HMMT 二月 2010 · GEN2 赛 · 第 1 题

1 . 5 1 0 . 5 0 0 . 5 1 1 . 5 2 From the picture above, we see that the union of the half-disks will be a quarter-circle with radius 2, and therefore area π . To prove that this is the case, we first prove that the boundary of every half-disk intersects the quarter-circle with radius 2, and then that the half-disk is internally tangent to the quarter-circle at that point. This is sufficient because it is clear from the diagram that we need not worry about covering the interior of the quarter-circle. Let O be the origin. For a given half-disk D , label the vertex on the y -axis A and the vertex on y the x -axis B . Let M be the midpoint of line segment AB . Draw segment OM , and extend it until it intersects the curved boundary of D . Label the intersection point C . This construction is shown in y the diagram below. 2 C A M D y O 2 B We first prove that C lies on the quarter-circle, centered at the origin, with radius 2. Since M is the midpoint of AB , and A is on the y -axis, M is horizontally halfway between B and the y -axis. Since General Test, Part 2 O and B are on the x -axis (which is perpendicular to the y -axis), segments OM and M B have the same length. Since M is the midpoint of AB , and AB = 2, OM = 1. Since D is a half-disk with y radius 1, all points on its curved boundary are 1 away from its center, M . Then C is 2 away from the origin, and the quarter-circle consists of all points which are 2 away from the origin. Thus, C is an intersection of the half-disk D with the positive quarter-circle of radius 2. y It remains to show that the half-disk D is internally tangent to the quarter-circle. Since OC is a y radius of the quarter-circle, it is perpendicular to the tangent of the quarter-circle at C . Since M C is a radius of the half-disk, it is perpendicular to the tangent of the half-disk at C . Then the tangents lines of the half-disk and the quarter-circle coincide, and the half-disk is tangent to the quarter-circle. It is obvious from the diagram that the half-disk lies at least partially inside of the quarter-circle, the half-disk D is internally tangent to the quarter-circle. y Then the union of the half-disks is be a quarter-circle with radius 2, and has area π .