Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: A cube of cheese $C=\{(x, y, z)| 0 \le x, y, z \le 1\}$ is cut along the planes $x=y, y=z$ and $z=x$ . How many pieces are there? 
(No cheese is moved until all three cuts are made.)
$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 9$

Solution: The cut $x = y$ separates the cube into points with $x < y$ and points with $x > y$ , and analogous results apply for the other cuts. Thus, which piece a particular point is in depends only on the relative sizes of its coordinates $x$ $y$ , and $z$ - for example, all points with the ordering $x < y < z$ are in the same piece. Thus, as there are $3! = 6$ possible orderings, there are $6$ pieces, which is answer $\boxed{6}$

Answer: B 6
Problem: Let $a, b, c, d$ be real numbers. Suppose that all the roots of $z^4+az^3+bz^2+cz+d=0$ are complex numbers lying on a circle
in the complex plane centered at $0+0i$ and having radius $1$ . The sum of the reciprocals of the roots is necessarily
$\textbf{(A)}\ a \qquad \textbf{(B)}\ b \qquad \textbf{(C)}\ c \qquad \textbf{(D)}\ -a \qquad \textbf{(E)}\ -b$

Solution: Let's denote the roots of the polynomial as $z_1, z_2, z_3, z_4$ . We know that the magnitudes of these 4 roots are 1 as given in the problem statement. Therefore, we have $z_1 \overline{z_1}, z_2 \overline{z_2}, z_3 \overline{z_3}, z_4 \overline{z_4} = 1$ . We want to find $\frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} + \frac{1}{z_4}$ which is $\overline{z_1} + \overline{z_2} + \overline{z_3} + \overline{z_4}$ . Remember that these are the roots of polynomials. Whenever complex numbers are the roots of a polynomial with real coefficients, we know they come in complex-conjugate pairs. Therefore, $\overline{z_1} + \overline{z_2} + \overline{z_3} + \overline{z_4} = z_1 + z_2 + z_3 + z_4$ . However, by Vieta's, $z_1 + z_2 + z_3 + z_4 = -a$ . Thus, the answer is $\boxed{01}$

Answer: D 01
Problem: Consider the sequence of numbers defined recursively by $t_1=1$ and for $n>1$ by $t_n=1+t_{(n/2)}$ when $n$ is even 
and by $t_n=\frac{1}{t_{(n-1)}}$ when $n$ is odd. Given that $t_n=\frac{19}{87}$ , the sum of the digits of $n$ is
$\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 17 \qquad \textbf{(C)}\ 19 \qquad \textbf{(D)}\ 21 \qquad \textbf{(E)}\ 23$

Solution: If $n$ is even, then $t_{(n/2)}$ would be negative, which is not possible. Therefore, $n$ is odd. With this function, backwards thinking is the key. If $t_x < 1$ , then $x$ is odd, and $t_{(x-1)} = \frac{1}{t_{x}}$ . Otherwise, you keep on subtracting 1 and halving x until $t_\frac{x}{2^{n}} < 1$ .
We can use this logic to go backwards until we reach $t_1 = 1$ , like so:
$t_n=\frac{19}{87}\\\\t_{n-1} = \frac{87}{19}\\\\t_{\frac{n-1}{2}} = \frac{68}{19}\\\\t_{\frac{n-1}{4}} = \frac{49}{19}\\\\t_{\frac{n-1}{8}} = \frac{30}{19}\\\\t_{\frac{n-1}{16}} = \frac{11}{19}\\\\t_{\frac{n-1}{16} - 1} = \frac{19}{11}\\\\t_{\frac{\frac{n-1}{16} - 1}{2}} = \frac{8}{11}\\\\t_{\frac{\frac{n-1}{16} - 1}{2} - 1} = \frac{11}{8}\\\\t_{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2}} = \frac{3}{8}\\\\t_{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2} - 1} = \frac{8}{3}\\\\t_{\frac{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{2}} = \frac{5}{3}\\\\t_{\frac{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4}} = \frac{2}{3}\\\\t_{\frac{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4} - 1} = \frac{3}{2}\\\\t_{\frac{\frac{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4} - 1}{2}} = \frac{1}{2}\\\\t_{\frac{\frac{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4} - 1}{2} - 1} = 2\\\\t_{\frac{\frac{\frac{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4} - 1}{2} - 1}{2}} = t_1 = 1 \Rightarrow \frac{\frac{\frac{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4} - 1}{2} - 1}{2} = 1 \Rightarrow n = 1905$ , so the answer is $\boxed{15}$

Answer: A 15
Problem: $[x-(y-z)] - [(x-y) - z] =$
$\textbf{(A)}\ 2y \qquad \textbf{(B)}\ 2z \qquad \textbf{(C)}\ -2y \qquad \textbf{(D)}\ -2z \qquad \textbf{(E)}\ 0$

Solution: The expression becomes $(x-y+z)-(x-y-z) = x-y+z-x+y+z = 2z$ , which is $\boxed{2}$

Answer: B 2
Problem: $\triangle ABC$ has a right angle at $C$ and $\angle A = 20^\circ$ . If $BD$ $D$ in $\overline{AC}$ ) is the bisector of $\angle ABC$ , then $\angle BDC =$
$\textbf{(A)}\ 40^\circ \qquad \textbf{(B)}\ 45^\circ \qquad \textbf{(C)}\ 50^\circ \qquad \textbf{(D)}\ 55^\circ\qquad \textbf{(E)}\ 60^\circ$

Solution: Since $\angle C = 90^{\circ}$ and $\angle A = 20^{\circ}$ , we have $\angle ABC = 70^{\circ}$ . Thus $\angle DBC = 35^{\circ}$ . It follows that $\angle BDC = 90^{\circ} - 35^{\circ} = 55^{\circ}$ , which is $\boxed{55}$

Answer: D 55
Problem: Using a table of a certain height, two identical blocks of wood are placed as shown in Figure 1.  Length $r$ is found to be $32$ inches.  After rearranging the blocks as in Figure 2, length $s$ is found to be $28$ inches.  How high is the table?
[asy] size(300); defaultpen(linewidth(0.8)+fontsize(13pt)); path table = origin--(1,0)--(1,6)--(6,6)--(6,0)--(7,0)--(7,7)--(0,7)--cycle; path block = origin--(3,0)--(3,1.5)--(0,1.5)--cycle; path rotblock = origin--(1.5,0)--(1.5,3)--(0,3)--cycle; draw(table^^shift((14,0))*table); filldraw(shift((7,0))*block^^shift((5.5,7))*rotblock^^shift((21,0))*rotblock^^shift((18,7))*block,gray); draw((7.25,1.75)--(8.5,3.5)--(8.5,8)--(7.25,9.75),Arrows(size=5)); draw((21.25,3.25)--(22,3.5)--(22,8)--(21.25,8.25),Arrows(size=5)); unfill((8,5)--(8,6.5)--(9,6.5)--(9,5)--cycle); unfill((21.5,5)--(21.5,6.5)--(23,6.5)--(23,5)--cycle); label("$r$",(8.5,5.75)); label("$s$",(22,5.75)); [/asy]
$\textbf{(A) }28\text{ inches}\qquad\textbf{(B) }29\text{ inches}\qquad\textbf{(C) }30\text{ inches}\qquad\textbf{(D) }31\text{ inches}\qquad\textbf{(E) }32\text{ inches}$

Solution: Let $h$ $l$ , and $w$ represent the height of the table and the length and width of the wood blocks, respectively, in inches. From Figure 1, we have $l+h-w=32$ , and from Figure 2, $w+h-l=28$ . Adding the equations gives $2h=60 \implies h=30$ , which is $\boxed{30}$

Answer: C 30
Problem: The population of the United States in $1980$ was $226,504,825$ . The area of the country is $3,615,122$ square miles. There are $(5280)^{2}$ square feet in one square mile. Which number below best approximates the average number of square feet per person?
$\textbf{(A)}\ 5,000\qquad \textbf{(B)}\ 10,000\qquad \textbf{(C)}\ 50,000\qquad \textbf{(D)}\ 100,000\qquad \textbf{(E)}\ 500,000$

Solution: With about $230$ million people and under $4$ million square miles, there are about $60$ people per square mile. Since a square mile is about $(5000 \ \text{ft})^{2} = 25$ million square feet, that gives approximately $\frac{25}{60}$ of a million square feet per person. $\frac{25}{60}$ is approximately half, so the answer is approximately half a million, which is $\boxed{500,000}$

Answer: E 500,000
Problem: In $\triangle ABC, AB = 13, BC = 14$ and $CA = 15$ . Also, $M$ is the midpoint of side $AB$ and $H$ is the foot of the altitude from $A$ to $BC$ . 
The length of $HM$ is
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair H=origin, A=(0,6), B=(-4,0), C=(5,0), M=B+3.6*dir(B--A); draw(B--C--A--B^^M--H--A^^rightanglemark(A,H,C)); label("A", A, NE); label("B", B, W); label("C", C, E); label("H", H, S); label("M", M, dir(M)); [/asy]
$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 6.5\qquad \textbf{(C)}\ 7\qquad \textbf{(D)}\ 7.5\qquad \textbf{(E)}\ 8$

Solution: Warning: this solution is very intensive in calculation. Please do NOT try this on the test!
Let's start by finding $AH$ . By Heron's Formula, $s=\frac{13+14+15}{2}=21, [ABC]=\sqrt{21*(21-13)(21-14)(21-15)}=84$ . Using the area formula $A=0.5bh$ $AH=12$ . Now using the Pythagorean Theorem, $BH=5, HC=9$
Now $AM=MB=6.5$ . Using Stewart's Theorem on $\triangle{ABH}$ , letting $HM=x$
$13*6.5*6.5+13x^2=12*6.5*12+5*6.5*5$ (remember that Stewart's Theorem is $man+dad=bmb+cnc$ ).
Thus $x=6.5$ or $x=-6.5$ (reject this solution since $x$ is positive). Thus $HM=6.5$ . Select $\boxed{6.5}$

Answer: B 6.5
Problem: A parabola $y = ax^{2} + bx + c$ has vertex $(4,2)$ . If $(2,0)$ is on the parabola, then $abc$ equals
$\textbf{(A)}\ -12\qquad \textbf{(B)}\ -6\qquad \textbf{(C)}\ 0\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ 12$

Solution: Consider the quadratic in completed square form: it must be $y=a(x-4)^{2}+2$ . Now substitute $x=2$ and $y=0$ to give $a=-\frac{1}{2}$ . Now expanding gives $y=-\frac{1}{2}x^{2}+4x-6$ , so the product is $-\frac{1}{2} \cdot 4 \cdot -6  = 3 \cdot 4 = 12$ , which is $\boxed{12}$

Answer: E 12
Problem: In $\triangle ABC, AB = 8, BC = 7, CA = 6$ and side $BC$ is extended, as shown in the figure, to a point $P$ so that $\triangle PAB$ is similar to $\triangle PCA$ . The length of $PC$ is
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, P=(1.5,5), B=(8,0), C=P+2.5*dir(P--B); draw(A--P--C--A--B--C); label("A", A, W); label("B", B, E); label("C", C, NE); label("P", P, NW); label("6", 3*dir(A--C), SE); label("7", B+3*dir(B--C), NE); label("8", (4,0), S); [/asy]
$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 8\qquad \textbf{(C)}\ 9\qquad \textbf{(D)}\ 10\qquad \textbf{(E)}\ 11$

Solution: Since we are given that $\triangle{PAB}\sim\triangle{PCA}$ , we have $\frac{PC}{PA}=\frac{6}{8}=\frac{PA}{PC+7}$
Solving for $PA$ in $\frac{PC}{PA}=\frac{6}{8}=\frac{3}{4}$ gives us $PA=\frac{4PC}{3}$
We also have $\frac{PA}{PC+7}=\frac{3}{4}$ . Substituting $PA$ in for our expression yields $\frac{\frac{4PC}{3}}{PC+7}=\frac{3}{4}$
Which we can further simplify to $\frac{16PC}{3}=3PC+21$
$\frac{7PC}{3}=21$
$PC=9\implies\boxed{9}$

Answer: C 9
Problem: A drawer in a darkened room contains $100$ red socks, $80$ green socks, $60$ blue socks and $40$ black socks. 
A youngster selects socks one at a time from the drawer but is unable to see the color of the socks drawn. 
What is the smallest number of socks that must be selected to guarantee that the selection contains at least $10$ pairs? 
(A pair of socks is two socks of the same color. No sock may be counted in more than one pair.)
$\textbf{(A)}\ 21\qquad \textbf{(B)}\ 23\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 30\qquad \textbf{(E)}\ 50$

Solution: Solution by e_power_pi_times_i
Suppose that you wish to draw one pair of socks from the drawer. Then you would pick $5$ socks (one of each kind, plus one). Notice that in the worst possible situation, you will continue to draw the same sock, until you get $10$ pairs. This is because drawing the same sock results in a pair every $2$ of that sock, whereas drawing another sock creates another pair. Thus the answer is $5+2\cdot(10-1) = \boxed{23}$

Answer: B 23
Problem: A park is in the shape of a regular hexagon $2$ km on a side. Starting at a corner, 
Alice walks along the perimeter of the park for a distance of $5$ km. 
How many kilometers is she from her starting point?
$\textbf{(A)}\ \sqrt{13}\qquad \textbf{(B)}\ \sqrt{14}\qquad \textbf{(C)}\ \sqrt{15}\qquad \textbf{(D)}\ \sqrt{16}\qquad \textbf{(E)}\ \sqrt{17}$

Solution: We imagine this problem on a coordinate plane and let Alice's starting position be the origin. We see that she will travel along two edges and then go halfway along a third. Therefore, her new $x$ -coordinate will be $1 + 2 + \frac{1}{2} = \frac{7}{2}$ because she travels along a distance of $2 \cdot \frac{1}{2} = 1$ km because of the side relationships of an equilateral triangle, then $2$ km because the line is parallel to the $x$ -axis, and the remaining  distance is $\frac{1}{2}$ km because she went halfway along and because of the logic for the first part of her route. For her $y$ -coordinate, we can use similar logic to find that the coordinate is $\sqrt{3} + 0 - \frac{\sqrt{3}}{2} =  \frac{\sqrt{3}}{2}$ . Therefore, her distance is \[\sqrt{\left(\frac{7}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{49}{4} + \frac{3}{4}} = \sqrt{\frac{52}{4}} = \sqrt{13},\] giving an answer of $\boxed{13}$

Answer: A 13
Problem: It is desired to construct a right triangle in the coordinate plane so that its legs are parallel 
to the $x$ and $y$ axes and so that the medians to the midpoints of the legs lie on the lines $y = 3x + 1$ and $y = mx + 2$ . The number of different constants $m$ for which such a triangle exists is
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 3\qquad \textbf{(E)}\ \text{more than 3}$

Solution: In any right triangle with legs parallel to the axes, one median to the midpoint of a leg has slope $4$ times that of the other. This can easily be shown with coordinates: any triangle of this sort may be labelled with right angle at $P(a,b)$ , other vertices $Q(a,b+2c)$ and $R(a-2d,b)$ , and thus midpoints $(a,b+c)$ and $(a-d,b)$ , so that the slopes are $\frac{c}{2d}$ and $\frac{2c}{d} = 4(\frac{c}{2d})$ , thus showing that one is $4$ times the other as required.
Thus in our problem, $m$ is either $3 \times 4 = 12$ or $3 \div 4 = \frac{3}{4}$ . In fact, both are possible, and each for infinitely many triangles. We shall show this for $m=12$ , and the argument is analogous for $m=\frac{3}{4}$ . Take any right triangle with legs parallel to the axes and a hypotenuse with slope $12 \div 2 = 6$ , e.g. the triangle with vertices $(0,0)$ $(1,0)$ , and $(1,6)$ . Then quick calculations show that the medians to the legs have slopes $12$ and $3$ . Now translate the triangle (without rotating it) so that its medians intersect at the point where the lines $y=12x+2$ and $y=3x+1$ intersect. This forces the medians to lie on these lines (since their slopes are determined, and now we force them to go through a particular point; a slope and a point uniquely determine a line). Finally, for any central dilation of this triangle (a larger or smaller triangle with the same centroid and sides parallel to this one's sides), the medians will still lie on these lines, showing the "infinitely many" part of the result.
Hence, to sum up, $m$ can in fact be both $12$ or $\frac{3}{4}$ , which is exactly $2$ values, i.e. $\boxed{2}$

Answer: C 2
Problem: $ABCDE$ is a regular pentagon. $AP, AQ$ and $AR$ are the perpendiculars dropped from $A$ onto $CD, CB$ extended and $DE$ extended, 
respectively. Let $O$ be the center of the pentagon. If $OP = 1$ , then $AO + AQ + AR$ equals
[asy] defaultpen(fontsize(10pt)+linewidth(.8pt)); pair O=origin, A=2*dir(90), B=2*dir(18), C=2*dir(306), D=2*dir(234), E=2*dir(162), P=(C+D)/2, Q=C+3.10*dir(C--B), R=D+3.10*dir(D--E), S=C+4.0*dir(C--B), T=D+4.0*dir(D--E); draw(A--B--C--D--E--A^^E--T^^B--S^^R--A--Q^^A--P^^rightanglemark(A,Q,S)^^rightanglemark(A,R,T)); dot(O); label("$O$",O,dir(B)); label("$1$",(O+P)/2,W); label("$A$",A,dir(A)); label("$B$",B,dir(B)); label("$C$",C,dir(C)); label("$D$",D,dir(D)); label("$E$",E,dir(E)); label("$P$",P,dir(P)); label("$Q$",Q,dir(Q)); label("$R$",R,dir(R)); [/asy]
$\textbf{(A)}\ 3\qquad \textbf{(B)}\ 1 + \sqrt{5}\qquad \textbf{(C)}\ 4\qquad \textbf{(D)}\ 2 + \sqrt{5}\qquad \textbf{(E)}\ 5$

Solution: To solve the problem, we compute the area of regular pentagon $ABCDE$ in two different ways. First, we can divide regular pentagon $ABCDE$ into five congruent triangles. [asy] unitsize(2 cm);  pair A, B, C, D, E, O, P, Q, R;  A = dir(90); B = dir(90 - 360/5); C = dir(90 - 2*360/5); D = dir(90 - 3*360/5); E = dir(90 - 4*360/5); O = (0,0); P = (C + D)/2; Q = (A + reflect(B,C)*(A))/2; R = (A + reflect(D,E)*(A))/2;  draw((2*R - E)--D--C--(2*Q - B)); draw(A--P); draw(A--Q); draw(A--R); draw(B--A--E); draw((O--B),dashed); draw((O--C),dashed); draw((O--D),dashed); draw((O--E),dashed);  label("$A$", A, N); label("$B$", B, dir(0)); label("$C$", C, SE); label("$D$", D, SW); label("$E$", E, W); dot("$O$", O, NE); label("$P$", P, S); label("$Q$", Q, dir(0)); label("$R$", R, W); label("$1$", (O + P)/2, dir(0)); [/asy]
If $s$ is the side length of the regular pentagon, then each of the triangles $AOB$ $BOC$ $COD$ $DOE$ , and $EOA$ has base $s$ and height 1, so the area of regular pentagon $ABCDE$ is $5s/2$
Next, we divide regular pentagon $ABCDE$ into triangles $ABC$ $ACD$ , and $ADE$
[asy] unitsize(2 cm);  pair A, B, C, D, E, O, P, Q, R;  A = dir(90); B = dir(90 - 360/5); C = dir(90 - 2*360/5); D = dir(90 - 3*360/5); E = dir(90 - 4*360/5); O = (0,0); P = (C + D)/2; Q = (A + reflect(B,C)*(A))/2; R = (A + reflect(D,E)*(A))/2;  draw((2*R - E)--D--C--(2*Q - B)); draw(A--P); draw(A--Q); draw(A--R); draw(B--A--E); draw(A--C,dashed); draw(A--D,dashed);  label("$A$", A, N); label("$B$", B, dir(0)); label("$C$", C, SE); label("$D$", D, SW); label("$E$", E, W); dot("$O$", O, dir(0)); label("$P$", P, S); label("$Q$", Q, dir(0)); label("$R$", R, W); label("$1$", (O + P)/2, dir(0)); [/asy] Triangle $ACD$ has base $s$ and height $AP = AO + 1$ . Triangle $ABC$ has base $s$ and height $AQ$ . Triangle $ADE$ has base $s$ and height $AR$ . Therefore, the area of regular pentagon $ABCDE$ is also \[\frac{s}{2} (AO + AQ + AR + 1).\] Hence, \[\frac{s}{2} (AO + AQ + AR + 1) = \frac{5s}{2},\] which means $AO + AQ + AR + 1 = 5$ , or $AO + AQ + AR = \boxed{4}$

Answer: C 4
Problem: $ABCDE$ is a regular pentagon. $AP, AQ$ and $AR$ are the perpendiculars dropped from $A$ onto $CD, CB$ extended and $DE$ extended, 
respectively. Let $O$ be the center of the pentagon. If $OP = 1$ , then $AO + AQ + AR$ equals
[asy] defaultpen(fontsize(10pt)+linewidth(.8pt)); pair O=origin, A=2*dir(90), B=2*dir(18), C=2*dir(306), D=2*dir(234), E=2*dir(162), P=(C+D)/2, Q=C+3.10*dir(C--B), R=D+3.10*dir(D--E), S=C+4.0*dir(C--B), T=D+4.0*dir(D--E); draw(A--B--C--D--E--A^^E--T^^B--S^^R--A--Q^^A--P^^rightanglemark(A,Q,S)^^rightanglemark(A,R,T)); dot(O); label("$O$",O,dir(B)); label("$1$",(O+P)/2,W); label("$A$",A,dir(A)); label("$B$",B,dir(B)); label("$C$",C,dir(C)); label("$D$",D,dir(D)); label("$E$",E,dir(E)); label("$P$",P,dir(P)); label("$Q$",Q,dir(Q)); label("$R$",R,dir(R)); [/asy]
$\textbf{(A)}\ 3\qquad \textbf{(B)}\ 1 + \sqrt{5}\qquad \textbf{(C)}\ 4\qquad \textbf{(D)}\ 2 + \sqrt{5}\qquad \textbf{(E)}\ 5$

Solution: Now, we know that angle $D$ has measure $\frac{180 \cdot 3}{5} = 108$ . Since \[\sin 54 = \frac{OP}{DO} = \frac{1}{DO}, DO = \frac{1}{\sin 54}\] \[\tan 54 = \frac{OP}{DP} = \frac{1}{DP}, DP = \frac{1}{\tan 54}\] Therefore, $AB = 2DP = \frac{2}{\tan 54}$ \[\sin 72 = \frac{AQ}{AB} = AQ \tan 54 \cdot \frac{1}{2}, AQ = \frac{2 \sin 72}{\tan 54}\] Therefore, $AO + AQ + AR = AO + 2AQ = \frac{1}{\sin 54}+\frac{4 \sin 72}{\tan 54} = \frac{1}{\sin 54} + 8 \sin 36 \cos 54 = \frac{1}{\cos 36} + 8-8\cos^2(36)$ . Recalling that $\cos 36 = \frac{1 + \sqrt{5}}{4}$ gives a final answer of $\boxed{4}$

Answer:  4
Problem: The number of real solutions $(x,y,z,w)$ of the simultaneous equations $2y = x + \frac{17}{x}, 2z = y + \frac{17}{y}, 2w = z + \frac{17}{z}, 2x = w + \frac{17}{w}$ is
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ 4\qquad  \textbf{(D)}\ 8\qquad \textbf{(E)}\ 16$

Solution: Consider the cases $x>0$ and $x<0$ , and also note that by AM-GM, for any positive number $a$ , we have $a+\frac{17}{a} \geq 2\sqrt{17}$ , with equality only if $a = \sqrt{17}$ . Thus, if $x>0$ , considering each equation in turn, we get that $y \geq \sqrt{17}, z \geq \sqrt{17}, w \geq \sqrt{17}$ , and finally $x \geq \sqrt{17}$
Now suppose $x > \sqrt{17}$ . Then $y - \sqrt{17} = \frac{x^{2}+17}{2x} - \sqrt{17} = (\frac{x-\sqrt{17}}{2x})(x-\sqrt{17}) < \frac{1}{2}(x-\sqrt{17})$ , so that $x > y$ . Similarly, we can get $y > z$ $z > w$ , and $w > x$ , and combining these gives $x > x$ , an obvious contradiction.
Thus we must have $x \geq \sqrt{17}$ , but $x \ngtr \sqrt{17}$ , so if $x > 0$ , the only possibility is $x = \sqrt{17}$ , and analogously from the other equations we get $x = y = z = w = \sqrt{17}$ ; indeed, by substituting, we verify that this works.
As for the other case, $x < 0$ , notice that $(x,y,z,w)$ is a solution if and only if $(-x,-y,-z,-w)$ is a solution, since this just negates both sides of each equation and so they are equivalent. Thus the only other solution is $x = y = z = w = -\sqrt{17}$ , so that we have $2$ solutions in total, and therefore the answer is $\boxed{2}$

Answer: B 2
Problem: If $2x+1=8$ , then $4x+1=$
$\mathrm{(A)\ } 15 \qquad \mathrm{(B) \ }16 \qquad \mathrm{(C) \  } 17 \qquad \mathrm{(D) \  } 18 \qquad \mathrm{(E) \  }19$

Solution: We have \begin{align*}2x+1 = 8 &\iff 2x = 7 \\ &\iff x = \frac{7}{2},\end{align*} so \begin{align*}4x+1 &= 4\left(\frac{7}{2}\right)+1 \\ &= 2(7)+1 \\ &= \boxed{15}

Answer: A 15
Problem: If $2x+1=8$ , then $4x+1=$
$\mathrm{(A)\ } 15 \qquad \mathrm{(B) \ }16 \qquad \mathrm{(C) \  } 17 \qquad \mathrm{(D) \  } 18 \qquad \mathrm{(E) \  }19$

Solution: From $2x = 7$ (as above), we can directly compute \begin{align*}4x &= 2(2x) \\ &= 2(7) \\ &= 14,\end{align*} so $4x+1 = 14+1 = \boxed{15}$

Answer: A 15
Problem: In right $\triangle ABC$ with legs $5$ and $12$ , arcs of circles are drawn, one with center $A$ and radius $12$ , the other with center $B$ and radius $5$ . They intersect the hypotenuse in $M$ and $N$ . Then $MN$ has length
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(12,7), C=(12,0), M=12*dir(A--B), N=B+B.y*dir(B--A); real r=degrees(B); draw(A--B--C--cycle^^Arc(A,12,0,r)^^Arc(B,B.y,180+r,270)); pair point=incenter(A,B,C); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$M$", M, dir(point--M)); label("$N$", N, dir(point--N)); label("$12$", (6,0), S); label("$5$", (12,3.5), E);[/asy]
$\mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }\frac{13}{5} \qquad \mathrm{(C) \  } 3 \qquad \mathrm{(D) \  } 4 \qquad \mathrm{(E) \  }\frac{24}{5}$

Solution: Firstly, the Pythagorean theorem gives \begin{align*}AB &=\sqrt{AC^2+BC^2} \\ &= \sqrt{12^2+5^2} \\ & =\sqrt{144+25} \\ &=\sqrt{169} \\ &= 13.\end{align*} Also, $AM = AC = 12$ and $BN = BC = 5$ since they are both radii of the respective circles. Thus $MB = AB-AM = 13-12 = 1$ , and so $MN = BN-BM = 5-1 = \boxed{4}$

Answer: D 4
Problem: A large bag of coins contains pennies, dimes and quarters. There are twice as many dimes as pennies and three times as many quarters as dimes. An amount of money which could be in the bag is
$\mathrm{(A)\ } $306 \qquad \mathrm{(B) \ }  $333 \qquad \mathrm{(C)\ } $342 \qquad \mathrm{(D) \  }  $348 \qquad \mathrm{(E) \  }  $360$

Solution: If there are $x$ pennies in the bag, then there are $2x$ dimes and $3(2x) = 6x$ quarters. Since pennies are $$0.01$ , dimes are $$0.10$ , and quarters are $$0.25$ , the total amount of money in the bag is \[$ \left(0.01x+(0.10)(2x)+(0.25)(6x)\right) = $1.71x.\] Therefore, the possible amounts of money are precisely the integer multiples of $$1.71$
Since the answer choices are all integer numbers of dollars, we multiply by $100$ to deduce that the answer must be an integer multiple of $$171$ . The only such multiple among the answer choices is $$(2 \cdot 171) = \boxed{342}$

Answer: C 342
Problem: An arbitrary circle can intersect the graph of $y = \sin x$ in
$\mathrm{(A) \ } \text{at most }2\text{ points} \qquad \mathrm{(B) \ }\text{at most }4\text{ points} \qquad \mathrm{(C) \  } \text{at most }6\text{ points} \qquad \mathrm{(D) \ } \text{at most }8\text{ points}$ $\mathrm{(E) \  }\text{more than }16\text{ points}$

Solution: Consider a circle whose center lies on the positive $y$ -axis and which passes through the origin. As the radius of this circle becomes arbitrarily large, its curvature near the $x$ -axis becomes almost flat, and so it can intersect the curve $y = \sin x$ arbitrarily many times (since the $x$ -axis itself intersects the curve infinitely many times). Hence, in particular, we can choose a radius sufficiently large that the circle intersects the curve at $\boxed{16}$

Answer: E 16