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Geometry Difficulty 5.2 AIME, harder Find the answer

Let RR be the region in the Cartesian plane of points (x,y)(x, y) satisfying x0,y0x \geq 0, y \geq 0, and x+y+x+y5x+y+\lfloor x\rfloor+\lfloor y\rfloor \leq 5. Determine the area of RR.

A number or a short expression. Spacing and $ signs are ignored.

Solution

We claim that a point in the first quadrant satisfies the desired property if the point is below the line x+y=3x+y=3 and does not satisfy the desired property if it is above the line. To see this, for a point inside the region, x+y<3x+y<3 and x+yx+y<3\lfloor x\rfloor+\lfloor y\rfloor \leq x+y<3 However, x+y\lfloor x\rfloor+\lfloor y\rfloor must equal to an integer. Thus, x+y2\lfloor x\rfloor+\lfloor y\rfloor \leq 2. Adding these two equations, x+y+x+y<5x+y+\lfloor x\rfloor+\lfloor y\rfloor<5, which satisfies the desired property. Conversely, for a point outside the region, x+y+{x}+{y}=x+y>3\lfloor x\rfloor+\lfloor y\rfloor+\{x\}+\{y\}=x+y>3 However, {x}+{y}<2\{x\}+\{y\}<2. Thus, x+y>1\lfloor x\rfloor+\lfloor y\rfloor>1, so x+y2\lfloor x\rfloor+\lfloor y\rfloor \geq 2, implying that x+y+x+y>5x+y+\lfloor x\rfloor+\lfloor y\rfloor>5. To finish, RR is the region bounded by the x -axis, the y -axis, and the line x+y=3x+y=3 is a right triangle whose legs have length 3. Consequently, RR has area 92\frac{9}{2}.

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.