Difference between revisions of "Spiral similarity"
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− | A '''spiral similarity''' is the [[ | + | ==Basic information== |
+ | A spiral similarity is a plane transformation composed of a rotation of the plane and a dilation of the plane having the common center. The order in which the composition is taken is not important. | ||
+ | |||
+ | Any two directly similar figures are related either by a translation or by a spiral similarity (directly similar figures are similar and have the same orientation). | ||
+ | |||
+ | The transformation is linear and transforms any given object into an object homothetic to given. | ||
+ | |||
+ | On the complex plane, any spiral similarity can be expressed in the form <math>T(x) = x_0+k (x-x_0),</math> where <math>k</math> is a complex number. The magnitude <math>|k|</math> is the dilation factor of the spiral similarity, and the argument <math>\arg(k)</math> is the angle of rotation. | ||
+ | |||
+ | The spiral similarity is uniquely defined by the images of two distinct points. It is easy to show using the complex plane. | ||
+ | [[File:Spiral center.png|450px|right]] | ||
+ | [[File:Spiral center 3.png|450px|right]] | ||
+ | Let <math>A' = T(A), B' = T(B),</math> with corresponding complex numbers <math>a', a, b',</math> and <math>b,</math> so | ||
+ | <cmath>a' = T(a) = x_0 + k (a - x_0), b' = T(b) = x_0+ k (b-x_0) \implies</cmath> | ||
+ | <cmath>k = \frac {T(b) - T(a)}{b-a} = \frac {b' - a' }{b - a},</cmath> | ||
+ | <cmath>x_0=\frac {ab' - ba' }{a-a'+b' -b}, a' - a \ne b' - b.</cmath> | ||
+ | For any points <math>A, B, A',</math> and <math>B'</math> the center of the spiral similarity taking <math>AB</math> to <math>A'B'</math> point <math>x_{0}</math> is also the center of a spiral similarity taking <math>A'B</math> to <math>AB'.</math> This fact explain existance of Miquel point. | ||
+ | |||
+ | <i><b>Case 1</b></i> Any line segment <math>AB</math> can be mapped into any other <math>A'B'</math> using the spiral similarity. Notation is shown on the diagram. | ||
+ | <math>P = AB \cap A'B'.</math> | ||
+ | |||
+ | <math>\Omega</math> is circle <math>AA'P, \omega </math> is circle <math>BB'P, x_0 = \Omega \cap \omega, x_0 \neq P,</math> | ||
+ | |||
+ | <math>C </math> is any point of <math>AB, \theta </math> is circle <math>CPx_0, C' = \theta \cap A'B'</math> is the image <math>C</math> under spiral symilarity centered at <math>x_0.</math> <cmath>\triangle AA'x_0 \sim \triangle BB'x_0 \sim \triangle CC'x_0.</cmath> | ||
+ | |||
+ | <math>|k| = \frac {A'B'}{AB} = \frac {A'x_0}{Ax_0} = \frac {B'x_0}{Bx_0} = \frac {C'x_0}{Cx_0}</math> is the dilation factor, | ||
+ | |||
+ | <math>\arg(k) =\angle APA'=\angle Ax_0A' =\angle Bx_0B' =\angle Cx_0C'</math> is the angle of rotation. | ||
+ | |||
+ | <i><b>Case 2</b></i> Any line segment <math>AB</math> can be mapped into any other <math>BB'</math> using the spiral similarity. Notation is shown on the diagram. <math>B = AB \cap BB', \Omega</math> is circle <math>ABB</math> (so circle is tangent to <math>BB'), \omega </math> is circle tangent to <math>AB, x_0 = \Omega \cap \omega, x_0 \neq B, C </math> is any point of <math>AB, \theta </math> is circle <math>CBx_0,</math> | ||
+ | <math>C' = \theta \cap BB'</math> is the image <math>C</math> under spiral symilarity centered at <math>x_0.</math> | ||
+ | <cmath>\triangle ABx_0 \sim \triangle BB'x_0 \sim \triangle CC'x_0.</cmath> | ||
+ | <math>|k| = \frac {BB'}{AB}</math> is the dilation factor, | ||
+ | |||
+ | <math>\angle Ax_0B = \arg(k)</math> is the angle of rotation. | ||
+ | |||
+ | ==Simple problems== | ||
+ | ===Explicit spiral similarity=== | ||
+ | [[File:1934 Pras.png|370px|right]] | ||
+ | Given two similar right triangles <math>ABC</math> and <math>A'B'C, k = \frac {AC}{BC},</math> | ||
+ | <math>\angle ACB = 90^\circ, D = AA' \cap BB'.</math> Find <math>\angle ADB</math> and <math>\frac {AA'}{BB'}.</math> | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | The spiral similarity centered at <math>C</math> with coefficient <math>k</math> and the angle of rotation <math>90^\circ</math> maps point <math>B</math> to point <math>A</math> and point <math>B'</math> to point <math>A'.</math> | ||
+ | |||
+ | Therefore this similarity maps <math>BB'</math> to <math>AA' \implies</math> | ||
+ | <cmath>\frac {AA'}{BB'} = k, \angle ADB = 90^\circ.</cmath> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ===Hidden spiral symilarity=== | ||
+ | [[File:1932a Pras.png|400px|right]] | ||
+ | [[File:1932b Pras.png|400px|right]] | ||
+ | Let <math>\triangle ABC</math> be an isosceles right triangle <math>(AC = BC).</math> Let <math>S</math> be a point on a circle with diameter <math>BC.</math> The line <math>\ell</math> is symmetrical to <math>SC</math> with respect to <math>AB</math> and intersects <math>BC</math> at <math>D.</math> Prove that <math>AS \perp DS.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Denote <math>\angle SBC = \alpha, \angle SCB = \beta = 90^\circ - \alpha,</math> | ||
+ | <cmath>\angle SCA = \alpha, \angle BSC = 90^\circ, k = \frac {SC}{SB} = \cot \beta.</cmath> | ||
+ | Let <math>SC</math> cross perpendicular to <math>BC</math> in point <math>B</math> at point <math>D'.</math> | ||
+ | |||
+ | Then <math>\frac {BC}{BD'} = \cot \beta.</math> | ||
+ | |||
+ | Points <math>D</math> and <math>D'</math> are simmetric with respect <math>AB,</math> so <math>BD = BD' \implies k = \frac {SC}{SB} = \frac {BC}{BD}.</math> | ||
+ | |||
+ | The spiral symilarity centered at <math>S</math> with coefficient <math>k</math> and the angle of rotation <math>90^\circ</math> maps <math>B</math> to <math>C</math> and <math>D</math> to point <math>D_0</math> such that <cmath>k \cdot BD_0 = BC = AC, \angle D_0CS = \angle DBS \implies D_0 = A.</cmath> | ||
+ | |||
+ | Therefore <math>\angle ASC = \angle DSB \implies</math> | ||
+ | <cmath>\angle ASD = \angle ASC - \angle DSC = \angle DSB - \angle DSC = \angle BSC = 90^\circ.</cmath> | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ===Linearity of the spiral symilarity=== | ||
+ | [[File:1933 Pras.png|450px|right]] | ||
+ | <math>\triangle ABF \sim \triangle BCD \sim \triangle CAE.</math> Points <math>D,E,F</math> are outside <math>\triangle ABC.</math> | ||
+ | |||
+ | Prove that the centroids of triangles <math>\triangle ABC</math> and <math>\triangle DEF</math> are coinsite. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let <math>\vec y = T(\vec x),</math> where <math>T</math> be the spiral similarity with the rotation angle <math>\angle BAF= \angle CBD = \angle ACE</math> and <math>k = \frac {|AF|}{|AB|} = \frac {|DB|}{|BC|} = \frac {|EC|}{|CA|}.</math> | ||
+ | |||
+ | A vector has two parameters, modulo and direction. It is not tied to a center of the spiral similarity. Therefore | ||
+ | <cmath>\vec AF = T(\vec AB), \vec BD = T(\vec BC), \vec CE = T(\vec CA).</cmath> | ||
+ | <cmath>\vec AB + \vec BC + \vec CA = \vec 0.</cmath> | ||
+ | We use the property of linearity and get | ||
+ | <cmath>\vec AF + \vec BD + \vec CE = k(\vec AB + \vec BC + \vec CA) = k \cdot \vec 0 = \vec 0.</cmath> | ||
+ | Let <math>G</math> be the centroid of <math>\triangle ABC</math> so <math>\vec GA + \vec GB + \vec GC = \vec 0 \implies </math> | ||
+ | |||
+ | <math>\vec GD + \vec GE + \vec GF = \vec 0 \implies G</math> is the centroid of the <math>\triangle DEF.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ===Construction of a similar triangle=== | ||
+ | [[File:1940 Pras.png|380px|right]] | ||
+ | Let triangle <math>\triangle ABC</math> and point <math>A'</math> on sideline <math>BC</math> be given. Construct <math>\triangle A'B'C' \sim \triangle ABC</math> where <math>B'</math> lies on sideline <math>AC</math> and <math>C'</math> lies on sideline <math>AB.</math> | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | Let <math>T(X)</math> be the spiral symilarity centered at <math>A'</math> with the dilation factor <math>k = \frac {AB}{AC}</math> and rotation angle <math>\alpha = \angle BAC, A_1 = T(A), B_1 = T(B).</math> | ||
+ | |||
+ | <math>A_1B_1 = T(AB)</math> so image of any point <math>C' \in AB</math> lies on <math>A_1B_1.</math> | ||
+ | <cmath>B' \in AC \implies B' = A_1B_1 \cap AC.</cmath> | ||
+ | The spiral symilarity <math>T^{-1}(X)</math> centered at <math>A'</math> with the dilation factor <math>k^{-1} = \frac {AC}{AB}</math> and rotation angle <math>-\alpha</math> maps <math>B'</math> into <math>C'</math> and <math>\angle B'A'C' = \alpha, \frac {A'C'}{A'B'} = k^{-1} = \frac {AC}{AB}</math> therefore the found triangle <math>\triangle A'B'C' \sim \triangle ABC</math> is the desired one. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ===Center of the spiral symilarity for similar triangles=== | ||
+ | [[File:5 133 Pras.png|400px|right]] | ||
+ | [[File:5 133aa Pras.png|400px|right]] | ||
+ | Let triangle <math>\triangle ABC</math> and point <math>C' (C' \ne C, C' \ne B)</math> on sideline <math>BC</math> be given. <math>\triangle A'B'C' \sim \triangle ABC</math> where <math>B'</math> lies on sideline <math>AB</math> and <math>A'</math> lies on sideline <math>AC.</math> The spiral symilarity <math>T</math> maps <math>\triangle ABC</math> into <math>\triangle A'B'C'.</math> | ||
+ | Prove | ||
+ | |||
+ | a) <math>\angle AB'A' = \angle BC'B' = \angle CA'C'.</math> | ||
+ | |||
+ | b) Center of <math>T</math> is the First Brocard point of triangles <math>\triangle ABC</math> and <math>\triangle A'B'C'.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | a) Let <math>\tau(X)</math> be the spiral symilarity centered at <math>C'</math> with the dilation factor <math>k = \frac {AC}{BC}</math> and rotation angle <math>\gamma = \angle ACB, A_1 = \tau(A), B_1 = \tau(B).</math> | ||
+ | <math>A' = A_1B_1 \cap AC, B' = \tau^{-1}(A').</math> | ||
+ | |||
+ | Denote <math>\varphi = \angle BC'B' \implies \angle CC'A' = 180^\circ - \varphi - \gamma \implies </math> | ||
+ | <math>\angle CA'C' = 180^\circ - \gamma - \angle CC'A' = \varphi.</math> Similarly <math>\angle AB'A' = \varphi.</math> | ||
+ | |||
+ | b) It is well known that the three circumcircles <math>AA'B', BB'C',</math> and <math> CC'A' </math> have the common point (it is <math>D</math> in the diagram). | ||
+ | |||
+ | Therefore <math>AB'DA'</math> is cyclic and <math>\angle AB'A' = \angle ADA' =\varphi.</math> | ||
+ | |||
+ | Similarly, <math>\angle BDB' = \angle CDC' = \varphi.</math> | ||
+ | |||
+ | <cmath>\angle A'DC' = 180^\circ - \gamma, \angle A'DC = 180^\circ - \gamma - \varphi \implies </cmath> | ||
+ | <cmath>\angle ADC = \angle A'DC' = 180^\circ - \gamma.</cmath> | ||
+ | <math>\angle CAD + \angle ACD = 180^\circ - \angle ADC = \gamma = \angle ACD + \angle BCD \implies \angle CAD = \angle BCD = \psi.</math> | ||
+ | |||
+ | Similarly, <math>\angle CAD = \angle ABD = \angle BCD = \psi.</math> | ||
+ | |||
+ | Therefore, <math>D</math> is the First Brocard point of <math>\triangle ABC.</math> | ||
+ | |||
+ | <math>AB'DA'</math> is cyclic <math>\implies \angle A'B'D = \angle A'AD = \psi.</math> Similarly, <math>\angle B'C'D = \angle C'A'D = \psi.</math> | ||
+ | |||
+ | Therefore <math>D</math> is the First Brocard point of <math>\triangle A'B'C',</math> and <math>\triangle A'DC' \sim \triangle ADC, \triangle A'DB' \sim \triangle ADB.</math> | ||
+ | |||
+ | Therefore the spiral symilarity <math>T</math> maps <math>\triangle ABC</math> into <math>\triangle A'B'C'</math> has the center <math>D,</math> the angle of the rotation <math>\varphi.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ===Spiral similarity in rectangle=== | ||
+ | [[File:Rectangle perp.png|430px|right]] | ||
+ | Let rectangle <math>ABCD</math> be given. Let point <math>E \in AC, DE \perp AC.</math> | ||
+ | |||
+ | Let points <math>F</math> and <math>G</math> be the midpoints of segments <math>AE</math> and <math>BC,</math> respectively. | ||
+ | |||
+ | Prove that <math>DF \perp FG.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let <math>H</math> be the midpoint <math>DE \implies FH = \frac {AD}{2} = CG,</math> | ||
+ | |||
+ | <math>FH || AD ||BC || CG \implies FHCG</math> is a parallelogram <math>\implies CH || GF.</math> | ||
+ | |||
+ | <cmath>\angle DAE = \angle CDE, \angle AED = \angle DEC = 90^\circ \implies</cmath> | ||
+ | <cmath>\triangle AED \sim \triangle DEC \implies</cmath> | ||
+ | <math>DF</math> and <math>CH</math> are corresponding medians of <math>\triangle AED</math> and <math>\triangle DEC.</math> | ||
+ | |||
+ | There is a spiral similarity <math>T</math> centered at <math>E</math> with rotation angle <math>90^\circ</math> that maps <math>\triangle AED</math> to <math>\triangle DEC.</math> Therefore <cmath>T(DF) = CH \implies DF \perp CH \implies DF \perp FG.</cmath> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ===Common point for 6 circles=== | ||
+ | [[File:1949 Pras.png|450px|right]] | ||
+ | Let <math>\triangle ABC</math> and point <math>A'</math> on sideline <math>BC</math> be given. <math>\triangle A'B'C' \sim \triangle ABC</math> where <math>B'</math> lies on sideline <math>AC</math> and <math>C'</math> lies on sideline <math>AB.</math> | ||
+ | |||
+ | Denote <math>D = BB' \cap CC', E = AA' \cap CC', F = BB' \cap AA'.</math> | ||
+ | |||
+ | Prove that circumcircles of triangles <math>\triangle ABF, \triangle A'B'F, \triangle BCD,</math> | ||
+ | <math>\triangle B'C'D, \triangle ACE, \triangle A'C'E</math> have the common point. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <math>\triangle A'B'C' \sim \triangle ABC</math> so there is the spiral symilarity <math>T</math> taking <math>\triangle ABC</math> to <math>\triangle A'B'C', T(ABC) = A'B'C'.</math> Denote <math>O</math> the center of <math>T.</math> | ||
+ | <cmath>A'B' = T(AB), A'B = \tau(AB'), F = AA' \cap BB' \implies</cmath> | ||
+ | the center of <math>\tau</math> is the secont crosspoint of circumcircles of <math>\triangle ABF</math> and <math>\triangle A'B'F,</math> but this center is point <math>O,</math> so these circles contain point <math>O</math>. Similarly for another circles. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ===Three spiral similarities=== | ||
+ | [[File:1947 Pras.png|450px|right]] | ||
+ | Let triangle <math>\triangle ABC</math> be given. The triangle <math>\triangle ACE</math> is constructed using a spiral similarity of <math>\triangle ABC</math> with center <math>A</math>, angle of rotation <math>\angle BAC</math> and coefficient <math>\frac {AC}{AB}.</math> | ||
+ | |||
+ | A point <math>D</math> is centrally symmetrical to a point <math>B</math> with respect to <math>C.</math> | ||
+ | |||
+ | Prove that the spiral similarity with center <math>E</math>, angle of rotation <math>\angle ACB</math> and coefficient <math>\frac {BC}{AC}</math> taking <math>\triangle ACE</math> to <math>\triangle CDE.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <cmath>\angle ECD = 180^\circ - \angle ACB - \angle ABC = \angle BAC.</cmath> | ||
+ | <math>EC = BC \cdot \frac {AC}{AB}, CD = BC = AC \cdot \frac {BC}{AC} \implies \frac {CD}{EC}=\frac {AB}{AC} \implies \triangle CDE \sim \triangle ACE \implies \angle DEC = \angle CEA.</math> | ||
+ | |||
+ | <i><b>Corollary</b></i> | ||
+ | |||
+ | Three spiral similarities centered on the images of the vertices of the given triangle <math>\triangle ABC</math> and with rotation angles equal to the angles of <math>\triangle ABC</math> take <math>\triangle ABC</math> to <math>\triangle FDC</math> centrally symmetric to <math>\triangle ABC</math> with respect to <math>C.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ===Superposition of two spiral similarities=== | ||
+ | Let <math>T</math> be the spiral similarity centered at <math>A</math> with angle <math>\alpha</math> and coefficient <math>\frac {1}{|k|}, Y = T(X), \frac {XA}{YA} = k.</math> | ||
+ | |||
+ | Let <math>t</math> be spiral similarity centered at <math>B</math> with angle <math>\alpha</math> and coefficient <math>k, Z = t(Y), \frac {BZ}{BY} = k.</math> | ||
+ | |||
+ | Let <math>t(T(O)) = O.</math> | ||
+ | |||
+ | Prove: | ||
+ | a) <math>XO = ZO, AO = BO, O</math> is the crosspoint of bisectors <math>AB</math> and <math>XZ.</math> | ||
+ | |||
+ | b)<math>\angle XOZ = 2 \alpha, \angle AOB = 2 \angle AXY.</math> | ||
+ | |||
+ | <i><b>Algebraic proof</b></i> | ||
+ | |||
+ | We use the complex plane <math>x = \vec X, y = \vec Y, z = \vec Z, a = \vec A,b = \vec B, o = \vec O.</math> | ||
+ | <cmath>y = T(x) = a + \frac {e^{i \alpha}}{k}(x - a), z = t(y) = b + k e^{i \alpha}(y - b) \implies </cmath> | ||
+ | <cmath>z = e^{2 i \alpha}x + b(1 - k e^{i \alpha}) + a e^{i \alpha}(k - e^{i \alpha}) = o + e^{2 i \alpha}(x - o).</cmath> | ||
+ | Let <math>a = (-1,0), b = (1,0).</math> Then <cmath>o = i(\cot \alpha - \frac {k}{\sin \alpha}) \implies XO = ZO, AO = BO, \angle XOZ = 2 \alpha, \angle AOB = 2 \angle AXY.</cmath> | ||
+ | <i><b>Geometric proof</b></i> | ||
+ | [[File:1938 Pras 1.png|410px|right]] | ||
+ | Denote <math>\angle XAY = \alpha, \angle AXY = \beta, \angle AYX = \gamma = 180^\circ - \alpha - \beta.</math> | ||
+ | |||
+ | Then <math>k = \frac {XA}{YA} = \frac {\sin \gamma}{\sin \beta}, \triangle BZY \sim \triangle AXY.</math> | ||
+ | |||
+ | Let <math>H</math> be the midpoint <math>AB, O</math> be the point on bisector <math>AB</math> such that <math>\angle AOH = \beta, C</math> be the point on bisector <math>AB</math> such that <math>\angle OAC = \alpha.</math> Then | ||
+ | |||
+ | <math>\triangle AOC = \triangle BOC \sim \triangle AXY \sim \triangle BZY \implies </math> | ||
+ | <cmath>k = \frac {OA}{CA}, k^{-1} = \frac {CB}{OB},\angle CBO =\alpha \implies</cmath> | ||
+ | <cmath>T(O) = C, t(C) = O, t(T(O)) = O, \angle AOB = 2 \beta.</cmath> | ||
+ | <cmath>T(XO) = YC \implies \frac {XO}{YC} = k; t(YC) = ZO \implies \frac {ZO}{YC} = k \implies</cmath> | ||
+ | <cmath>XO = ZO, \angle (XO)(YC) = \angle (YC)(ZO) = \alpha \implies \angle (XO)(ZO) = \angle XOZ = 2 \alpha.</cmath> | ||
+ | <math>XO = ZO, AO = BO \implies O</math> is the crosspoint of bisectors <math>AB</math> and <math>XZ.</math> | ||
+ | |||
+ | <i><b>Corollary</b></i> | ||
+ | |||
+ | There is another pare of the spiral similarities centered at <math>X</math> and <math>Z</math> with angle <math>\beta,</math> coefficients <math>k' = \frac {XA}{XY} = \frac {\sin \gamma}{\sin \alpha}</math> and <math>k'^{-1}, Y = T'(A), B = t'(Y).</math> | ||
+ | |||
+ | In this case <math>t'(T'(O)) = O.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ===Spiral similarity for circles=== | ||
+ | [[File:1928 Pras.png|400px|right]] | ||
+ | [[File:1927 Pras.png|400px|right]] | ||
+ | 1. Let circle <math>\omega</math> cross circle <math>\Omega</math> at points <math>A</math> and <math>B.</math> Point <math>C</math> lies on <math>\omega.</math> | ||
+ | |||
+ | Spiral similarity <math>T</math> centered at <math>A</math> maps <math>\omega</math> into <math>\Omega: \Omega = T(\omega), C' = T(C).</math> | ||
+ | |||
+ | Prove that points <math>C, C',</math> and <math>B</math> are collinear. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Arcs <math>\overset{\Large\frown} {AC} = \overset{\Large\frown} {ABC'}, \angle ABC = \frac {\overset{\Large\frown} {AC}}{2},</math> | ||
+ | <cmath>\angle ABC' = \frac {360^\circ -\overset{\Large\frown} {ABC'}}{2} = 180 ^\circ - \frac {\overset{\Large\frown} {ABC'}}{2} = 180 ^\circ - \angle ABC' \implies</cmath> | ||
+ | <cmath>B \in CC'.</cmath> | ||
+ | |||
+ | <i><b>Corollary</b></i> | ||
+ | |||
+ | Let <math>C \in \omega, C' \in \Omega,</math> points <math>C, C',</math> and <math>B</math> be collinear. | ||
+ | |||
+ | Then exist the spiral similarity <math>T</math> centered at <math>A</math> such that <math>T(\omega) = \Omega, T(C) = C'.</math> | ||
+ | |||
+ | |||
+ | 2. Let circle <math>\omega</math> cross circle <math>\Omega</math> at points <math>A</math> and <math>B.</math> | ||
+ | |||
+ | Points <math>C</math> and <math>D</math> lie on <math>\omega, C' = BC \cap \Omega, D' = BD \cap \Omega.</math> | ||
+ | |||
+ | Let <math>EA</math> be the tangent to <math>\omega, FA</math> be the tangent to <math>\Omega.</math> | ||
+ | |||
+ | Prove that angle between tangents is equal angle between lines <math>CD</math> and <math>C'D'.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | There is the spiral similarity <math>T</math> centered at <math>A</math> such that <cmath>T(\omega) = \Omega, T(C) = C', T(D) = D'.</cmath> | ||
+ | Therefore <math>T(CD) = C'D', T(EA) = FA \implies</math> angles between these lines are the same. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ===Remarkable point for spiral similarity=== | ||
+ | [[File:Point X.png|410px|right]] | ||
+ | [[File:Point Xx.png|410px|right]] | ||
+ | Circles <math>\omega</math> and <math>\Omega</math> centered at points <math>O</math> and <math>O',</math> respectively intersect at points <math>A</math> and <math>B.</math> Points <math>C \in \omega, C' \in \Omega,</math> and <math>A</math> are collinear. Point <math>X</math> is symmetrical to <math>A</math> with respect to the midpoint <math>OO'</math> point <math>M.</math> | ||
+ | Prove: | ||
+ | |||
+ | a) <math>\angle ABX = 90^\circ,</math> b) <math>CX = C'X.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | a) <math>AB \perp OO', OO'</math> cross <math>AB</math> in midpoint <math>\implies BX || OO' \implies \angle ABX = 90^\circ.</math> | ||
+ | |||
+ | b) <math>OM = O'M, AM = MX \implies AOXO'</math> is parallelogram <math>\implies AO = XO' = CO, OX = AO' = C'O'.</math> | ||
+ | |||
+ | Denote <math>\angle OAC = \angle OCA = \alpha, \angle O'AC' = \angle O'C'A = \beta,</math> | ||
+ | <cmath>\angle XO'A = \angle XOA = \varphi \implies</cmath> | ||
+ | |||
+ | <cmath>\angle CAC' = 180^\circ = \alpha + (180^\circ - \varphi) + \beta \implies \alpha + \beta = \varphi.</cmath> | ||
+ | <cmath>\angle COX = 360^\circ - (180^\circ - 2 \alpha) - \varphi = 180^\circ + \alpha - \beta.</cmath> | ||
+ | <cmath>\angle XO'C' = \varphi + \angle AO'C' = \varphi + (180^\circ - 2 \beta) = 180^\circ + \alpha - \beta.</cmath> | ||
+ | <cmath>\angle XO'C' = \angle COX \implies \triangle COX = \triangle XO'C' \implies CX = C'X.</cmath> | ||
+ | |||
+ | <i><b>Corollary</b></i> | ||
+ | |||
+ | Let points <math>D \in \omega, D' \in \Omega,</math> and <math>A</math> be collinear. Then <math>DX = D'X.</math> | ||
+ | |||
+ | Therefore <math>X</math> is the crosspoint of the bisectors <math>CC'</math> and <math>DD'.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ===Remarkable point for pair of similar triangles=== | ||
+ | [[File:Similar triangles.png|400px|right]] | ||
+ | [[File:Similar triangles A.png|400px|right]] | ||
+ | Let <math>\triangle ACD \sim \triangle AC'D'.</math> | ||
+ | |||
+ | Let the points <math>O</math> and <math>O'</math> be the circumcenters of <math>\omega (\triangle ACD)</math> and <math>\Omega (\triangle AC'D').</math> | ||
+ | |||
+ | Let point <math>M</math> be the midpoint of <math>OO'.</math> | ||
+ | |||
+ | The point <math>X</math> is symmetric to <math>A</math> with respect point <math>M.</math> | ||
+ | |||
+ | Prove: | ||
+ | |||
+ | a) point <math>X</math> be the crosspoint of the bisectors <math>CC'</math> and <math>DD'.</math> | ||
+ | |||
+ | b) <math>\angle ABX = 90^\circ.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <math>OM = MO', AM = MX \implies AOXO'</math> is parallelogram <math>\implies</math> | ||
+ | <cmath>CO = AO = XO', C'O' = AO' = XO.</cmath> | ||
+ | Denote <math>\angle ADC = \angle AD'C' = \alpha \implies \overset{\Large\frown} {AC} = \overset{\Large\frown} {AC'} = 2\alpha \implies</math> | ||
+ | |||
+ | <cmath>\angle AOC = \angle AO'C' = 2 \alpha \implies</cmath> | ||
+ | <cmath>\angle COX = 360^\circ - \angle AOX - 2 \alpha = \angle C'O'X \implies</cmath> | ||
+ | <cmath>\triangle XOC = \triangle C'O'X \implies CX = C'X.</cmath> | ||
+ | <cmath>\angle CXC' = \angle CXO + \angle OXO' + \angle O'XC'</cmath> | ||
+ | <cmath>\angle CXC' = \angle CXO + (180^\circ - \angle AOX) + \angle OCX</cmath> | ||
+ | <cmath>\angle CXC' = (180^\circ - \angle AOX) + (180^\circ - \angle COX).</cmath> | ||
+ | <cmath>\angle CXC' = 360^\circ - \angle AOX - \angle COX) = \angle AOC = 2 \alpha.</cmath> | ||
+ | Similarly, <math>DX = D'X, \angle DXD' = 2 \angle ACD.</math> | ||
+ | |||
+ | The statement that <math>\angle ABX = 90^\circ</math> was proved in the previous section. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ===Remarkable point’s problems=== | ||
+ | [[File:Point 1A.png|240px|right]] | ||
+ | [[File:Point 2A.png|240px|right]] | ||
+ | [[File:Point X tangents M.png|240px|right]] | ||
+ | [[File:Point X 3 tangents new.png|240px|right]] | ||
+ | |||
+ | <i><b>Problem 1</b></i> | ||
+ | |||
+ | Let a convex quadrilateral <math>ABCD</math> be given, <math>E = AC \cap BD.</math> | ||
+ | |||
+ | Let <math>M</math> and <math>N</math> be the midpoints of <math>AC</math> and <math>BD,</math> respectively. | ||
+ | |||
+ | Circumcircles <math>ABE</math> and <math>CDE</math> intersect a second time at point <math>K.</math> | ||
+ | |||
+ | Prove that points <math>M, K, N,</math> and <math>E</math> are concyclic. | ||
+ | |||
+ | <i><b>Problem 2</b></i> | ||
+ | |||
+ | Let triangle <math>\triangle ABC</math> be given. | ||
+ | |||
+ | Let point <math>D</math> lies on sideline <math>BC.</math> | ||
+ | |||
+ | Denote the circumcircle of the <math>\triangle ABD</math> as <math>\omega</math>, the circumcircle of the <math>\triangle ACD</math> as <math>\Omega</math>. | ||
+ | |||
+ | Let <math>X</math> be the circumcenter of <math>\Omega.</math> | ||
+ | |||
+ | |||
+ | Let circle <math>ACD</math> cross sideline <math>AB</math> at point <math>E.</math> | ||
+ | |||
+ | Let the circumcircle of the <math>\triangle BCE</math> cross <math>\omega</math> at point <math>F.</math> | ||
+ | |||
+ | Prove that <math>\angle BFX = 90^\circ.</math> | ||
+ | |||
+ | <i><b>Problem 3</b></i> | ||
+ | |||
+ | The circles <math>\omega</math> and <math>\Omega</math> are crossed at points <math>A</math> and <math>B,</math> points <math>C \in \Omega, D \in \omega.</math> | ||
+ | |||
+ | Let <math>AC</math> be the tangent to <math>\omega, AD</math> be the tangent to <math>\Omega.</math> | ||
+ | |||
+ | Point <math>E</math> is symmetric <math>A</math> with respect to <math>B.</math> | ||
+ | |||
+ | Prove that points <math>A, C, D,</math> and <math>E</math> are concyclic. | ||
+ | |||
+ | <i><b>Problem 4</b></i> | ||
+ | |||
+ | The circles <math>\omega</math> and <math>\Omega</math> are crossed at points <math>A</math> and <math>B,</math> points <math>C \in \Omega, D \in \omega.</math> | ||
+ | |||
+ | Let <math>AC</math> be the tangent to <math>\omega, AD</math> be the tangent to <math>\Omega.</math> | ||
+ | |||
+ | Points <math>E \in \omega</math> and <math>F \in \Omega</math> lye on bisector of the angle <math>\angle CAD.</math> | ||
+ | |||
+ | Points <math>E' \in \omega</math> and <math>F' \in \Omega</math> lye on external bisector of the angle <math>\angle CAD.</math> | ||
+ | |||
+ | Prove that <math>DE, CF,</math> and bisector <math>E'F'</math> are tangent to the circle <math>\theta = BEF,</math> | ||
+ | |||
+ | bisector <math>EF</math> is tangent to the circle <math>\Theta =BE'F'.</math> | ||
+ | |||
+ | <i><b>Solutions</b></i> | ||
+ | [[File:Solutions 1 and 2.png|480px|right]] | ||
+ | [[File:Point X tangents A.png|300px|right]] | ||
+ | [[File:Point X 3 tangents n.png|300px|right]] | ||
+ | Solutons are clear from diagrams. | ||
+ | |||
+ | In each case we use remarcable point <math>X,</math> as the point of bisectors crossing. | ||
+ | |||
+ | <i><b>Solution 1</b></i> | ||
+ | |||
+ | We use bisectors <math>AC</math> and <math>BD</math>. | ||
+ | <cmath>\angle XME = \angle XNE = \angle XKE = 90^\circ.</cmath> | ||
+ | The points <math>X, M, K, N,</math> and <math>E</math> are concyclic. | ||
+ | |||
+ | <i><b>Solution 2</b></i> | ||
+ | <cmath>AX = CX, DX = EX.</cmath> | ||
+ | We use bisectors of <math>AC</math> and <math>DE.</math> <math>\angle BFX = 90^\circ.</math> | ||
+ | |||
+ | <i><b>Solution 3</b></i> | ||
+ | |||
+ | We use bisectors of <math>AC</math> and <math>AD</math>. | ||
+ | |||
+ | <math>X</math> is the circumcenter of <math>\triangle ACD, \angle ABX = 90^\circ.</math> | ||
+ | |||
+ | Circle <math>ACD</math> is symmetric with respect diameter <math>BX.</math> | ||
+ | |||
+ | Point <math>E</math> is symmetric to <math>A \in ACD</math> with respect diameter <math>BX.</math> | ||
+ | |||
+ | Therefore <math>E \in ACD.</math> | ||
+ | |||
+ | <i><b>Solution 4</b></i> | ||
+ | |||
+ | Let <math>X = EE' \cap FF' \implies XE = XF, XE' = XF', XB \perp AB.</math> | ||
+ | |||
+ | Let <math>Y = AB \cap \theta, Z = AB \cap \Theta, M</math> be midpoint <math>EF, M'</math> be midpoint <math>E'F'</math>. | ||
+ | |||
+ | <math> \angle EAE' = 90^\circ.</math> We need prove that <math>X \in \theta</math> and <math>X \in \Theta.</math> | ||
+ | |||
+ | Denote <math>\angle E'AD = \alpha \implies \angle EAD = \angle EAC = 90^\circ - \alpha \implies \angle CAF' = \alpha.</math> | ||
+ | |||
+ | The angle between a chord and a tangent is half the arc belonging to the chord. | ||
+ | |||
+ | <math>AD</math> is tangent to <math>\Omega \implies \overset{\Large\frown} {AF'} = 2 \alpha \implies \angle ACF' = \alpha.</math> | ||
+ | |||
+ | <math>\angle AF'C = 180^\circ - 2\alpha, \angle CF'F = \angle CAF = 90^\circ - \alpha = \angle ACF \implies</math> | ||
+ | |||
+ | <math>FF'</math> is diameter <math>\Omega.</math> Similarly, <math>EE'</math> is diameter <math>\omega.</math> | ||
+ | <cmath>\angle XEF = \angle E'EA = \alpha, \angle EFX = \angle AFF' = \alpha.</cmath> | ||
+ | <math>\angle EBX = 90^\circ - \angle EBA = 90^\circ - \angle EE'A =\alpha \implies X \in \theta \implies ZX</math> is tangent to <math>\theta.</math> | ||
+ | |||
+ | Let <math>T</math> be the spiral similarity <math>T(\omega) = \theta</math> centered at <math>B.</math> | ||
+ | |||
+ | Points <math>A, F,</math> and <math>E</math> are collinear <math>\implies T(A) = F.</math> Points <math>E', X,</math> and <math>E</math> are collinear <math>\implies T(E') = X.</math> | ||
+ | |||
+ | Therefore <math>T(D) = E \implies DE</math> is tangent to <math>\theta.</math> | ||
+ | |||
+ | Similarly, <math>CF</math> is tangent to <math>\theta.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ===Japan Mathematical Olympiad Finals 2018 Q2=== | ||
+ | [[File:Japan.png|300px|right]] | ||
+ | [[File:Japan pro.png|300px|right]] | ||
+ | Given a scalene <math>\triangle ABC,</math> let <math>D</math> and <math>E</math> be points on lines <math>AB</math> and <math>AC,</math> respectively, so that <math>CA = CD, BA = BE.</math> | ||
+ | |||
+ | Let <math>\omega</math> be the circumcircle of <math>\triangle ADE</math> and <math>P</math> the reflection of <math>A</math> across <math>BC.</math> | ||
+ | |||
+ | Lines <math>PD</math> and <math>PE</math> meet <math>\omega</math> again at <math>X</math> and <math>Y,</math> respectively. | ||
+ | |||
+ | Prove that <math>BX</math> and <math>CY</math> intersect on <math>\omega.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let <math>H</math> be the orthocenter of <math>\triangle ABC.</math> | ||
+ | |||
+ | <math>AB = BE \implies</math> Point <math>E</math> is symmetrical to point <math>A</math> with respect to height <math>BH.</math> | ||
+ | |||
+ | Point <math>D</math> is symmetrical to point <math>A</math> with respect to height <math>CH \implies</math> | ||
+ | |||
+ | <math>AH = EH = DH \implies \omega</math> is centered at <math>H \implies</math> | ||
+ | |||
+ | <math>\omega</math> is symmetrical with respect to heightline <math>AHP \implies</math> | ||
+ | |||
+ | <math>X</math> is symmetrical to point <math>E</math> with respect to height <math>AH,</math> | ||
+ | |||
+ | <math>Y</math> is symmetrical to point <math>D</math> with respect to height <math>AH \implies</math> | ||
+ | |||
+ | <cmath>\angle XAD = \angle EAY = \angle CAY.</cmath> | ||
+ | |||
+ | The isosceles triangles <math>\triangle ABE \sim ACD \implies</math> | ||
+ | <cmath>\frac {AB}{AC} = \frac {AE}{AD} = \frac {AX}{AY} \implies \triangle AXB \sim \triangle AYC \implies</cmath> | ||
+ | |||
+ | a) <math>\angle AXB = \angle AYC \implies A, X, F = BX \cap CY, Y</math> are concyclic. | ||
+ | |||
+ | b) <math>A</math> is the spiral center that maps <math>AXB</math> to <math>AYC \implies</math> maps <math>XB</math> to <math>YC.</math> | ||
+ | |||
+ | Therefore <math>F, A, X, Y</math> are concyclic and <math>F, A, B, C</math> are concyclic. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Latest revision as of 00:29, 16 December 2023
Contents
- 1 Basic information
- 2 Simple problems
- 2.1 Explicit spiral similarity
- 2.2 Hidden spiral symilarity
- 2.3 Linearity of the spiral symilarity
- 2.4 Construction of a similar triangle
- 2.5 Center of the spiral symilarity for similar triangles
- 2.6 Spiral similarity in rectangle
- 2.7 Common point for 6 circles
- 2.8 Three spiral similarities
- 2.9 Superposition of two spiral similarities
- 2.10 Spiral similarity for circles
- 2.11 Remarkable point for spiral similarity
- 2.12 Remarkable point for pair of similar triangles
- 2.13 Remarkable point’s problems
- 2.14 Japan Mathematical Olympiad Finals 2018 Q2
Basic information
A spiral similarity is a plane transformation composed of a rotation of the plane and a dilation of the plane having the common center. The order in which the composition is taken is not important.
Any two directly similar figures are related either by a translation or by a spiral similarity (directly similar figures are similar and have the same orientation).
The transformation is linear and transforms any given object into an object homothetic to given.
On the complex plane, any spiral similarity can be expressed in the form where
is a complex number. The magnitude
is the dilation factor of the spiral similarity, and the argument
is the angle of rotation.
The spiral similarity is uniquely defined by the images of two distinct points. It is easy to show using the complex plane.
Let with corresponding complex numbers
and
so
For any points
and
the center of the spiral similarity taking
to
point
is also the center of a spiral similarity taking
to
This fact explain existance of Miquel point.
Case 1 Any line segment can be mapped into any other
using the spiral similarity. Notation is shown on the diagram.
is circle
is circle
is any point of
is circle
is the image
under spiral symilarity centered at
is the dilation factor,
is the angle of rotation.
Case 2 Any line segment can be mapped into any other
using the spiral similarity. Notation is shown on the diagram.
is circle
(so circle is tangent to
is circle tangent to
is any point of
is circle
is the image
under spiral symilarity centered at
is the dilation factor,
is the angle of rotation.
Simple problems
Explicit spiral similarity
Given two similar right triangles and
Find
and
Solution
The spiral similarity centered at with coefficient
and the angle of rotation
maps point
to point
and point
to point
Therefore this similarity maps to
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Hidden spiral symilarity
Let be an isosceles right triangle
Let
be a point on a circle with diameter
The line
is symmetrical to
with respect to
and intersects
at
Prove that
Proof
Denote
Let
cross perpendicular to
in point
at point
Then
Points and
are simmetric with respect
so
The spiral symilarity centered at with coefficient
and the angle of rotation
maps
to
and
to point
such that
Therefore
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Linearity of the spiral symilarity
Points
are outside
Prove that the centroids of triangles and
are coinsite.
Proof
Let where
be the spiral similarity with the rotation angle
and
A vector has two parameters, modulo and direction. It is not tied to a center of the spiral similarity. Therefore
We use the property of linearity and get
Let
be the centroid of
so
is the centroid of the
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Construction of a similar triangle
Let triangle and point
on sideline
be given. Construct
where
lies on sideline
and
lies on sideline
Solution
Let be the spiral symilarity centered at
with the dilation factor
and rotation angle
so image of any point
lies on
The spiral symilarity
centered at
with the dilation factor
and rotation angle
maps
into
and
therefore the found triangle
is the desired one.
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Center of the spiral symilarity for similar triangles
Let triangle and point
on sideline
be given.
where
lies on sideline
and
lies on sideline
The spiral symilarity
maps
into
Prove
a)
b) Center of is the First Brocard point of triangles
and
Proof
a) Let be the spiral symilarity centered at
with the dilation factor
and rotation angle
Denote
Similarly
b) It is well known that the three circumcircles and
have the common point (it is
in the diagram).
Therefore is cyclic and
Similarly,
Similarly,
Therefore, is the First Brocard point of
is cyclic
Similarly,
Therefore is the First Brocard point of
and
Therefore the spiral symilarity maps
into
has the center
the angle of the rotation
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Spiral similarity in rectangle
Let rectangle be given. Let point
Let points and
be the midpoints of segments
and
respectively.
Prove that
Proof
Let be the midpoint
is a parallelogram
and
are corresponding medians of
and
There is a spiral similarity centered at
with rotation angle
that maps
to
Therefore
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Common point for 6 circles
Let and point
on sideline
be given.
where
lies on sideline
and
lies on sideline
Denote
Prove that circumcircles of triangles
have the common point.
Proof
so there is the spiral symilarity
taking
to
Denote
the center of
the center of
is the secont crosspoint of circumcircles of
and
but this center is point
so these circles contain point
. Similarly for another circles.
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Three spiral similarities
Let triangle be given. The triangle
is constructed using a spiral similarity of
with center
, angle of rotation
and coefficient
A point is centrally symmetrical to a point
with respect to
Prove that the spiral similarity with center , angle of rotation
and coefficient
taking
to
Proof
Corollary
Three spiral similarities centered on the images of the vertices of the given triangle and with rotation angles equal to the angles of
take
to
centrally symmetric to
with respect to
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Superposition of two spiral similarities
Let be the spiral similarity centered at
with angle
and coefficient
Let be spiral similarity centered at
with angle
and coefficient
Let
Prove:
a) is the crosspoint of bisectors
and
b)
Algebraic proof
We use the complex plane
Let
Then
Geometric proof
Denote
Then
Let be the midpoint
be the point on bisector
such that
be the point on bisector
such that
Then
is the crosspoint of bisectors
and
Corollary
There is another pare of the spiral similarities centered at and
with angle
coefficients
and
In this case
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Spiral similarity for circles
1. Let circle cross circle
at points
and
Point
lies on
Spiral similarity centered at
maps
into
Prove that points and
are collinear.
Proof
Arcs
Corollary
Let points
and
be collinear.
Then exist the spiral similarity centered at
such that
2. Let circle cross circle
at points
and
Points and
lie on
Let be the tangent to
be the tangent to
Prove that angle between tangents is equal angle between lines and
Proof
There is the spiral similarity centered at
such that
Therefore
angles between these lines are the same.
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Remarkable point for spiral similarity
Circles and
centered at points
and
respectively intersect at points
and
Points
and
are collinear. Point
is symmetrical to
with respect to the midpoint
point
Prove:
a) b)
Proof
a) cross
in midpoint
b) is parallelogram
Denote
Corollary
Let points and
be collinear. Then
Therefore is the crosspoint of the bisectors
and
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Remarkable point for pair of similar triangles
Let
Let the points and
be the circumcenters of
and
Let point be the midpoint of
The point is symmetric to
with respect point
Prove:
a) point be the crosspoint of the bisectors
and
b)
Proof
is parallelogram
Denote
Similarly,
The statement that was proved in the previous section.
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Remarkable point’s problems
Problem 1
Let a convex quadrilateral be given,
Let and
be the midpoints of
and
respectively.
Circumcircles and
intersect a second time at point
Prove that points and
are concyclic.
Problem 2
Let triangle be given.
Let point lies on sideline
Denote the circumcircle of the as
, the circumcircle of the
as
.
Let be the circumcenter of
Let circle cross sideline
at point
Let the circumcircle of the cross
at point
Prove that
Problem 3
The circles and
are crossed at points
and
points
Let be the tangent to
be the tangent to
Point is symmetric
with respect to
Prove that points and
are concyclic.
Problem 4
The circles and
are crossed at points
and
points
Let be the tangent to
be the tangent to
Points and
lye on bisector of the angle
Points and
lye on external bisector of the angle
Prove that and bisector
are tangent to the circle
bisector is tangent to the circle
Solutions
Solutons are clear from diagrams.
In each case we use remarcable point as the point of bisectors crossing.
Solution 1
We use bisectors and
.
The points
and
are concyclic.
Solution 2
We use bisectors of
and
Solution 3
We use bisectors of and
.
is the circumcenter of
Circle is symmetric with respect diameter
Point is symmetric to
with respect diameter
Therefore
Solution 4
Let
Let be midpoint
be midpoint
.
We need prove that
and
Denote
The angle between a chord and a tangent is half the arc belonging to the chord.
is tangent to
is diameter
Similarly,
is diameter
is tangent to
Let be the spiral similarity
centered at
Points and
are collinear
Points
and
are collinear
Therefore is tangent to
Similarly, is tangent to
vladimir.shelomovskii@gmail.com, vvsss
Japan Mathematical Olympiad Finals 2018 Q2
Given a scalene let
and
be points on lines
and
respectively, so that
Let be the circumcircle of
and
the reflection of
across
Lines and
meet
again at
and
respectively.
Prove that and
intersect on
Proof
Let be the orthocenter of
Point
is symmetrical to point
with respect to height
Point is symmetrical to point
with respect to height
is centered at
is symmetrical with respect to heightline
is symmetrical to point
with respect to height
is symmetrical to point
with respect to height
The isosceles triangles
a) are concyclic.
b) is the spiral center that maps
to
maps
to
Therefore are concyclic and
are concyclic.
vladimir.shelomovskii@gmail.com, vvsss