tag:blogger.com,1999:blog-6933544261975483399.post7799896982387802202..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1155: Right Triangle, Altitude, Incenter, Incircle, AreaAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-6933544261975483399.post-50170039385997063722015-11-05T12:58:22.180-08:002015-11-05T12:58:22.180-08:00Easily BO1_|_CO, BO2_|_AO, and let BO1, BO2 inters...Easily BO1_|_CO, BO2_|_AO, and let BO1, BO2 intersect AC at M, N respectively. By symmetry tr.ANO1 and ABO1 are congruent, and so are CO2M and CO2B. Now, if MO2 and NO1 intersect at X, that is the orthocenter of tr BMN with <MBN=45, since <BO1O=(<A+ABH)/2 =45 gives <BO1N=90, and so is <BO2M, hence OO1XO2 is a parallelogram, and S[OO1XO2]=2S[O1XO2] ( 1 ). <br />MNO2O1 is cyclic, tr. O1XO2 and MXN are similar, their similitude ratio is O1O2/MN, but since <O1MO2=45, we get O1O2/MN=1/sqrt{2}, i.e. S[MXN]/S[O1XO2]=(MN/O1O2)^2=2; with (1) we got S[MXN]=S[OO1XO2] and indeed S[AO1B]+S[BO2C]=S[AOC], doneStan Fulgerhttps://www.facebook.com/stan.fulgernoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-36350815204946176282015-10-26T19:09:49.221-07:002015-10-26T19:09:49.221-07:00Dear Antonio - from the data given is it possible ...Dear Antonio - from the data given is it possible to find the 3 sides of the Tr. ?<br /><br />Rgds<br /><br />Sumith Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-39963605354992363702015-10-21T22:31:45.663-07:002015-10-21T22:31:45.663-07:00See Problem 36:
Area Tr. AOC = Area Tr. AO1B + A...See Problem 36: <br />Area Tr. AOC = Area Tr. AO1B + Area Tr. CO2B or 13 = 5 + Area Tr. CO2B or Area Tr. CO2B = 8 Sq. UnitsAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-43933842103854279342015-10-21T22:21:33.786-07:002015-10-21T22:21:33.786-07:00Let r1 be the radius of O1 and r2 of O2 and r of. ...Let r1 be the radius of O1 and r2 of O2 and r of. O. <br /><br />Let the respective areas be S1, S2 and S<br /><br />From similar Tr.s r1= rc/b and r2 = ra/b<br /><br />So S1 = rc^2 / 2b and S2 = ra^ 2/2b and S = rb/2<br /><br />Now since a^2 + c^2 = b^2 we have<br /><br />S1(2b/r) + S2(2b/r) = S (2b/r)<br /><br />Simplifying S1 + S 2 = S<br /><br />So 5 + S2 = 13 and S = 8<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-58578886356060318842015-10-21T16:01:16.468-07:002015-10-21T16:01:16.468-07:00Let S(XYZ) denote area of triangle XYZ
Let r1, r2,...Let S(XYZ) denote area of triangle XYZ<br />Let r1, r2, r3 are radii of incircles of triangles AHB, BHC and ABC<br />Note that triangles AHB, BHC and ABC are similar ( case AA)<br />Triangles AO1B, AOC and BO2C are similar ( case AA)<br />S(AO1B)/S(AOC)= (AB/AC).(AO1/AO)<br />Since Triangle ABH similar to ABC so we have AO1/AO= r1/r= AB/AC<br />So S(AO1B)/S(AOC)= (AB/AC)^2<br />Similarly we have S(BO2C)/S(AOC)= (BC/AC) ^2<br />Since AC^2=AB^2+BC^2 => S(AOC)=S(AO1B)+S(BO2C)= 13<br />So S(BO2C)= 13-5=8<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com