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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 786.
http://img7.imageshack.us/img7/479/problem786.pngDraw lines per sketch.Let AC=3, AH=x => BH=6/5We have BH^2=AH.HC …( relation in right triangle)Or 36/25=x(3-x)We get 2 solutions : x=3/5 or x=12/5For x=3/5 , HD=2/5 and HE=7/5So tan(beta)=1/3 and tan(alpha)=7/6DF/BF=tan(alpha-beta)= 3/5Due to symmetric , we will get the same result for x=12/5
Let AB=a, BC=b, AC=c, BH=h. Since ab=ch, h/c=2/5, ab:c^2 = ab:(a^2+b^2) = 2:5Solving it, we have a:b = 1:2 or a:b = 2:1WLOG, let a:b = 1:2. Let vecBC = 2i, vecBA=j. Using vector, we havevecBD=2/3 (i+j)vecBE=1/3 (4i+j)Therefore, angle DBC = 45 degreesalso, tan(angle EBC)=1/4Hence, DF/BF= tan(angle DBE)= tan(45 - angle EBC)= [1 - 1/4] / [1 + 1/4]= 3/5
let AC=30, then BH=12AB=6sqrt(5), BC=12sqrt(5), AH=6,HD=4, BD=4sqrt(10), BE=6sqrt(10)the area of triangle BDE is 60then DF=2sqrt(10)DF/BF=sqrt(3)/3
To Anonymous (Problem 786) your solution is not correct.
Let AH = p, BD = d, BE = e and BF = h and BF = gAH. HC = BH^2 sop(b-p) = 4b^2/25Solving the quadratic p = b/5 or 4b/5Take p = b/5; c^2 = b^2/5 and a^2 = 4b^2/5Hence a/c = 2 and so BD bisects the right angle at BHD = b/3 -b/5 = 2b/15 so d^2 = 8b^2/45Similarly from right Tr. BHE e^2 = 17b^2/45From the calculations using pure geometry alone become a bit unwieldy We can use the fact that eh = 2b^2/15 by noting that the Tr.s ABD and EBD have equal areas Also since BFDH is cyclic h/(2b/5) = (b/3)/e = (e-g)/ (2b/3)But the easiest way is to use Trigonometry as Jacob has done above so that h/g = tan DBF = (1-1/3)/
Correction tanDBF = (1-1/4)/(1+1/4)= 3/5