1975 AMC 12 Problem 28
Attempt Problem 28 of the 1975 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 1975 AMC 12 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
28.
In triangle shown in the adjoining figure, is the midpoint of side and Points and are taken on and respectively, and lines and intersect at If then equals
not enough information given to solve the problem
Answer: A
Small hint:
Represent by vectors from
Big hint:
If then equate the coefficients of a point on and on
Solution:
Put at the origin and write the position vectors of as If then and so Write Because also lies on the median its and coefficients are equal: Hence Along and so
Therefore, the correct answer is A.
Problem 28 in other years
- 1950 AMC 12
- 1951 AMC 12
- 1952 AMC 12
- 1953 AMC 12
- 1954 AMC 12
- 1955 AMC 12
- 1956 AMC 12
- 1957 AMC 12
- 1958 AMC 12
- 1959 AMC 12
- 1960 AMC 12
- 1961 AMC 12
- 1962 AMC 12
- 1963 AMC 12
- 1964 AMC 12
- 1965 AMC 12
- 1966 AMC 12
- 1967 AMC 12
- 1968 AMC 12
- 1969 AMC 12
- 1970 AMC 12
- 1971 AMC 12
- 1972 AMC 12
- 1973 AMC 12
- 1974 AMC 12
- 1976 AMC 12
- 1977 AMC 12
- 1978 AMC 12
- 1979 AMC 12
- 1980 AMC 12
- 1981 AMC 12
- 1982 AMC 12
- 1983 AMC 12
- 1984 AMC 12
- 1985 AMC 12
- 1986 AMC 12
- 1987 AMC 12
- 1988 AMC 12
- 1989 AMC 12
- 1990 AMC 12
- 1991 AMC 12
- 1992 AMC 12
- 1993 AMC 12
- 1994 AMC 12
- 1995 AMC 12
- 1996 AMC 12
- 1997 AMC 12
- 1998 AMC 12
- 1999 AMC 12