2017 AMC 10A Problems/Problem 10
Contents
[hide]Problem
Joy has thin rods, one each of every integer length from
cm through
cm. She places the rods with lengths
cm,
cm, and
cm on a table. She then wants to choose a fourth rod that she can put with these three to form a quadrilateral with positive area. How many of the remaining rods can she choose as the fourth rod?
Solution 1
The triangle inequality generalizes to all polygons, so and
yields
. Now, we know that there are
numbers between
and
exclusive, but we must subtract
to account for the 2 lengths already used that are between those numbers, which gives
Solution 2
The quadrilateral inequality states that , where
and
and the side lengths, and
is the largest one. Now, we can just plug these values in. We have
cases. First if this other side length (we will call it
) is the largest side length we have:
, so there are
, so
values. Now, we have case
which is if
is the largest side length. We can say that
, so we have the lower bound to be
and the upper bound to be
, so we have
values for this case. When adding these
cases we get a total of
Video Solution
~savannahsolver
See Also
2017 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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All AMC 10 Problems and Solutions |
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