Furman University Wylie Mathematics Tournament/1996 Senior Exam/Solutions

Problem 1

Using the rational root theorem, we see that $x=2$ is a root. We use synthetic division to factor the equation as: $(x-2)(3x^{2}+x-10).$ The roots of the second factor are $\frac{5}{3}$ and $-2.$ $\frac{5}{3}=.6.$ The answer is 6.

Problem 2

Multiplying the denominator by $1-2i$ gives us that the expression is $2+3i,$ the coefficient is thus 3.

Problem 3

With the first cut, we must remove a $8 \times 8$ square leaving us with an $8 \times 5$ rectangle. We now remove a $5 \times 5$ square leaving us with a $3 \times 5$ rectangle. We now remove a $3 \times 3$ square and are left with a $2 \times 3$ rectangle. We now remove a $2 \times 2$ square and are left with a $1 \times 2$ rectangle. From this, we remove two $1 \times 1$ squares. We have 6 total squares.

Problem 4

Problem 5

We have that $2b^{2}+3b+7=159 \Longrightarrow b=8.$ So therefore $3(8^{2})+4(8)+5=229$

Problem 6

We have that $\frac{8}{3}(1+p)=\frac{10}{2}.$ The closest to $p$ is 87.

Problem 7

The area of the squares is $4^{2}+13^{2}+15^{2}=410.$ 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 $\frac{\log{x}}{\log{3}} = p$ and $\frac{\log{x}}{\log{7}}=q.$ Also, notice that we are looking for the value of $\frac{\log{x}}{\log{21}}$ and $\log{21}=\log{3}+\log{7}.$ From this we get the value we are looking for as $\frac{1}{1/p+1/q}.$

Problem 9

We use the Principle of Inclusion Exclusion. There are $\left[\frac{1000}{12}\right]+\left[\frac{1000}{18}\right]=83+55=138.$ We must now the subtract the number of numbers divisible by their LCM which is 36 which is $\left[\frac{1000}{36}\right]=27.$ The answer is 138-27=111.

Problem 10

There are $6(7)/2=21$ 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 $\frac{10!}{2!3!5!}=2520$ arrangements.

Problem 12

We have:

$a_{1}=2(1)+3(2)=8$ $b_{1}=1-2(2)=-3$ $a_{2}=2(8)+3(-3)=7$ $b_{2}=8-2(-3)=14$ $a_{3}=2(7)+3(14)=56 \equiv 0 \mod{4}$

Problem 13

The roots of $p(x)$ are 1 and -2. These are the values of $r$ we can use on $q(x).$ So we have $a+b+c=-1$ and $4a-2b+c=8.$ 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

$\binom{3}{1}\left(\frac 12\right)^{3} = 3/8$

Case 2: We have 1+1+1+2. This has probability of $\binom{4}{3}(\frac 12)^{4} = 1/4.$

Case 3: We have 1+1+1+1+1 which has a probability of $\binom{5}{5}(\frac 12)^{5} = 1/32.$

The total probability is $3/8 + 1/4 + 1/32 = \frac{21}{32}.$

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 $\frac{1}{9}\pi+2(\frac 1{2 \sqrt{3}})$ and $a+b=4.$

Problem 16

Problem 17

We use trial and error for each of the $n$ 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 $n$ that works.

Problem 18

Note that $\sin{\frac{23}{3}\pi} = \sin{\frac{5}{3}\pi}.$ $\arccos \left(\frac{5}{3}\pi\right) = \frac{5}{6}\pi$ and 11 is our answer.

Problem 19

There are $\binom{10}{0} + \binom{10}{1} + \cdots + \binom{10}{10} = 2^{10} {}$ 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 $B.$ There are $\binom{7}{0} + \binom{7}{1} + \cdots + \binom{7}{7} = 2^{7}$ possible subsets. The number of subsets with atleast one element of $B$ is therefore 1024-128=896.

Problem 20

With triangle inequality, we have that $c+19>17 \Longrightarrow c>2.$ Also, using the fact that the triangle is acute, $c^{2}<17^{2}+19^{2}\Longrightarrow c<26.$ So we have $2<c<26$ which gives us 23 integer values for $c.$

Problem 21

Let $y=8^{x}.$ The equation becomes $2y+32/y=65 \Longrightarrow 2y^{2}-65y+32 \Longrightarrow y=1/2, 32 \Longrightarrow x=5/3,-1/3$ and the sum of the solutions of $x$ 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 $((x+2y)+(3z+5w))^{8}.$ Since we are looking for the coefficient of $xy^{2}$ and $z^{3}w^{2},$ we basically need to look at $\displaystyle \binom{8}{4} (x + 2y)^3 (3z + 5w)^5$ (after using the binomial theorem and looking at different possibilities). Using the binomial theorem, the coefficient we have is $\binom{8}{5}\binom{3}{2}(2y)^{2}x \binom{5}{2}(3z)^{3}(5w)^{2}$ and the coefficient is $56 \cdot 3 \cdot 4 \cdot 10 \cdot 27 \cdot 25 \mod{10000}=6000.$ The answer is 6.

Problem 24

Problem 25

Using the law of cosines, we have $17^{2}=25^{2}+26^{2}-2(25)(26)(\cos{\alpha}) \Longrightarrow \cos{\alpha}=\frac{253}{13 \cdot 25}.$ Now, using Heron’s Formula, we see that the area of the triangle is 204, so $\frac{1}{2}\cdot 25 \cdot 26 \cdot \sin{\alpha}=204 \Longrightarrow \sin{\alpha}=\frac{204}{15 \cdot 13}.$ The result follows.

Problem 26

$0\leq \sin{\theta}\leq 1$ which is at an interval of length 2. But because a parabola contains two roots for each value of $y,$ the interval length is 1.

Problem 27

We have $2\sin{2\theta}\cos{2\theta}=\cos{2\theta}\Longrightarrow \sin{2\theta}=1/2 \Longrightarrow \theta=\pi/24, 25\pi/24$ (keeping in mind of the signs of cosine and sine. From there, we can get our answer by adding $2\pi$ 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 $r.$ Drawing $r$ to the tangents of the circle, we form a 30-60-90 triangle and the line of length 2 can be written as $2r+2r/\sqrt{3}+2r/\sqrt{3}.$ Thus $r=2\sqrt{3}-3$ and the answer is -1.

Problem 29

We can simplify the inequality to $\frac{x^{2}+3x-7}{x+3}>3 \Longrightarrow x^{2}>16 \Longrightarrow |x|<4.$ 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 $x_{4}=12$ 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 $84 \cdot 2=168.$

Problem 34

Problem 35

Notice that $CF=CE$ and are the altitudes of the equilateral triangles. Thus, $CF=CE=\sqrt{3}.$ Now, $\triangle AFE$ is equilateral (using similar triangles), so $FE=1$ . It follows that $[\triangle CFE]=\sqrt{11}/4$ and our answer is 4.

Problem 36

$\sqrt{2}+\sqrt{6-2\sqrt{8}}=\sqrt{2}+\sqrt{6-4\sqrt{2}}=\sqrt{2}+\sqrt{(2-\sqrt{2})^{2}}=2$

Problem 37

Using similar triangles, we see that $PC=1$ and thus both of the triangles are 30-60-90. The area that we are looking for is $[\triangle{ACO}]-[\triangle{PBC}]-[POA]=\sqrt{3}/2-\sqrt{3}/8-1/6\pi=\frac{9\sqrt{3}-4\pi}{24}$ . 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 $\alpha$ . Using Heron's formula, the area of the triangle is 84. Now use sine area to find out the sine of $\alpha$ . The result follows.

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