 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Take The Midpoints, Get Equilateral? (Posted on 2007-08-01) The circumcircle of the hexagon PQRSTU has a radius of length 1. It is known that PQ = RS = TU = 1.

Let J, K and L correspond respectively to the mid points of the sides QR, ST and UP.

Is it always the case that the triangle JKL is equilateral?

 See The Solution Submitted by K Sengupta Rating: 3.0000 (1 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Solution | Comment 3 of 4 | `Let cis(theta) denote cos(theta) + i sin(theta).`
`WOLOG let the vertices be`
`    P = cis(alpha+30)    Q = cis(alpha+90)    R = cis(-[beta+90])    S = cis(-[beta+30])    T = cis(-30)    U = cis(30)`
`The midpoints then become`
`    J = (Q + R)/2    K = (S + T)/2    L = (U + P)/2`
`It is easy, but tedious to show that`
`    (K - J) cis(60) = L - J`
`Therefore, JKL is an equilateral triangle.`
` `

 Posted by Bractals on 2007-08-02 03:48:07 Please log in:

 Search: Search body:
Forums (1)