1956 AMC 12 Problem 21

Attempt Problem 21 of the 1956 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1956 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

21.

If each of two intersecting lines intersects a hyperbola and neither line is tangent to the hyperbola, then the possible number of points of intersection with the hyperbola is:
22
22 or 33
22 or 44
33 or 44
2,2, 3,3, or 44
Answer: E
Concepts:hyperbolaline-conic intersectionscounting distinct points
Small hint:
A non-tangent line that intersects a hyperbola can meet it in either one or two points
Big hint:
The two lines may share one hyperbola point, since the lines themselves intersect
Solution:
A line can meet a hyperbola in either one finite point, when its direction is parallel to an asymptote, or two finite points. By choosing the two lines so that their finite intersection sets on the hyperbola are disjoint, they can therefore contribute 1+1=2,1+1=2, 1+2=3,1+2=3, or 2+2=42+2=4 distinct intersections. If their own intersection lies on the hyperbola, one point is shared instead.
Thus, 2,2, 3,3, or 44 are possible, and the correct answer is E.
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