tag:blogger.com,1999:blog-6933544261975483399.post6342985182194497807..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1441 Intersecting Circles, Perpendicular Bisector, Parallel LinesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-7425723587341672062019-12-19T03:06:52.734-08:002019-12-19T03:06:52.734-08:00We use the results of problem n°1440 applied to bo...We use the results of problem n°1440 applied to both lines (O,C,D) and (O,E,F), and consider alignements (A,C,H) and (B,E,M) to which we apply Reim’s theorem (see problem n°72).<br />Thus: MH//CE.<br />Similarly we obtain alignements (A,E,N) and (B,C,G) resulting in GN//CE.<br />Gewürtztraminer68https://www.blogger.com/profile/00329980114313015297noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-73452281881333676672019-07-20T05:35:31.195-07:002019-07-20T05:35:31.195-07:00If Point's
Q(R,0) and O(0,0)
Define:
(x-r)²+y...If Point's<br />Q(R,0) and O(0,0)<br /><br />Define:<br />(x-r)²+y²=R²<br />x²+y²=r²<br /><br />On Point O, can define two lines<br />y=a*x (Line on Point's OCD) and y=b*x (Line on Point's OEF)<br /><br />So:<br />(x-R)²+y²=R²<br />y=a*x<br />=> Point D [R/(1+a²),(a*R/(1+a²))]<br /><br />(x-R)²+y²=R²<br />y=b*x<br />=> Point F [R/(1+b²),(b*R/(1+b²))]<br /><br />x²+y²=R²<br />y=a*x<br />=> Point C [r²/(1+a²),(a*r²/(1+a²))]<br /><br />x²+y²=R²<br />y=b*x<br />=> Point B [r²/(1+b²),(b*r²/(1+b²))]<br /><br /><br />Solve for the Slope k (germ. Steigung) at the Line on the Point's CE<br />k(1)=a-b (i)<br /><br /><br />On Point's CD can construction a Normal line, who will Intersect the (Big) Circle (-R,0)<br /><br />n: y=-(1/a)*(x-(R+r²)/(2*(1+a²)))+a*(R+r²)/(2(1+a²))<br /><br />=> The Intersect Point's:<br />G [(p+q)/(2(1+a²)),(a*(r²-R)-p/a)/(2(1+a²))] and H [(q-p)/(2(1+a²)),(a*(r²-R)+p/a)/(2(1+a²))]<br /><br /><br />The same procedure can be define the Point's M and N.<br />...<br />=> The Intersect Point's:<br />M [(u+v)/(2(1+b²)),(b*(r²-R)-u/b)/(2(1+b²))] and N [(v-u)/(2(1+b²)),(b*(r²-R)+u/b)/(2(1+b²))]<br /><br />In wich:<br />p=a*√(R²(4a²+3)+r²(2R-r²)) , q=2a²R+R+r²<br /><br />u=b*√(R²(4b²+3)+r²(2R-r²)) , v=2b²R+R+r²<br /><br />__________<br /><br />Now it can define a new Lines with Point's G and N, M and H.<br />The Slope from the new Lines, a solve will be show:<br /><br />k(2)=a-b, line with Point's G and N (ii)<br />k(3)=a-b, line with Point's M and H (iii)<br />__________________________<br /><br />So:<br />For the three Lines, the Slope is a same. <br />A prove for the Parallelity lines.<br /><br /><br /><br /><br /><br /><br /><br />Ludwig Mercknoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-71961959617219279112019-07-14T18:35:42.089-07:002019-07-14T18:35:42.089-07:00From previous problem, A,C,H co line, M,E,B co lin...From previous problem, A,C,H co line, M,E,B co line.<br />Connect ACH, connect BEM. <br />Connect AB; <br />so angle AHM = angle ABM; <br />also angle CEM = angle ABE (ACEM on the same circle O), <br />so CE//MH<br /><br />Let GN and OF cross at P;<br />OA=OB, so curve OA=curve OB;<br />curve AG= curve GD; curve BN=curve NF; so angle(OA)+angle(OB)+angel(AG)+angle(GD)+angle(DMHF)+angle(NF)+angle(BN)=180; <br />so angle NPF = angle(NF)+angle(OA)+angle(AG) = 90-1/2 angle(DMHF) =90- 1/2angleFOD = angle CEO.<br /> So CE//GN;<br />So CE//GN//MH<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-75958725771960328352019-07-13T19:28:03.140-07:002019-07-13T19:28:03.140-07:00https://photos.app.goo.gl/VcpDVKKDqwxshkUg9
Defin...https://photos.app.goo.gl/VcpDVKKDqwxshkUg9<br /><br />Define points I, J, K, L, P, S,T, X and Y as per attached sketch<br />Note that I, J are midpoints of OD and CD<br />L and K are midpoints of OF and EF<br />QXJI and QYKL are rectangles<br />We have OJ= ½(OC+OD)= 1/2OC + OI<br />And JI= OJ-OI= 1/2OC= QX<br />Similarly, KL= ½ OE=1/2OC= QY<br />Since QX=QY => Chord GH= Chord MN=> Arc GH= Arc MN<br />And Arc GM=Arc NH => GN//MH<br />Let OS ⊥GN , we have quadrilaterals OSGJ and OSNK are cyclic <br />We have ∠ (COS)= ∠ (SGJ)= u=∠ (SNK)= ∠ (SOP)<br />So ∠ (COS)=u=∠ (SOE) => OS is the angle bisector of ∠ (TOP) and ∠ (COE)=> CE//GN <br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com