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Guest Message by DevFuse

Perhaps check it again

Member Since 02 Dec 2013
Offline Last Active Today, 02:35 AM

Topics I've Started

Six sixes to make 1,000 (modified version)

25 October 2014 - 06:26 PM

Form as many horizontal-style arithmetical expressions as possible that equal 1,000.


In order to limit the total number of possible solutions, the following rules of mine are in place:



Use exactly six sixes.


Use no more than one pair of parentheses.


Use no more than one "/" symbol.


Use no more than two decimal points.


Use no more than one minus (subtraction) sign.


Concatenation is allowed.


No other symbols/characters/operations are allowed.



Total number of equilateral triangles

15 August 2014 - 06:08 PM

*             *              *             *    


       *              *             *             *


               *             *             *            *


                      *              *


              *              *




The 16 points above lie in a plane on an equilateral triangular lattice.


Certain sets of three points of the figure correspond to the vertices of equilateral triangles.


Suppose you were to form all of the equilateral triangles possible, such that, for any given


equilateral triangle, it must have its three vertices coincide with three of the 16 points.



How many total equilateral triangles can be formed this way in the figure?



Minimizing the number of terms of the square of a polynomial

23 July 2014 - 05:42 AM

Suppose a, b, c, and d  belong to the set of nonzero integers.


Let  P(x)  =  (x+ ax3 + bx2 + cx + d)2.



Determine one of the sets of values of a, b, c, and d, such that when P(x) is

expanded into individual terms of an 8th degree polynomial, that polynomial

will have the fewest number of nonzero terms possible.




Write P(x) in its expanded form.

concave octagon with the least number of internal diagonals

13 July 2014 - 07:03 PM

Consider concave octagons in the plane that are non-self-intersecting.


What is the minimum number of diagonals possible for one of these octagons

if it is required that each of the diagonals lie entirely within the octagon?

Max. # of convex/concave hexagons possible using 6 points

25 June 2014 - 07:57 PM

In the plane you get to choose six points.


By choosing these six points in one of an infinite number of appropriate configurations,

what is the maximum number of combined convex and concave (but not self-intersecting)

hexagons that can be formed, if each of the six points is to be one of the vertices for every

hexagon so formed?


For reference sake, the points could be labeled A, B, C, D, E, and, F.


I am not asking for any coordinates.  However, if you want to volunteer a set of them to

illustrate your work/solution, then that would be fine.






Here is an example with four points:  A, B, C, and, D.


Place them in the plane.  You choose where.


What is the maximum number of combined convex and concave (but not self-intersecting)

quadrilaterals that can be formed, if each of the four points is to be one of the vertices for

every quadrilateral so formed?


If you do not choose all of the points to be distinct and/or you place at least three of them

on the same lime, you won't get any quadrilaterals formed.


If you place the four points so that they are the vertices of a convex quadrilateral, then

you will get one total (convex) quadrilateral.


If you place three of the points as vertices of a triangle and the fourth point as an interior

point of that triangle, then three total (concave) quadrilaterals can be formed.


"Three" is the answer, unless there is some other general configuration for four points that

has been overlooked with a higher total number of quadrilaterals.