2010 AMC 12A Problems/Problem 25
Contents
Problem
Two quadrilaterals are considered the same if one can be obtained from the other by a rotation and a translation. How many different convex cyclic quadrilaterals are there with integer sides and perimeter equal to 32?
Solution 1
It should first be noted that given any quadrilateral of fixed side lengths, there is exactly one way to manipulate the angles so that the quadrilateral becomes cyclic.
Proof. Given a quadrilateral where all sides are fixed, we can construct the diagonal
. When
is the minimum allowed by the triangle inequality, one of the angles
or
will be degenerate and measure
, so opposite angles will sum to less than
. When
is the maximum allowed, one of the angles will be degenerate and measure
, so opposite angles will sum to more than
. Thus, since the sum of opposite angles increases continuously as
is lengthened from the minimum to the maximum values, there is a unique value of
somewhere in the middle such that the sum of opposite angles is exactly
.
Denote ,
,
, and
as the integer side lengths of the quadrilateral. Without loss of generality, let
.
Since , the Triangle Inequality implies that
.
We will now split into cases.
Case :
(
side lengths are equal)
Clearly there is only way to select the side lengths
, and no matter how the sides are rearranged only
unique quadrilateral can be formed.
Case :
or
(
side lengths are equal)
If side lengths are equal, then each of those side lengths can only be integers from
to
except for
(because that is counted in the first case). Obviously there is still only
unique quadrilateral that can be formed from one set of side lengths, resulting in a total of
quadrilaterals.
Case :
(
pairs of side lengths are equal)
and
can be any integer from
to
, and likewise
and
can be any integer from
to
. However, a single set of side lengths can form
different cyclic quadrilaterals (a rectangle and a kite), so the total number of quadrilaterals for this case is
.
Case :
or
or
(
side lengths are equal)
If the equal side lengths are each
, then the other
sides must each be
, which we have already counted in an earlier case. If the equal side lengths are each
, there is
possible set of side lengths. Likewise, for side lengths of
there are
sets. Continuing this pattern, we find a total of
sets of side lengths. (Be VERY careful when adding up the total for this case!) For each set of side lengths, there are
possible quadrilaterals that can be formed, so the total number of quadrilaterals for this case is
.
Case :
(no side lengths are equal)
Using the same counting principles starting from
and eventually reaching
, we find that the total number of possible side lengths is
. There are
ways to arrange the
side lengths, but there is only
unique quadrilateral for
rotations, so the number of quadrilaterals for each set of side lengths is
. The total number of quadrilaterals is
.
And so, the total number of quadrilaterals that can be made is .
Solution 2
As with solution we would like to note that given any quadrilateral we can change its angles to make a cyclic one.
Let be the sides of the quadrilateral.
There are ways to partition
. However, some of these will not be quadrilaterals since they would have one side bigger than the sum of the other three. This occurs when
. For
,
. There are
ways to partition
. Since
could be any of the four sides, we have counted
degenerate quadrilaterals. Similarly, there are
,
for other values of
. Thus, there are
non-degenerate partitions of
by the hockey stick theorem. However, for
or
, each quadrilateral is counted
times,
for each rotation. Also, for
, each quadrilateral is counted twice. Since there is
quadrilateral for which
, and
for which
, there are
quads for which
or
. Thus there are
total quadrilaterals.
See also
2010 AMC 12A (Problems • Answer Key • Resources) | |
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