2000 STATE HIGH SCHOOL MATHEMATICS CONTEST

SOLUTIONS

(D)

Since the lengths of all of the sides of

are one-half the lengths of the sides of

, the area of

is one-fourth the area of

, making the area of quadrilateral MNRQ three-fourths the are of the original triangle.

2.	(D)	Graphically we see that the function g(x) is a typical absolute value graph, shifted to the right so that the minimum point is while f(x) is a line passing through the origin. The slope of the right-hand side of the absolute value function is 1, so it is clear that the linear function will intersect the absolute value function twice in and only if k is between 0 and 1.
Algebraically, if . If . The first implication forces and the second forces or . Since we want two points of intersection, both must be true at the same time, making .

(B)

For any p we have

. Since p is prime, its only factors are 1 and p, so

and

. Solving for y and x we get

and

. Thus, for any given p there is only one ordered pair (x,y).

4.	(C)	Since the coordinates of point T are , we know that the length of segment OT is Segment OF, being a tangent to the circle also has this length. Let the length of the radius of the circle be r, so by the Law of Cosines we have . We also know that . Combining and substituting the values we have gives us	.
		. Simplifying we get . This implies that , so . Of course, if you happen to know the tangent of or the half angle formula for the tangent of , you could use this and the length of OT to immediately determine the length of side TC.

(A)

There are a total of

possible outcomes to this experiment. Of these there are

in which each face occurs only once. Thus the probability is

, making

(D)

Solve for r in the first equation and substitute into the second, yielding

. This simplifies to

and the only real solution to this is

, making

. Thus

(B)

. Together these yield

(B)

Since the 100% does nothing to change the size, we must use some combination of the others. Hence

. The smallest possible value for m is 2 and fortunately

gives us the right power of 3 and cancels the 4's.

(D)

. Similarly with the subset of M numbers

. Subtracting we get the other subset of M-N numbers with mean

10.	(D)	In the figure, since QX bisects the angle at Q. Since XY was constructed parallel to PQ, . Now the exterior angle XYR to has measure equal to the sum of the two interior angles and . Hence both and are congruent to those indicated in the figure. We now know that .
		We also know that angle-bisectors cut the opposite side of a triangle into segments with lengths proportional to the lengths of the adjacent sides, so we know that . Thus n must divide .

11.

(B)

Your first impulse, like mine, might be to complete the square to get the expression

somewhere in the problem. While this works, it is easier to substitute in two values for x and force them to be equal, since the function is to be constant. Any two values will work, but we want to use ones for which we know the sine and cosine. All multiples of

result in 1 + K, so we need to put in one value that is not a multiple of

. Try

. This gives us

. So now we know that

, so

. This does in fact give us the function

12.

(D)

The diagonals of a rhombus are perpendicular-bisectors of each other. Using this we can call the longer diagonal L and the shorter S. The area of a rhombus is half the product of the diagonals, so

. We also know that

. Solving this system of equations yields

. There are several other ways to work this problem, and each method gives the same answer in a different form. Equivalent answers are

and

13.

(A)

In n had Property T, the sum of the n subsets would just be the sum of the integers from 1 to 3n, or

. This in turn would equal

, so

. This means that 3 must divide m, but then that would force 9 to divide 3(3n+1), which is impossible.

14.

(C)

If x is the length of the two congruent sides and y the length of the third side of the triangle, the perimeter is

and the area is

. Solving for y in the first equation and substituting into the second we get

. Since we are not interested in the exact solutions, only the number of real solutions to this equation, a graph of the function

shows that there are two real solutions in the possible range from zero to eight.

15.

(E)

The grid shows the "distance" between each pair of 5-tuples. The only 5-tuple with the correct distances is E. Distance here means the sum of the absolute value of the difference of each coordinate from in one 5-tuple from another.

16.

(B)

The amounts that Alice (or Bob) can win are 1, 2, 4, 8, 16, 32, 64, and 128 dollars. In Alice won all of the games, she would have $255. Let A be the amount Alice won and B the amount Bob won. Clearly

and we are told that

. Solving this system of equations, we see that Alice actually won $150 while Bob only won $105. Alice had to win the last game, since the sum of all the others only totals $127. Similarly, Bob had to win the $64 game since the total of the first five is only $63. At this point we know that Alice only has to win $22 more and Bob $41. So Alice cannot win game 6 with $32, Bob must. Now Alice still has $22 to win from 1,2,4,8,16, so she must with game 5 for $16. Then Bob must win $9 more, so he must with the 4^th game form $8. Now Alice must win $6 more, so she must win game 2 and 3 for $2 and $4. So Alice wins games 2, 3, 5, and 8.

17.

(C)

First we see the

, and

. It certainly appears that there is a pattern here and it looks like

. You would probably assume this was the case on a timed test, but an induction argument proves that this is in fact the case. First

. Next assume

, but recall, that by definition

. Now we need to show that

. By definition,

, so

. By the induction hypothesis,

, and

. We see now that

. Now we know that

. Counting, we see that there are 20 twos.

18.	(A)	Since the two tangent lines are parallel, we can use alternate interior angles to show that . From this we see that . Segment EB will be perpendicular to AE, since angle AEB is inscribed in a semicircle. From the figure we see that , , and . Subtracting equation (2)from equation (3), and substituting, we get .
		Now simplify this last equation to get . This last equation, together with equation (1) forms a system of equation which, when solved simultaneously, give .

19.

(B)

Suppose you are looking at the tetrahedron ABCD from a point directly above the vertex D and that all edges and medians shown are projected perpendicularly down to the plane containing the points A, B, and C. Since

, we know that

, and that

. Likewise, we know that

. Triangle GDH is isosceles with congruent sides of length

and included angle measuring 120 degrees. From this we find that GH has length

20.

(E)

Since F is on the perpendicular-bisector of segment AE, we know that |AF| = |EF|, making triangle AFE isosceles and angles FAC and FEC congruent. First in

we can use the Law of Sines to get

. In

, again using the Law of Sines, we have

. If we let the sides of the square be 1 unit, then |CE| = 1 and

, so we can combine the two Law of Sines equations into

. From this we see that

, so

and