## Monday, October 15, 2012

### Problem 814: Two Equilateral Triangles with a common Vertex, Angle

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

Click the figure below to see the complete problem 814.

1. Observe that triangle BCF is congruent to triangle BAD by rotation.
Let angle BCF = x, then angle BAD = x,
Angle AGF = angle ACF - angle CAG = (60+x)-(x-60)=120

2. @W Fung: The proof is forced. We do not know anything about angles FBC and DBA or sides (CF) and (AD) so that we can say that triangle (BAD) is obtained from the triangle (BCF) using rotation.

1. To Rusu:

I skipped the short proof of congruence as i believe it is too obvious.

BA = BC ; BD = BF ; angle ABD = 60 - angle DBC = angle FBC
So by S-A-S, triangle ABD and CBF are congruent. Hence the rotation at point B exists.

3. Rotation clockwise 60° about B.

B→B
C→A
F→D

---
⇒ ∠BCF = ∠BAG
⇒ B,A,G,C concyclic
⇒ ∠AGC = 180° - ∠ABC = 120°
⇒ ∠AGF = 120°

4. solution to the problem 814( without rotation), by Michael Tsourakakis from Greece

angle x+angle ω=60 and angle y+angle ω=60, so,angle x=angle y
The triangles ABD, BCF ,are equal ,why have, AB=BC ,BD=BF and angle x=angle y Therefore,angle BDA=angle BFC .So, the quadrilateral BCDF ,is , cyclic quadrilateral. Therefore,angleFGD=angleDBF=60. So, angle AGF=120
Image: http://img4.imageshack.us/img4/4129/56463515.png

5. http://www.mathematica.gr/forum/viewtopic.php?f=20&t=31981&p=148019

6. < ABD = < CBF so Tr.s ABD and CBF are congruent SAS

Hence < ADB = < BFG .= say @

So < AGF = < ADF + DFG = @+60 + 60-@ = 120

Sumith Peiris
Moratuwa
Sri Lanka

7. Further it follows that BGDF and ABCG are cyclic