 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Perpendicular Concurrency Concern (Posted on 2010-08-08) ABCDEF is a convex, but not necessarily regular, hexagon with AB = BC; CD = DE; EF = FA and < ABC + < CDE + < EFA = 300o.

Prove that the perpendiculars from A, C and E respectively to FB, BD and DF are concurrent.

 No Solution Yet Submitted by K Sengupta No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Solution | Comment 2 of 3 | Let the perpendiculars from A to BF and from C to BD intersect at O.

Let the position vectors of points A, B, ...relative to O, be a, b, ...
with magnitudes a, b, ...

Let M be the mid-point of AC, with position vector 0.5(a + c).

AB = BC, so MB is perpendicular to AC, giving  [b - 0.5(a + c)].(c - a) = 0

This can be simplified to                                      b.c - b.a = 0.5(c2 - a2)

Using the other two sides in a similar way gives    d.e - d.c = 0.5(e2 - c2)

and                                                                   f.a - f.e = 0.5(a2 - e2)

Adding these three equation:       f.a - b.a + b.c - d.c + d.e - f.e = 0

a.(f - b) + c.(b - d) + e.(d - f) = 0

Since OA is perpendicular to BF and OC is perpendicular to BD, the first two terms are both zero. This leaves e.(d - f) = 0, proving that OE is perpendicular to DF and therefore that the three perpendiculars all pass through the same point, O.

This property seems to be independent of the sum of angles B, D and F, so the constraint that B + D + F = 300 degrees seems unnecessary (or have I missed something?).

 Posted by Harry on 2010-08-27 23:41:22 Please log in:
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