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These are Posts that have been published by Kitonum

It is known that the trigonometric functions of an integer number of degrees may be expressed by radicals if the number of degrees is divisible by 3. Simple code finds all these values ​​in the range 0 to 90 degrees:

 [sin(`0`^`o`)=0,`   cos`(`0`^`o`)=1,`   tan`(`0`^`o`)=0,`   cot`(`0`^`o`)=infinity];

for n from 3 to 87 by 3 do


Clock:=proc(H, M)  # H and M - time in hours (0 <= H <= 23) and minutes (0<=M<60)

local A1,A2, A, B, B1, C1, C, E, F,alpha,t, T, T1, T2, P, G;

uses plottools, plots; 


A1:=(irem(H,12)+M/60)*30;  A2:=M*6;

A:=circle([0,0], thickness=5);

P:=disk([0,0], 1, color=grey);

Here are two procedures associated with a clock with hands. I think they are interesting not only for fun, but will be useful for teachers in schools in the preparation of such tasks.

The first procedure called  ClockHandsAngle  finds the angle between the hands of the clock at any time.

The second procedure called  TimeFromAngle , for a given angle finds all times in a given time range. The first procedure  is...

Consider two sets in the Euclidean plane, each consisting of 4 points.

First set:  A(0, 0),  B(3, 4),  C(12, 4),  E(4, -1)

Second set:  F(0, -8),  G(12, -4),  H(9, -8),  K(4, -9)

It is easy to check that the set of all pairwise distances between the points of each of the given sets (6 numbers for each set ) are the same. At the same time it is obvious that there is no any...

We assume that the length of a match is 1, then the perimeter of a polygon is equal to the number of matches N. If a match can be located at arbitrary angles to each other, then at a given perimeter of the area can take on any value between zero and the area of a regular polygon (for even number of matches) . For an odd number of matches the lower bound equals to the area of an equilateral triangle of side 1. For any given area within these boundaries will be infinitely many solutions.

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