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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. | 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 tur 2 klass 10 Problem 6== | ||
+ | [[File:2024 final 10 6.png|350px|right]] | ||
+ | A point <math>P</math> lies on one of medians of triangle <math>\triangle ABC</math> in such a way that <math>\angle PAB = \angle PBC = \angle PCA.</math> Prove that there exists a point <math>Q</math> on another median such that <math>\angle QBA = \angle QCB = \angle QAC.</math> (A.Zaslavsky) | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | 1. Denote <math>a = BC, b = AC, c = AB. </math> | ||
+ | It is known that barycentric coordinates are | ||
+ | <cmath>P = (a^2 b^2 : b^2 c^2 : c^2 a^2), P \in AD \implies b^2 c^2 = c^2 a^2 \implies AC = BC.</cmath> | ||
+ | |||
+ | 2. Denote <math>\omega = \odot APB, \Omega = \odot APC. </math> | ||
+ | |||
+ | <math>\angle ABP = \angle PAC \implies AC </math> is tangent to <math>\omega.</math> | ||
+ | |||
+ | <math>\angle PAC = \angle PCB \implies BC</math> is tangent <math>\Omega.</math> | ||
+ | |||
+ | <math>PC</math> is the radical axes of <math>\omega</math> and <math>\Omega,</math> the power of a point <math>D</math> with respect to a circle <math>\Omega</math> is <math>CD^2</math> so the power of a point <math>D</math> with respect to a circle <math>\omega</math> is <math>CD^2.</math> | ||
+ | |||
+ | <math>BD^2 = CD^2,</math> so <math>BD</math> is tangent to <math>\omega \implies \angle ABC = \angle CAB \implies AC = BC.</math> | ||
+ | |||
+ | <math>AC = BC,</math> so point <math>Q</math> symmetrical to <math>P</math> with respect to the <math>C-</math>median satisfies the conditions. <math>\blacksquare</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | |||
+ | ==2024 tur 2 klass 10 Problem 7== | ||
+ | [[File:2024 final 10 7.png|380px|right]] | ||
+ | Let <math>ABC</math> be a triangle with <math>\angle A = 60^\circ, AD, BE,</math> and <math>CF</math> be its bisectors, <math>P, Q</math> be the projections of <math>A</math> to <math>EF</math> and <math>BC</math> respectively, and <math>R</math> be the second common point of the circle <math>\omega = \odot DEF</math> with <math>AD.</math> | ||
+ | |||
+ | Prove that points <math>P, Q, R</math> are collinear. (K.Belsky) | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Denote <math>a = BC, b = AC, c = AB, I -</math> the incenter of <math>\triangle ABC,</math> | ||
+ | <math>L=FE \cap AD, R' -</math> the midpoint of <math>AI.</math> | ||
+ | |||
+ | It is known ([[Bisector | Division of bisector]]) that | ||
+ | <cmath>\frac {AI}{DI} = \frac {b+c}{a}, \frac {DI}{IL} = 1+ \frac {2a}{b+c} \implies \frac{1}{IL} = \frac{1}{DI} + \frac{2}{AI} \implies</cmath> <cmath>2 DI (\frac{AI}{2} - IL) =AI \cdot IL \implies 2 DI \cdot LR' = AI \cdot IL.</cmath> | ||
+ | |||
+ | <cmath>DL \cdot LR' = (DI + IL) \cdot LR' = \frac {AI \cdot IL}{2} + IL \cdot (\frac{AI}{2} - IL) = AI \cdot IL - IL^2 = AL \cdot IL.</cmath> | ||
+ | <math>\angle BAC = 60^\circ \implies \angle BIC = \angle FIE = 120^\circ \implies AEIF </math> is cyclic. | ||
+ | |||
+ | Therefore <math>AL \cdot IL = LE \cdot LF = DL \cdot LR' \implies EDFR'</math> is cyclic <math>\implies R = R'.</math> | ||
+ | [[File:2024 final 10 7a.png|430px|right]] | ||
+ | Let <math>AG \perp AD, G \in BC, H = AP \cap BC, Q' = RP \cap BC.</math> | ||
+ | |||
+ | It is known that points <math>F, E,</math> and <math>G</math> are collinear, | ||
+ | <cmath>\angle AGB = \frac{|\angle ACB - \angle ABC|}{2}, \angle AGE = \angle BGE.</cmath> | ||
+ | |||
+ | <math>AP \perp EF, AQ \perp BG \implies AG</math> is the diameter of <math>\odot APQG \implies</math> | ||
+ | <cmath>\overset{\Large\frown} {QP} = \overset{\Large\frown} {AP} \implies \angle DAP = \angle QAP, AP = PH.</cmath> | ||
+ | <cmath>AR = RI, AP = PH \implies RP || IH \implies \frac {DH}{HQ'} = \frac {DI}{IR} = \frac {2a}{b+c}.</cmath> | ||
+ | <math>AH</math> is the bisector of <math>\angle DAQ \implies \frac {DH}{HQ} = \frac {AD}{AQ}.</math> | ||
+ | |||
+ | Bisector <math>AD = \frac {2 b c \cos 30^\circ}{b+c}.</math> | ||
+ | |||
+ | Altitude <math>AQ = \frac {2[ABC]}{BC}= \frac {bc \sin 60^\circ}{a} \implies \frac {AD}{AQ} = \frac {2a}{b+c} = \frac {DH}{HQ} = \frac {DH}{HQ'} \implies Q = Q'. \blacksquare</math> | ||
+ | |||
+ | Note that the point <math>R</math> is a Feuerbach point of <math>\triangle ABC</math> since both the inscribed circle and the Euler circle pass through it. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==2024 tur 2 klass 9 Problem 7== | ||
+ | [[File:Incircle and secants.png|400px|right]] | ||
+ | Let triangle <math>\triangle ABC</math> and point <math>P</math> on the side <math>BC</math> be given. Let <math>P'</math> be such point on the side <math>BC</math> that <math>BP = P'C.</math> The cross points of segments <math>AP</math> and <math>AP'</math> with the incircle <math>\omega</math> of <math>\triangle ABC</math> form a convex quadrilateral <math>EFE'F'.</math> | ||
+ | |||
+ | Find the locus of crosspoints of diagonals <math>EFE'F'.</math> (D.Brodsky) | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | 1. Particular case of [[Projective geometry (simplest cases) | Fixed point]] . | ||
+ | |||
+ | 2. Denote <math>p_a = \frac{b+c-a}{2}, p_b = \frac{a-b+c}{2}, p_c = \frac{a+b-c}{2}, m = \frac {CP}{BP}, \mu = \frac {PF'}{AF'}.</math> | ||
+ | <cmath>PD = \frac {m p_b - p_c} {m+1}, AD' = p_a \implies</cmath> | ||
+ | <cmath>x(x+y) = p_a^2, z(z+y) = PD^2, x+y+z = AP, \mu = \frac {z}{x+y}.</cmath> | ||
+ | We perform simple transformations and get: | ||
+ | <cmath>\mu^2 p_a^2 - \mu (AP^2 - p_a^2 - PD^2) + PD^2 = 0.</cmath> | ||
+ | We use Stewart's theorem and get: | ||
+ | <cmath> AP^2 = \frac {AC^2}{1+m} + \frac {m AB^2}{1+m} - \frac {m BC^2}{(1+m)^2} \implies</cmath> | ||
+ | <cmath>(\mu p_a)^2 - 2 (\mu p_a) \frac {m p_b + p_c}{m+1} + \frac {(m p_b - p_c)^2} {(m+1)^2} = 0.</cmath> | ||
+ | <cmath>\mu = \frac {(\sqrt{m p_b} \pm \sqrt{p_c})^2}{(m+1) p_a}.</cmath> | ||
+ | Similarly <cmath>\nu = \frac {\eta + \zeta}{\xi}= \frac {(\sqrt{m p_c} \pm \sqrt{p_b})^2}{(m+1) p_a}.</cmath> | ||
+ | Therefore <math>\frac {\nu + \mu}{2} = \frac {a}{2 p_a}</math> not depends from <math>m.</math> | ||
+ | |||
+ | Let <math>M</math> be the midpoint of <math>BC, AM</math> is the median of <math>\triangle ABC</math> and <math>\triangle PAP'.</math> | ||
+ | |||
+ | The line <math>FF'</math> cross the median of <math>\triangle PAP'</math> at point <math>G'</math> such that <math>\frac {MG'}{G'A} = \frac {\nu + \mu}{2} = \frac {p_b + p_c}{2 p_a} = \frac {a}{2 p_a}.</math> | ||
+ | |||
+ | So point <math>G'</math> is fixed and this point lyes on <math>EE' \implies G = G'</math>. | ||
+ | |||
+ | Therefore the locus of crosspoints of diagonals <math>EFE'F'</math> is point <math>G.</math> | ||
+ | |||
+ | <i><b>Corollary</b></i> | ||
+ | |||
+ | Let line <math>MQ||D'D'', Q \in AC</math>. Then <math>\frac {D''Q}{AD''} = \frac {a}{2 p_a} \implies 2 D''Q = BC.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==2024 tur 2 klass 9 Problem 5== | ||
+ | [[File:2024 final 9 5.png|350px|right]] | ||
+ | Let <math>ABC</math> be an isosceles triangle <math>(AC = BC), O</math> be its circumcenter,<math>H</math> be the orthocenter, and <math>P</math> be a point inside the triangle such that <math>\angle APH = \angle BPO = 90^\circ.</math> | ||
+ | |||
+ | Prove that <math>\angle PAC = \angle PBA = \angle PCB.</math> (A.Zaslavsky) | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Denote <math>E -</math> the midpoint <math>BC, M-</math> the midpoint <math>AB, D -</math> the foot from <math>B</math> to <math>\overline{AC}, \alpha = \angle ABC, \omega = \odot ADHM, \Omega = \odot BEOM, \Theta = \odot BHC, \theta = \odot AB</math> tangent to <math>\overline{BC}.</math> | ||
+ | |||
+ | <math>AH = BH, BO = CO , \angle BAH = 90^\circ - \alpha = \angle BCO \implies \triangle ABH \sim \triangle BCO.</math> | ||
+ | There is a spiral similarity <math>T</math> centered at point <math>X</math> that maps <math>\triangle ABH</math> into <math>\triangle BCO.</math> | ||
+ | |||
+ | The coefficient of similarity <math>k = \frac {BC}{AB} = \frac {1}{2 \cos \alpha},</math> rotation angle equal <math>180^\circ - \alpha.</math> | ||
+ | <cmath>B = T(A), B = AB \cap BC \implies X \in \theta.</cmath> | ||
+ | <cmath>C = T(B), B = AB \cap BC \implies X \in \Theta</cmath> | ||
+ | <math>(\angle BCH = \angle ABH,</math> so <math>AB</math> is tangent to <math>\Theta).</math> [[Spiral similarity | Basic information]] | ||
+ | <cmath>E = T(M), B = AM \cap BE \implies X \in \Omega.</cmath> | ||
+ | <cmath>AB = T(CA), CO = T(BH), D = AC \cap BH, M = AB \cap CO \implies M = T(D).</cmath> | ||
+ | <cmath>A = AD \cap BM \implies X \in \omega.</cmath> | ||
+ | <cmath>P = \theta \cap \Theta \implies P = X.</cmath> | ||
+ | <cmath>P = T(P) \implies \triangle BPC = T(\triangle APB) \implies \triangle BPC \sim \triangle APB \implies \angle PAB = \angle PCB.</cmath> | ||
+ | <cmath>\triangle OBP = T(\triangle HAP) \implies \triangle OBP \sim \triangle HAP \implies \angle OBP = \angle HAP .</cmath> | ||
+ | <cmath>\angle PAC = \alpha - \angle HAP - \angle BAH = \alpha - \angle OBP - \angle OBC = \angle PBA.\blacksquare</cmath> | ||
+ | <cmath>\triangle ADM \sim \triangle ABC \implies \angle ADM = \alpha = \angle APM, \angle EPM = 180^\circ - \alpha \implies</cmath> | ||
+ | Points <math>A,P,</math> and <math>E</math> are collinear, so <math>P \in A-</math> median of <math>\triangle ABC.</math> | ||
+ | |||
+ | <math>\angle PAB = \angle PCB, \angle ABC = \angle BAC \implies PC</math> is <math>B-</math> symmedian of <math>\triangle ABC.</math> | ||
+ | |||
+ | <math>\angle APH = 90^\circ \implies P</math> is <math>A-</math> Humpty point. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==2024 tur 2 klass 9 Problem 4== | ||
+ | [[File:2024 final 9 4d.png|400px|right]] | ||
+ | For which <math>n > 0</math> it is possible to mark several different points and several different circles on the plane in such a way that: | ||
+ | |||
+ | - exactly <math>n</math> marked circles pass through each marked point; | ||
+ | |||
+ | - exactly <math>n</math> marked points lie on each marked circle; | ||
+ | |||
+ | - the center of each marked circle is marked? (P.Puchkov) | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | Case <math>n = 1.</math> Circles centered at <math>A</math> and <math>B</math> with radii <math>R = |\vec {AB}|.</math> | ||
+ | |||
+ | Case <math>n = 2, \vec {AC} = \vec {BC'}, |\vec {AB} = \vec {AC}|, \vec {AC}</math> is not paralel to <math>\vec {AB}.</math> | ||
+ | |||
+ | Four circles are centered at points <math>A, B, C,</math> and <math>C'.</math> Each radius is equal <math>R.</math> | ||
+ | |||
+ | Case <math>n = 3, \vec {CD} = \vec {AD''} = \vec {C'D'} = \vec {BD'''}, \vec {CD}</math> is not paralel to <math>\vec {AB}</math> or <math>\vec {AC}, |\vec {AB}| = \vec {CD}|</math> | ||
+ | |||
+ | Eight circles centered at <math>A, B, C, C', D, D',D''</math> and <math>D'''</math> have radii <math>R.</math> | ||
+ | |||
+ | Case <math>n = 4, \vec {DE} = \vec {D'E'} = ...,|\vec {DE}| = R,..</math> | ||
+ | |||
+ | <i><b>Answer </b></i> For all <math>n.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==2024 tur 2 klass 9 Problem 3== | ||
+ | [[File:2024 final 9 3.png|450px|right]] | ||
+ | Let <math>(P,P')</math> and <math>(Q,Q')</math> be two pairs of points isogonally conjugated with respect to a triangle <math>ABC,</math> and <math>R</math> be the common point of lines <math>PQ</math> and <math>P'Q'.</math> | ||
+ | Prove that the pedal circles of points <math>P,Q,</math> and <math>R</math> are coaxial. (L.Shatunov, V.Shelomovskii) | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | 1. Let <math>P'</math> be the isogonal conjugate of a point <math>P</math> with respect to a triangle <math>\triangle ABC.</math> Then circle centered at the midpoint <math>PP'</math> is the common pedal circle of points <math>P</math> and <math>P'.</math> ([[Isogonal conjugate | Circumcircle of pedal triangles]]) So center <math>O</math> is the midpoint <math>PP'</math> and center <math>U</math> is the midpoint <math>QQ'.</math> | ||
+ | |||
+ | 2. Denote <math>R' = PQ' \cap P'Q.</math> Then <math>R'</math> is the isogonal conjugate of a point <math>R</math> with respect to <math>\triangle ABC.</math> So center <math>V</math> is the midpoint <math>RR'.</math> ([[Isogonal conjugate | Two pares of isogonally conjugate points]]) | ||
+ | |||
+ | 3. The Gauss line (or Gauss–Newton line) is the line joining the midpoints of the three diagonals of a complete quadrilateral <math>PQ'P'Q</math> ([[Gauss line]]).So points <math>U,O,</math> and <math>V</math> are collinear as was to be proven. <math>\blacksquare</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | == 2024 tur 2 klass 8 Problem 4== | ||
+ | [[File:2024 final 8 4.png|300px|right]] | ||
+ | [[File:2024 final 8 4a.png|300px|right]] | ||
+ | A square with sidelength <math>1</math> is cut from the paper. Construct a segment with length <math>\frac{1}{2024}</math> using at most <math>13</math> folds. No instruments are available, it is allowed only to fold the paper and to mark the common points of folding lines. (M.Evdokimov) | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | Main idea: <math>\frac{1}{2024} = \frac{1}{4} \cdot \left [ \frac{1}{22} - \frac{1}{23} \right ].</math> | ||
+ | <cmath>\frac {a}{b} = \frac{x}{x+c} \implies x = \frac {a \cdot c}{b - a}.</cmath> | ||
+ | <cmath>EG = EA - GA = c \cdot \left [ \frac {1}{b/a - 1} - \frac {1}{(b+FB)/a - 1} \right ].</cmath> | ||
+ | Let <math>c = \frac{1}{2}, a = \frac{1}{64}, b = \frac{45}{64}, FB = \frac{2}{64}.</math> | ||
+ | <cmath>EG = \frac{1}{2} \cdot \left [ \frac {1}{45 - 1} - \frac {1}{47 - 1} \right] = \frac{1}{2024}.</cmath> | ||
+ | We perform <math>1</math> horizontal fold of the sheet. We get line <math>AD (AC = \frac{1}{2}).</math> We perform | ||
+ | |||
+ | <math>6</math> vertical folds of the sheet. We get <math>62</math> vertical lines at a distance of <math>\frac{1}{64}</math> from each other. | ||
+ | |||
+ | Point <math>F</math> is the lower left corner of the sheet, point <math>B</math> is the lower point of the second vertical line, point <math>C</math> is the lower point of the <math>47^{th}</math> line, point <math>D</math> is the point at the intersection of the horizontal line and the <math>46^{th}</math> vertical line. | ||
+ | |||
+ | Points <math>E</math> and <math>G</math> are at the intersection of the lines <math>BD</math> and <math>FD</math> and the <math>47^{th}</math> vertical line. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | == 2024 tur 2 klass 8 Problem 2== | ||
+ | [[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) | ||
+ | |||
+ | <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> 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>\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 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> | ||
+ | <math>\angle PDH = \angle MDH \implies</math> points <math>M,P,</math> and <math>D</math> are collinear. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
==2024, Problem 23== | ==2024, Problem 23== | ||
Line 18: | Line 233: | ||
<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> | ||
Line 28: | Line 243: | ||
<cmath>O'D' = O'C, OC = OD \implies \triangle OCD \sim \triangle O'D'C \implies OC||O'D'.</cmath> | <cmath>O'D' = O'C, OC = OD \implies \triangle OCD \sim \triangle O'D'C \implies OC||O'D'.</cmath> | ||
Let <math>F= CD \cup OO' \implies \frac {FO}{FO'} = \frac {OC}{O'D'} \implies Q = F.</math> | Let <math>F= CD \cup OO' \implies \frac {FO}{FO'} = \frac {OC}{O'D'} \implies Q = F.</math> | ||
+ | [[File:2024 23 3.png|350px|right]] | ||
<cmath>\angle LCB' = \alpha = \angle B'BL' \implies LC || L'B.</cmath> | <cmath>\angle LCB' = \alpha = \angle B'BL' \implies LC || L'B.</cmath> | ||
Similarly, <math>AL || CL' \implies \triangle DLC \sim \triangle CL'D' \implies \frac {LC}{L'D'} = \frac {DC}{CD'} = \frac {OC}{O'D'}.</math> | Similarly, <math>AL || CL' \implies \triangle DLC \sim \triangle CL'D' \implies \frac {LC}{L'D'} = \frac {DC}{CD'} = \frac {OC}{O'D'}.</math> | ||
Line 34: | Line 250: | ||
Therefore <math>Q \in LL'.</math> Similarly, if <math>P \in \overset{\Large\frown} {B'A'}</math> then <math>Q \in KK'.</math> | Therefore <math>Q \in LL'.</math> Similarly, if <math>P \in \overset{\Large\frown} {B'A'}</math> then <math>Q \in KK'.</math> | ||
+ | |||
+ | <i><b>Claim</b></i> | ||
+ | |||
+ | Points <math>D, C,</math> and <math>D'</math> are collinear. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <math>S</math> is the midpoint of arc <math>\overset{\Large\frown}{A'B'} \implies \angle SAC = \angle SBC.</math> | ||
+ | Denote <math>\angle CAP = \alpha, \angle CBP = \beta, \angle SAC = \angle SBC = \varphi.</math> | ||
+ | <cmath>D \in \omega \implies \angle PDC = \alpha, \angle PCD = \pi - \alpha - \varphi.</cmath> | ||
+ | <cmath>D' \in \omega' \implies \angle PD'C = \beta, \angle PCD' = \pi - \beta - \varphi.</cmath> | ||
+ | <cmath>S \in \Omega \implies \angle SAP + \angle SBP = \alpha + \beta + 2 \varphi = \pi.</cmath> | ||
+ | Therefore <math>\angle PCD + \angle PCD' = \pi \implies </math> points <math>D, C,</math> and <math>D'</math> are collinear. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==One-to-one mapping of the circle== | ||
+ | [[File:2024 23 AA.png|350px|right]] | ||
+ | Let a circle <math>\Omega,</math> two fixed points <math>A</math> and <math>B</math> on it and a point <math>C</math> inside it be given. | ||
+ | Then there is a one-to-one mapping of the circle <math>\Omega</math> onto itself, based on the following two theorems. | ||
+ | |||
+ | 1. Let a circle <math>\Omega,</math> two fixed points <math>A</math> and <math>B</math> on <math>\Omega,</math> and a point <math>C</math> inside <math>\Omega</math> be given. | ||
+ | |||
+ | Let an arbitrary point <math>Q \in \Omega</math> be given. | ||
+ | |||
+ | Let <math>A' = AC \cap \Omega, A' \ne A, B' = BC \cap \Omega, B' \ne B. S</math> is the midpoint of the arc <math>A'B', D = A'Q \cap BS, E = B'Q \cap AS.</math> | ||
+ | |||
+ | Denote <math>\omega = \odot AEC, \omega' = \odot BCD, P = \omega \cap \omega'.</math> Prove that <math>P \in \Omega.</math> | ||
+ | |||
+ | 2. Let a circle <math>\Omega,</math> two fixed points <math>A</math> and <math>B</math> on <math>\Omega,</math> and a point <math>C</math> inside <math>\Omega</math> be given. | ||
+ | |||
+ | Let an arbitrary point <math>P \in \Omega</math> be given. | ||
+ | |||
+ | Let <math>A' = AC \cap \Omega, A' \ne A, B' = BC \cap \Omega, B' \ne B. S</math> is the midpoint of the arc <math>A'B'.</math> | ||
+ | |||
+ | Denote <math>\omega = \odot ACP, \omega' = \odot BCP, D = BS \cap \omega' , E = \omega \cap AS.</math> | ||
+ | |||
+ | Denote <math>Q = A'D \cap B'E.</math> Prove that <math>Q \in \Omega.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <math>1. C = AA' \cap BB', D = A'Q \cap BS, E = B'Q \cap AS \implies </math> | ||
+ | |||
+ | Points <math>D,C,E </math> are collinear. | ||
+ | <cmath>\angle APC = \angle SEC, \angle BPC = \angle SDC.</cmath> | ||
+ | <cmath>\angle APC + \angle BPC + \angle ASC = \angle SEC + \angle SDC + \angle DSE = 180 ^\circ \implies P \in \Omega.</cmath> | ||
+ | |||
+ | 2. Points <math>D, C,</math> and <math>E</math> are collinear (see Claim in <i><b>2024, Problem 23</b></i>). | ||
+ | |||
+ | We use Pascal's theorem for points <math>A,B',S,A',B</math> and crosspoints <math>C,D,E</math> and get <math>Q \in \Omega.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==2024, Problem 22== | ||
+ | [[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,</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> | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | Denote <math>M</math> the midpoint <math>AB, D</math> the point on the line <math>CO, DO = MO, \alpha = \angle CBM, b = OM.</math> | ||
+ | |||
+ | <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}, 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 = b \sin \alpha.</cmath> | ||
+ | Denote <math>Q = AB \cap DP \implies BQ = \frac {BP}{\cos \alpha} = b \tan \alpha = MB.</math> | ||
+ | |||
+ | So point <math>Q</math> is the fixed point (<math>P</math> not depends from angle <math>\alpha, \angle BPQ = 90^\circ ).</math> | ||
+ | |||
+ | Therefore point <math>P</math> lies on the circle with diameter <math>BQ</math> (except points <math>B</math> and <math>Q.)</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==2024, Problem 21== | ||
+ | [[File:2024 21 0.png|350px|right]] | ||
+ | [[File:2024 21 1.png|350px|right]] | ||
+ | A chord <math>PQ</math> of the circumcircle of a triangle <math>ABC</math> meets the sides <math>BC, AC</math> at points <math>A', B',</math> respectively. The tangents to the circumcircle at <math>A</math> and <math>B</math> meet at point <math>X,</math> and the tangents at points <math>P</math> and <math>Q</math> meets at point <math>Y.</math> The line <math>XY</math> meets <math>AB</math> at point <math>C'.</math> | ||
+ | |||
+ | Prove that the lines <math>AA', BB',</math> and <math>CC'</math> concur. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | WLOG, <math>P \in \overset{\Large\frown} {AC}.</math> | ||
+ | Denote <math>\Omega = \odot ABC, Z = BB' \cap AA', D = AQ \cap BP.</math> | ||
+ | |||
+ | Point <math>D</math> is inside <math>\Omega.</math> | ||
+ | |||
+ | We use Pascal’s theorem for quadrilateral <math>APQB</math> and get <math>D \in XY.</math> | ||
+ | |||
+ | We use projective transformation which maps <math>\Omega</math> to a circle and that maps the point <math>D</math> to its center. | ||
+ | |||
+ | From this point we use the same letters for the results of mapping. Therefore the segments <math>AQ</math> and <math>BP</math> are the diameters of <math>\Omega, C'D \in XY || AP \implies C'</math> is the midpoint <math>AB.</math> | ||
+ | |||
+ | <math>AB|| PQ \implies AB || B'A' \implies C' \in CZ \implies</math> | ||
+ | |||
+ | preimage <math>Z</math> lies on preimage <math>CC'.\blacksquare</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==2024, Problem 20== | ||
+ | [[File:2024 20.png|350px|right]] | ||
+ | Let a triangle <math>ABC,</math> points <math>D</math> and <math>E \in BD</math> be given, <math>F = AD \cap CE.</math> | ||
+ | Points <math>D', E'</math> and <math>F'</math> are the isogonal conjugate of the points <math>D, E,</math> and <math>F,</math> respectively, with respect to <math>\triangle ABC.</math> | ||
+ | |||
+ | Denote <math>R</math> and <math>R'</math> the circumradii of triangles <math>\triangle DEF</math> and <math>\triangle D'E'F',</math> respectively. | ||
+ | |||
+ | Prove that <math>\frac {[DEF]}{R} = \frac {[D'E'F']}{R'},</math> where <math>[DEF]</math> is the area of <math>\triangle DEF.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Denote <math>\angle BAC = \alpha, \angle ABC = \beta, \angle ACB = \gamma,</math> | ||
+ | <cmath>\angle EDF = \Theta, \angle E'D'F' = \theta, \angle DEF = \Psi, \angle D'E'F' = \psi,</cmath> | ||
+ | <cmath>\angle BAD' = \angle CAF = \varphi_A, \angle CBD' = \angle ABD = \varphi_B, \angle BCE' = \angle ACE = \varphi_C.</cmath> | ||
+ | It is easy to prove that | ||
+ | <math>\frac {[DEF]}{R} = \frac {[D'E'F']}{R'}</math> is equivalent to <math>DE \cdot \sin \Theta \cdot \sin \Psi = D'E' \cdot \sin \theta \cdot \sin \psi.</math> | ||
+ | <cmath>\Theta = \alpha - \varphi_A + \varphi_B, \theta = \beta - \varphi_B + \varphi_A, \Psi = \beta - \varphi_B + \varphi_C, \psi = \gamma - \varphi_C + \varphi_B \implies</cmath> | ||
+ | <cmath>\theta = 180^\circ - \gamma - \Theta, \psi = 180^\circ - \alpha - \Psi.</cmath> | ||
+ | By applying the law of sines, we get | ||
+ | <cmath>\frac {BE}{\sin \varphi_C} = \frac {BC}{\sin \Psi}, \frac {BD}{\sin (\alpha - \varphi_A)} = \frac {AB}{\sin \Theta}, \frac {BC}{\sin \alpha} = \frac {AB}{\sin \gamma}.</cmath> | ||
+ | <cmath>ED = BD - BE \implies DE \cdot \sin \Theta \cdot \sin \Psi = \frac {AB}{\sin \gamma} \left ( \sin \gamma \cdot \sin (\alpha - \varphi_A) \cdot \sin \Psi - \sin \alpha \cdot \sin \varphi_C \cdot \sin \Theta \right )</cmath> | ||
+ | |||
+ | <cmath>\frac {BE'}{\sin (\gamma - \varphi_C)} = \frac {BC}{\sin \psi}, \frac {BD'}{\sin \varphi_A} = \frac {AB}{\sin \theta}.</cmath> | ||
+ | <cmath>D'E' = BE' - BD' \implies D'E' \cdot \sin \theta \cdot \sin \psi = \frac {AB}{\sin \gamma} (\sin \alpha \cdot \sin (\gamma - \varphi_C) \cdot \sin \theta - \sin \gamma \cdot \sin \varphi_A \cdot \sin \psi ).</cmath> | ||
+ | We need to prove that | ||
+ | <cmath>\sin \gamma \cdot \sin (\alpha - \varphi_A) \cdot \sin \Psi - \sin \alpha \cdot \sin \varphi_C \cdot \sin \Theta = \sin \alpha \cdot \sin (\gamma - \varphi_C) \cdot \sin \theta - \sin \gamma \cdot \sin \varphi_A \cdot \sin \psi</cmath> | ||
+ | We make the transformations: | ||
+ | <cmath>\sin \alpha \left[ \sin \varphi_C \cdot \sin \Theta + \sin (\gamma - \varphi_C) \cdot \sin (\gamma + \Theta) \right] = \sin \gamma \left[ \sin (\alpha - \varphi_A) \cdot \sin \Psi + \sin \varphi_A \cdot \sin (\alpha + \Psi) \right]</cmath> | ||
+ | |||
+ | <cmath>\sin \alpha \left[ \sin \varphi_C \cdot \sin \Theta + (\sin \gamma \cos \varphi_C - \cos \gamma \sin \varphi_C) \cdot (\sin \gamma \cos \Theta + \cos \gamma \sin \Theta \right] =</cmath> | ||
+ | <cmath>= \sin \gamma \left[ \sin \alpha \cos \varphi_A \cdot \sin \Psi - \cos \alpha \sin \varphi_A \cdot \sin \Psi + \sin \varphi_A \cdot \sin \alpha \cos \Psi + \sin \varphi_A \cdot \cos \alpha \sin \Psi \right]</cmath> | ||
+ | <cmath>\sin \alpha \left[ \sin \varphi_C \cdot \sin \Theta + \sin^2 \gamma \cos \varphi_C \cos \Theta - \cos ^2 \gamma \sin \varphi_C \cdot \sin \Theta + \cos \gamma \sin \Theta + \sin \gamma \cos \gamma (\cos \varphi_C \sin \Theta - \sin \varphi_C \cos \Theta) \right] =</cmath> | ||
+ | <cmath>= \sin \gamma \cdot \sin \alpha \left[\cos \varphi_A \cdot \sin \Psi + \sin \varphi_A \cdot \cos \Psi \right]</cmath> | ||
+ | |||
+ | <cmath>\sin \alpha \left[ \sin^2 \gamma \cdot \cos (\Theta - \varphi_C) + \sin \gamma \cdot \cos \gamma \cdot \sin (\Theta - \varphi_C)\right] = \sin \gamma \cdot \sin \alpha \cdot \sin (\Psi + \varphi_A)</cmath> | ||
+ | |||
+ | <cmath>\sin \alpha \cdot \sin \gamma \cdot \sin (\gamma + \Theta - \varphi_C) = \sin \gamma \cdot \sin \alpha \cdot \sin (\beta + \varphi_A - \varphi_B + \varphi_C).</cmath> The last statement is obvious. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==2024, Problem 19== | ||
+ | [[File:2024 19 4.png|250px|right]] | ||
+ | [[File:2024 19 2.png|250px|right]] | ||
+ | [[File:2024 19 3.png|250px|right]] | ||
+ | A triangle <math>ABC,</math> its circumcircle <math>\Omega</math>, and its incenter <math>I</math> are drawn on the plane. | ||
+ | |||
+ | Construct the circumcenter <math>O</math> of <math>\triangle ABC</math> using only a ruler. | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | We successively construct: | ||
+ | |||
+ | - the midpoint <math>D = BI \cap \Omega</math> of the arc <math>AB,</math> | ||
+ | |||
+ | - the midpoint <math>E = CI \cap \Omega</math> of the arc <math>AC,</math> | ||
+ | |||
+ | - the polar <math>H'H''</math> of point <math>H \in DE,</math> | ||
+ | |||
+ | - the polar <math>G'G''</math> of point <math>G \in DE,</math> | ||
+ | |||
+ | - the polar <math>F = H'H'' \cap G'G''</math> of the line <math>DE,</math> | ||
+ | |||
+ | - the tangent <math>FD || AC</math> to <math>\Omega,</math> | ||
+ | |||
+ | - the tangent <math>FE || AB</math> to <math>\Omega,</math> | ||
+ | |||
+ | - the trapezium <math>ACDF,</math> | ||
+ | |||
+ | - the point <math>K = AF \cap CD,</math> | ||
+ | |||
+ | - the point <math>L = AD \cap CF,</math> | ||
+ | |||
+ | - the midpoint <math>M = AC \cap KL</math> of the segment <math>AB,</math> | ||
+ | |||
+ | - the midpoint <math>M'</math> of the segment <math>AC,</math> | ||
+ | |||
+ | - the diameter <math>DM</math> of <math>\Omega,</math> | ||
+ | |||
+ | - the diameter <math>EM'</math> of <math>\Omega,</math> | ||
+ | |||
+ | - the circumcenter <math>O = DM \cap EM'.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==2024, Problem 18== | ||
+ | [[File:2024 18 1.png|390px|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''' | ||
+ | ==2024, Problem 17== | ||
+ | [[File:2024 17 1.png|400px|right]] | ||
+ | Let <math>\triangle ABC</math> be not isosceles triangle, <math>\omega</math> be its incircle. | ||
+ | |||
+ | Let <math>D, E,</math> and <math>F</math> be the points at which the incircle of <math>\triangle ABC</math> touches the sides <math>BC, CA,</math> and <math>AB,</math> respectively. | ||
+ | |||
+ | Let <math>K</math> be the point on ray <math>EF</math> such that <math>EK = AB.</math> | ||
+ | |||
+ | Let <math>L</math> be the point on ray <math>FE</math> such that <math>FL = AC.</math> | ||
+ | |||
+ | The circumcircles of <math>\triangle BFK</math> and <math>\triangle CEL</math> intersect <math>\omega</math> again at <math>Q</math> and <math>P,</math> respectively. | ||
+ | |||
+ | Prove that <math>BQ, CP,</math> and <math>AD</math> are concurrent. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <math>\frac {KE} {FL} = \frac {AB}{AC},</math> so points <math>Q,P,</math> and <math>G = FE \cap BC</math> are collinear (see [[Symmetry | Symmetry and incircle]] for details). | ||
+ | |||
+ | Therefore lines <math>BQ, CP,</math> and <math>AD</math> are concurrent (see [[ Symmetry | Symmetry and incircle A]] for details.) | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==2024, Problem 16== | ||
+ | [[File:2024 16 1.png|300px|right]] | ||
+ | Let <math>AA', BB',</math> and <math>CC'</math> be the bisectors of a triangle <math>\triangle ABC.</math> | ||
+ | |||
+ | The segments <math>BB'</math> and <math>A'C'</math> meet at point <math>D.</math> Let <math>E</math> be the projection of <math>D</math> to <math>AC.</math> | ||
+ | |||
+ | Points <math>P</math> and <math>Q</math> on the sides <math>AB</math> and <math>BC,</math> respectively, are such that <math>EP = PD, EQ = QD.</math> | ||
+ | |||
+ | Prove that <math>\angle PDB' = \angle EDQ.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <math>\triangle PDQ = \triangle PEQ (DQ = EQ, DP = PF, PQ</math> is the common side) <math>\implies</math> | ||
+ | |||
+ | <math>PQ \perp DE, F = PQ \cap DE</math> is the midpoint <math>DE \implies</math> | ||
+ | |||
+ | <math>G = BB' \cap PQ</math> is the midpoint of <math>DB'.</math> | ||
+ | <cmath>\frac{BG}{BB'} =\frac {BQ}{BC} = \frac {BP}{BA} = \frac {BD}{BI}.</cmath> | ||
+ | (see [[Bisector | Division of bisector]] for details.) | ||
+ | |||
+ | So <math>DQ || CC', PD || AA'.</math> Denote <math>\angle ACC' = \angle BCC' = \gamma, \angle A'AC = \alpha, B'BC = \beta.</math> | ||
+ | <cmath>\angle PDB' = \angle AIB' = \angle BB'C - \angle IAC = 180^\circ - \beta - 2 \gamma - \alpha = 90^\circ - \gamma.</cmath> | ||
+ | <cmath>\angle QDE = 90^\circ - \angle DQP = 90^\circ - \gamma = \angle PDB'.</cmath> | ||
+ | |||
+ | Another solution see [[Isogonal_conjugate | 2024_Sharygin_olimpiad_Problem_16]] | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==2024, Problem 15== | ||
+ | [[File:2024 15.png|390px|right]] | ||
+ | The difference of two angles of a triangle is greater than <math>90^\circ.</math> Prove that the ratio of its circumradius and inradius is greater than <math>4.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Suppose, <math>\angle BAC = \alpha = \angle ACB + 90^\circ = \gamma + 90^\circ.</math> | ||
+ | |||
+ | Let <math>\Omega = \odot ABC, C'</math> be the point on <math>\Omega</math> opposite <math>C, B'</math> be the midpoint of arc <math>\overset{\Large\frown} {AC}.</math> | ||
+ | Then <cmath>\angle CAC' = 90^\circ \implies \angle BAC' = \angle ACB = \gamma = \angle BCC' \implies</cmath> | ||
+ | <cmath>\overset{\Large\frown} {AB} = \overset{\Large\frown} {BC'} \implies BB' || CC'.</cmath> | ||
+ | <cmath>BO = CO \implies \angle CBO = \angle OCB = \angle ACB = \gamma.</cmath> | ||
+ | <cmath>\angle ABC = \beta = 180^\circ - \gamma - \alpha = 90^\circ - 2 \gamma.</cmath> | ||
+ | <cmath>\angle OBB' = \frac {\beta}{2} + \gamma = 45^\circ.</cmath> | ||
+ | <cmath>BO = OB' = R \implies \angle BB'O = 45^\circ, \angle BOB' = 90^\circ.</cmath> | ||
+ | Incenter <math>I</math> triangle <math>\triangle ABC</math> lies on <math>BB',</math> therefore <math>OI \ge \frac {R}{\sqrt {2}}.</math> | ||
+ | |||
+ | We use the Euler law <math>OI^2 = R^2 - 2 Rr \implies \frac {2r}{R} = 1 - \frac {OI^2}{R^2} \le \frac{1}{2} \implies \frac{R}{r} \ge 4.</math> | ||
+ | |||
+ | If <math>R = 4r</math> then <math>OI \perp BB' \implies \frac {BI}{IB'} = 1 \implies \frac {BC + AB}{AC} = 2, BI = \frac{R}{\sqrt{2}}.</math> | ||
+ | <cmath>\sin CBI = \frac {r}{BI} = \frac{r \sqrt{2}}{R} = \frac{1}{2\sqrt{2}} \implies \sin \beta = \frac{\sqrt{7}}{4} \implies</cmath> | ||
+ | <cmath>AC = 2R \sin \beta = R \frac {\sqrt{7}}{2},AB = AC (1 - \frac{1}{\sqrt {7}}), BC = AC (1 + \frac{1}{\sqrt {7}}).</cmath> | ||
+ | |||
+ | If <math>\angle BAC' > \angle ACB \implies \frac {OI}{R}</math> increases so <math>\frac {r}{R}</math> decreases. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==2024, Problem 14== | ||
+ | [[File:2024 14 1.png|390px|right]] | ||
+ | The incircle <math>\omega</math> of a right-angled triangle <math>\triangle ABC</math> touches the circumcircle <math>\theta</math> of its medial triangle at point <math>F.</math> Let <math>OE</math> be the tangent to <math>\omega</math> from the midpoint <math>O</math> of the hypothenuse <math>AB</math> distinct from <math>AB.</math> Prove that <math>CE = CF.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let <math>\Omega</math> and <math>I</math> be the circumcircle and the incenter of <math>\triangle ABC, D = \omega \cap AB.</math> | ||
+ | |||
+ | Let <math>Q</math> be nine-point center of <math>\triangle ABC, G</math> be the point at <math>OB</math> such that <math>OG = DO, K \in DI, KQ ||AB.</math> | ||
+ | |||
+ | Denote <math>r = ID, R = AO,a = BC, b = AC,c= AB, \beta = \angle ABC.</math> | ||
+ | |||
+ | <math>\triangle ABC</math> is the right-angled triangle, so <math>Q</math> is the midpoint <math>CO,</math> | ||
+ | |||
+ | <cmath>2R = c, r = \frac {a+b-c}{2},FQ = \frac{R}{2}.</cmath> | ||
+ | Let <math>h(X)</math> be the result of the homothety of the point <math>X</math> centered in <math>C</math> with the coefficient <math>2.</math> Then | ||
+ | <cmath>h(\theta) = \Omega, F' = h(F) \in \Omega, O = h(Q) \implies FQ || F'O.</cmath> | ||
+ | <cmath>DO = EO = GO \implies \angle DEG = 90^\circ.</cmath> | ||
+ | <cmath>\angle DOE = 2 \angle DOI = 2 \angle EGD \implies \angle DOI = \angle EGD, IO || EG.</cmath> | ||
+ | WLOG, <math>a > b \implies DO = \frac {a-b}{2} = \tan IOD = \frac{r}{DO} = \frac{a+b-c}{a-b} = \frac{\cos\beta + \sin \beta -1}{\cos\beta - \sin \beta}.</math> | ||
+ | |||
+ | Let <math>H</math> be the foot from <math>C</math> to <math>\overline{AB}</math>. | ||
+ | <math>CH = a \sin \beta = c \sin \beta \cos \beta, BH = a \cos \beta, BG = AD = \frac{c+b-a}{2}.</math> | ||
+ | <cmath>\tan \angle CGH = \frac {CH}{BH - BG} = \frac{2 \sin \beta \cos \beta}{2 \cos^2 \beta +\cos \beta - \sin \beta -1} =</cmath> | ||
+ | <cmath>= \frac{(\sin \beta + \cos \beta -1)(\sin \beta + \cos \beta +1)}{(\cos \beta - \sin \beta)(\sin \beta + \cos \beta +1)} = \frac{\cos\beta + \sin \beta -1}{\cos\beta - \sin \beta} = \tan IOD.</cmath> | ||
+ | Therefore points <math>C,E,</math> and <math>G</math> are collinear. | ||
+ | <cmath>\psi = \angle BCG = \angle AGC - \angle ABC \implies</cmath> | ||
+ | <cmath>\tan \psi = \frac {\tan \angle AGC - \tan \angle ABC}{1 + \tan \angle AGC \cdot \tan \angle ABC} = \frac {c - a}{c-b}.</cmath> | ||
+ | <cmath>\sin 2\psi = \frac {2\tan \psi}{1 + \tan^2 \psi} = \frac {2(c-a)(c-b)}{c(3c - 2b - 2a)},</cmath> | ||
+ | <cmath>\angle AOF' = \angle KQI, \sin \angle KQI = \frac{CH/2 -r}{QF - r}.</cmath> | ||
+ | <cmath>4c(QF - r) = c(c -2a -2b +2c) = c(3c -2a -2b),</cmath> | ||
+ | <cmath>2c(CH - 2r) = 2(ab -ac -bc +c^2) = 2(c - a)(c - b),</cmath> | ||
+ | <cmath>\sin \angle KQI = \frac {2(c-a)(c-b)}{c(3c - 2b - 2a)}= \sin 2 \psi \implies \angle KQI = 2 \psi.</cmath> | ||
+ | <cmath>\angle ACF' = \frac {\angle AOF'}{2} = \frac {\angle KQI}{2} = \psi = \angle BCG \implies \angle FCI = \angle ECI \implies CF = CE.</cmath> | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==2024, Problem 12== | ||
+ | [[File:2024 12.png|390px|right]] | ||
+ | The bisectors <math>AE, CD</math> of a <math>\triangle ABC</math> with <math>\angle B = 60^\circ</math> meet at point <math>I.</math> | ||
+ | |||
+ | The circumcircles of triangles <math>ABC, DIE</math> meet at point <math>P.</math> | ||
+ | |||
+ | Prove that the line <math>PI</math> bisects the side <math>AC.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Denote <math>M</math> the midpoint <math>AC, \omega = \odot DIE,</math> | ||
+ | <cmath>\varphi = \angle CIM, \phi = \angle DIP, f(x) = \frac {\sin x}{\sin(120^\circ -x)}.</cmath> | ||
+ | <cmath>\angle AIC = 90^\circ + \frac {\angle ABC}{2} = 120^\circ = 180^\circ - \angle ABC \implies B \in \omega.</cmath> | ||
+ | In triangles <math>\triangle CIM</math> and <math>\triangle AIM</math>, by applying the law of sines, we get | ||
+ | <cmath>\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}.</cmath> | ||
+ | |||
+ | We use the formulas for circle <math>\omega</math> and get <cmath>\frac {PD}{\sin \angle PID} = \frac {PE}{\sin \angle PIE} \implies f(\phi) = \frac {\sin \phi} {\sin (120^\circ -\phi)} = \frac {PD}{PE}.</cmath> | ||
+ | |||
+ | <cmath>\angle BDI = \angle AEC \implies \angle ADC = 180^\circ - \angle AEC.</cmath> | ||
+ | In triangles <math>\triangle ADC</math> and <math>\triangle AEC</math>, by applying the law of sines, we get | ||
+ | <cmath> \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}.</cmath> | ||
+ | |||
+ | <cmath>\angle BEP = \angle BDP, \angle BCP = \angle BAP \implies \triangle APD \sim \triangle CPE \implies \frac {AD}{CE} = \frac {PD}{PE}.</cmath> | ||
+ | |||
+ | Therefore <math>f(\phi) = f(\varphi).</math> The function <math>f</math> increases monotonically on the interval <math>(0, \frac {2 \pi}{3}).</math> | ||
+ | |||
+ | This means <math>\phi = \varphi</math> and points <math>P,I,</math> and <math>M</math> are collinear. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==2024, Problem 9== | ||
+ | [[File:2024 9.png|370px|right]] | ||
+ | Let <math>ABCD (AD || BC</math> be a trapezoid circumscribed around a circle <math>\omega,</math> centered at <math>O</math> which touches the sides <math>AB, BC, CD,</math> and <math>AD</math> at points <math>P, Q, R, S,</math> respectively. | ||
+ | |||
+ | The line passing trough <math>P</math> and parallel to the bases of trapezoid meets <math>QR</math> at point <math>X.</math> | ||
+ | |||
+ | Prove that <math>AB, QS,</math> and <math>DX</math> concur. | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | Solution 1. <math>AP = AS \implies \angle AOS = \frac {\overset{\Large\frown} {PS}}{2} = \angle PQS \implies PQ||AO.</math> | ||
+ | |||
+ | <cmath>OD \perp OC, QR \perp OC \implies OD || QX. AD ||PX \implies</cmath> | ||
+ | |||
+ | <math>E = AP \cap QS</math> is the center of similarity of triangles <math>\triangle PQX</math> and <math>\triangle AOD.</math> | ||
+ | |||
+ | Solution 2. <math>\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}.</math> | ||
+ | |||
+ | Denote <math>EA = x, AP = AS = y, BP = BQ = z.</math> | ||
+ | |||
+ | <cmath>\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</cmath> | ||
+ | |||
+ | <cmath>\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.</cmath> | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==2024, Problem 8== | ||
+ | [[File:2024 8.png|390px|right]] | ||
+ | Let <math>ABCD</math> be a quadrilateral with <math>\angle B = \angle D</math> and <math>AD = CD.</math> | ||
+ | |||
+ | The incircle of <math>\triangle ABC</math> touches the sides <math>BC</math> and <math>AB</math> at points <math>E</math> and <math>F</math> respectively. | ||
+ | |||
+ | The midpoints of segments <math>AC, BD, AE,</math> and <math>CF</math> are points <math>M,X,Y,Z.</math> | ||
+ | |||
+ | Prove that points <math>M,X,Y,Z.</math> are concyclic. | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | <cmath>ZM || AB, YM || BC \implies \angle YMZ = \angle ABC = \angle ADC = \alpha.</cmath> | ||
+ | <cmath>2 \vec {XY} = \vec {DA} + \vec {BE}, 2 \vec {XZ} = \vec {DC} + \vec {BF}.</cmath> | ||
+ | <math>\vec {DC}</math> is the rotation of <math>\vec {DA}</math> around a point <math>D</math> through an angle <math>\alpha.</math> | ||
+ | |||
+ | <math>\vec {BF}</math> is the rotation of <math>\vec {BE}</math> around a point <math>B</math> through an angle <math>\alpha.</math> | ||
+ | |||
+ | So <math>\vec {XZ}</math> is the rotation of <math>\vec {XE}</math> around a point <math>X</math> through an angle <math>\alpha.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==2024, Problem 2== | ||
+ | [[File:2024 2.png|330px|right]] | ||
+ | [[File:2024 2a.png|330px|right]] | ||
+ | [[File:2024 2b.png|330px|right]] | ||
+ | Three distinct collinear points are given. Construct the isosceles triangles such that these points are their circumcenter, incenter and excenter (in some order). | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | Let <math>M</math> be the midpoint of the segment connecting the incenter and excenter. It is known that point <math>M</math> belong the circumcircle. | ||
+ | Construction is possible if a circle with diameter IE (incenter – excenter) intersects a circle with radius OM (circumcenter – M). Situation when <math>E</math> between <math>I</math> and <math>O</math> is impossible. | ||
+ | |||
+ | Denote points <math>A, B, C</math> such that <math>B \in AC</math> and <math>AB \le BC.</math> | ||
+ | |||
+ | Suppose point <math>A</math> is circumcenter, so <math>B</math> is incenter. <math>M</math> is midpoint BC. The vertices of the desired triangle are located at the intersection of a circle with center <math>A</math> and radius <math>AM</math> with <math>\omega</math> and a line <math>AB.</math> | ||
+ | |||
+ | Suppose point <math>C</math> is circumcenter, so <math>B</math> is incenter. <math>M</math> is midpoint <math>AB.</math> The vertices of the desired triangle are located at the intersection of a circle with center <math>A</math> and radius <math>AM</math> with <math>\omega</math> and a line <math>AB.</math> | ||
+ | |||
+ | Suppose point <math>B</math> is circumcenter, so <math>A</math> is incenter. <math>M</math> is midpoint <math>AB.</math> Suppose <math>3 AB < BC.</math> The vertices of the desired triangle are located at the intersection of a circle with center <math>B</math> and radius <math>BM</math> with <math>\omega</math> and a line <math>AB.</math> | ||
+ | |||
+ | If <math>3 AB \ge BC</math> there is not desired triangle. | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==The problem from MGTU== | ||
+ | [[File:2024 olimp november pr 9.png|390px|right]] | ||
+ | The lateral face of the regular triangular pyramid <math>SABC</math> is inclined to the plane of the base <math>ABC</math> at an angle of <math>\alpha = \arctan \frac{3}{4}.</math> Points <math>D, E, F</math> are the midpoints of the sides of the <math>\triangle ABC.</math> Triangle <math>\triangle DEF</math> is the lower base of a right prism. The edges of the upper base of the prism intersect the lateral edges of the pyramid <math>SABC</math> at points <math>K, L, N.</math> The area of the total surface of the polyhedron with vertices <math>D, E, F, K, L, N</math> is equal to <math>53 \sqrt{3}.</math> Find the side of <math>\triangle ABC.</math> | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | Denote <math>AB = BC = AC = a, O</math> is the center of <math>\triangle ABC, G = DF \cap AO, GK \perp ABC, K \in AS, M = SD \cap KN.</math> | ||
+ | <cmath>OD = \frac{ CD}{3} = \frac {a \sqrt {3}}{6}, SO = OD \cdot \tan \alpha = \frac {a \sqrt {3}}{8}.</cmath> | ||
+ | <cmath>AG = \frac {AE}{2}, AO = \frac{2AE}{3} \implies \frac {KG}{SO} = \frac{AG}{AO} = \frac{3}{4} \implies KG = \frac {3 a \sqrt{3}}{32} = h.</cmath> | ||
+ | <cmath>MD = \frac{KG}{\sin \alpha} = \frac{5}{3} KG = \frac{5h}{3}.</cmath> | ||
+ | The area of the total surface of the polyhedron with vertices <math>D, E, F, K, L, N</math> is | ||
+ | <cmath>Area = [DEF] + [KLN] + 3[DFK]+ 3[DKN].</cmath> | ||
+ | <cmath>[ABC] = \frac {a^2 \sqrt{3}}{4} = 4[DEF] =4 s \implies s = \frac {a^2 \sqrt{3}}{16}.</cmath> | ||
+ | <cmath>\frac{KN}{AB} = \frac {SK}{SA} = \frac{GO}{AO} = \frac {1}{4} \implies [KLN] =\frac{[ABC]}{16} = \frac {s}{4} \implies [DEF] + [KLN] = \frac {5s}{4}.</cmath> | ||
+ | <cmath>[DFK] = \frac{KG \cdot DF}{2} = \frac{ah}{4}, ah = \frac{3}{2} s.</cmath> | ||
+ | <cmath>[DKN] = \frac{MD \cdot KN}{2} = \frac{5ah}{24} \implies 3[DFK]+ 3[DKN] = \frac{11ah}{8} = \frac{33s}{16}.</cmath> | ||
+ | <cmath>Area = \frac {5s}{4} + \frac{33s}{16} = \frac{53s}{16} = 53 \sqrt{3} \implies s = 16 \sqrt {3} = \frac {a^2 \sqrt{3}}{16} \implies a = 16.</cmath> | ||
+ | ==The trapezoid problem from MGTU== | ||
+ | [[File:2024 olimp november pr 7.png|470px|right]] | ||
+ | Points <math>M</math> and <math>N</math> are the midpoints of bases <math>AD</math> and <math>BC</math> of trapezoid <math>ABCD.</math> | ||
+ | |||
+ | Denote <math>\alpha</math> the angle between lines <math>MN</math> and <math>AC, \cos \alpha = \frac{11}{16}.</math> | ||
+ | |||
+ | Find the area of trapezoid <math>ABCD</math> if <math>MN = 2, BD = 6.</math> | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | <cmath>AF||MN||CG \implies AG = CF = AM+NC \implies FB=DG \implies</cmath> | ||
+ | <cmath>[ABCD] = [AFCG] = AC \cdot CG \cdot \sin \alpha = 2 CH \cdot CG \cdot \sin \alpha.</cmath> | ||
+ | By applying the Law of Cosines on <math>\triangle CHG, HG = \frac {BD}{2} = 3,</math> we get | ||
+ | <cmath>CH^2 + CG^2- 2 CH \cdot CG \cos \alpha = GH^2 \implies CH^2 - \frac {11}{4} CH - 5 = 0 \implies CH = 4.</cmath> | ||
+ | <cmath>\sin ^2 \alpha = 1 - \cos^2 \alpha = \frac {135}{16^2} \implies [ABCD] = 2 \cdot 4 \cdot 2 \frac {\sqrt{135}}{16} = \sqrt{135}.</cmath> |
Latest revision as of 14:29, 17 November 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 10 Problem 6
- 2 2024 tur 2 klass 10 Problem 7
- 3 2024 tur 2 klass 9 Problem 7
- 4 2024 tur 2 klass 9 Problem 5
- 5 2024 tur 2 klass 9 Problem 4
- 6 2024 tur 2 klass 9 Problem 3
- 7 2024 tur 2 klass 8 Problem 4
- 8 2024 tur 2 klass 8 Problem 2
- 9 2024, Problem 23
- 10 One-to-one mapping of the circle
- 11 2024, Problem 22
- 12 2024, Problem 21
- 13 2024, Problem 20
- 14 2024, Problem 19
- 15 2024, Problem 18
- 16 2024, Problem 17
- 17 2024, Problem 16
- 18 2024, Problem 15
- 19 2024, Problem 14
- 20 2024, Problem 12
- 21 2024, Problem 9
- 22 2024, Problem 8
- 23 2024, Problem 2
- 24 The problem from MGTU
- 25 The trapezoid problem from MGTU
2024 tur 2 klass 10 Problem 6
A point lies on one of medians of triangle in such a way that Prove that there exists a point on another median such that (A.Zaslavsky)
Proof
1. Denote It is known that barycentric coordinates are
2. Denote
is tangent to
is tangent
is the radical axes of and the power of a point with respect to a circle is so the power of a point with respect to a circle is
so is tangent to
so point symmetrical to with respect to the median satisfies the conditions.
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2024 tur 2 klass 10 Problem 7
Let be a triangle with and be its bisectors, be the projections of to and respectively, and be the second common point of the circle with
Prove that points are collinear. (K.Belsky)
Proof
Denote the incenter of the midpoint of
It is known ( Division of bisector) that
is cyclic.
Therefore is cyclic
Let
It is known that points and are collinear,
is the diameter of is the bisector of
Bisector
Altitude
Note that the point is a Feuerbach point of since both the inscribed circle and the Euler circle pass through it.
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2024 tur 2 klass 9 Problem 7
Let triangle and point on the side be given. Let be such point on the side that The cross points of segments and with the incircle of form a convex quadrilateral
Find the locus of crosspoints of diagonals (D.Brodsky)
Solution 1. Particular case of Fixed point .
2. Denote We perform simple transformations and get: We use Stewart's theorem and get: Similarly Therefore not depends from
Let be the midpoint of is the median of and
The line cross the median of at point such that
So point is fixed and this point lyes on .
Therefore the locus of crosspoints of diagonals is point
Corollary
Let line . Then
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2024 tur 2 klass 9 Problem 5
Let be an isosceles triangle be its circumcenter, be the orthocenter, and be a point inside the triangle such that
Prove that (A.Zaslavsky)
Proof
Denote the midpoint the midpoint the foot from to tangent to
There is a spiral similarity centered at point that maps into
The coefficient of similarity rotation angle equal so is tangent to Basic information Points and are collinear, so median of
is symmedian of
is Humpty point.
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2024 tur 2 klass 9 Problem 4
For which it is possible to mark several different points and several different circles on the plane in such a way that:
- exactly marked circles pass through each marked point;
- exactly marked points lie on each marked circle;
- the center of each marked circle is marked? (P.Puchkov)
Solution
Case Circles centered at and with radii
Case is not paralel to
Four circles are centered at points and Each radius is equal
Case is not paralel to or
Eight circles centered at and have radii
Case
Answer For all
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2024 tur 2 klass 9 Problem 3
Let and be two pairs of points isogonally conjugated with respect to a triangle and be the common point of lines and Prove that the pedal circles of points and are coaxial. (L.Shatunov, V.Shelomovskii)
Solution
1. Let be the isogonal conjugate of a point with respect to a triangle Then circle centered at the midpoint is the common pedal circle of points and ( Circumcircle of pedal triangles) So center is the midpoint and center is the midpoint
2. Denote Then is the isogonal conjugate of a point with respect to So center is the midpoint ( Two pares of isogonally conjugate points)
3. The Gauss line (or Gauss–Newton line) is the line joining the midpoints of the three diagonals of a complete quadrilateral (Gauss line).So points and are collinear as was to be proven.
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2024 tur 2 klass 8 Problem 4
A square with sidelength is cut from the paper. Construct a segment with length using at most folds. No instruments are available, it is allowed only to fold the paper and to mark the common points of folding lines. (M.Evdokimov)
Solution
Main idea: Let We perform horizontal fold of the sheet. We get line We perform
vertical folds of the sheet. We get vertical lines at a distance of from each other.
Point is the lower left corner of the sheet, point is the lower point of the second vertical line, point is the lower point of the line, point is the point at the intersection of the horizontal line and the vertical line.
Points and are at the intersection of the lines and and the vertical line.
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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.
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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.
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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
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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
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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
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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.
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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
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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
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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.
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The problem from MGTU
The lateral face of the regular triangular pyramid is inclined to the plane of the base at an angle of Points are the midpoints of the sides of the Triangle is the lower base of a right prism. The edges of the upper base of the prism intersect the lateral edges of the pyramid at points The area of the total surface of the polyhedron with vertices is equal to Find the side of
Solution
Denote is the center of The area of the total surface of the polyhedron with vertices is
The trapezoid problem from MGTU
Points and are the midpoints of bases and of trapezoid
Denote the angle between lines and
Find the area of trapezoid if
Solution
By applying the Law of Cosines on we get