## Wednesday, July 18, 2012

### Problem 785: Equilateral Triangle, Parallel to a side, Perpendicular to a side

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

Click the figure below to see the complete problem 785.

1. Let K be the mid-point of AC, such that BK is the altitude.

Note that triangles ADE and EFC are both equilateral,
and also DBFE is a parallelogram, so
AE = DE = BF
EC = EF = DB

Thus
a + d + f
= d + EC + AE + f
= d + DB + BF + f
= GB + BH
= 2*MK + 2*KN
= 2*(MK + KN)
= 2x

2. Let AE = b, then AD = b, EC = a-b, FC = a-b
Then
NC = CFcos60 = (a-b-f)/2
by considering the length of AC = AM + MN + NC,
a = (b-d)/2 + x + (a-b-f)/2
x = (a+d+f)/2
q.e.d.

3. Triangles ADE and CFE are equilateral
So DE + EF = AE + EC = a
Draw DJ, FK each perpendicular to AC
x = ME + EN = (MJ + JE)+(EK + KN)
= [(GD + DE) + (EF + FH)]cos 60°
= [GD + (DE + EF) + FH]/2
= (d + a + f)/2

4. Let AG = p so that AM = p/2
Let CH = q so that CN = q/2

Since AED and CEF are equilateral
p+d+q+f = a ....(1)
p/2 + d + q/2+f = x
So 2x = p+2d+q+2f....(2)

Comparing (1) and (2) the result follows easily

Sumith Peiris
Moratuwa
Sri Lanka

5. https://photos.app.goo.gl/hq74dAKX1uzHKGqG3
define points G1, D1, F1 and H1 per attached sketch
we have DB=EF=FC
so DB+BF=FC+BF=a
GG1=d/2
HH1=f/2
So x= GG1+DD1+FF1+HH1= DG/2+DB/2+BF/2+FH/2
X= (a+d+f)/2