Proposed Problem
Click the figure below to see the complete problem 416 about Triangle, Cevians, Trisectors, Area.
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Complete Problem 416
Level: High School, SAT Prep, College geometry
Friday, January 8, 2010
Problem 416: Triangle, Cevians, Trisectors, Area
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1) AB' = 1/3 AC => SABB' = 1/3 S = SBCC' = SACC'
ReplyDelete2) Draw BB", CC" perpendicular to AA'
tr ADB and DB'CA' have same area ( - SADB')
(AD∙BB")/2 = (AA'∙CC")/2 - (AD∙CC"/3)/2 (1)
(draw // from B'to AA' give us altitude for tr ADB')
From area ABA' and AA'C get BB" = 2 CC" (2)
substitute (2) at (1)
(AD∙BB")/2 = (AA'BB")/4 - (AD∙BB")/12
AD/2 = AA'/4 - AD/12 => 6AD/12 + AD/12 = AA'/4
7AD/3 = AA' = AD = 3/7 AA'
from
AA' ∙ BB" = 2/3 S => SADB = 2/7 S
at the same way SAFC = SAEC = 2/7 S
3)S1 = S - ( 2/7 S + 2/7 S + 2/7 S ) = S/7
corecting
ReplyDeleterow 9 at 2) is 7AD/3 = AA' => AD = 3/7 AA'
This seems to be taken (or say inspired) a problem posted in a different webside:
ReplyDeletehttp://www.8foxes.com/Home/216
http://8foxes.blogspot.com/2009/12/fox-216.html
It was in their first page. You should have given a reference.
To Anonymous:
ReplyDeleteThanks for your comment.
Problem 416 is for any triangle. The problem that you mention on 8foxes is for an equilateral triangle.
E is the midpoint of BD !
ReplyDelete