MathDB

5

Part of 2018 Iranian Geometry Olympiad

Problems(3)

equal angles starting with a parallelogram ABCD with <DAC=90^o

Source: Iranian Geometry Olympiad 2018 IGO Intermediate p5

9/19/2018
Suppose that ABCDABCD is a parallelogram such that ∠DAC=90o\angle DAC = 90^o. Let HH be the foot of perpendicular from AA to DCDC, also let PP be a point along the line ACAC such that the line PDPD is tangent to the circumcircle of the triangle ABDABD. Prove that ∠PBA=∠DBH\angle PBA = \angle DBH.
Proposed by Iman Maghsoudi
geometryparallelogramequal anglescircumcircle
2018 IGO Elementary Level P5

Source:

9/20/2018
There are some segments on the plane such that no two of them intersect each other (even at the ending points). We say segment ABAB breaks segment CDCD if the extension of ABAB cuts CDCD at some point between CC and DD. [asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(4cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -5.267474904743955, xmax = 11.572179069738377, ymin = -10.642621257034536, ymax = 4.543526642434019; /* image dimensions */
/* draw figures */ draw((-4,-2)--(1.08,-2.03), linewidth(2)); draw(shift((-2.1866176795507295,-2.0107089507113147))*scale(0.21166666666666667)*(expi(pi/4)--expi(5*pi/4)^^expi(3*pi/4)--expi(7*pi/4))); /* special point */ draw((-0.16981767035094117,3.225314210196242)--(-2.1866176795507295,-2.0107089507113147), linewidth(2) + linetype("4 4")); draw((-0.16981767035094117,3.225314210196242)--(-0.8194002739586808,1.538865607509914), linewidth(2)); label("AA",(-1.2684397405642523,3.860690076971137),SE*labelscalefactor,fontsize(16)); label("BB",(-1.9211395070170559,2.002590777612728),SE*labelscalefactor,fontsize(16)); label("CC",(-4.971261820527631,-1.6571211388676117),SE*labelscalefactor,fontsize(16)); label("DD",(1.08925640451367566,-1.6571211388676117),SE*labelscalefactor,fontsize(16)); /* dots and labels */ dot((-4,-2),dotstyle); dot((1.08,-2.03),dotstyle); dot((-0.16981767035094117,3.225314210196242),dotstyle); dot((-0.8194002739586808,1.538865607509914),dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]
a)a) Is it possible that each segment when extended from both ends, breaks exactly one other segment from each way? [asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(4cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.8, xmax = 8.68, ymin = -10.32, ymax = 3.64; /* image dimensions */
/* draw figures */ draw((-2.56,1.24)--(-0.36,1.4), linewidth(2)); draw((-3.32,-2.68)--(-1.24,-3.08), linewidth(2)); draw(shift((-2.551651190956802,-2.8277593863544612))*scale(0.17638888888888887)*(expi(pi/4)--expi(5*pi/4)^^expi(3*pi/4)--expi(7*pi/4))); /* special point */ draw(shift((-0.8889576602618603,1.3615303519809556))*scale(0.17638888888888887)*(expi(pi/4)--expi(5*pi/4)^^expi(3*pi/4)--expi(7*pi/4))); /* special point */ draw((-2.551651190956802,-2.8277593863544612)--(-0.8889576602618603,1.3615303519809556), linewidth(2) + linetype("4 4")); draw((-1.4097008194020806,0.049476186483185636)--(-1.8514772275312024,-1.0636149148218605), linewidth(2)); /* dots and labels */ dot((-2.56,1.24),dotstyle); dot((-0.36,1.4),dotstyle); dot((-3.32,-2.68),dotstyle); dot((-1.24,-3.08),dotstyle); dot((-1.4097008194020806,0.049476186483185636),dotstyle); dot((-1.8514772275312024,-1.0636149148218605),dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]
b)b) A segment is called surrounded if from both sides of it, there is exactly one segment that breaks it.\\ (e.g. segment ABAB in the figure.) Is it possible to have all segments to be surrounded?
[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(7cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -10.70976151557872, xmax = 18.64292748469251, ymin = -16.354300717041443, ymax = 9.136192362141452; /* image dimensions */
/* draw figures */ draw((1.0313140845297686,0.748205038977829)--(-1.3,-4), linewidth(2.8)); draw((-5.780195085389632,-2.13088646583346)--(-2.549994860479401,-2.13088646583346), linewidth(2.8)); draw((4.121070821400425,-3.816208322308361)--(1.78,-1.88), linewidth(2.8)); draw(shift((-0.38228674372374466,-2.13088646583346))*scale(0.21166666666666667)*(expi(pi/4)--expi(5*pi/4)^^expi(3*pi/4)--expi(7*pi/4))); /* special point */ draw((-2.549994860479401,-2.13088646583346)--(-0.38228674372374466,-2.13088646583346), linewidth(2.8) + linetype("4 4")); draw(shift((0.32979226045261084,-0.6805897691262632))*scale(0.21166666666666667)*(expi(pi/4)--expi(5*pi/4)^^expi(3*pi/4)--expi(7*pi/4))); /* special point */ draw((4.121070821400425,-3.816208322308361)--(0.32979226045261084,-0.6805897691262632), linewidth(2.8) + linetype("4 4")); draw((-3.6313140845297687,-8.74820503897783)--(3.600422205681574,5.980726991931396), linewidth(2.8) + linetype("2 2")); label("AA",(-0.397698406272906,1.754593418658662),SE*labelscalefactor,fontsize(16)); label("BB",(-2.6377720405041316,-3.266261278756151),SE*labelscalefactor,fontsize(16)); /* dots and labels */ dot((1.0313140845297686,0.748205038977829),linewidth(6pt) + dotstyle); dot((-1.3,-4),linewidth(6pt) + dotstyle); dot((-5.780195085389632,-2.13088646583346),linewidth(6pt) + dotstyle); dot((-2.549994860479401,-2.13088646583346),linewidth(6pt) + dotstyle); dot((4.121070821400425,-3.816208322308361),linewidth(6pt) + dotstyle); dot((1.78,-1.88),linewidth(6pt) + dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy] Proposed by Morteza Saghafian
IranIGO2018 igogeometry
incenters are concyclic, started with a cyclic ABCD, vietnamese IGO proposal

Source: Iranian Geometry Olympiad 2018 IGO Advanced p5

9/19/2018
ABCDABCD is a cyclic quadrilateral. A circle passing through A,BA,B is tangent to segment CDCD at point EE. Another circle passing through C,DC,D is tangent to ABAB at point FF. Point GG is the intersection point of AE,DFAE,DF, and point HH is the intersection point of BEBE, CFCF. Prove that the incenters of triangles AGFAGF, BHFBHF, CHECHE, DGEDGE lie on a circle.
Proposed by Le Viet An (Vietnam)
geometryincenterConcycliccyclic quadrilateral