Problem 3 - Entrance Test

Let's set up a coordinate system. Let A = (0,0), B = (1,0), C = (1,1), D = (0,1). Since M is the midpoint of AB, M = ( (0+1)/2, (0+0)/2 ) = (1/2, 0). Since N is the midpoint of BC, N = ( (1+1)/2, (0+1)/2 ) = (1, 1/2). Next, we find the equations of lines CM and DN. Line CM passes through C(1,1) and M(1/2,0). Slope of CM = (1-0) / (1-1/2) = 1 / (1/2) = 2. Equation of CM: y - 0 = 2(x - 1/2) => y = 2x - 1. Line DN passes through D(0,1) and N(1,1/2). Slope of DN = (1/2-1) / (1-0) = (-1/2) / 1 = -1/2. Equation of DN: y - 1 = -1/2 (x - 0) => y = -1/2 x + 1. Point P is the intersection of CM and DN. We set the y-values equal: 2x - 1 = -1/2 x + 1 2x + 1/2 x = 1 + 1 5/2 x = 2 x = 4/5. Now find the y-coordinate of P using either equation: y = 2(4/5) - 1 = 8/5 - 5/5 = 3/5. So, the coordinates of point P are (4/5, 3/5). We need to find the area of triangle APD. The vertices are A(0,0), P(4/5, 3/5), D(0,1). We can choose AD as the base of the triangle. The length of base AD is the distance between (0,0) and (0,1), which is 1. Since AD lies along the y-axis, the height of the triangle with respect to this base is the perpendicular distance from P to the y-axis, which is the x-coordinate of P. Height = 4/5. Area of triangle APD = 1/2 * base * height = 1/2 * 1 * (4/5) = 2/5.