Furman University Wylie Mathematics Tournament/1996 Senior Exam/Solutions
Contents
[hide]- 1 Problem 1
- 2 Problem 2
- 3 Problem 3
- 4 Problem 4
- 5 Problem 5
- 6 Problem 6
- 7 Problem 7
- 8 Problem 8
- 9 Problem 9
- 10 Problem 10
- 11 Problem 11
- 12 Problem 12
- 13 Problem 13
- 14 Problem 14
- 15 Problem 15
- 16 Problem 16
- 17 Problem 17
- 18 Problem 18
- 19 Problem 19
- 20 Problem 20
- 21 Problem 21
- 22 Problem 22
- 23 Problem 23
- 24 Problem 24
- 25 Problem 25
- 26 Problem 26
- 27 Problem 27
- 28 Problem 28
- 29 Problem 29
- 30 Problem 30
- 31 Problem 31
- 32 Problem 32
- 33 Problem 33
- 34 Problem 34
- 35 Problem 35
- 36 Problem 36
- 37 Problem 37
- 38 Problem 38
- 39 Problem 39
- 40 Problem 40
- 41 Tiebreaker 1
- 42 Tiebreaker 2
- 43 See also
Problem 1
Using the rational root theorem, we see that is a root. We use synthetic division to factor the equation as:
The roots of the second factor are
and
The answer is 6.
Problem 2
Multiplying the denominator by gives us that the expression is
the coefficient is thus 3.
Problem 3
With the first cut, we must remove a square leaving us with an
rectangle. We now remove a
square leaving us with a
rectangle. We now remove a
square and are left with a
rectangle. We now remove a
square and are left with a
rectangle. From this, we remove two
squares. We have 6 total squares.
Problem 4
Problem 5
We have that So therefore
Problem 6
We have that The closest to
is 87.
Problem 7
The area of the squares is Now after using Heron’s Formula, we have that the area of the triangle is 24. Thus, the total area is 434.
Problem 8
Using the change of base formula, we have and
Also, notice that we are looking for the value of
and
From this we get the value we are looking for as
Problem 9
We use the Principle of Inclusion Exclusion. There are We must now the subtract the number of numbers divisible by their LCM which is 36 which is
The answer is 138-27=111.
Problem 10
There are squares of length 1. There are 8 squares of length 2, and there are 3 squares of length 3. The answer is 21+8+3=32.
Problem 11
There are arrangements.
Problem 12
We have:
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Problem 13
The roots of are 1 and -2. These are the values of
we can use on
So we have
and
From these, we have that $a-b=3$ and <math> c=2$ (Error compiling LaTeX. Unknown error_msg) and the answer is 5.
Problem 14
There are three possible cases to get a score of 5.
Case 1: We have 1+2+2. There are three flips of the coin and we must have one head and two tails. Thus the probability is
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Case 2: We have 1+1+1+2. This has probability of
Case 3: We have 1+1+1+1+1 which has a probability of
The total probability is
Problem 15
The area of the intersection is the area of a square of sidelength 1 that is INSIDE the circle. After drawing in the radii of the circle, we see that this area is two 30-60-90 triangles (with one of the legs of length 1 and the hypotenuse as the radius). The total area is and
Problem 16
Problem 17
We use trial and error for each of the gons by splitting them into triangles and using the law of cosines to find out the sidelength of each of the sides. 7 is the smallest
that works.
Problem 18
Note that
and 11 is our answer.
Problem 19
There are total subsets that can be made. Now, we can find out the total number of subjects that can be made that DO NOT contain any elements of
There are
possible subsets. The number of subsets with atleast one element of
is therefore 1024-128=896.
Problem 20
With triangle inequality, we have that Also, using the fact that the triangle is acute,
So we have
which gives us 23 integer values for
Problem 21
Let The equation becomes
and the sum of the solutions of
is 4/3.
Problem 22
The perfect cubes are 1,8,27,64,125. Thus, our sum is 7(1)+19(2)+37(3)+61(4)=7+38+111+244=400. Our answer is thus (e).
Problem 23
We can expand the expression as such Since we are looking for the coefficient of
and
we basically need to look at
(after using the binomial theorem and looking at different possibilities). Using the binomial theorem, the coefficient we have is
and the coefficient is
The answer is 6.
Problem 24
Problem 25
Using the law of cosines, we have Now, using Heron’s Formula, we see that the area of the triangle is 204, so
The result follows.
Problem 26
which is at an interval of length 2. But because a parabola contains two roots for each value of
the interval length is 1.
Problem 27
We have (keeping in mind of the signs of cosine and sine. From there, we can get our answer by adding
to each of these base angles.
Problem 28
The line that goes through the two opposite vertices of the hexagon (in which the centers of the two circles are collinear to that line) is of length 2 (from equilateral triangles). Let the radius of each of the circles be Drawing
to the tangents of the circle, we form a 30-60-90 triangle and the line of length 2 can be written as
Thus
and the answer is -1.
Problem 29
We can simplify the inequality to The length of this interval is 8 and thus 40% of the given interval.
Problem 30
Problem 31
Problem 32
Trial and error shows that works.
Problem 33
The area of the triangle (after using Heron’s Formula) is 84. It is obvious that the minimum area of the rectangle is when the altitude and the base of the triangle are the sidelengths of the rectangle. Thus the minimum area is
Problem 34
Problem 35
Notice that and are the altitudes of the equilateral triangles. Thus,
Now,
is equilateral (using similar triangles), so
. It follows that
and our answer is 4.
Problem 36
Problem 37
Using similar triangles, we see that and thus both of the triangles are 30-60-90. The area that we are looking for is
. The answer is 5.
Problem 38
Problem 39
Shoelace Theorem gives us the area as 5/8, thus the answer is 13.
Problem 40
Use law of cosines to find out the cosine of the angle . Using Heron's formula, the area of the triangle is 84. Now use sine area to find out the sine of
. The result follows.