tag:blogger.com,1999:blog-6933544261975483399.post4877523922687868126..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1132: Triangle, Excircle, Tangency Point, Parallel, MidpointAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-27408271491253746582018-10-24T00:28:35.676-07:002018-10-24T00:28:35.676-07:00Considering usual triangle notations
we know AC1EB...Considering usual triangle notations<br />we know AC1EB1 are concyclic (since m(AC1E)=m(AB1E)=90) ------(1)<br />Since AC2//B2C1 => m(C2AB)=m(BC1B2)=B/2 -----(2)<br />Also in the triangle C1A1B1, m(C1B1A1)=B/2-----------(3)<br />From (2) & (3), C2 lies on the circle (1)<br />Similarly it can be shown that B2 also lie on the same circle (1)<br />Hence m(AC2B2)=m(AC1B2)=B/2 => m(AC3B3)=B and m(AB3C3)=C<br />=> C3B3//BC -------(4)<br />Connect C2E and m(C2EA)=m(C2B2A)=C/2<br />We also know that m(BEA)=C/2 => C2,B & E are collinear <br />Hereafter,it can be easily found that C3 is the circum-center of AC2B -----(5)<br />From (4) & (5) it can be deduced that C3 and B3 are the mid-points of AB & AC respectively and the result followsAnonymoushttps://www.blogger.com/profile/11901058471570620532noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-32454320083760894572018-10-22T07:55:49.273-07:002018-10-22T07:55:49.273-07:00< B1AB2 = <AB1A 1 = < A1C1B1 = C/2, so A...< B1AB2 = <AB1A 1 = < A1C1B1 = C/2, so AC1B1B2 is con-cyclic<br /><br />Similarly AB1C1C2 is con-cyclic<br /><br />Hence AB2B1C1C2 are all con-cyclic<br /><br />So B/2 = < AC1B2 = < AC2B2 = < C2B2C1 = B2A1C, so BC // B3C3<br /><br />If the diagonals of parallelogram AC2A1B2 cut at X, AX = XA1<br /><br />Hence applying the mid point theorem to Tr. ABC, XC3 = A1B/2 and XB3 = A1C/2<br /><br />Therefore B3C3 =BC/2<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-88018770625113711212016-02-20T14:44:04.512-08:002016-02-20T14:44:04.512-08:00Let <BC1A1=a and CB1A1=b.
<BA1C1=<B2A1C=...Let <BC1A1=a and CB1A1=b. <br /><BA1C1=<B2A1C=a and <A1C1B1=b. <br />Since AC2 parallel to A1C1, <AC2B1=<B2A1B1=<AC1B1=a+b so AC2C1B1 is a circle<br />Similarly, <AB2C1=<C2A1C1=<AB1C1=a+b so AB2B1C1 is a circle.<br />AC1EB1 is circle because <AC1E=<AB1E=90<br />All six points A,C2,C1,E,B1,B2 are cyclic<br />Triangles AC3C2 and A1BC1 are isosceles so <AC3B3=2a=<ABC and C3B3 parallel to BC.<br />Triangles A1BC1 and A1C2C1 are isosceles so EBC2 is a line with <AC2E=<AC1E=90<br />Because AC2B=90 and triangle AC3C2 isosceles, C3A=C3C2=C3B and B3C3=BC/2<br />Ivan Bazarovnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-80325096813236495152015-07-12T20:37:05.293-07:002015-07-12T20:37:05.293-07:00http://s11.postimg.org/mjwfzd603/pro_1132.png
Let...http://s11.postimg.org/mjwfzd603/pro_1132.png<br /><br />Let 2p= perimeter of triangle ABC<br />Let AB2 and AC2 cut BC at G and H<br />We have AB1=AC1= p and triangles A1CB1, ACG, BA1C1 and ABH are isosceles<br />So GA1=AB1= p and HA1=AC1= p<br />So A is the midpoint of HG<br />Since A1C2//AB2 and A1B2//AH => C2 and B2 are midpoints of AH and AG<br />C3B3 =1/2 BC<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com