Geometry Problem

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Level: Mathematics Education, High School, Honors Geometry, College.

This entry contributed by Ajit Athle.

Click the figure below to see the complete problem 926.

## Saturday, September 21, 2013

### Geometry Problem 926: Two Equilateral Triangles, Midpoint, Perpendicular, Metric Relations

Labels:
equilateral,
metric relations,
midpoint,
perpendicular,
triangle

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Let P on BC be the feet of A-altitude for tr ABC, then by the spiral similarity that carries ABC into DFE, we have <PAG= <FBC, so APGB is cyclic, therefore <AGB=<APB=90°.

ReplyDeleteLet T(M,θ,k) be a transformation that rotate a point P, with angle θ clockwise about M, with a scale factor k. i.e. ∠PMP'=θ and P'M=k×PM.

ReplyDeleteConsider T(M,90°,√3), then

A→B

D→F

∠AMD = ∠BMF = 90°−∠BMD

Thus ΔAMD→ΔBMF. Hence, AD→BF.

i.e. AD⊥BF and BF = AD×√3.

Triangles AMD and BMF are similar ..( case SAS)

ReplyDeleteand MB/MA=MF/MD= sqrt(3)

So BMF is the image of AMD in the hemothethy and rotation transformation center M, ratio= sqrt(3) , angle of rotation=90

in this transformation correspondent angles are preserved

so AD ⊥BF and BF=AD.sqrt(3)

If AB = 2b and DF = 2a then,

ReplyDeleteDM = a and AM = b

BM =√3b and FM = √3a

So triangles AMD ≈ BMF since < AMD = < BMF and AM/BM = DM/MF

Hence < MAD = MBG and so AMGB is concyclic and < AGB = 90

Further BF/AD = MF/MD = √3

Sumith Peiris

Moratuwa

Sri Lanka