<html><body bgcolor="#FFFFFF"><div>On Jul 20, 2010, at 1:07 AM, Mathew Kurian <<a href="mailto:bluejamesbond@gmail.com">bluejamesbond@gmail.com</a>> wrote:<br><br></div><div></div><blockquote type="cite"><div><span class="Apple-style-span" style="font-family: arial, sans-serif; font-size: 13px; border-collapse: collapse; white-space: nowrap; -webkit-border-horizontal-spacing: 2px; -webkit-border-vertical-spacing: 2px; ">in this example given by Paul Ojanen ...</span><div><span class="Apple-style-span" style="font-family: arial, sans-serif; font-size: 13px; border-collapse: collapse; "><br> </span></div><div><span class="Apple-style-span" style="font-family: arial, sans-serif; font-size: 13px; border-collapse: collapse; ">The triangle is not being consistent by keeping its center point on the center of BACKGROUND.</span></div></div></blockquote><div><br></div>Yes it is.  The problem is that what you think of as the "center" of an equilateral triangle (where the three angle bisectors meet) isn't quite the same as what 2htdp/image considers the "center" (the center of the bounding rectangle).  However, if you figure out (by trigonometry) where the former "center" is, and overlay an invisible circle centered at that point, then when you rotate the whole complex, that point will appear unmoving.<div><br></div><div>Steve Bloch</div><div><br><blockquote type="cite"><div> <div><span class="Apple-style-span" style="-webkit-tap-highlight-color: rgba(26, 26, 26, 0.296875); -webkit-composition-fill-color: rgba(175, 192, 227, 0.230469); -webkit-composition-frame-color: rgba(77, 128, 180, 0.230469); font-size: 15px; border-collapse: collapse; font-family: arial, sans-serif; ">Hopefully, this makes sense.</span><br></div> </div></blockquote><div><br></div>Ditto :-)</div><div><br></div><div><br></div></body></html>