Problem:
A transformation of the first quadrant of the coordinate plane maps each point (x,y) to the point (xβ,yβ). The vertices of quadrilateral ABCD are A=(900,300), B=(1800,600),C=(600,1800), and D=(300,900). Let k be the area of the region enclosed by the image of quadrilateral ABCD. Find the greatest integer that does not exceed k.
Solution:
Notice that AB is contained in the line whose equation is 3y=x. The image of a point on AB must therefore satisfy 3y2=x2. Because the coordinates of the image points must be positive, the image Aβ²Bβ² of AB is contained in the line y3β=x. In a similar fashion, it follows that the image Cβ²Dβ² of CD is contained in the line y=x3β. An equation for line AD is x+y=1200, so the image of AD is contained in the first-quadrant part of the circle x2+y2=1200. In a similar fashion, it follows that the image of BC is contained in the first-quadrant part of the circle x2+y2=2400. Thus the area of the region enclosed by the image of quadrilateral ABCD is 360ΞΈβ(Ο(OBβ²)2βΟ(OAβ²)2), where O is the origin and ΞΈ is the degree measure of the angle formed by OAβ² and ODβ². Notice that ΞΈ=tanβ13ββtanβ13β1β=30, because the slope of ODβ² is 3β and the slope of OAβ² is 3β1β. Hence the area of the region enclosed by the image of ABCD is k=121β(2400β1200)Ο=100Ο, and the greatest integer that does not exceed k is 314β.
