Page 22 - RMGO 3

22 Thanos KALOGERAKIS

Clearly, a1 = a2 = a3 and b1 = b2 = b3, that implies triangles KAB and LAB
b
b
b
c
c
c
are isosceles and therefore K and L are lying on the perpendicular bisector of AB.
Finally we draw KL which intersects AB at the desired point M, the midpoint
of AB.
A construction with neither a compass nor a straightedge

Problem 7 (Figure 13). Given, as much cardboard squares as needed, of sides a,
b and a unit segment on the plane, construct segments with lengths:
√ √ √ √
17, 37, 65, 101.

Figure 13

Solution. We place the a-side and b-side squares, as in the Figure 14.
√ √ √ √
2
It’s easy to prove that c 1 = 2, c 2 = 5, c 3 = 10, . . . , c n = n + 1.

Figure 14