## Wednesday, May 30, 2012

### Problem 758: Equilateral Triangle, Cevians, Trisection of sides, Congruence, Perpendicular, 90 Degrees

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

This entry contributed by Ajit Athle.
Click the figure below to see the problem 758 details.

1. Let vec(BA)=a, vec(BC)=b.

By Menelaus' theorem in triangle BEC and line AFD,
EF/FC * CD/DB * BA/AE = EF/FC * 2/1 * 3/1 = 1
so EF/FC = 1/6.

vec(BE) = 2/3 a
vec(BF) = [6 vec(BE) + vec(BC)]/7 = (4a+b)/7
vec(CE) = vec(BE) - vec(BC) = (2a-3b)/3

Now let |a|=|b|=k, then
a.a = k^2
b.b = k^2
a.b = k*k*cos(60) = k^2 / 2

Therefore,
vec(BF).vec(CE) = (4a+b).(2a-3b)/9
= [8 a.a - 10 a.b - 3 b.b] / 9
= [8 - 5 - 3]k^2 / 9
= 0

Hence, BF perp CE.

2. Haven't found the synthetic solution yet but here's my analytical attempt: WLOG assume the midpoint of AC as the origin (0,0) and let A:(-1.0) & B:(1.0) hence C:(0,√3). We can now determine D to be (1/3,2/√3) & E:(-2/3,1/√3). Slope CE=-√3 /5 and
eqn CE: 5y =-√3x + √3. Similarly, slope AD = √3/2 and eqn. AD: 2y =√3x + √3. Solve the two eqns. simultaneously to obtain F as (-3/7,2√3/7). Therefore, slope BF = 5/√3 and finally (slope BF * slope CE)= -1. Hence BF _l_ CE.
Will keep looking for a synthetic solution.

3. http://henrik-geometry.webs.com/758.htm

4. ∆ABD≡∆CAE
∴ ∠AEC = ∠ADB and so B,E,F,D are concyclic
∴ ∠AFE = 60°
Let M be the midpoint of BE
∆BMD is equilateral
Since ME = MB = MD, ∆EDB is right angled at D
B,D,E,F being concyclic,
we have ∠BFE = ∠BDE = 90°

5. Tr.s CAE & ABD are congruent. So Tr.s AEF & ABD are similar since Tr.s AEF & CAE are similar. Hence BEFD is con cyclic.
Now consider Tr. BDE where BE = 2. BD & the included < is 60. So Tr. BDE is a 30-60-90 Tr. and the result follows.

Sumith Peiris
Moratuwa
Sri Lanka

6. Let the side of triangle ABC be 3
Extend BF to meet AC at G
Apply Ceva's to triangle ABC
=> AE/EB.BD/DC.CG/GA=1
=>1/2.1/2.CG/GA=1
=>CG=4GA
Since CG+GA=3 => CG = 12/5 and GA = 3/5 -------(1)
Consider triangle AEC
AE =1, AC =3 and m(A) = 60
so EC^2=1+9-2.1.3.cos60
=> EC = Sqrt(7)
Apply Menulas to triangle EBC
=> AB/AE.EF/FC.CD/DB= 1
=> 3/1.EF/FC.2/1=1
=>EF=FC/6
Since EF+FC=EC=Sqrt(7) => EF=1/Sqrt(7) and FC = 6/Sqrt(7) ------(2)
Consider the triangle GBC
BC=1, CG=12/5 and m(C) = 60
=> BG^2 = 9+144/25-2.3.12/5.COS60
=> BG = sQRT(189/25) = 3Sqrt(21)/5 -------------(3)
Now apply Menulas to triangle GBC
=> CA/AG.GF/FB.BD.DC=1
=>3/(3/5).GF/FB.1/2=1
=> GF=2/5FB
Since GF+FB=BG=3Sqrt(21)/5 => FB = 3Sqrt(3/7)------------(4)
We observe that
FB^2+FC^2 = 27/7+36/7 = 63/7 = 9 = BC^2
or
FB^2+EF^2= 27/7+1/7=28/7=4=BE^2
Hence EF Perpendicular to EC