A certain convex hexagon is equilateral but
not equiangular. Its opposite sides are also parallel, and separated by distances of 27, 31 and 38.
Find the area of this hexagon.
It seems that no such figure exists.
Starting with some construction, let O be a point at, say, (20,20).
Then C1, C2, and C3 are circles centred on O with radii 13.5, 15.5, and 19 respectively.
We know that one side is tangent to C1 at, say, (20,6.5), namely T. The tangent line passes through C2 at, say A, around (12.4, 6.5) and through C3 at, say, B, around (33.4,6.5), so line AB is somewhat offset comparative to O.
By some easy geometry we now have the side length of the desired figure: sqrt((31/2)^2-(27/2)^2)=sqrt(58) for the shorter part AT and
sqrt((38/2)^2-(27/2)^2)=sqrt(715)/2 for the longer part BT, totalling
sqrt((31/2)^2-(27/2)^2)+sqrt((38/2)^2-(27/2)^2)= 1/2(2sqrt(58) + sqrt(715)), somewhat over 20.99.
With compasses construct circles of the same radius on A and B respectively. It helps to reflect line AB to A'B' about O at this stage.
We know that the next side of the desired figure starts at either A or B. We need a line which both meets AB and is tangent to C2.
Assume we start at B, then since the next lside must be tangent to C2 at some point, and end on the circle on B, that end point will be very close to A', making the desired figure impossible to complete.
So the side that meets meets AB and is tangent to C2 starts at A, and accordingly is tangent to C2 at A. This gives us a new point D at around (-5.9,16.8). Note that this point is fixed and also provides both ends of the next line we want.
We can therefore add line B'D immediately to complete half of the desired figure. Unfortunately, though this line is very close to the desired length, it is somewhat shorter, and is far from tangent to A'B'at B'. Ths is even more obvious if a circle of the required radius is constructed on B'; the circles do not meet at D, but overlap somewhat.
The area of the polygon completed in this manner is just under 1001; the side lengths are 4*20.99, and 2*20.87.
If we ignore the requirement of tangency, the figure can more easily be drawn by simply noting the crossing points of the circles of required length on A,B,A', and B'. In this case, the sides are of equal and correct length, and the area is almost exactly 1003. However, perpendiculars to the lines AD and B'D reveal that the distance requirement of the parallel lines is not met: O to AD is very close, but O to B'D is 18.79 rather than 19.
Edited on September 13, 2020, 1:13 am
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Posted by broll
on 2020-09-09 03:29:24 |
Some thoughts
