See complete Problem 122
Marion Walter's Theorem, Area of triangle and hexagon. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Monday, June 16, 2008
Elearn Geometry Problem 122
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SAC'F = SFCA" (1)
ReplyDeleteSAEB' = SEBA" (2) ( Both: S/3 - common area )
from (1), and the fact h1 = h/3, h2 = h/2
(h alt of ABA",h1 of AC'F, h2 of FCA")
=>
AF = 6/10 AA", FA" = 4/10 AA"
SAFC' = SFCA" = 2/15 S
from (2)
=>
AE = 3/4 AA"
SAFC = 1/3 - 2/15
SAFC = 1/5 S (3)
SECA" = 1/4 ∙1/3
SECA" = 1/12
SFEC = 2/15 - 1/12
SFEC = 1/20 (4)
SAGFEC = 1/20 + 1/5 + 1/20
SAGFEC = 3/10
for three sides of ABC
SAGFEC + SAGHMB + SBMDEC = 9/10
SFGHMDE = 10/10 - 9/10
SFGHMDE = 1/10
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Solution to problem 122.
ReplyDeleteLet K be the center of gravity, and AL, BN, CP the medians of triangle ABC. Let S be the area of ABC. Applying Ceva’s theorem to the cevians AA”, BN and CC’, we have
(AN/NC).(CA”/A”B).(BC’/C’A)=(1/1).(1/2).(2/1)=1.
Thus F lies on the median BN. Similarly D and H lie on the medians AL and CP, respectively.
For cevians AA’, BN and CC”, we have
(AN/NC).(CA’/A’B).(BC”/C”A)=(1/1).(2/1).(1/2)=1,
so M lies on the median BN, and similarly G and E lie on the medians AL and CP, respectively.
Now we have the same situation of problem 121, where it was proved that S(KEF) = S/60. The hexagonal central region is the union of six triangles KDE, KEF, KFG, KGH, KHM, and KMD, each one if them with area S/60.
Hence S1 = 6.(S/60) = S/10.