Difference between revisions of "Sharygin Olympiads, the best"
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<math>O</math> is the circumcenter of <math>\triangle APC, O'</math> is the circumcenter of <math>\triangle BPC.</math> | <math>O</math> is the circumcenter of <math>\triangle APC, O'</math> is the circumcenter of <math>\triangle BPC.</math> | ||
− | Let <math>K</math> and <math>L</math> be the midpoints of the arcs <math>\overset{\Large\frown}{CB'}</math> of <math>\theta.</math> | + | Let <math>K</math> and <math>L</math> be the midpoints of the arcs <math>\overset{\Large\frown}{CB'}</math> of <math>\theta,D = AL \cap \omega.</math> |
− | Let <math>K'</math> and <math>L'</math> be the midpoints of the arcs <math>\overset{\Large\frown}{CA'}</math> of <math>\theta'.</math> | + | Let <math>K'</math> and <math>L'</math> be the midpoints of the arcs <math>\overset{\Large\frown}{CA'}</math> of <math>\theta', D' = BL' \cap \omega'.</math> |
These points not depends from position of point <math>P.</math> | These points not depends from position of point <math>P.</math> | ||
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<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> | ||
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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''' | '''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== | ==2024, Problem 22== | ||
[[File:2023 22 2.png|350px|right]] | [[File:2023 22 2.png|350px|right]] | ||
[[File:2024 22.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> | Find the locus of touching points <math>P</math> of the ellipse with the line <math>BC.</math> | ||
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<cmath>\angle CBO = 90^\circ \implies \angle COB = \alpha, MB = b \tan \alpha,</cmath> | <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}, | + | <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> | 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>\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>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>\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> | Denote <math>Q = AB \cap DP \implies BQ = \frac {BP}{\cos \alpha} = b \tan \alpha = MB.</math> | ||
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'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
==2024, Problem 21== | ==2024, Problem 21== | ||
[[File:2024 21 0.png|350px|right]] | [[File:2024 21 0.png|350px|right]] | ||
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preimage <math>Z</math> lies on preimage <math>CC'.\blacksquare</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''' |
Latest revision as of 18:06, 16 May 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
- 1 2024, Problem 23
- 2 One-to-one mapping of the circle
- 3 2024, Problem 22
- 4 2024, Problem 21
- 5 2024, Problem 20
- 6 2024, Problem 19
- 7 2024, Problem 18
- 8 2024, Problem 17
- 9 2024, Problem 16
- 10 2024, Problem 15
- 11 2024, Problem 14
- 12 2024, Problem 12
- 13 2024, Problem 9
- 14 2024, Problem 8
- 15 2024, Problem 2
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
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
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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.)
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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
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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.
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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.
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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
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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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