## Saturday, September 21, 2013

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

Geometry Problem
Post your solution in the comment box below

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

This entry contributed by Ajit Athle.
Click the figure below to see the complete problem 926.

1. 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°.

2. Let 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.

Consider T(M,90°,√3), then
A→B
D→F
∠AMD = ∠BMF = 90°−∠BMD

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

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

DM = 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