $CC'$ be the bisectors of a triangle $\triangle ABC.$

The segments $BB'$ and $A'C'$ meet at point $D.$ Let $E$ be the projection of $D$ to $AC.$

Points $P$ and $Q$ on the sides $AB$ and $BC,$ respectively, are such that $EP = PD, EQ = QD.$

Prove that $\angle PDB' = \angle EDQ.$

Proof

$\triangle PDQ = \triangle PEQ (DQ = EQ, DP = PF, PQ$ is the common side) $\implies$

$PQ \perp DE, F = PQ \cap DE$ is the midpoint $DE \implies$

$G = BB' \cap PQ$ is the midpoint of $DB'.$ $\frac{BG}{BB'} =\frac {BQ}{BC} = \frac {BP}{BA} = \frac {BD}{BI}.$ (see Division of bisector for details.)

So $DQ || CC', PD || AA'.$ Denote $\angle ACC' = \angle BCC' = \gamma, \angle A'AC = \alpha, B'BC = \beta.$ $\angle PDB' = \angle AIB' = \angle BB'C - \angle IAC = 180^\circ - \beta - 2 \gamma - \alpha = 90^\circ - \gamma.$ $\angle QDE = 90^\circ - \angle DQP = 90^\circ - \gamma = \angle PDB'.$

Another solution see 2024_Sharygin_olimpiad_Problem_16

vladimir.shelomovskii@gmail.com, vvsss

2024, Problem 12

The bisectors $AE, CD$ of a $\triangle ABC$ with $\angle B = 60^\circ$ meet at point $I.$

The circumcircles of triangles $ABC, DIE$ meet at point $P.$

Prove that the line $PI$ bisects the side $AC.$

Proof

Denote $M$ the midpoint $AC, \omega = \odot DIE,$ $\varphi = \angle CIM, \phi = \angle DIP, f(x) = \frac {\sin x}{\sin(120^\circ -x)}.$ $\angle AIC = 90^\circ + \frac {\angle ABC}{2} = 120^\circ = 180^\circ - \angle ABC \implies B \in \omega.$ In triangles $\triangle CIM$ and $\triangle AIM$ , by applying the law of sines, we get $\frac {IM}{\sin \angle ACI} = \frac {CM}{\sin \varphi}, \frac {IM}{\sin \angle CAI} = \frac {AM}{\sin (120^\circ -\varphi)} \implies f(\varphi) = \frac {\sin \varphi} {\sin (120^\circ -\varphi)} = \frac {\sin \angle ACI}{\sin \angle CAI}.$

We use the formulas for circle $\omega$ and get $\frac {PD}{\sin \angle PID} = \frac {PE}{\sin \angle PIE} \implies f(\phi) = \frac {\sin \phi} {\sin (120^\circ -\phi)} = \frac {PD}{PE}.$

$\angle BDI = \angle AEC \implies \angle ADC = 180^\circ - \angle AEC.$ In triangles $\triangle ADC$ and $\triangle AEC$ , by applying the law of sines, we get $\frac {AD}{\sin \angle ACD} = \frac {AC}{\sin \angle ADC} = \frac {AC}{\sin \angle AEC} \implies \frac {AD}{CE} = \frac {\sin \angle ACI}{\sin \angle CAI}.$

$\angle BEP = \angle BDP, \angle BCP = \angle BAP \implies \triangle APD \sim \triangle CPE \implies \frac {AD}{CE} = \frac {PD}{PE}.$

Therefore $f(\phi) = f(\varphi).$ The function $f$ increases monotonically on the interval $(0, \frac {2 \pi}{3}).$

This means $\phi = \varphi$ and points $P,I,$ and $M$ are collinear.

vladimir.shelomovskii@gmail.com, vvsss

2024, Problem 9

Let $ABCD (AD || BC$ be a trapezoid circumscribed around a circle $\omega,$ centered at $O$ which touches the sides $AB, BC, CD,$ and $AD$ at points $P, Q, R, S,$ respectively.

The line passing trough $P$ and parallel to the bases of trapezoid meets $QR$ at point $X.$

Prove that $AB, QS,$ and $DX$ concur.

Solution

Solution 1. $AP = AS \implies \angle AOS = \frac {\overset{\Large\frown} {PS}}{2} = \angle PQS \implies PQ||AO.$

$OD \perp OC, QR \perp OC \implies OD || QX. AD ||PX \implies$

$E = AP \cap QS$ is the center of similarity of triangles $\triangle PQX$ and $\triangle AOD.$

Solution 2. $\triangle ODS \sim \triangle QXG \implies \frac {SD}{GX} = \frac {SO}{GQ} = \frac {1}{2} \cdot \frac {SQ}{GQ} = \frac {1}{2} \cdot \frac {AB}{PB}.$

Denote $EA = x, AP = AS = y, BP = BQ = z.$

$\triangle AES \sim BEQ \implies \frac {x}{x+y+z}= \frac {y}{z} \implies xz = xy + y^2 + yz \implies 2xz = xy + y^2 + yz + xz \implies$

$\frac {x}{x+y}= \frac {y+z}{2z} \implies \frac {AS}{AG} = \frac {1}{2} \cdot \frac {AB}{PB} \implies \triangle ESD \sim \triangle EGX \implies E \in DX.$ vladimir.shelomovskii@gmail.com, vvsss

2024, Problem 8

Let $ABCD$ be a quadrilateral with $\angle B = \angle D$ and $AD = CD.$

The incircle of $\triangle ABC$ touches the sides $BC$ and $AB$ at points $E$ and $F$ respectively.

The midpoints of segments $AC, BD, AE,$ and $CF$ are points $M,X,Y,Z.$

Prove that points $M,X,Y,Z.$ are concyclic.

Solution

$ZM || AB, YM || BC \implies \angle YMZ = \angle ABC = \angle ADC = \alpha.$ $2 \vec {XY} = \vec {DA} + \vec {BE}, 2 \vec {XZ} = \vec {DC} + \vec {BF}.$ $\vec {DC}$ is the rotation of $\vec {DA}$ around a point $D$ through an angle $\alpha.$

$\vec {BF}$ is the rotation of $\vec {BE}$ around a point $B$ through an angle $\alpha.$

So $\vec {XZ}$ is the rotation of $\vec {XE}$ around a point $X$ through an angle $\alpha.$

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 $M$ be the midpoint of the segment connecting the incenter and excenter. It is known that point $M$ belong the circumcircle. Construction is possible if a circle with diameter IE (incenter – excenter) intersects a circle with radius OM (circumcenter – M). Situation when $E$ between $I$ and $O$ is impossible.

Denote points $A, B, C$ such that $B \in AC$ and $AB \le BC.$

Suppose point $A$ is circumcenter, so $B$ is incenter. $M$ is midpoint BC. The vertices of the desired triangle are located at the intersection of a circle with center $A$ and radius $AM$ with $\omega$ and a line $AB.$

Suppose point $C$ is circumcenter, so $B$ is incenter. $M$ is midpoint $AB.$ The vertices of the desired triangle are located at the intersection of a circle with center $A$ and radius $AM$ with $\omega$ and a line $AB.$

Suppose point $B$ is circumcenter, so $A$ is incenter. $M$ is midpoint $AB.$ Suppose $3 AB < BC.$ The vertices of the desired triangle are located at the intersection of a circle with center $B$ and radius $BM$ with $\omega$ and a line $AB.$

If $3 AB \ge BC$ there is not desired triangle.

vladimir.shelomovskii@gmail.com, vvsss

@@ Line 39: / Line 39: @@
 [[File:2023 22 2.png|350px|right]]
 [[File:2024 22.png|350px|right]]
-A segment <math>AB</math> is given. Let <math>C</math> be an arbitrary point of the perpendicular bisector to <math>AB, O</math> be the point on the circumcircle of <math>\triangle ABC</math> opposite to <math>C,</math> and an ellipse centered at <math>O</math> touche <math>AB, BC, CA.</math>
+A segment <math>AB</math> is given. Let <math>C</math> be an arbitrary point of the perpendicular bisector to <math>AB,</math>
+<math>O</math> be the point on the circumcircle of <math>\triangle ABC</math> opposite to <math>C,</math> and an ellipse centered at <math>O</math> touche <math>AB, BC, CA.</math>
 Find the locus of touching points <math>P</math> of the ellipse with the line <math>BC.</math>
@@ Line 48: / Line 49: @@
 <cmath>\angle CBO = 90^\circ \implies \angle COB = \alpha,  MB = b \tan \alpha,</cmath>
-<cmath>CB = \frac {b \sin \alpha}{\cos^2 \alpha}, CO = \frac {b} {\cos^2 \alpha}, CN = b \left (1 +  \frac {1} {\cos^2 \alpha} \right ).</cmath>
+<cmath>CB = \frac {b \sin \alpha}{\cos^2 \alpha}, CO = \frac {b} {\cos^2 \alpha}, CD = b \left (1 +  \frac {1} {\cos^2 \alpha} \right ).</cmath>
 In order to find the ordinate of point <math>P,</math> we perform an affine transformation (compression along axis <math>AB)</math> which will transform the ellipse <math>MPD</math> into a circle with diameter <math>MD.</math> The tangent of the <math>CP</math> maps into the tangent of the <math>CE, E = \odot CBO \cap \odot MD, PF \perp CO.</math>
 <cmath>\angle OEF = \angle ECO \implies OF = OE \sin \angle OEF = OE  \sin \angle ECO = b \cos^2 \alpha.</cmath>
 <cmath>CP = \frac {CF}{\sin \alpha} = \frac {b}{\sin \alpha}\left ( \frac {1} {\cos^2 \alpha} - \cos^2 \alpha \right ) = b \sin \alpha  \left ( \frac {1}{\cos^2 \alpha } + 1 \right).</cmath>
 <cmath>\frac {CP}{CD} = \sin \alpha , \angle PCD = 90^\circ - \alpha \implies \angle CPD = 90^\circ.</cmath>
-<cmath>BP = CP - CB = \sin \alpha </cmath>
+<cmath>BP = CP - CB = b \sin \alpha.</cmath>
 Denote <math>Q = AB \cap DP \implies BQ = \frac {BP}{\cos \alpha} = b \tan \alpha = MB.</math>
@@ Line 61: / Line 62: @@
 '''vladimir.shelomovskii@gmail.com, vvsss'''
 ==2024, Problem 21==
 [[File:2024 21 0.png|350px|right]]

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Difference between revisions of "Sharygin Olympiads, the best"

Revision as of 10:07, 31 March 2024

Contents

2024, Problem 23

2024, Problem 22

2024, Problem 21

2024, Problem 19

2024, Problem 18

2024, Problem 16

2024, Problem 12

2024, Problem 9

2024, Problem 8

2024, Problem 2