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

Labels:
area,
hexagon,
similarity,
triangle,
trisection

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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.