choose the point in the feasible region farthest from point Either one, when substituting into the function, yields
19.
, which is .
.
Since
is a positive integer,
.
must be in the form of
for
some positive integer
The two smallest possible value of respectively.
Sum of the two smallest possible value of
where is a positive integers are and
20. From , we know that . From the first inequality:
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Since
must be an integer, it follows that
. Similarly, from the
second inequality:
And it follows that it gives us
And since
, which is
.
, we find that
. We now have a system of three equations. Solving . From this, we find that
21. The domain of is defined when .
. Applying the domain of
that square roots must be positive, we get at the domain for
, which is defined when
and the fact
. Simplify this to arrive . Repeat this process for
12
to get a domain of . For , since
square roots are positive, we can exclude the negative values of the previous domain to arrive at
as the domain of
. We now arrive at a
domain with a single number that defines , however, since we are looking for the largest value for for which the domain of
is nonempty, we must continue until
to get a domain of
we arrive at a domain that is empty. We continue with
. Solve for to get
. Since square roots cannot be negative,
.
this is the last nonempty domain. We add to get
22. First, notice that there must be four rays emanating from
that intersect the , the number of
four corners of the square region. Depending on the location of the corner-most point that is
rays distributed among these four triangular sectors will vary. We start by finding
-ray partitional (let this point be the
bottom-left-most point). We first draw the four rays that intersect the vertices. At this point, the triangular sectors with bases as the sides of the square that the point is closest to both do not have rays dividing their areas. Therefore, their heights are equivalent since their areas are equal. The remaining other two triangular sectors, each sector with sectors into
rays are divided among the
rays, thus dividing these two
triangles of equal areas. Let the distance from this corner point to the
closest side be and the side of the square be . From this, we get the equation
. Solve for to get
. Therefore, point
is
of
the side length away from the two sides it is closest to. By moving another
-ray partitional point. We can continue moving
to the right,
we also move one ray from the right sector to the left sector, which determines
right and up to derive
the set of points that are points each
-ray partitional. In the end, we get a square grid of
apart from one another. Since this grid ranges from a distance of
from the same side, we have a
grid, a total of
-ray partitional, we must find
from one side to
-ray partitional points. To find the overlap from the the distance from the corner-most Since the
-ray partitional point to the sides closest to it.
grid, each point
apart from
grid,
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-ray partitional points form a
each other, we can deduce that the -ray partitional points form a
each point apart from each other. To find the overlap points, we must find the
and
which are
and
. Therefore, the overlapping
common divisors of
points will form grids with points , , , and Since the grid with points
away from each other respectively.
away from each other includes the other points, we can
grid, which .
disregard the other grids. The total overlapping set of points is a has
23. Answer: (C)
points. Subtract from to get
Lemma) if , then
The tedious algebra is left to the reader. (it is not bad at all) Well, let us consider the cases where each of those step is definite (evaluate).
is never
So, we have
,
--- (exception -> case 2)
--- (exception -> case 3) --- (exception -> case 4)
--- (exception -> case 5)
If it is not any of the above 5 cases, then
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if
(--- exception -> case 6), then , ,
Hence, it is possible maximum of and minimum is 1.
2 possible combination of are and . Verification is left upto the
reader. Right now, (C) is the most possible answer out of those 5. Case 2) Case 3) Case 5)
, then
= 1, which is in the range.
, hence
Case 6) Since , ,
Case 4) , this is quite an annoying special case. In this case, ,
is not define.
In this case, and Hence,
and
.
. Once,
you work out this system, you will get no solution with Thus, answer is (C).
24. Answer:
Given, a 14-9-7-12 quadrilateral ( which has an in-circle). Find the largest possible in-radius.
Solution:
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