 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Three Squares make a Triangle (Posted on 2019-12-23) Take three unit squares made of paper.
Select one of the diagonals of each square and rearrange the squares so the three selected diagonals form an equilateral triangle with side length equal to the square root of 2.
The outline of all the paper now forms an irregular hexagon (three angles of which are 90 degrees).
Find the area of this irregular hexagon, and find the percentage of area which is one layer of paper, vs two layers, vs three layers.
What is the shape of the region which is three layers?

 No Solution Yet Submitted by Larry No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Solution | Comment 2 of 3 | Let the vertecies of the equilateral triangle be A, B, C.  AB=BC=AC=sqrt(2) and the area of ABC equals sqrt(3)/2.

Let A*, B* and C* be the vertecies of the squares laying outside ABC, outside of sides BC, AC, and AB respectively.
Then the hexagon consists of points A,C*,B,A*,C,B* in cyclic order.  All its edges are 1 unit and has angles alternating between 90 and 150 degrees.
Then areas of ABC*, area(AB*C), and area(A*BC) are all 1/2.  Then the area of the hexagon is 3/2 + sqrt(3)/2.

Let A', B' and 'C be the vertecies of the squares laying inside ABC, opposite of sides BC, AC, and AB respectively.  Then the four squares are A*BA'C, B*CB'A, and C*AC'B.  A'B, A'C, B'A, B'C, C'A, C'B all have unit length.

The concave quadrilateral ACA'B is the portion of triangle ABC not covered by square A*BA'C.  BAB'C and CBC'A are similar quadrilaterals.  Each of these quadrilaterals has an area of sqrt(3)/2 - 1/2.

Let AB' and BA' intersect at C".  Then ABC" is a 15-15-150 isosceles triangle.  This triangle is the portion of ABC covered by only square A*BA'C.
ABC" can be split into two 15-75-90 triangles.  A 15-75-90 right triangle has sides in a 1:2+sqrt(3):sqrt(6)+sqrt(2) ratio.
Scaling this to fit a longer leg of sqrt(2)/2 then means that the area of ABC" equals 1-sqrt(3)/2 and AC" = BC" = sqrt(3)-1.
Points B" and C" forming triangles AB"C and A"BC are derived the same way and have the same properties.

The union of ABC*, AB*C, A*BC, ABC", AB"C and A"BC is the area of the hexagon which is covered by just one of the squares.  That total area is 9/2 - (3/2)*sqrt(3)

The small hexagon A'C"B'A"C'B" is the area of the hexagon which is covered by all three of the squares.  Angles B"A'C", C"B'A", and A"C'B" are right angles and angles C'B"A', A'C"B', and B"A'C" are all 150 degrees.  All the sides are congruent by symmetry.  Therefore hexagon A'C"B'A"C'B" is similar to the main hexagon AC*BA*CB*.

The main hexagon has sides of unit length.  The length of A'B", one of the sides of the small hexagon, can be found by deducting B"C from A'C.  Then A'B" = 2-sqrt(3); which then makes the area of A'C"B'A"C'B" equal to (2-sqrt(3))^2*(3/2 + sqrt(3)/2) = 9/2 - (5/2)*sqrt(3).

The union concave quadrilaterals AB"A'C", BC"B'A" and CA"C'B" is the area of the hexagon which is covered by exactly two of the squares.  This area can be found by subtracting the other areas from the total hexagon.  That total is (3/2 + sqrt(3)/2) - (9/2-(3/2)*sqrt(3)) - (9/2-(5/2)*sqrt(3)) = -15/2 + (9/2)*sqrt(3).

The portion of the main hexagon covered by just one square is (9/2 - (3/2)*sqrt(3)) / (3/2 + sqrt(3)/2) = 6-3*sqrt(3) = 0.803848
The portion of the main hexagon covered by exactly two squares is (-15/2 + (9/2)*sqrt(3)) / (3/2 + sqrt(3)/2) = -12+7*sqrt(3) = 0.124356
The portion of the main hexagon covered by all three squares is (9/2 - (5/2)*sqrt(3)) / (3/2 + sqrt(3)/2) = 7-4*sqrt(3) = 0.071797

 Posted by Brian Smith on 2019-12-23 17:39:52 Please log in:

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