: 2 1/3. Let the square be ABCD. Let the vertices of the rhombus be P on AB, Q on AD, and R on BC. We require the locus of the fourth vertex S of the rhombus. Suppose P is a distance x from B. We may take x  1/2 is just the reflection of the locus for x  BR, then the two extreme positions are (1) QR parallel to AB, giving S on the line AB, (2) Q at D, giving S a distance x from the line AB. So as x varies from 0 to 1 we get a right-angled triangle sides 1, 1 and √2 and area 1/2. However, we can also have BR > AQ. This gives points below the line AB. The extreme position is with R at C. Suppose QD = y. Then 1 + y 2 = x 2 + (1 - y) 2 , so y = x 2 /2. This gives S a distance y below the line AB. This gives an additional area of 1/6 (by calculus - integrate x 2 /2 from 0 to 1; I do not see how to do it without). The triangle and the curvilinear triangle together form a curvilinear triangle area 1/2 + 1/6 = 2/3. There is an identical triangle formed by reflection in MN. Thus the total area is 1 + 2/3 + 2/3 = 2 1/3. Thanks to Robert Hill and John Jones for pointing out that the original solution missed out the two triangles. 1st ASU 1967 (C) John Scholes jscholes@kalva.demon.co.uk 19 Nov 1998
1st ASU 1967 Problem 9 ABCD is a unit square. One vertex of a rhombus lies on side AB, another on side BC, and a third on side AD. Find the area of the set of all possible locations for the fourth vertex of the rhombus. Solution