tag:blogger.com,1999:blog-6933544261975483399.post3825524780615474487..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 1104: Right Triangle, Incircle, Circumcircle, Inscribed Circle, Radius, TangentAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-45058056636297410552015-11-09T09:29:47.930-08:002015-11-09T09:29:47.930-08:00Draw OL // BC where L is on O1M
Apply Pythagoras ...Draw OL // BC where L is on O1M<br /><br />Apply Pythagoras to right Tr. O1OL<br /><br />(b/2 - x)^2 = (x-c/2)^2 + (a/2-x)^2 from which we have upon simplification using b^2=c^2 + a^2, <br /><br />x = a+c-b which is anyway = to 2r since BDOE is a square of side r and AF = c-r and FC = a-r<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-44617116353564443402015-04-02T14:47:58.960-07:002015-04-02T14:47:58.960-07:00http://s25.postimg.org/59a5jt4qn/pro_1104.png
T...http://s25.postimg.org/59a5jt4qn/pro_1104.png<br /> This is a special case of general Sangaku problem and<br />the general solution is <br />x= r+ 2.d.(s-a).(s-c)/(b.s)…………..(1)<br />see sketch below for detail<br />http://s25.postimg.org/vvmm8s8xr/General_Sangaku_problem.png<br />In our case we have d= ½. b. <br />replace it in (1) and simplify we get<br />x= r+ (s-a).(s-c)/s ………(2)<br />in right triangle ABC (s-a)= ½(-a+b+c) and (s-c)= ½(a+b-c)<br />so (s-a).(s-c)= ¼.(b^2- (a-c)^2)<br />replace b^2= a^2+c^2 in above and simplify<br />we get (s-a).(s-c)= ½. ac= area of triangle ABC= s.r<br />so (s-a)(s-c)/s= r<br /> and x= 2.r= a+c-b<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-38211951179510281092015-04-02T12:57:41.852-07:002015-04-02T12:57:41.852-07:00Referring to lines " Also <BHM = <BMH...Referring to lines " Also <BHM = <BMH = ½ HO1M = 45 deg.<br />Follows HBMO1 is a square.<br />So <DOH = <EOM = 45 deg and H, O, M are collinear."<br />Note that <DOH = <EOM = 45 deg if H, O, M are collinnear . Please provide explanation.<br />Note that the solution never use given data " circle O, tangent to circle Q" . Please check it.<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-41886287346918641072015-03-30T20:31:06.168-07:002015-03-30T20:31:06.168-07:00HBMO1 is a rectangle.
HD = EM, OD = OE imply ΔHDO ...HBMO1 is a rectangle.<br />HD = EM, OD = OE imply ΔHDO and ΔMEO are congruent.<br />So <DHO = <EMO.<br />Also <BHM = <BMH = ½ HO1M = 45 deg.<br />Follows HBMO1 is a square.<br />So <DOH = <EOM = 45 deg and H, O, M are collinear.<br />Implies HBMO1 is a square and O is its center.<br />Hence x = O1M = 2 OE =2r = b + c – a.<br />Pravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.com