tag:blogger.com,1999:blog-6933544261975483399.post583082250803659096..comments2018-03-20T06:39:05.874-07:00Comments on Go Geometry: Geometry Problem 1340: Triangle, Incenter, Concentric Circles, Isosceles Triangles, CongruenceAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-6933544261975483399.post-38126047179755526372017-08-12T07:31:38.404-07:002017-08-12T07:31:38.404-07:00In Circle 1, ED, HM and FG are equidistant from th...In Circle 1, ED, HM and FG are equidistant from the Centre O and are hence equal.<br /><br />If the tangent point on ED is X and on HM is Y then EX = HY and because AX = AY, AH = AE<br /><br />Similarly BF = BD and CM = CG<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-74240513353069253622017-08-09T15:26:38.138-07:002017-08-09T15:26:38.138-07:00Problem 1340
Suppose that a recorded circle tange...Problem 1340<br />Suppose that a recorded circle tangents to P, Q, R on the sides AC, CB, BA respectively.<br />Then ER=RD=HP=PM=GQ=QF(OP=OR=OQ). But AR=AP so AH=AE.<br />Similar BD=BF and CM=CG.<br />APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL OF KORYDALLOS PIRAEUS GREECE<br />apostolis manoloudishttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-39679218579253187982017-08-09T15:05:25.732-07:002017-08-09T15:05:25.732-07:00I will include your conclusion DE = FG = HM. Thank...I will include your conclusion DE = FG = HM. Thanks.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-70096508988762636462017-08-09T13:39:38.490-07:002017-08-09T13:39:38.490-07:00If tr AEH is isosceles then tr ADM is also isoscel...If tr AEH is isosceles then tr ADM is also isosceles since HEDM forms a cyclic quad. This is what we'll show.<br /><br />Let the tangent point on AB be X and the tangent on AC be Y.<br />1. tr OXD is congruent to tr OYM since OM = OD (larg radius) and OX=OY (small radius) and they are right triangles. So XD = YM<br />2. Similarly tr AXO is congruent to AYO by 2 sides (1 common and the radius) and being a right tr. so AX = AY.<br />3. Add those together AD = AM so tr. ADM is isosceles.<br />4. As noted from the cyclic quad tr AEH must be also.<br /><br />This can be repeated for the other 2 sides.<br /><br /><br />Benjamin Leishttps://www.blogger.com/profile/10974191081762367425noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-58807930504087712142017-08-09T13:33:52.559-07:002017-08-09T13:33:52.559-07:00Let the Radius of Circle 1 be R and the incircle o...Let the Radius of Circle 1 be R and the incircle of Trianlge ABC be r<br />drop _|_lars from O to AB, AC and BC. Denote the points as X, Y & Z<br />We have OX=OY=OZ=r<br />Join O to D and consider the right trianlge OXD.<br />We have XD^2+OX^2 = R^2<br />=> XD^2 = R^2-r^2 --------------(1)<br />Similary join OM and consider the right triangle OYM<br />we have YM^2=R^2-r^2--------------(2)<br />From (1) and (2) we can say DX = MY and also we have AX = AY (Tangents to the incircle)<br />=> DA = MA ----------(3)<br />As HA.AM = EA.AD<br />=> HA=EA (HAE is isosceles)<br />Also We can say the chords AM,ED & GF are equal in lengthAnonymousnoreply@blogger.com