tag:blogger.com,1999:blog-6933544261975483399.post7848338491740724406..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1402: Triangle, Three Tangent Circles, Parallel Diameter, Collinear PointsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger9125tag:blogger.com,1999:blog-6933544261975483399.post-38182027935291730212019-11-10T05:48:11.744-08:002019-11-10T05:48:11.744-08:00Problem solve with Hyperbola (Adriaan van Roomen)....Problem solve with Hyperbola (Adriaan van Roomen).<br /><br />Big Circle Radius: R<br />Small Circle Radius: r<br /><br />Let the big Circle Midpont A(0,0) => F1 (Hyperbola Foci)<br />Let the small Circle Midpoint C(R+r,0) => F2 (Hyperbola Foci)<br /><br />1.)<br />We know the Hyperbola have the linear eccentricity length:<br />2*e=R+r<br /><br />The Hyperbola Center (e,0).<br /><br />2.)<br />The Hyperola curve will be construction on the Intersections Point with the big and Small Circle, so:<br /><br />(x-e)²/a² - y²/b² = 1 (hyp.)<br />And: e-a=r => a=(R+r)/2 , b=√(R*r) | Because: a²+b²=e²<br /><br />3.)<br />On the Hyperbola curve can be construction a other Circle (only touch Circle to the two Circle (F1 , F2)) with the Radius w and Center B(k,l).<br /><br />-> k²+l² = (R+w)²<br />(k-e)²/a² - l²/b² = 1 (Point (k,l) on hyp.)<br /><br />=> Solve (k,l) <=> (R; r; w) or B(k,l) (Point B)<br /><br />=> Solve Intersect Point's (D and E) from the Circles F1 and F2, to Point B(k,l)<br />x²+y²=R²<br />(x-k)²+(y-l)²=w²<br />-> Solve Point D<br /><br />(x-(R+r))² + y²=r²<br />(x-k)²+(y-l)²=w²<br />-> Solve Point E<br /><br />4.)<br />Construction with the Parallel Line to Ox,<br />so: y=l<br /><br />And now it can win two other Points, with Intersect of Line and the 3'rd Circle (k,l).<br /><br />(x-k)²+(y-l)²=w²<br />y=l<br /><br />Solve x1 and x2: x1=k+w , x2=k-w, who x1 > x2, and x1=Point H, x2=Point G<br /><br />5.)<br />Construction two Lines on Points x1 and F1, x2 and F2.<br />The Intersect Point give the Point M<br /><br />6.)<br />Point D, M and E can construction three Lines, who will Prove the three Lines are the same.Ludwig Mercknoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-12847231940380511522019-10-19T11:01:11.238-07:002019-10-19T11:01:11.238-07:00https://photos.app.goo.gl/cRCLTb68yREoFXGg7
note ...https://photos.app.goo.gl/cRCLTb68yREoFXGg7<br /><br />note that triangles DBG and DAF are isosceles => F, D, G are collinear<br />similarly F, E, H are colinear <br />Per Pappus theorem applied to colinear points G,B,H and A, F, C .<br />Pappus line formed by D,M,E will be colinear <br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-47151018701439810792019-10-18T11:06:40.280-07:002019-10-18T11:06:40.280-07:00I try to translate the solution of Florin to Engli...I try to translate the solution of Florin to English . see below<br /><br />The triangles GMH and CMA are similar,<br />whence GH / AC = GM / MC = MH / MA<br />and how the angles GMA = CMH result in GA // HC<br />that is, ACHG parallelogram, from where<br />GH = AC, GM = MC, MH = MA.<br />Let BM intersect with AC at point P.<br />As M is the midpoint of the diagonals it results<br />BM = MP, BG = BH, AP = PC.<br />Let now be the point N the intersection of DE with BP.<br />Then from the cross-sectional theorem in the triangle BAC it follows that<br />NP / NB * AC = (DA / DB) * PC + (EC / EB) * AP = AP * (DA / DB + EC / EB) =<br />= AP * (DA + EC) / DB = DA + EC = AF + FC = AC. From here NP / NB = 1,<br />That is, NP = NB and hence the points M = N.<br />Florin Popa, Comănești, Romania.<br /> <br /><br />In my opinion , there are some errors in above solution.<br />GA is not parallel to HC and ACHG is not a parallelogram<br /><br />Peter Tran<br /> Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-30199519092823279452018-12-09T11:45:09.098-08:002018-12-09T11:45:09.098-08:00ACHG is a paralelogram if radius of circle B is ar...ACHG is a paralelogram if radius of circle B is arithmetic average of radius of the circle A and the circle C. In all the other cases ACGH trapezus. <br /><br />Erina,NJAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-56817690420610602272018-12-03T11:03:41.495-08:002018-12-03T11:03:41.495-08:00Also F,D,G and F,E,H are collinear. Problem would ...Also F,D,G and F,E,H are collinear. Problem would have been more challenging if FD and FE extended meets circle "b" at G and H. Then we have to prove GH is diameter and also parallel to AC. Pradyumna Agashehttps://www.blogger.com/profile/10300531209692781145noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-39416727727680965982018-11-29T11:28:38.978-08:002018-11-29T11:28:38.978-08:00ACGH is not a paralelogram ?ACGH is not a paralelogram ?c.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-65883818331198418562018-11-27T12:08:49.660-08:002018-11-27T12:08:49.660-08:00Mark Δ(delta) the line that goes through the point...Mark Δ(delta) the line that goes through the points D and E.<br />Δ ∩ AC =Q; Δ ∩ GH = P; Δ ∩ AB=D₀ <br />1. ΔAMC ͠ ΔMGH; AM/MH=CM/MG=AC/GH=(R₁+R₃/2R₂)<br />2. ΔCEQ ͠ ΔBPE; (CQ/BP=CE/BE X/Y+R₂= R₃/R₂)<br />3. ΔCMQ ͠ ΔMPG; (CQ/PG=CM/MG X/Y= R₁+R₃/2R₂)<br />4. ΔAQD₀ ͠ ΔPBD₀; (AQ/PB=AD₀/D₀B R₁+R₃+X/R₂+Y= Z/R₁+R₂-Z)<br />By solving the system:<br /><br />X= (R₁+R₃)R₃/R₁-R₃<br />y= 2R₂*R₃/R₁-R₃<br />Z=R₁<br />AD₀=R₁=AD D₀ Ξ D<br /><br />E,M,D are colinear<br /><br />Erina, NJAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-52493944176930668672018-11-27T11:25:08.345-08:002018-11-27T11:25:08.345-08:00It is the same thing to be proved M, F, intersecti...It is the same thing to be proved M, F, intersections of internal tg are collinearc.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-17207140079837376012018-11-26T08:15:46.398-08:002018-11-26T08:15:46.398-08:00Triunghiurile GMH și CMA sunt similare ,
de unde...Triunghiurile GMH și CMA sunt similare , <br />de unde GH/AC = GM/MC = MH/MA <br />și cum unghiurile GMA = CMH rezultă GA // HC <br />adică ACHG paralelogram, de unde <br />GH = AC, GM = MC, MH = MA.<br />Fie BM intersectat cu AC în punctul P. <br />Cum M este punctul de mijloc al diagonalelor rezultă <br />BM = MP, BG =BH, AP=PC.<br />Fie acum punctul N intersecția lui DE cu BP.<br />Atunci din teorema transversalei în triunghiul BAC rezultă că <br />NP/NB *AC = (DA/DB)*PC + (EC/EB )*AP = AP*(DA/DB + EC/EB) =<br />= AP*(DA+EC)/DB = DA + EC = AF + FC =AC . De aici NP/NB =1, <br />Adică NP = NB și de aici punctele M=N.<br />Florin Popa, Comănești , România.<br />Anonymousnoreply@blogger.com