tag:blogger.com,1999:blog-6933544261975483399.post3645872120891920513..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1288 Triangle, 30-50-100 Degrees, Area, Inradius, Metric Relations, MeasurementAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-52408611808741090932016-11-28T08:12:36.985-08:002016-11-28T08:12:36.985-08:00Draw a perpendicular from B to AC and meet it at D...Draw a perpendicular from B to AC and meet it at D<br />Consider the 30-60-90 triangle ABD<br />Since AB = c, BD = c/2, AD = Sqrt(3)c/2 <br />Hence Area of ABC = bc/4 ---------- (1)<br /><br />Draw an Angle Bisector for m(AB,BC) and let it meet AC at E<br />From Angle bisector theorem, CE = ab/a+c and AE = bc/a+c<br />Since triangle BEC is isosceles (50,50,80), BE = CE = ab/a+c ---------(2)<br />If you observe, triangle ABC is similar to AEB<br />Hence BE/c=a/b => BE = ac/b --------------(3)<br />From (2) and (3) ab/a+c = ac/b<br />hence a = b^2-c^2/c -------(4)<br />As we know semi-perimeter for triangle ABC, S = a+b+c/2<br />substituting value of 'a' from (4), we get S = b^2+bc/2c <br />We know, Area ABC in terms of S and r = S.r = bc/4 (from (1))<br />=> (b^2+bc/2c ).r = bc/4<br />=> r= c^2/2(b+c) ----------- (5)<br /><br /><br />Now consider the right triangle BED<br />We know BD = c/2 , BE = b^2-c^2/2b (Substitute for 'a' in ac/b)<br />and ED = AD - AE = Sqrt(3)bc-2c^2/2b<br />Applying pythogorus, we get BE^2 = BD^2+ED^2<br />=> (b^2-c^2/b)^2 = (Sqrt(3)bc-2c^2/2b)^2 + c^2/4<br />Simplifying, we get b^3+Sqrt(3)c^3=3b^2c^2 ----------------(6)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-74691371053033298182016-11-25T12:35:34.845-08:002016-11-25T12:35:34.845-08:00https://goo.gl/photos/mJcfykipaSNAr2en6
Draw CD ...https://goo.gl/photos/mJcfykipaSNAr2en6<br /><br />Draw CD such that angle BCD= 50 . D is on AB extended (see sketch).<br />1. S(ABC) =1/2.AC.AB.sin(30)= b.c/4<br />2. Note that angle BDC= 50 and tri CBD is isosceles => BD=BC= a<br />Triangle ABC and ACD are similar => AC/AD=AB/AC= c/b => AD= b^2/c<br />BD= AD-AB= b^2/c-c= a<br />AB/AC=BC/CD=c/b => CD=a.b/c= b(b^2/c^2-1)<br />Inradius r= S(ABC)/2p where 2p= a+b+c= b+b^2/c<br /> . r=b.c/4p= bc/2/(b+b^2/c) => r=c^2/(2(b+c))<br />3. Apply cosine formula to triangle ACD<br />CD^2=AC^2+AD^2-2.AC.AD.cos(30)…..(1)<br />Replace CD=b(b^2/c^2-1) ; AD= b^2/c in (1) and simplifying we have<br /> . (b^2/c^2-1)^2=1+b^2/c^2-(b/c).sqrt(3)<br />This equation will simplify to b^3+sqrt(3).c^3= 3.b.c^2 <br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-12792058994787365532016-11-25T04:48:09.478-08:002016-11-25T04:48:09.478-08:00Problem 1288
Draw BK perpendicular in AC,then...Problem 1288<br />Draw BK perpendicular in AC,then BK=AB/2=c/2 and AK=√3c/2.<br />Let area triangle ABC is (ABC)=(AC.BK)/2=bc/4.<br />Let s=(a+b+c)/2 then (ABC)=sr.So r=(ABC)/s=bc/4s, suffices to show that bc/4s=c^2/2(b+c) or b/(a+b+c)=c/(b+c) or b^2=c(a+c).<br />In extension of the AB passes point D such that BD=BC=a,then triangle ABC is similar with triangle ACD so AB/AC=AC/AD or c/b=b/(a+c) or b^2=c(a+c).So r=(c^2)/(2(b+c)).<br />APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE<br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.com