Difference between revisions of "Spiral similarity"
(→Basic information) |
m (→Explicit spiral similarity) |
||
(38 intermediate revisions by one other user not shown) | |||
Line 36: | Line 36: | ||
==Simple problems== | ==Simple problems== | ||
− | ===Explicit spiral | + | ===Explicit spiral similarity=== |
[[File:1934 Pras.png|370px|right]] | [[File:1934 Pras.png|370px|right]] | ||
Given two similar right triangles <math>ABC</math> and <math>A'B'C, k = \frac {AC}{BC},</math> | Given two similar right triangles <math>ABC</math> and <math>A'B'C, k = \frac {AC}{BC},</math> | ||
Line 43: | Line 43: | ||
<i><b>Solution</b></i> | <i><b>Solution</b></i> | ||
− | The spiral | + | 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 | + | Therefore this similarity maps <math>BB'</math> to <math>AA' \implies</math> |
<cmath>\frac {AA'}{BB'} = k, \angle ADB = 90^\circ.</cmath> | <cmath>\frac {AA'}{BB'} = k, \angle ADB = 90^\circ.</cmath> | ||
Line 142: | Line 142: | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''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=== | ===Common point for 6 circles=== | ||
[[File:1949 Pras.png|450px|right]] | [[File:1949 Pras.png|450px|right]] | ||
Line 157: | Line 179: | ||
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. | 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''' | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Revision as of 23:29, 15 December 2023
Contents
[hide]- 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
vladimir.shelomovskii@gmail.com, vvsss
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 vladimir.shelomovskii@gmail.com, vvsss
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
vladimir.shelomovskii@gmail.com, vvsss
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.
vladimir.shelomovskii@gmail.com, vvsss
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
vladimir.shelomovskii@gmail.com, vvsss
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
vladimir.shelomovskii@gmail.com, vvsss
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.
vladimir.shelomovskii@gmail.com, vvsss
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
vladimir.shelomovskii@gmail.com, vvsss
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
vladimir.shelomovskii@gmail.com, vvsss
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.
vladimir.shelomovskii@gmail.com, vvsss
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
vladimir.shelomovskii@gmail.com, vvsss
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.
vladimir.shelomovskii@gmail.com, vvsss
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