tag:blogger.com,1999:blog-6933544261975483399.post1050335178358386329..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 786: Right Triangle, Altitude, Hypotenuse, Cevian, Perpendicular, Ratio, Metric relationsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-6933544261975483399.post-12621339960988494872016-02-24T01:48:23.450-08:002016-02-24T01:48:23.450-08:00Correction tanDBF = (1-1/4)/(1+1/4)= 3/5Correction tanDBF = (1-1/4)/(1+1/4)= 3/5Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-68734379833394515622016-02-24T01:42:07.723-08:002016-02-24T01:42:07.723-08:00Let AH = p, BD = d, BE = e and BF = h and BF = g
...Let AH = p, BD = d, BE = e and BF = h and BF = g<br /><br />AH. HC = BH^2 so<br /><br />p(b-p) = 4b^2/25<br /><br />Solving the quadratic p = b/5 or 4b/5<br /><br />Take p = b/5; c^2 = b^2/5 and a^2 = 4b^2/5<br /><br />Hence a/c = 2 and so BD bisects the right angle at B<br /><br />HD = b/3 -b/5 = 2b/15 so d^2 = 8b^2/45<br /><br />Similarly from right Tr. BHE e^2 = 17b^2/45<br /><br />From the calculations using pure geometry alone become a bit unwieldy <br /><br />We can use the fact that eh = 2b^2/15 by noting that the Tr.s ABD and EBD have equal areas <br /><br />Also since BFDH is cyclic h/(2b/5) = (b/3)/e = (e-g)/ (2b/3)<br /><br />But the easiest way is to use Trigonometry as Jacob has done above so that h/g = tan DBF = (1-1/3)/Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-39657699053753025782012-08-04T12:18:14.637-07:002012-08-04T12:18:14.637-07:00To Anonymous (Problem 786) your solution is not co...To Anonymous (Problem 786) your solution is not correct.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-48665702591806320082012-08-04T08:30:19.058-07:002012-08-04T08:30:19.058-07:00let AC=30, then BH=12
AB=6sqrt(5), BC=12sqrt(5), A...let AC=30, then BH=12<br />AB=6sqrt(5), BC=12sqrt(5), AH=6,HD=4, BD=4sqrt(10), BE=6sqrt(10)<br />the area of triangle BDE is 60<br />then DF=2sqrt(10)<br />DF/BF=sqrt(3)/3Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-40056459686279757572012-07-18T18:30:11.075-07:002012-07-18T18:30:11.075-07:00Let AB=a, BC=b, AC=c, BH=h.
Since ab=ch, h/c=2/5...Let AB=a, BC=b, AC=c, BH=h. <br /><br />Since ab=ch, h/c=2/5, <br />ab:c^2 = ab:(a^2+b^2) = 2:5<br /><br />Solving it, we have <br />a:b = 1:2 or a:b = 2:1<br /><br />WLOG, let a:b = 1:2. <br /><br />Let vecBC = 2i, vecBA=j. <br /><br />Using vector, we have<br />vecBD=2/3 (i+j)<br />vecBE=1/3 (4i+j)<br /><br />Therefore, angle DBC = 45 degrees<br />also, tan(angle EBC)=1/4<br /><br />Hence, <br />DF/BF<br />= tan(angle DBE)<br />= tan(45 - angle EBC)<br />= [1 - 1/4] / [1 + 1/4]<br />= 3/5Jacob Hanoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-49054594168695390952012-07-18T16:48:36.229-07:002012-07-18T16:48:36.229-07:00http://img7.imageshack.us/img7/479/problem786.png
...http://img7.imageshack.us/img7/479/problem786.png<br /><br />Draw lines per sketch.<br />Let AC=3, AH=x => BH=6/5<br />We have BH^2=AH.HC …( relation in right triangle)<br />Or 36/25=x(3-x)<br />We get 2 solutions : x=3/5 or x=12/5<br />For x=3/5 , HD=2/5 and HE=7/5<br />So tan(beta)=1/3 and tan(alpha)=7/6<br />DF/BF=tan(alpha-beta)= 3/5<br />Due to symmetric , we will get the same result for x=12/5Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com