Difficulty: Hard
Correct Answer: 535
Explanation:
Introduction / Context:
This problem combines general counting of intersections with corrections for special configurations where many lines meet at the same point. Normally, any two non parallel lines intersect in exactly one point, but here several lines share point A and point B. We must correctly count distinct intersection points while avoiding overcounting the intersections that occur at A and B. This is a common type of advanced combinatorics question in geometry.
Given Data / Assumptions:
Total number of lines in the plane = 37.
13 of these lines pass through point A.
11 of these lines pass through point B.
No line passes through both A and B.
No two lines are parallel.
Except for the lines through A or B, no three lines pass through a single point.
Concept / Approach:
If no special configurations existed, any two lines would intersect at a unique point, and the total number of intersection points would be 37C2. However, at point A, the 13 lines intersect all together, so the naive count 37C2 includes many repeated counts of the same geometric point A. Similarly at point B. We first compute 37C2 and then subtract the extra counts at A and B. At point A, the number of pairs of lines passing through A is 13C2, all meeting at the single point A. This should contribute just 1 distinct intersection, not 13C2. The same logic holds for point B with 11C2 pairs.
Step-by-Step Solution:
Step 1: Compute the total number of pairwise intersections if all lines were in general position.
Step 2: Total naive intersections = 37C2 = 37 * 36 / 2 = 666.
Step 3: At point A, there are 13 lines, so the number of pairs among them is 13C2 = 13 * 12 / 2 = 78.
Step 4: All these 78 pairs meet at the same point A, but we want to count that single point only once.
Step 5: Hence, the overcount at A is 78 - 1 because we should have counted only one intersection point there.
Step 6: Overcount at A = 78 - 1 = 77.
Step 7: Similarly, at point B there are 11 lines, so pairs = 11C2 = 11 * 10 / 2 = 55.
Step 8: All these 55 pairs meet at the same point B, but this should be counted as a single intersection point.
Step 9: Overcount at B = 55 - 1 = 54.
Step 10: Total overcount = 77 + 54 = 131.
Step 11: Required number of distinct intersection points = 666 - 131 = 535.
Verification / Alternative check:
To cross-check, imagine replacing the clusters of lines at A and B with equivalent configurations where each pair intersection is distinct except one. For 13 lines at A, if they were in general position, we would have 78 distinct intersection points. But when they all pass through A, these 78 points collapse into one. This is exactly a loss of 77 distinct intersection points. Similarly, for 11 lines at B, 55 potential intersection points collapse into one, a loss of 54. Subtracting these losses from the naive count 666 gives 535. This conceptual view matches the detailed step-by-step subtraction and confirms the correctness.
Why Other Options Are Wrong:
The value 525 suggests subtracting an incorrect amount of overcount. Likewise 545 and 555 correspond to different incorrect corrections that do not match the actual multiplicities of intersections at A and B. If someone forgets to adjust for one of the special points or miscomputes 13C2 or 11C2, they may get one of these values. Only 535 correctly reflects the elimination of repeated counting at both special points A and B.
Common Pitfalls:
The most common error is to compute only 37C2 and stop, forgetting that many line pairs meet at exactly the same points A and B. Another pitfall is to subtract 13C2 and 11C2 without adding back 1 for each common point, effectively removing those intersections entirely instead of collapsing them into one. Some students also confuse conditions like no three lines concurrent and no two parallel. Careful reading and a clear understanding of overcounting corrections are essential.
Final Answer:
The total number of distinct intersection points determined by the 37 lines is 535.
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