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:

15