See complete Problem 61
Triangle, Trisection of Sides. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Monday, May 19, 2008
Elearn Geometry Problem 61
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See complete Problem 61
Triangle, Trisection of Sides. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
hi good day.uhm you had a great project. i just wnt to ask where did u get the values 3/20? thank you
ReplyDeleteIt can be easily demonstrated that line AMF meeting AC in F1 is a median of the triangle ABC. Likewise AGD and CEH extended are also medians of Tr. ABC.
ReplyDeleteNow we apply Menelaus's theorem repetitively to various triangles and their transversals. First we can see that A'M/MA*AF1/F1C*3/1=1 or A'M/MA*1/1*3/1=1 or A'M/MA=1/3 or A'M/AA'=1/4.
AE/EA"*A"B/BC*CB"/B"A=1 or AE/EA"*2/3*1/2=1 or AE/EA"=3/1. Thus, A'H/HA*1/2*3/1=1 or A'H/HA=2/3 or A'H/AA'=2/5 or AH/AA'=3/5
Now, HM = AA' - AH - MA' = AA'(1-3/5-1/4)=3/20*AA'(This the genesis of the magic fraction "3/20"). Similarly, it is possible to prove that: MD = 3/20 * CC", DE = 3/20 * BB", EF = 3/20*AA", FG = 3/20*CC', GH = 3/20*BB'.
Hence, P=(3/20)(AA'+AA"+BB'+BB"+CC'+CC")
Ajit: ajitathle@gmail.com
FA" = 4/10 AA"
ReplyDeleteEA" = 1/4 AA" ( see P122, P123 )
EF = 4/10 - 1/4
EF = 3/20 AA"
in the same way others
Due to the “symmetry” of the construction of the figure with regards to the three angles and the three sides of ΔABC, proving, in sufficiently general terms, that any of the sides of the hexagon MDEFGH pertaining to any of the 6 cevians trisecting the sides of the triangle respects the 3/20 ratio of the proposed formula will suffice for our purpose.
ReplyDeleteIn our case, we are going to prove that MD = 3/20 x CC”. See graph at https://drive.google.com/file/d/1rK-xPuW7WBcHXyQVW4_b__XsGnsl7oF3/view?usp=sharing.
First, note that due to the trisection of the sides from the angles, we have: A’C’’ // A”C // AC and also A”B” // A’B’ // AB and B’C’ // B”C” // BC (magenta dotted lines on the graph).
Therefore, there is literally a “maze” of parallelograms (such as, for our purpose for segment MD, AC”A’B’, A’CB”C”, A’KJC” (J and K to be introduced later), BC”B”A” and A”B”PQ (P and Q to be introduced later)) and triangles congruent to ΔABC everywhere.
Part I
Let us consider parallelogram AC”A’B’ and point J (in red) the intersection of its diagonals AA’ and B’C” (J is also on A”C’). Therefore J is middle of AA’.
If likewise K (in red) is the intersection of diagonals CC” and A’B” of parallelogram A’CB”C” (like J, K is also on A’C’), then A’KJC” is a parallelogram (because A’B”B’C” is also a parallelogram and J is middle of B’C” and K middle of A’B”).
And so M is on A’J and C”K (diagonals of A’KJC”), therefore M is middle of JA’. As J is middle of AA’, then M is on AA’ and AM = 3/4 x AA’ (1).
Likewise, with parallelograms BC”B”A” and A”B”PQ, with points P and Q in blue on the graph chosen like J and J previously, we find that AE = 3/4 x AA” (2).
From (1) and (2), we get ME // A’A”// BC and ME = 3/4 x A’A” = 1/4 x BC (3).
Likewise, it can be shown that MG // AB and MG = 1/4 x AB and EG // AC and EG = 1/4 x AC.
So ΔGME (in green dotted line on the graph) ≅ ΔABC (SSS) and the ratio of their sides is 1/4.
Part II
Look at ΔEDM : it is congruent with ΔBDC (B,D and E on same line as well as C, D and M, and EM = 1/4 x BC). Therefore, MD = 1/4 x CD. Therefore CM = CD+MD = 5/4 x CD.
As CM = 3/4 x CC” (like (1) and (2) above), therefore MD = 3/20 x CC” (4).
Part III
The proof leading to (4) above can be repeated with suitable parallelograms for any of the segments of the hexagon MDEFGH of the figure.
Therefore P = 3/20 x (AA’ + AA” + BB’+ BB” + CC’ + CC”) , QED.
Graph of solution
ReplyDelete