Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: How many distinguishable rearrangements of the letters in $CONTEST$ have both the vowels first? (For instance, $OETCNST$ is one such arrangement, but $OTETSNC$ is not.)
$\mathrm{(A)\ } 60 \qquad \mathrm{(B) \ }120 \qquad \mathrm{(C) \  } 240 \qquad \mathrm{(D) \  } 720 \qquad \mathrm{(E) \  }2520$

Solution: We consider the vowels and consonants separately. There are $2$ vowels ( $O$ and $E$ ), giving $2! = 2$ choices for the first two letters; similarly, there are $5$ consonants ( $C$ $N$ $S$ , and two $T$ s), which would give $5! = 120$ possible choices for letters $3$ to $7$ , except that since the two $T$ s are indistinguishable, this actually counts each order exactly twice. Therefore the number of possible orderings of the consonants is $\frac{120}{2} = 60$ , giving a total of $2 \cdot 60 = \boxed{120}$ possible rearrangements.

Answer: B 120
Problem: Pegs are put in a board $1$ unit apart both horizontally and vertically. A rubber band is stretched over $4$ pegs as shown in the figure, forming a quadrilateral. Its area in square units is
[asy] int i,j; for(i=0; i<5; i=i+1) { for(j=0; j<4; j=j+1) { dot((i,j)); }} draw((0,1)--(1,3)--(4,1)--(3,0)--cycle, linewidth(0.7));[/asy]
$\mathrm{(A)\ } 4 \qquad \mathrm{(B) \ }4.5 \qquad \mathrm{(C) \  } 5 \qquad \mathrm{(D) \  } 5.5 \qquad \mathrm{(E) \  }6$

Solution: [asy] int i,j; for(i=0; i<5; i=i+1) { for(j=0; j<4; j=j+1) { dot((i,j)); }} draw((0,1)--(1,3)--(4,1)--(3,0)--cycle, linewidth(0.7)); draw((0,0)--(4,0)--(4,3)--(0,3)--cycle); label("$A$",(0,3),NW); label("$B$",(4,3),NE); label("$C$",(4,0),SE); label("$D$",(0,0),SW); label("$E$",(1,3),N); label("$F$",(4,1),E); label("$G$",(3,0),S); label("$H$",(0,1),W); [/asy] We draw in the rectangle bounding the given quadrilateral and label the points as shown. The area of rectangle $ABCD$ is $(3)(4) = 12$ , while the areas of the triangles $AEH$ $EBF$ $FCG$ , and $GDH$ are, respectively, \begin{align*}&\text{area of } \triangle AEH = \frac{1}{2}(1)(2) = 1, \\ &\text{area of } \triangle EBF = \frac{1}{2}(3)(2) = 3, \\ &\text{area of } \triangle FCG = \frac{1}{2}(1)(1) = \frac{1}{2}, \text{ and} \\ &\text{area of } \triangle GDH = \frac{1}{2}(3)(1) = \frac{3}{2}.\end{align*} Hence the area of the given quadrilateral $EFGH$ is $12-\left(1+3+\frac{1}{2}+\frac{3}{2}\right) = \boxed{6}$

Answer: E 6
Problem: Pegs are put in a board $1$ unit apart both horizontally and vertically. A rubber band is stretched over $4$ pegs as shown in the figure, forming a quadrilateral. Its area in square units is
[asy] int i,j; for(i=0; i<5; i=i+1) { for(j=0; j<4; j=j+1) { dot((i,j)); }} draw((0,1)--(1,3)--(4,1)--(3,0)--cycle, linewidth(0.7));[/asy]
$\mathrm{(A)\ } 4 \qquad \mathrm{(B) \ }4.5 \qquad \mathrm{(C) \  } 5 \qquad \mathrm{(D) \  } 5.5 \qquad \mathrm{(E) \  }6$

Solution: The number of lattice points (i.e. pegs on the board) strictly inside the quadrilateral is $5$ and the number of lattice points on its boundary is $4$ . Therefore, by Pick's theorem , its area is $5+\frac{4}{2}-1 = \boxed{6}$

Answer: E 6
Problem: Exactly three of the interior angles of a convex polygon are obtuse. What is the maximum number of sides of such a polygon?
$\mathrm{(A)\ } 4 \qquad \mathrm{(B) \ }5 \qquad \mathrm{(C) \  } 6 \qquad \mathrm{(D) \  } 7 \qquad \mathrm{(E) \  }8$

Solution: Suppose that such a polygon has $n$ sides. Let the three obtuse angle measures, in degrees, be $o_1$ $o_2$ , and $o_3$ and the $(n-3)$ acute angle measures, again in degrees, be $a_1,a_2,a_3, \dotsc a_{n-3}$
Since $90 < o_i < 180$ for each $i$ , we have \[3(90) = 270 < o_1+o_2+o_3 < 3(180) = 540,\] and similarly, since $0 < a_i < 90$ for each $i$ \[0 < a_1+a_2+a_3+\dotsb+a_{n-3} < 90(n-3) = 90n-270.\] It follows that \[270+0 < o_1+o_2+o_3+a_1+a_2+a_3+\dotsb+a_{n-3} < 540+90n-270,\] and recalling that the sum of the interior angle measures of an $n$ -gon is $180(n-2) = 180n-360$ , this reduces to $270 < 180n-360 < 90n+270$ . Hence \[\frac{540}{180} < n < \frac{270+360}{90} \iff 3 < n < 7,\] so an upper bound is $n \leq 6$ , and it is easy to check that this bound can be attained by e.g. a convex hexagon with a right angle, $2$ acute angles, and $3$ obtuse angles, as shown below:
HrKCfF2ETNSQSy3uGlkg hexagonsurvey1.gif
Accordingly, the maximum possible number of sides of such a polygon is $\boxed{6}$

Answer: C 6
Problem: If $\usepackage{gensymb} A = 20 \degree$ and $\usepackage{gensymb} B = 25 \degree$ , then the value of $\left(1+\tan A\right)\left(1+\tan B\right)$ is
$\mathrm{(A)\ } \sqrt{3} \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \  } 1+\sqrt{2} \qquad \mathrm{(D) \  } 2\left(\tan A+\tan B\right) \qquad \mathrm{(E) \  }\text{none of these}$

Solution: Noting that $\usepackage{gensymb} 25 \degree + 20 \degree = 45 \degree$ , we apply the angle sum formula \[\tan(A+B) = \frac{\tan A+\tan B}{1-\tan A\tan B},\] giving \begin{align*}1 &= \tan 45^{\circ} \\ &= \tan(A+B) \\ &= \frac{\tan A+\tan B}{1-\tan A\tan B},\end{align*} so \[\tan A + \tan B = 1-\tan A\tan B.\] Hence \begin{align*}(1+\tan A)(1+\tan B) &= 1+\tan A+\tan B+\tan A\tan B \\ &= 1+\left(1-\tan A\tan B\right)+\tan A\tan B \\ &= \boxed{2}

Answer: B 2
Problem: If $\usepackage{gensymb} A = 20 \degree$ and $\usepackage{gensymb} B = 25 \degree$ , then the value of $\left(1+\tan A\right)\left(1+\tan B\right)$ is
$\mathrm{(A)\ } \sqrt{3} \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \  } 1+\sqrt{2} \qquad \mathrm{(D) \  } 2\left(\tan A+\tan B\right) \qquad \mathrm{(E) \  }\text{none of these}$

Solution: Expanding in terms of sines and cosines, we obtain \begin{align*}\left(1+\tan A\right)\left(1+\tan B\right) &= \left(\frac{\cos A}{\cos A}+\frac{\sin A}{\cos A}\right)\left(\frac{\cos B}{\cos B}+\frac{\sin B}{\cos B}\right) \\ &=\frac{\left(\sin A+\cos A\right)\left(\sin B+\cos B\right)}{\cos A\cos B} \\ &= \frac{\sin A\sin B+\cos A\cos B+\sin A\cos B+\sin B\cos A}{\cos A\cos B}.\end{align*}
Recalling the angle sum identities \begin{align*}&\cos(A-B) = \sin A\sin B+\cos A\cos B \text{ and} \\ &\sin(A+B)=\sin A\cos B+\sin B\cos A,\end{align*} this reduces to \[\frac{\cos(A-B)+\sin(A+B)}{\cos A\cos B}.\]
Now, using the product-to-sum formula \[\cos A\cos B = \frac{1}{2}\left(\cos(A-B)+\cos(A+B)\right),\] we can simplify the denominator, yielding \[\left(1+\tan A\right)\left(1+\tan B\right) = \frac{\cos(A-B)+\sin(A+B)}{\frac{1}{2}\left(\cos(A-B)+\cos(A+B)\right)}.\]
Finally, since $\usepackage{gensymb} A+B = 45 \degree$ , we have $\sin(A+B) = \cos(A+B)$ , so \begin{align*}\left(1+\tan A\right)\left(1+\tan B\right) &= \frac{\cos(A-B)+\sin(A+B)}{\frac{1}{2}\left(\cos(A-B)+\sin(A+B)\right)} \\ &= \frac{1}{\left(\frac{1}{2}\right)} \\ &= \boxed{2}

Answer: B 2
Problem: If $\usepackage{gensymb} A = 20 \degree$ and $\usepackage{gensymb} B = 25 \degree$ , then the value of $\left(1+\tan A\right)\left(1+\tan B\right)$ is
$\mathrm{(A)\ } \sqrt{3} \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \  } 1+\sqrt{2} \qquad \mathrm{(D) \  } 2\left(\tan A+\tan B\right) \qquad \mathrm{(E) \  }\text{none of these}$

Solution: As in Solution 2, we rewrite the expression as \[\frac{\cos 20^{\circ} \cos 25^{\circ} + \cos 20^{\circ} \sin 25^{\circ} + \sin 20^{\circ} \cos 25^{\circ} + \sin 20^{\circ} \sin 25^{\circ}}{\cos 20^{\circ} \cos 25^{\circ}},\] and hence as \[1 + \frac{\cos 20^{\circ} \sin 25^{\circ} + \sin 20^{\circ} \cos 25^{\circ} + \sin 20^{\circ} \sin 25^{\circ}}{\cos 20^{\circ} \cos 25^{\circ}}.\] Using the angle sum identities \begin{align*}&\cos(A-B) = \sin A\sin B+\cos A\cos B \text{ and} \\ &\sin(A+B)=\sin A\cos B+\sin B\cos A,\end{align*} we obtain \begin{align*}&\cos 20^{\circ} \sin 25^{\circ} + \sin 20^{\circ} \cos 25^{\circ} = \sin\left(20^{\circ}+25^{\circ}\right) = \sin 45^{\circ} \text{ and} \\ &\cos 20^{\circ} \cos 25^{\circ} - \sin 20^{\circ} \sin 25^{\circ} = \cos\left(20^{\circ}+25^{\circ}\right) = \cos 45^{\circ}.\end{align*} Therefore the expression becomes \begin{align*}1+\frac{\sin 45^{\circ} + \sin 20^{\circ} \sin 25^{\circ}}{\cos 45^{\circ} + \sin 20^{\circ} \sin 25^{\circ}} &= 1+1 \qquad \text{(since } \sin 45^{\circ} = \cos 45^{\circ}\text{)} \\ &= \boxed{2}

Answer: B 2
Problem: If $\usepackage{gensymb} A = 20 \degree$ and $\usepackage{gensymb} B = 25 \degree$ , then the value of $\left(1+\tan A\right)\left(1+\tan B\right)$ is
$\mathrm{(A)\ } \sqrt{3} \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \  } 1+\sqrt{2} \qquad \mathrm{(D) \  } 2\left(\tan A+\tan B\right) \qquad \mathrm{(E) \  }\text{none of these}$

Solution: As in Solutions 2 and 3, the expression becomes \[\left(\frac{\cos A+\sin A}{\cos A}\right)\left(\frac{\cos B+\sin B}{\cos B}\right).\]
Now, using the identity $\cos^2 A + \sin^2 A = 1$ and the double-angle identity $\sin(2A) = 2\sin A\cos A$ , we observe that \begin{align*}\left(\cos A + \sin A\right)^2 &= \cos^2 A + \sin^2 A + 2\sin A \cos A \\ &= 1 + 2\sin A \cos A \\ &= 1 + \sin(2A).\end{align*}
Since $A$ and $B$ are acute, we have $\sin A,\cos A,\sin B,\cos B > 0$ , so $\cos A + \sin A > 0$ and $\cos B + \sin B > 0$ . Hence, taking the positive square root of both sides in the above identity, the expression becomes \[\left(\frac{\sqrt{1+\sin(2A)}}{\cos A}\right)\left(\frac{\sqrt{1+\sin(2B)}}{\cos B}\right).\] Recalling the further identity $\sin A = \cos\left(90^{\circ}-A\right)$ , together with the half-angle identity \[\cos\left(\frac{A}{2}\right) = \sqrt{\frac{1+\cos A}{2}} \qquad \text{for } 0^{\circ} \leq A \leq 180^{\circ},\] we finally obtain \begin{align*}\left(1+\tan A\right)\left(1+\tan B\right) &= 2\left(\frac{\sqrt{\frac{1+\cos\left(90^{\circ}-2A\right)}{2}}}{\cos A}\right)\left(\frac{\sqrt{\frac{1+\cos\left(90^{\circ}-2B\right)}{2}}}{\cos B}\right) \\ &= 2\left(\frac{\cos\left(45^{\circ}-A\right)}{\cos A}\right)\left(\frac{\cos\left(45^{\circ}-B\right)}{\cos B}\right) \\ &= 2\left(\frac{\cos 25^{\circ} \cos 20^{\circ}}{\cos 20^{\circ} \cos 25^{\circ}}\right) = \boxed{2}

Answer: B 2
Problem: Diagonal $DB$ of rectangle $ABCD$ is divided into three segments of length $1$ by parallel lines $L$ and $L'$ that pass through $A$ and $C$ and are perpendicular to $DB$ . The area of $ABCD$ , rounded to the one decimal place, is
[asy] defaultpen(linewidth(0.7)+fontsize(10)); real x=sqrt(6), y=sqrt(3), a=0.4; pair D=origin, A=(0,y), B=(x,y), C=(x,0), E=foot(C,B,D), F=foot(A,B,D); real r=degrees(B); pair M1=F+3*dir(r)*dir(90), M2=F+3*dir(r)*dir(-90), N1=E+3*dir(r)*dir(90), N2=E+3*dir(r)*dir(-90); markscalefactor=0.02; draw(B--C--D--A--B--D^^M1--M2^^N1--N2^^rightanglemark(A,F,B)^^rightanglemark(N1,E,B)); pair W=A+a*dir(135), X=B+a*dir(45), Y=C+a*dir(-45), Z=D+a*dir(-135); label("A", A, NE); label("B", B, NE); label("C", C, dir(0)); label("D", D, dir(180)); label("$L$", (x/2,0), SW); label("$L^\prime$", C, SW); label("1", D--F, NW); label("1", F--E, SE); label("1", E--B, SE); clip(W--X--Y--Z--cycle);[/asy]
$\mathrm{(A)\ } 4.1 \qquad \mathrm{(B) \ }4.2 \qquad \mathrm{(C) \  } 4.3 \qquad \mathrm{(D) \  } 4.4 \qquad \mathrm{(E) \  }4.5$

Solution: Let $E$ be the point of intersection of $L$ and $\overline{BD}$ . Then, because $AE$ is the altitude to the hypotenuse of right triangle $ABD$ , triangles $ADE$ and $BAE$ are similar, giving \[\frac{AE}{BE} = \frac{ED}{EA},\] and so \begin{align*}AE &= \sqrt{BE \cdot ED} \\ &= \sqrt{(1+1)(1)} \\ &= \sqrt{2}.\end{align*} Thus, taking $BD$ and $AE$ as the base and perpendicular height, respectively, of triangle $ABD$ , we may compute its area as $\frac{1}{2}(3)\left(\sqrt{2}\right) = \frac{3\sqrt{2}}{2}$ . By symmetry, the area of the entire rectangle $ABCD$ is \[2\left(\frac{3\sqrt{2}}{2}\right) = 3\sqrt{2} \approx (3)(1.4) = \boxed{4.2}.\]

Answer: B 4.2
Problem: Six bags of marbles contain $18, 19, 21, 23, 25$ and $34$ marbles, respectively. One bag contains chipped marbles only. The other $5$ bags contain no chipped marbles. Jane takes three of the bags and George takes two of the others. Only the bag of chipped marbles remains. If Jane gets twice as many marbles as George, how many chipped marbles are there?
$\mathrm{(A)\ } 18 \qquad \mathrm{(B) \ }19 \qquad \mathrm{(C) \  } 21 \qquad \mathrm{(D) \  } 23 \qquad \mathrm{(E) \  }25$

Solution: Let George's bags contain a total of $x$ marbles, so Jane's bag contains $2x$ marbles. This means the total number of non-chipped marbles is $3x \equiv 0 \pmod{3}$ , while the total number of marbles is $18+19+21+23+25+34 = 140 \equiv 2 \pmod{3}$ , so the number of chipped marbles must also be congruent to $2-0 \equiv 2 \pmod{3}$
The answer choices are congruent modulo 3 to $0$ $1$ $0$ $2$ , and $1$ respectively, so the only possible number of chipped marbles among these is $23$ . Indeed, if Jane takes the bags containing $19$ $25$ , and $34$ marbles and George takes the remaining bags containing $18$ and $21$ marbles, then Jane will have a total of $19+25+34 = 78$ marbles, which is twice as many as George's $18+21 = 39$ marbles, as desired. Thus the answer is precisely $\boxed{23}$

Answer: D 23
Problem: Consider the graphs of $y = Ax^2$ and $y^2+3 = x^2+4y$ , where $A$ is a positive constant and $x$ and $y$ are real variables. In how many points do the two graphs intersect?
$\mathrm{(A) \ }\text{exactly }4 \qquad \mathrm{(B) \ }\text{exactly }2 \qquad$
$\mathrm{(C) \  }\text{at least }1,\text{ but the number varies for different positive values of }A \qquad$
$\mathrm{(D) \  }0\text{ for at least one positive value of }A \qquad \mathrm{(E) \ }\text{none of these}$

Solution: Substituting $y = Ax^2$ into the equation $y^2+3 = x^2+4y$ gives \begin{align*}\left(Ax^2\right)+3 = x^2+4\left(Ax^2\right) &\iff A^2x^4+3 = x^2+4Ax^2 \\ &\iff A^2x^4-\left(4A+1\right)x^2+3 = 0 \\ &\iff x^2 = \frac{4A+1 \pm \sqrt{4A^2+8A+1}}{2A^2} \\ &\text{(using the quadratic formula)}.\end{align*} Now observe that since $A$ is positive, $4A^2+8A+1$ is also positive, so the square root will always give two distinct real values. Moreover, \[\left(4A+1\right)^2 = 16A^2+8A+1 > 4A^2+8A+1,\] so $4A+1-\sqrt{4A^2+8A+1} > 0$ , meaning that both solutions for $x^2$ are positive. Hence both solutions will give $2$ distinct values of $x$ (the positive and negative square roots), and each of these will correspond to a distinct point of intersection of the graphs, so there are $2 \cdot 2 = \boxed{4}$ points of intersection.

Answer: A 4
Problem: Consider the graphs of $y = Ax^2$ and $y^2+3 = x^2+4y$ , where $A$ is a positive constant and $x$ and $y$ are real variables. In how many points do the two graphs intersect?
$\mathrm{(A) \ }\text{exactly }4 \qquad \mathrm{(B) \ }\text{exactly }2 \qquad$
$\mathrm{(C) \  }\text{at least }1,\text{ but the number varies for different positive values of }A \qquad$
$\mathrm{(D) \  }0\text{ for at least one positive value of }A \qquad \mathrm{(E) \ }\text{none of these}$

Solution: Firstly, note that $y = Ax^2$ is an upward-facing parabola (since $A > 0$ ) whose vertex is at the origin. We now manipulate the equation of the second graph as follows: \begin{align*}y^2+3 = x^2+4y &\iff y^2-4y+3-x^2 = 0 \\ &\iff y^2-4y+4-x^2 = 1 \\ &\iff \frac{(y-2)^2}{1}-\frac{(x-0)^2}{1} = 1,\end{align*} showing that it is a vertical (upward- and downward-opening) hyperbola with center $(0,2)$ and asymptotes $y=x+1$ and $y=-x+1$ . It therefore remains to consider graphically where the parabola will intersect the hyperbola.
On the lower branch of the hyperbola, the maximum point is $(0,2-1) = (0,1)$ , which is above the vertex of the parabola. Therefore, by continuity and the symmetry of both the parabola and the hyperbola in the $y$ -axis, there are always exactly $2$ intersection points here.
For the top branch, as it approaches the asymptote $y = x+1$ , its slope also approaches that of this asymptote, which is $1$ . However, for any upward-opening parabola, the slope approaches infinity as $x$ does, so no matter how small $A$ is (i.e. how 'flat' the parabola is), the parabola will eventually overtake the hyperbola, giving a point of intersection with positive $x$ -coordinate. As above, symmetry gives another point of intersection with negative $x$ -coordinate, so that there are $2$ intersection points with this branch too.
Thus there are a total of $2+2 = \boxed{4}$ intersection points.

Answer: A 4
Problem: A wooden cube with edge length $n$ units (where $n$ is an integer $>2$ ) is painted black all over. By slices parallel to its faces, the cube is cut into $n^3$ smaller cubes each of unit edge length. If the number of smaller cubes with just one face painted black is equal to the number of smaller cubes completely free of paint, what is $n$
$\mathrm{(A)\ } 5 \qquad \mathrm{(B) \ }6 \qquad \mathrm{(C) \  } 7 \qquad \mathrm{(D) \  } 8 \qquad \mathrm{(E) \  }\text{none of these}$

Solution: Observe that if we remove the outer layer of unit cubes from the entire cube, what remains is a smaller cube of side length $(n-2)$ , which contains all of the unpainted cubes and no others. This shows that there are exactly $(n-2)^3$ unpainted cubes. Similarly, taking one face of the cube and removing the outer edge leaves a square of side length $(n-2)$ containing all of the cubes on that face with exactly one face painted. Making the same argument for the other $5$ faces as well, we deduce that there are a total of $6(n-2)^2$ cubes with only one face painted.
Accordingly, we require \begin{align*}(n-2)^3 = 6(n-2)^2 &\iff n-2 = 6 \qquad \text{(as } n > 2\text{, so } n-2 \neq 0\text{)} \\ &\iff n = \boxed{8}

Answer: D 8
Problem: How many integers $x$ satisfy the equation \[\left(x^2-x-1\right)^{x+2} = 1?\]
$\mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \  } 4 \qquad \mathrm{(D) \  } 5 \qquad \mathrm{(E) \  }\text{none of these}$

Solution: We recall that for real numbers $a$ and $b$ , there are exactly $3$ ways in which we can have $a^b = 1$ , namely $a = 1$ $b = 0$ and $a \neq 0$ ; or $a = -1$ and $b$ is an even integer.
The first case therefore gives \begin{align*}x^2-x-1 = 1 &\iff x^2-x-2 = 0 \\&\iff (x-2)(x+1) = 0 \\&\iff x = 2 \text{ or } x = -1.\end{align*}
Similarly, the second case gives $x+2 = 0$ , i.e. $x = -2$ , and this indeed gives $x^2-x-1 = 4+2-1 = 5 \neq 0$ , so $x = -2$ is a further valid solution.
Lastly, for the third case, we have \begin{align*}x^2-x-1 = -1 &\iff x^2-x = 0 \\&\iff x(x-1) = 0 \\&\iff x = 0 \text{ or } x = 1,\end{align*} but $x = 1$ would give $x+2 = 3$ , which is odd, whereas $x = 0$ gives $x+2 = 2$ , which is even. Therefore, this case gives only one further solution, namely $x = 0$
Accordingly, the possible values of $x = -2$ $-1$ $0$ , or $2$ , yielding a total of $\boxed{4}$ solutions.

Answer: C 4
Problem: In a circle with center $O$ $AD$ is a diameter, $ABC$ is a chord, $BO = 5$ and $\angle ABO = \ \stackrel{\frown}{CD} \ = 60^{\circ}$ . Then the length of $BC$ is
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair O=origin, A=dir(35), C=dir(155), D=dir(215), B=intersectionpoint(dir(125)--O, A--C); draw(C--A--D^^B--O^^Circle(O,1)); pair point=O; label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$O$", O, dir(305)); label("$5$", B--O, dir(O--B)*dir(90)); label("$60^\circ$", dir(185), dir(185)); label("$60^\circ$", B+0.05*dir(-25), dir(-25));[/asy]
$\mathrm{(A)\ } 3 \qquad \mathrm{(B) \ }3+\sqrt{3} \qquad \mathrm{(C) \  } 5-\frac{\sqrt{3}}{2} \qquad \mathrm{(D) \  } 5 \qquad \mathrm{(E) \  }\text{none of the above}$

Solution: Since $\angle CAD$ is an angle inscribed in a $60{^\circ}$ arc, we obtain $\angle CAD =\frac{60^{\circ}}{2} = 30^{\circ}$ , so $\triangle ABO$ is a $30^{\circ}$ $60^{\circ}$ $90^{\circ}$ right triangle. This gives $AO = BO\sqrt{3} = 5\sqrt{3}$ and $AB = 2BO = 10$ , and now since $AD$ is a diameter, $AD = 2AO = 10\sqrt{3}$ . The fact that $AD$ is a diameter also means that $\angle ACD = 90^{\circ}$ , so $\triangle ACD$ is again a $30^{\circ}$ $60^{\circ}$ $90^{\circ}$ right triangle, yielding $CD = \frac{1}{2}AD = 5\sqrt{3}$ and $AC = \frac{\sqrt{3}}{2}AD = 15$ , which finally gives $BC = AC-AB = 15-10 = \boxed{5}$

Answer: D 5
Problem: A non-zero digit is chosen in such a way that the probability of choosing digit $d$ is $\log_{10}{(d+1)}-\log_{10}{d}$ . The probability that the digit $2$ is chosen is exactly $1/2$ the probability that the digit chosen is in the set
$\mathrm{(A)\ } \{2,3\} \qquad \mathrm{(B) \ }\{3,4\} \qquad \mathrm{(C) \  } \{4,5,6,7,8\} \qquad \mathrm{(D) \  } \{5,6,7,8,9\} \qquad \mathrm{(E) \  }\{4,5,6,7,8,9\}$

Solution: We have $\log_{10}{(d+1)}-\log_{10}{d} = \log_{10}{\left(\frac{d+1}{d}\right)}$ , so the probability of choosing $2$ is $\log_{10}{\left(\frac{3}{2}\right)}$ . The probability that the digit chosen is in the set must therefore be \begin{align*}2\log_{10}{\left(\frac{3}{2}\right)} = &\log_{10}{\left(\left(\frac{3}{2}\right)^2\right)} \\ = &\log_{10}{\left(\frac{9}{4}\right)} \\ = &\log_{10}{9}-\log_{10}{4} \\ = &\left(\log_{10}{9}-\log_{10}{8}\right)+\left(\log_{10}{8}-\log_{10}{7}\right)+\left(\log_{10}{7}-\log_{10}{6}\right) \\ &+\left(\log_{10}{6}-\log_{10}{5}\right)+\left(\log_{10}{5}-\log_{10}{4}\right),\end{align*} which, by definition, is the probability that the digit chosen is in the set $\boxed{4,5,6,7,8}$

Answer: C 4,5,6,7,8
Problem: The volume of a certain rectangular solid is $8$ cm , its total surface area is $32$ cm , and its three dimensions are in geometric progression. The sums of the lengths in cm of all the edges of this solid is
$\mathrm{(A)\ } 28 \qquad \mathrm{(B) \ }32 \qquad \mathrm{(C) \  } 36 \qquad \mathrm{(D) \  } 40 \qquad \mathrm{(E) \  }44$

Solution: As the dimensions are in geometric progression, let them be $\frac{b}{r}$ $b$ , and $br$ cm, so the volume is $\left(\frac{b}{r}\right)(b)(br) = b^3$ , giving $b^3 = 8$ and thus $b = 2$ . The surface area condition now yields \begin{align*}2\left(\frac{2}{r}\right)(2)+2(2)(2r)+2(2r)\left(\frac{2}{r}\right) = 32 &\iff \frac{8}{r}+8+8r = 32 \\ &\iff r+\frac{1}{r} = 3 \\ &\iff r^2-3r+1 = 0 \\ &\iff r = \frac{3 \pm \sqrt{5}}{2}.\end{align*}
Since \[\frac{3-\sqrt{5}}{2} = \frac{1}{\left(\frac{3+\sqrt{5}}{2}\right)},\] the two possible values of $r$ do not actually give different dimensions, but merely determine whether they are in increasing or decreasing order. Therefore, without loss of generality, we take $r = \frac{3+\sqrt{5}}{2}$ , giving the dimensions (in cm) as \begin{align*}&\frac{2}{\left(\frac{3+\sqrt{5}}{2}\right)}, 2, \text{ and } 2\left(\frac{3+\sqrt{5}}{2}\right) \\ &= \frac{4}{3+\sqrt{5}}, 2, \text{ and } 3+\sqrt{5} \\ &= 3-\sqrt{5}, 2, \text{ and } 3+\sqrt{5}.\end{align*}
As there are $4$ edges with each of these distinct lengths, it follows that the sum of all the edge lengths (in cm) is \begin{align*}4\left(3-\sqrt{5}+2+3+\sqrt{5}\right) &= 4(8) \\ &= \boxed{32}

Answer: B 32
Problem: The volume of a certain rectangular solid is $8$ cm , its total surface area is $32$ cm , and its three dimensions are in geometric progression. The sums of the lengths in cm of all the edges of this solid is
$\mathrm{(A)\ } 28 \qquad \mathrm{(B) \ }32 \qquad \mathrm{(C) \  } 36 \qquad \mathrm{(D) \  } 40 \qquad \mathrm{(E) \  }44$

Solution: Similarly to in Solution 1, we let the dimensions (in cm) be $b$ $br$ , and $br^2$ , so that the volume condition gives $8 = b^3r^3 = (br)^3$ and thus $br = 2$ . The surface area condition now becomes \begin{align*}2(b)(br)+2(br)\left(br^2\right)+2\left(br^2\right)(b) = 32 &\iff b^2r+b^2r^2+b^2r^3 = 16 \\&\iff br\left(b+br+br^2\right) = 16,\end{align*} so substituting $br = 2$ from above immediately gives \[b+br+br^2 = \frac{16}{2} = 8,\] and hence, without needing to actually compute the dimensions, we deduce that the sum of the edge lengths (in cm) is \[4b + 4br + 4br^2 = 4(8) = \boxed{32}.\]

Answer: B 32