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A Tilted Square

Stage: 4 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

Much of this solution uses the work of Andrew I and Elizabeth F. Well done.


For the square with opposite vertices of co-ordinates (5, 3) and (5, 7) respectively, you can obtain the other vertices (7, 5) and (3, 5) by creating 4 right-angled congruent triangles. The resulting figure is a square because: it has all its sides congruent, and all its angles right angles.

For the opposite corners of co-ordinates (5, 3) and (6, 6), using the same method, you get (4, 5) and (7, 4).

Veryfying that the four point sets from the problem are squares:

A. (8, 3); (7, 8); (2, 7); (3, 2) - Yes
B. (3, 3); (7, 4); (8, 8); (4; 7) - No
C. (16, 19); (18; 22); (21, 20); (19, 17) - Yes
D. (4, 20); (21, 19); (20, 2); (3, 3) - Yes

For the general case, using the drawing from the figure.

Let the co-ordinates of one corner be (a, b), and of the opposite be (c, d). Note that (x1, y1) and (x2, y2) are the co-ordinates of the other two corners. The geometrical problem has unique solution, that is obtained in the following manner:

Using the values m and p

x2 = c - p (i)
y2 = d - m (ii)
x1 = a + p (iii)
y1 = b + m (iv)
c = a +p - m (v)
d = b + p + m (vi)

(v) + (vi) gives
c + d = a + b + 2p
p = (c + d - a - b)/2

(v) - (vi) gives
c - d = a - b - 2m
m = (a - b + d - c)/2

Substituting these values for m and p iinto (i), (ii), (iii) and (iv) gives:

x1 = (a + c + d - b)/2
y1 = (a + b + d - c)/2
x2 = (a + b + c - d)/2
y2 = (b + c + d - a)/2