1995 AHSME Problems/Problem 4
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Contents
Problem
If is
of
,
is
of
, and
is
of
, then
Solution 1
We are given: ,
,
. We want M in terms of N, so we substitute N into everything:
Solution 2
Altenrately, picking an arbitrarly value for of
, we find that
.
We find that , meaning
, giving
.
Finally, since is
of
, we have
.
Thus, and
, so their ratio
This method does not prove that the answer must be constant, but it proves that if the answer is a constant, it must be .
See also
1995 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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All AHSME Problems and Solutions |
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