2019 AIME II Problems/Problem 10
Contents
Problem 10
There is a unique angle between
and
such that for nonnegative integers
, the value of
is positive when
is a multiple of
, and negative otherwise. The degree measure of
is
, where
and
are relatively prime integers. Find
.
Solution 1
Note that if is positive, then
is in the first or third quadrant, so
. Also notice that the only way
can be positive for all
that are multiples of
is when
, etc. are all the same value
. This happens if
, so
. Therefore, the only possible values of theta between
and
are
,
, and
. However
does not work since
is positive, and
does not work because
is positive. Thus,
.
.
Solution 2
As in the previous solution, we note that is positive when
is in the first or third quadrant. In order for
to be positive for all
divisible by
, we must have
,
,
, etc to lie in the first or second quadrants. We already know that
. We can keep track of the range of
for each
by considering the portion in the desired quadrants, which gives
at which point we realize a pattern emerging. Specifically, the intervals repeat every
after
. We can use these repeating intervals to determine the desired value of
since the upper and lower bounds will converge to such a value (since it is unique, as indicated in the problem). Let's keep track of the lower bound.
Initially, the lower bound is (at
), then increases to
at
. This then becomes
at
,
at
,
at
,
at
. Due to the observed pattern of the intervals, the lower bound follows a partial geometric series. Hence, as
approaches infinity, the lower bound converges to
-ktong
See Also
2019 AIME II (Problems • Answer Key • Resources) | ||
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Followed by Problem 11 | |
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