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A triangle ABC is formed by the lines x - 3y - 6 = 0, 3x - y + 6 = 0, and 3x + 4y - 24 = 0. Points P(alpha, 0) and Q(0, beta) always lie on or inside triangle ABC. If the range of alpha is [a, b] and the range of beta is [c, d], then which of the following is correct?

1. a + b = 4
2. c + d = 1
3. a + b = 2
4. c + d = 4

Correct answer: c + d = 4

Solution

Finding vertices: L1: x-3y-6=0, L2: 3x-y+6=0, L3: 3x+4y-24=0. V1 (L1 and L2): x-3y=6 and 3x-y=-6. From first: x=6+3y. Sub: 3(6+3y)-y=-6 => 18+9y-y=-6 => 8y=-24 => y=-3, x=6-9=-3. V1=(-3,-3). V2 (L1 and L3): x-3y=6 and 3x+4y=24. From first: x=6+3y. Sub: 3(6+3y)+4y=24 => 18+9y+4y=24 => 13y=6 => y=6/13, x=6+18/13=96/13. V2=(96/13, 6/13). V3 (L2 and L3): 3x-y=-6 and 3x+4y=24. Subtract: -5y=-30 => y=6, x=(-6+6)/3=0. V3=(0,6). For P(alpha,0) on x-axis inside triangle: substitute y=0 into constraints. L1: alpha-6>=0? No, we need alpha to be on correct side. Checking each line's sign for interior: take centroid = ((-3+96/13+0)/3, (-3+6/13+6)/3) = ((−39+96)/13/3, (39+6)/13/3) approx (57/39, 48/39) = (57/39, 48/39). For L1 at centroid: 57/39 - 3*48/39 - 6 = (57-144)/39 - 6 = -87/39 - 6 < 0. So interior satisfies x-3y-6 < 0. For P(alpha,0): alpha-6 < 0 => alpha < 6. For L2 at centroid: 3*57/39 - 48/39 + 6 = (171-48)/39+6 = 123/39+6 > 0. For P(alpha,0): 3*alpha+6 > 0 => alpha > -2. For L3 at centroid: 3*57/39 + 4*48/39-24 = (171+192)/39-24 = 363/39-24 < 0 (363/39~9.3 < 24). For P(alpha,0): 3*alpha-24 < 0 => alpha < 8. Combined: alpha in (-2, 6). But also P must be on or inside, so range of alpha is [-2, 6]. Thus a+b = -2+6 = 4. For Q(0,beta): L1: -3*beta-6 < 0 => beta > -2. L2: -beta+6 > 0 => beta < 6. L3: 4*beta-24 < 0 => beta < 6. Also need to check vertex constraints more carefully. At x=0 on L2: y=6; on L3: y=6. Range: beta in [-2,6]? No — at x=0, y=-2 gives L1: -(-6)-6=0 (on L1). So beta ranges from -2 to 6. But wait: V3=(0,6) is a vertex at x=0. At x=0, the triangle boundary: V1=(-3,-3), V3=(0,6), V2=(96/13,6/13). The y-axis enters the triangle. At x=0 and y=0: check if (0,0) is inside: L1: -6<0 yes, L2: 6>0 yes, L3: -24<0 yes. So (0,0) is inside. Q(0,beta) ranges: lower bound where y-axis crosses a triangle edge. L1 at x=0: -3y-6=0 => y=-2. L2 at x=0: -y+6=0 => y=6 (vertex V3). Upper bound is 6, lower bound is -2. So c=-2, d=6 => c+d=4.

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