tag:blogger.com,1999:blog-6933544261975483399.post3602440165983377419..comments2024-03-17T11:34:28.946-07:00Comments on Go Geometry (Problem Solutions): Problem 587: Triangle, Incenter, Incircle, Tangency Point, Midpoint, Distance, Half the DifferenceAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-6933544261975483399.post-89891661822314958632023-12-26T21:36:56.973-08:002023-12-26T21:36:56.973-08:00Let the tangency point of BC and AB be E & F ...Let the tangency point of BC and AB be E & F respectively<br />BE=a1, CE=a2, BF=c1, AF=c2<br />CE=BF =>a1=c1<br />AD=c2, CD=a2<br /><br />AM=AD+DM<br />AM=c2+x<br /><br />CM=CD-DM<br />CM=a2-x<br /><br />Since AM=CM, c2+x=a2-x<br />2x=a2-c2<br />2x=a2-c2+a1-c1 [a1=c1]<br />2x=a-c<br />x=(a-c)/2Marcohttps://www.blogger.com/profile/04632526355171968456noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-92108054637652259132019-04-29T08:36:02.660-07:002019-04-29T08:36:02.660-07:00Name tangency point on BC as P and the tangency po...Name tangency point on BC as P and the tangency point on AB as R <br /><br />Let MC = b<br />then DC = x + b<br />but,<br />DC = PC (tangents of the circle)<br />thus PC = x + b<br /><br />Since AM = MC = b (M is the midpoint of AC)<br />AD = AM - x<br /> = b - x<br />then AR = b - x (tangents of the circle)<br /><br />BR = AC - AR<br /> = c - b + x<br />BP = BC - PC<br /> = a - x - b<br /><br />Since BR = BP (tangents of the circle)<br />c - b + x = a - x - b<br />2x = a - c <br /> x = (a - c)/2Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-32026966724677575742011-03-04T04:18:18.463-08:002011-03-04T04:18:18.463-08:00i shall use the usual notation for the sides of th...i shall use the usual notation for the sides of the triangle and the semi perimeter.<br />AB = c, BC = a and AC = b and s = (1/2)*(a+b+c)<br />now by the tangency property of the in circle we have AD = s-a and AM = b/2 as M is the midpoint of AC. now DM = AM - AD = b/2 - (s-a) = a/2 - c/2 = (1/2)*(a-c). this implies that DM = (1/2)*(BC - AB).<br />Q. E. D.sourav mishranoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-18911755429064112842011-03-03T19:46:30.398-08:002011-03-03T19:46:30.398-08:00E,F tangency points of AB, AC with incircle, say.
...E,F tangency points of AB, AC with incircle, say.<br /><br />Then<br /><br />2MD<br />=CD-AD<br />=CF-AE<br />=(CF+BF)-(AE+BE)<br />=BC-AB<br />=a-cPravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-72539724678789701982011-03-02T11:33:22.890-08:002011-03-02T11:33:22.890-08:00Name E, F tgpoints on AB, BC
AD+x = DC-x => 2...Name E, F tgpoints on AB, BC<br />AD+x = DC-x => 2x = DC-AD = CF-AE = (CF+BF)-(AE+BE)<br />2x = BC-AB = a - c<br />x = (a - c)/2c .t . e. onoreply@blogger.com