## Monday, October 13, 2014

### Geometry Problem 1050: Regular Hexagon, Center, Any Point, Inside, Outside, Distance, Congruence, Metric Relations

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

Click the figure below to view the complete problem 1050.

1. Let each side of the hexagon be a.
Let M,N be the midpoints of AF,CD resp.
Required sum
= (2PM^2 + a^2 / 2) + (2PO^2 + 2OE^2) + (2PN^2 + a^2 / 2)
= 2(PM^2 + PN^2) + a^2 + 2(PO^2 + OE^2)
= 2( 2PO^2 + 1/2 MN^2) + a^2 + 2PO^2 + 2a^2
= 6PO^2 + MN^2 + 3a^2
= 6PO^2 + 3a^2 + 3a^2
= 6PO^2 + 6a^2

2. Let ∠POC=x, ∠POD=y. Let AB=OA=R, PO=d.

Using cosine law, we have
PA² = R² + d² − 2Rd cos(x+120°)
PB² = R² + d² − 2Rd cos(x+60°)
PC² = R² + d² − 2Rd cos(x)
PD² = R² + d² − 2Rd cos(y)
PE² = R² + d² − 2Rd cos(y+60°)
PF² = R² + d² − 2Rd cos(y+120°)

Note that
cos(x)+cos(x+60°)+cos(x+120°)
= 2 cos(x+60°) cos(60°) + cos(x+60°)
= 2 cos(x+60°)

cos(y)+cos(y+60°)+cos(y+120°)
= 2 cos(y+60°)
= 2 cos(180°−(x+60°))
= −2 cos(x+60°)

Thus,
cos(x)+cos(x+60°)+cos(x+120°) + cos(y)+cos(y+60°)+cos(y+120°) = 0

Hence,
PA²+PB²+PC²+PD²+PE²+PF² = 6R²+6d² = 6 AB²+6 PO²

3. Problem 1050
Is AB=BC=CD=DE=EF=FA=AO=BO=CO=DO=EO=FO.But PA^2+PB^2+PC^2+PD^2+PE^2+PF^2=(PA^2+PD^2)+(PB^2+PE^2)+(PC^2+PF^2)=
(2PO^2+2AO^2)+(2PO^2+2BO^2)+(2PO^2+2CO^2)=6PO^2+6AB^2.
APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

4. Considering the geometrical structure in a 3-D plane.
Join OF,OE and consider the triangles OEF & OEP

OEP is right angle triangle
=> PE^2 = PO^2 + OE^2 ---- (1)

In the triangle OEF, OE = OF and Angle EFO = 60 => OEF is equilateral triangle
Hence OE=OF=FE ------- (2)

Substituting (2) in (1) => PE^2 = PO^2 + FE^2

Considering symmetry, PE=PF=PA=PB=PC=PD & EF=FA=AB=BC=CD=DE
The result thus follows

5. Note that AD, BE and CF are concurent at O

Use Apollonius Theorem in each of the triangles PAD, PBE, PCF and add. The result follows easily

Sumith Peiris
Moratuwa
Sri Lanka