Geometry Problem. Post your solution in the comments box below.

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

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## Monday, October 13, 2014

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

Labels:
center,
congruence,
distance,
hexagon,
metric relations,
regular polygon

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Let each side of the hexagon be a.

ReplyDeleteLet 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

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

ReplyDeleteUsing 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

Problem 1050

ReplyDeleteIs 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

Considering the geometrical structure in a 3-D plane.

ReplyDeleteJoin 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