AMC 10 · 2013 · #21

The ceiling is 2C(30, 2) = 870, choice (E) — what you get by assuming every pair meets twice — and 810 sits just under it, short by 60 = 2 · 30, a small and checkable deficit. Spot-check both cases by hand with a = 0. Directrix y = 1 gives y = (1-x²)/2 and directrix y = 2 gives y = 1 - x²/4; setting them equal gives x² = -2, no real point, matching the nested case. Directrix y = -1 gives y = (x²-1)/2, and (1-x²)/2 = (x²-1)/2 gives x = ± 1, y = 0 — exactly two points, even though those two directrices are parallel. That single example shows why the split has to be by opening direction, not by parallel directrices. The distractors sit exactly on the plausible slips: (A) 720 = 2(435-75) discards all 5C(6, 2) = 75 parallel-directrix pairs, forgetting the opposite-side ones do meet; (D) 840 = 2(435-15) keeps only the b ＞ 0 half of the nested pairs; (E) 870 forgets the nested pairs entirely. One honest caveat: the 'no three concurrent' clause is load-bearing and is handed to you rather than proved. A direct check shows it is not literally true of this configuration — for instance (1/2,1/2) is at distance 1/√(2) from the focus and from each of the three directrices y = x+1, y = x-1, y = -x+2, so three of the parabolas pass through it. Counting the way the problem instructs, one fresh point per good pair, gives 810.