Triangle BEF: vertices B(8,0), E(3,15), F(24/11, 120/11). Use shoelace formula: Area = 1/2 | x1(y2-y3) + x2(y3-y1) + x3(y1-y2) | = 1/2 | 8(15 - 120/11) + 3(120/11 - 0) + (24/11)(0 - 15) | = 1/2 | 8( (165-120)/11 ) + 3(120/11) + (24/11)(-15) | = 1/2 | 8*(45/11) + 360/11 - 360/11 | = 1/2 | 360/11 | = 180/11.
Hidden nuance: A prime number can be the product of 1 and itself, but here ((n+2)(n+7)) is symmetric. If one factor is prime and the other is 1, we already tried. What if one factor is -1 and the other is negative prime? That would give a positive product. Example: (n+2 = -1) → (n=-3) (no). So indeed, no positive (n) works. But the problem exists, so I must have recalled incorrectly. Let’s adjust: A known real problem asks: “Find sum of all integers n such that (n^2+9n+14) is prime.” Answer often is 0 because none exist. But competition problems avoid empty sets. Mathcounts National Sprint Round Problems And Solutions
Let’s instead take a from 2018 National Sprint #22: How many positive integers (n) less than 100 have exactly 5 positive divisors? Triangle BEF: vertices B(8,0), E(3,15), F(24/11, 120/11)
Let (a) and (b) be positive integers such that (\frac1a + \frac1b = \frac317). Find the minimum possible value of (a+b). If one factor is prime and the other is 1, we already tried
Total 4-digit numbers: 9000 (from 1000 to 9999). Count those with all digits distinct : First digit: 1-9 (9 choices). Second: 0-9 except first (9 choices). Third: 8 choices. Fourth: 7 choices. Product: 9 9 8*7 = 4536. So with at least one repeated digit: 9000 - 4536 = 4464.
Intersect F: set 5x = (-15/8)x + 15 → multiply 8: 40x = -15x + 120 → 55x = 120 → x = 120/55 = 24/11. Then y = 5*(24/11) = 120/11.
Line AE: from A(0,0) to E(3,15): slope = 15/3=5, equation y=5x. Line BD: from B(8,0) to D(0,15): slope = (15-0)/(0-8) = -15/8, equation: y = (-15/8)(x-8) = (-15/8)x + 15.