Evaluate \beginaligned &\log_10(\tan1°)\ &\quad+\log_10(\tan2°)\ &\quad+\log_10(\tan3°)+\cdots\ &\quad+\log_10(\tan88°)\ &\quad+\log_10(\tan89°). \endaligned

(B) \frac12\log_10((\sqrt3)/(2))

1987 AMC 12 — Problems & Online Timed Exam

Timed

1:15:00

1.

$(1+x^2)(1-x^3)$ equals

$1-x^5$

$1-x^6$

$1+x^2-x^3$

$1+x^2-x^3-x^5$

$1+x^2-x^3-x^6$

Answer: D

Concepts:polynomial multiplication distributive property

Difficulty rating: 960

Small Hint:

Distribute each term of the first binomial across the second

Big Hint:

The product $x^2(-x^3)$ contributes the highest-degree term

Solution:

Expanding gives $\begin{aligned} (1+x^2)(1-x^3) &=1+x^2-x^3\\ &\qquad-x^5. \end{aligned}$

Thus the correct answer is D.

2.

As shown in the figure, a triangular corner with side lengths $DB=EB=1$ is cut from equilateral triangle $ABC$ of side length $3.$ The perimeter of the remaining quadrilateral $ADEC$ is

$6$

$6\frac12$

$7$

$7\frac12$

$8$

Answer: E

Concepts:equilateral triangle perimeter angle-side relationship

Difficulty rating: 1030

Small Hint:

Determine the angle between $BD$ and $BE$

Big Hint:

Start from the original perimeter, remove $BD$ and $BE,$ and add $DE$

Solution:

Since $\angle DBE=60^\circ$ and $DB=EB=1,$ triangle $DBE$ is equilateral, so $DE=1.$ The original perimeter is $9.$ Cutting off the corner removes two unit segments and adds one: $9-1-1+1=8.$

Thus the correct answer is E.

3.

How many primes less than $100$ have $7$ as the ones digit? (Assume the usual base $10$ representation.)

$4$

$5$

$6$

$7$

$8$

Answer: C

Concepts:prime numbers divisibility enumeration

Difficulty rating: 1000

Small Hint:

List every positive integer below $100$ whose ones digit is $7$

Big Hint:

For numbers below $100,$ trial division only requires primes through their square roots

Solution:

The primes are $7,\ 17,\ 37,\ 47,\ 67,\ 97.$ The other candidates are composite: $27,57,77,$ and $87.$ Hence there are $6$ such primes.

Thus the correct answer is C.

4.

$\frac{2^1+2^0+2^{-1}}{2^{-2}+2^{-3}+2^{-4}}$ equals

$6$

$8$

$\frac{31}{2}$

$24$

$512$

Answer: B

Concepts:negative exponents geometric scaling fractions

Difficulty rating: 1260

Small Hint:

Rewrite every negative power of $2$ as a reciprocal

Big Hint:

Compare corresponding numerator and denominator terms by a common factor

Solution:

The numerator is $2+1+\frac12=\frac72,$ while the denominator is $\frac14+\frac18+\frac1{16}=\frac7{16}.$ Their quotient is $\frac{\frac{7}{2}}{\frac{7}{16}}=8.$

Thus the correct answer is B.

5.

A student recorded the exact percentage frequency distribution for a set of measurements, as shown below. However, the student neglected to indicate $N,$ the total number of measurements. What is the smallest possible value of $N?$

measured value	percent frequency
$0$	$12.5$
$1$	$0$
$2$	$50$
$3$	$25$
$4$	$12.5$
	$100$

$5$

$8$

$16$

$25$

$50$

Answer: B

Concepts:percentage frequency divisibility sample size

Difficulty rating: 1150

Small Hint:

Express each nonzero percentage as a reduced fraction of the total

Big Hint:

Every resulting frequency must be a whole number

Solution:

Since $12.5\%=\frac{1}{8},$ the total $N$ must be divisible by $8.$ Taking $N=8$ gives frequencies $1,0,4,2,1,$ all whole numbers. Thus the smallest possible total is $8.$

Therefore the correct answer is B.

6.

In the $\triangle ABC$ shown, $D$ is some interior point, and $x,$ $y,$ $z,$ $w$ are the measures of angles in degrees. Solve for $x$ in terms of $y,$ $z,$ and $w.$

$w-y-z$

$w-2y-2z$

$180-w-y-z$

$2w-y-z$

$180-w+y+z$

Answer: A

Concepts:triangle angle sum angle subtraction interior point

Difficulty rating: 1290

Small Hint:

Name the two unlabelled angles at $A$ and $B$ inside triangle $ADB$

Big Hint:

Subtract the angle sum of $\triangle ADB$ from that of $\triangle ABC$

Solution:

Let the unlabelled angles of triangle $ADB$ at $A$ and $B$ be $\alpha$ and $\beta.$ Then $\alpha+\beta+w=180^\circ.$ The angle sum of triangle $ABC$ gives $(\alpha+y)+(\beta+z)+x=180^\circ.$ Subtracting yields $x+y+z=w,$ so $x=w-y-z.$

Thus the correct answer is A.

7.

If $a-1=b+2=c-3=d+4,$ which of the four quantities $a,$ $b,$ $c,$ $d$ is the largest?

$a$

$b$

$c$

$d$

no one is always largest

Answer: C

Concepts:equal expressions variable comparison linear relations

Difficulty rating: 890

Small Hint:

Set all four expressions equal to one common value

Big Hint:

Solve each variable as the common value plus or minus a constant

Solution:

If the common value is $k,$ then $\begin{aligned} a&=k+1,& b&=k-2,\\ c&=k+3,& d&=k-4. \end{aligned}$ Therefore $c$ is always the largest.

Thus the correct answer is C.

8.

In the figure the sum of the distances $AD$ and $BD$ is

between $10$ and $11$

$12$

between $15$ and $16$

between $16$ and $17$

$17$

Answer: C

Concepts:Pythagorean theorem distance estimation

Difficulty rating: 1260

Small Hint:

Use the $3$ - $4$ - $5$ right triangle to find $BD$

Big Hint:

The horizontal and vertical displacements from $A$ to $D$ are $10$ and $4$

Solution:

Triangle $BCD$ is a $3$ - $4$ - $5$ right triangle, so $BD=5.$ From $A$ to $D,$ the horizontal displacement is $13-3=10$ and the vertical displacement is $4,$ hence $AD=\sqrt{10^2+4^2}=\sqrt{116}.$ Since $10\lt\sqrt{116}\lt11,$ the sum $AD+BD$ is between $15$ and $16.$

Thus the correct answer is C.

9.

The first four terms of an arithmetic sequence are $a,$ $x,$ $b,$ $2x.$ The ratio of $a$ to $b$ is

$\frac14$

$\frac13$

$\frac12$

$\frac23$

$2$

Answer: B

Concepts:arithmetic sequence common difference ratio

Difficulty rating: 1230

Small Hint:

Write all four terms using $a$ and a common difference

Big Hint:

Use the equation relating the second term $x$ to the fourth term $2x$

Solution:

Let the common difference be $d.$ Then $x=a+d,\ b=a+2d,$ and $2x=a+3d.$ Thus $2(a+d)=a+3d,$ so $a=d$ and $b=3d.$ Therefore $\frac{a}{b}=\frac{1}{3}.$

Thus the correct answer is B.

10.

How many ordered triples $(a,b,c)$ of nonzero real numbers have the property that each number is the product of the other two?

$1$

$2$

$3$

$4$

$5$

Answer: D

Concepts:system of equations sign patterns product

Difficulty rating: 1670

Small Hint:

Translate the condition into $a=bc,\ b=ca,\ c=ab$

Big Hint:

Multiply the equations and use the nonzero condition to determine the possible magnitudes

Solution:

Multiplying $a=bc,\ b=ca,\ c=ab$ gives $abc=(abc)^2.$ Since none of the variables is zero, $abc=1.$ Also $abc=a^2=b^2=c^2,$ so each variable is $1$ or $-1.$ Product $1$ allows either no negative signs or exactly two, giving $1+3=4$ ordered triples.

Thus the correct answer is D.

11.

Let $c$ be a constant. The simultaneous equations $\begin{aligned} x-y&=2,\\ cx+y&=3. \end{aligned}$ have a solution $(x,y)$ inside Quadrant I if and only if

$c=-1$

$c\gt-1$

$c\lt\frac32$

$0\lt c\lt\frac32$

$-1\lt c\lt\frac32$

Answer: E

Concepts:linear system inequalities coordinate quadrants

Difficulty rating: 1500

Small Hint:

Solve the two equations for $x$ and $y$ in terms of $c$

Big Hint:

Quadrant I requires both coordinates to be strictly positive

Solution:

Solving the system gives $x=\frac5{c+1},\qquad y=\frac{3-2c}{c+1}.$ The condition $x\gt0$ requires $c\gt-1.$ With this positive denominator, $y\gt0$ is equivalent to $c\lt\frac{3}{2}.$ Hence $-1\lt c\lt\frac{3}{2}.$

Thus the correct answer is E.

12.

In an office, at various times during the day the boss gives the secretary a letter to type, each time putting the letter on top of the pile in the secretary’s in-box. When there is time, the secretary takes the top letter off the pile and types it. If there are five letters in all, and the boss delivers them in the order $1,$ $2,$ $3,$ $4,$ $5,$ which of the following could not be the order in which the secretary types them?

$1,$ $2,$ $3,$ $4,$ $5$

$2,$ $4,$ $3,$ $5,$ $1$

$3,$ $2,$ $4,$ $1,$ $5$

$4,$ $5,$ $2,$ $3,$ $1$

$5,$ $4,$ $3,$ $2,$ $1$

Answer: D

Concepts:stack order permutations logical feasibility

Difficulty rating: 1590

Small Hint:

Letters still in the pile always remain in increasing order from bottom to top

Big Hint:

After letter $4$ is typed, examine which smaller letter must be on top before $2$ can be typed

Solution:

To type $4$ first in choice D, letters $1,2,3,4$ must all have been delivered, leaving $3$ on top after $4$ is removed. Letter $5$ can then be delivered and typed, but $3$ is still above $2,$ so $2$ cannot be typed next. Thus $4,5,2,3,1$ is impossible. Each other listed order can be produced by interleaving deliveries and typings.

Therefore the correct answer is D.

13.

A long piece of paper $5$ cm wide is made into a roll for cash registers by wrapping it $600$ times around a cardboard tube of diameter $2$ cm, forming a roll $10$ cm in diameter. Approximate the length of the paper in meters. (Pretend the paper forms $600$ concentric circles with diameters evenly spaced from $2$ cm to $10$ cm.)

$36\pi$

$45\pi$

$60\pi$

$72\pi$

$90\pi$

Answer: A

Concepts:arithmetic sequence circumference unit conversion

Difficulty rating: 1490

Small Hint:

The $600$ diameters form an arithmetic sequence

Big Hint:

Use the average diameter to find the sum of all $600$ circumferences

Solution:

The average of the first and last diameters is $\frac{2+10}{2}=6\text{ cm}.$ Hence the total length is approximately $600(6\pi)=3600\pi$ cm, or $36\pi$ meters.

Thus the correct answer is A.

14.

$ABCD$ is a square and $M$ and $N$ are the midpoints of $BC$ and $CD,$ respectively. Then $\sin\theta=$

$\frac{\sqrt5}{5}$

$\frac35$

$\frac{\sqrt{10}}5$

$\frac45$

none of these

Answer: B

Concepts:vectors sine of angle midpoint geometry

Difficulty rating: 1530

Small Hint:

Assign coordinates to the square and write vectors $\overrightarrow{AM}$ and $\overrightarrow{AN}$

Big Hint:

Use the determinant formula for the sine of the angle between two vectors

Solution:

Take the square to have side $2,$ with $A=(0,0),\ M=(1,2),$ and $N=(2,1).$ Then $\begin{aligned} \sin\theta &=\frac{|\det((1,2),(2,1))|} {\sqrt5\sqrt5}\\ &=\frac{|1-4|}{5} =\frac35. \end{aligned}$

Thus the correct answer is B.

15.

If $(x,y)$ is a solution to the system $\begin{aligned} xy&=6,\\ x^2y+xy^2+x+y&=63, \end{aligned}$ find $x^2+y^2.$

$13$

$\frac{1173}{32}$

$55$

$69$

$81$

Answer: D

Concepts:symmetric expressions factoring system of equations

Difficulty rating: 1470

Small Hint:

Factor the second equation using $x+y$

Big Hint:

Use $x^2+y^2=(x+y)^2-2xy$

Solution:

The second equation becomes $\begin{aligned} xy(x+y)&+(x+y)\\ &=(xy+1)(x+y)\\ &=63. \end{aligned}$ Since $xy=6,$ we get $x+y=9.$ Therefore $\begin{aligned} x^2+y^2 &=(x+y)^2-2xy\\ &=81-12=69. \end{aligned}$

Thus the correct answer is D.

16.

A cryptographer devises the following method for encoding positive integers. First, the integer is expressed in base $5.$ Second, a $1$ -to- $1$ correspondence is established between the digits that appear in the expressions in base $5$ and the elements of the set $\{V,W,X,Y,Z\}.$ Using this correspondence, the cryptographer finds that three consecutive integers in increasing order are coded as $VYZ,$ $VYX,$ $VVW,$ respectively. What is the base- $10$ expression for the integer coded as $XYZ?$

$48$

$71$

$82$

$108$

$113$

Answer: D

Concepts:base representation carrying bijection

Difficulty rating: 1860

Small Hint:

Compare the final digits of $VYZ$ and $VYX$

Big Hint:

The change from $VYX$ to $VVW$ forces a base- $5$ carry

Solution:

Because $VYX$ is one more than $VYZ,$ the digit for $X$ is one more than the digit for $Z.$ The next increment changes $VYX$ to $VVW,$ so $X=4,\ W=0,$ and $Z=3.$ The carry changes $Y$ to $V,$ leaving $Y=1,\ V=2.$ Thus $\begin{aligned} XYZ_5&=413_5,\\ 413_5&=4\cdot25+5+3\\ &=108. \end{aligned}$

Thus the correct answer is D.

17.

In a mathematics competition, the sum of the scores of Bill and Dick equalled the sum of the scores of Ann and Carol. If the scores of Bill and Carol had been interchanged, then the sum of the scores of Ann and Carol would have exceeded the sum of the scores of the other two. Also, Dick’s score exceeded the sum of the scores of Bill and Carol. Determine the order in which the four contestants finished, from highest to lowest. Assume all scores were nonnegative.

Dick, Ann, Carol, Bill

Dick, Ann, Bill, Carol

Dick, Carol, Bill, Ann

Ann, Dick, Carol, Bill

Ann, Dick, Bill, Carol

Answer: E

Concepts:inequalities ordering algebraic comparison

Difficulty rating: 1830

Small Hint:

Represent the four scores by $A,B,C,D$ in name order

Big Hint:

Add and subtract the equality $A+C=B+D$ and the inequality $A+B\gt C+D$

Solution:

Let $A,B,C,D$ be the scores of Ann, Bill, Carol, and Dick. The conditions are $\begin{aligned} A+C&=B+D,\\ A+B&\gt C+D,\\ D&\gt B+C. \end{aligned}$ Adding the first two comparisons gives $A\gt D.$ Subtracting the equality from the inequality gives $B\gt C.$ Finally, $D\gt B+C\ge B.$ Hence $A\gt D\gt B\gt C.$

Thus the correct answer is E.

18.

It takes $A$ algebra books (all the same thickness) and $H$ geometry books (all the same thickness, which is greater than that of an algebra book) to completely fill a certain shelf. Also, $S$ of the algebra books and $M$ of the geometry books would fill the same shelf. Finally, $E$ of the algebra books alone would fill this shelf. Given that $A,$ $H,$ $S,$ $M,$ $E$ are distinct positive integers, it follows that $E$ is

$\frac{AM+SH}{M+H}$

$\frac{AM^2+SH^2}{M^2+H^2}$

$\frac{AH-SM}{M-H}$

$\frac{AM-SH}{M-H}$

$\frac{AM^2-SH^2}{M^2-H^2}$

Answer: D

Concepts:linear equations rates and mixtures algebraic elimination

Difficulty rating: 1980

Small Hint:

Let $a$ and $g$ be the two book thicknesses and normalize the shelf length to $1$

Big Hint:

Eliminate the geometry-book thickness from $Aa+Hg=1$ and $Sa+Mg=1$

Solution:

Let the shelf length be $1,$ and let $a,g$ be the algebra- and geometry-book thicknesses. Then $\begin{aligned} Aa+Hg&=1,\\ Sa+Mg&=1,\qquad Ea=1. \end{aligned}$ Multiplying the first equation by $M,$ the second by $H,$ and subtracting gives $(AM-SH)a=M-H.$ Hence $E=\frac1a=\frac{AM-SH}{M-H}.$

Thus the correct answer is D.

19.

Which of the following is closest to $\sqrt{65}-\sqrt{63}?$

$0.12$

$0.13$

$0.14$

$0.15$

$0.16$

Answer: B

Concepts:radical rationalization estimation inequalities

Difficulty rating: 1490

Small Hint:

Rationalize the difference of the two square roots

Big Hint:

Compare $\sqrt{65}+\sqrt{63}$ with $16$ to decide which side of $0.125$ the result lies on

Solution:

Rationalizing gives $\sqrt{65}-\sqrt{63}=\frac2{\sqrt{65}+\sqrt{63}}.$ The denominator is less than $16,$ because its square is $128+2\sqrt{4095}\lt256.$ Thus the value is greater than $\frac{2}{16}=0.125.$ Also each radical exceeds $7.5,$ so the value is less than $\frac{2}{15}<0.134.$ It is therefore closer to $0.13$ than to any other choice.

Thus the correct answer is B.

20.

Evaluate $\begin{aligned} &\log_{10}(\tan1^\circ)\\ &\quad+\log_{10}(\tan2^\circ)\\ &\quad+\log_{10}(\tan3^\circ)+\cdots\\ &\quad+\log_{10}(\tan88^\circ)\\ &\quad+\log_{10}(\tan89^\circ). \end{aligned}$

$0$

$\frac12\log_{10}\left(\frac{\sqrt3}{2}\right)$

$\frac12\log_{10}2$

$1$

none of these

Answer: A

Concepts:logarithm product rule complementary angles tangent

Difficulty rating: 1570

Small Hint:

Pair the term at angle $k^\circ$ with the term at angle $(90-k)^\circ$

Big Hint:

Use $\tan(90^\circ-\theta)=\cot\theta$ before combining logarithms

Solution:

For each $k=1,\ldots,44,$ $\tan k^\circ\tan(90^\circ-k^\circ)=1.$ Thus each paired sum of logarithms is $\log_{10}1=0.$ The remaining middle term is $\log_{10}(\tan45^\circ)=0.$ Hence the whole sum is $0.$

Thus the correct answer is A.

21.

There are two natural ways to inscribe a square in a given isosceles right triangle. If it is done as in Figure $1$ below, then one finds that the area of the square is $441\text{ cm}^2.$ What is the area (in $\text{cm}^2$ ) of the square inscribed in the same $\triangle ABC$ as shown in Figure $2$ below?

$378$

$392$

$400$

$441$

$484$

Answer: B

Difficulty rating: 1980

Small Hint:

Use Figure $1$ to determine the leg length of the isosceles right triangle

Big Hint:

In Figure $2,$ compare the side of the tilted square with the two equal leg segments cut off by its opposite side

Solution:

The first square has side $21,$ so the triangle’s legs have length $42.$ Let $s$ be the side of the tilted square. Its side on the hypotenuse is parallel to the opposite side joining the legs. That opposite side has endpoints $(0,k)$ and $(k,0),$ so $s=k\sqrt2.$ The distance between the parallel lines $x+y=k$ and $x+y=42$ is also $s,$ giving $s=\frac{42-k}{\sqrt2},\qquad k=\frac{s}{\sqrt2}.$ Hence $s=14\sqrt2,$ and its area is $s^2=392.$

Thus the correct answer is B.

22.

A ball was floating in a lake when the lake froze. The ball was removed (without breaking the ice), leaving a hole $24$ cm across at the top and $8$ cm deep. What was the radius of the ball (in centimeters)?

$8$

$12$

$13$

$8\sqrt3$

$6\sqrt6$

Answer: C

Concepts:sphere cross-section right triangle chord geometry

Difficulty rating: 1790

Small Hint:

Take a vertical cross-section through the center of the circular hole

Big Hint:

The half-chord is $12,$ while the center-to-ice distance is $r-8$

Solution:

In a vertical cross-section, half the $24$ -cm hole is a chord segment of length $12.$ If the sphere radius is $r,$ the distance from its center to the ice plane is $r-8.$ The resulting right triangle gives $(r-8)^2+12^2=r^2.$ Solving yields $16r=208,$ so $r=13.$

Thus the correct answer is C.

23.

If $p$ is a prime and both roots of $x^2+px-444p=0$ are integers, then

$1\lt p\le11$

$11\lt p\le21$

$21\lt p\le31$

$31\lt p\le41$

$41\lt p\le51$

Answer: D

Concepts:integer roots prime divisibility Vieta’s formulas

Difficulty rating: 2110

Small Hint:

For an integer root $x,$ rewrite the equation as $x^2=p(444-x)$

Big Hint:

Since $p$ is prime, write $x=np$ and factor $444$

Solution:

An integer root satisfies $x^2=p(444-x).$ Thus $x$ is divisible by $p,$ so write $x=np.$ Substitution gives $n(n+1)p=444=2^2\cdot3\cdot37.$ The consecutive product condition works only with $p=37,$ for which $n=3$ or $-4.$ The roots are $111$ and $-148,$ so $31\lt p\le41.$

Thus the correct answer is D.

24.

How many polynomial functions $f$ of degree $\ge1$ satisfy $f(x^2)=[f(x)]^2=f(f(x))?$

$0$

$1$

$2$

finitely many but more than $2$

infinitely many

Answer: B

Concepts:polynomial degree leading coefficient functional equation

Difficulty rating: 2310

Small Hint:

Compare the degrees and leading coefficients of the three polynomials

Big Hint:

After proving $f$ is a monic quadratic, compare coefficients in $f(x^2)=[f(x)]^2$

Solution:

Let $f$ have degree $n\ge1$ and leading coefficient $a.$ The three expressions have degrees $2n,2n,n^2,$ so $n^2=2n$ and therefore $n=2.$ Comparing leading coefficients in $f(x^2)=[f(x)]^2$ gives $a=a^2,$ hence $a=1.$

Write $f(x)=x^2+bx+c.$ Then $x^4+bx^2+c=(x^2+bx+c)^2.$ The cubic coefficient forces $b=0,$ and then the quadratic coefficient forces $c=0.$ Thus the only candidate is $f(x)=x^2,$ which indeed satisfies all three expressions. There is exactly one function.

Thus the correct answer is B.

25.

$ABC$ is a triangle: $A=(0,0),$ $B=(36,15),$ and both the coordinates of $C$ are integers. What is the minimum area $\triangle ABC$ can have?

$\frac12$

$1$

$\frac32$

$\frac{13}{2}$

there is no minimum

Answer: C

Concepts:coordinate geometry determinant area greatest common divisor

Difficulty rating: 2170

Small Hint:

Write $C=(u,v)$ and use the determinant formula for triangle area

Big Hint:

Find the smallest positive value of $|36v-15u|$ using $\gcd(36,15)$

Solution:

For $C=(u,v),$ the area is $\frac12|36v-15u|.$ Since $\gcd(36,15)=3,$ the smallest possible positive value of the determinant is at least $3.$ It is attained, for example, by $u=7,\ v=3,$ since $36(3)-15(7)=3.$ Therefore the minimum area is $\frac{3}{2}.$

Thus the correct answer is C.

26.

The amount $2.5$ is split into two nonnegative real numbers uniformly at random, for instance, into $2.143$ and $0.357,$ or into $\sqrt3$ and $2.5-\sqrt3.$ Then each number is rounded to its nearest integer, for instance, $2$ and $0$ in the first case above, $2$ and $1$ in the second. What is the probability that the two integers sum to $3?$

$\frac14$

$\frac25$

$\frac12$

$\frac35$

$\frac34$

Answer: B

Concepts:continuous probability rounding interval length

Difficulty rating: 1980

Small Hint:

Let the first number be $x$ uniformly distributed on $[0,2.5]$

Big Hint:

Find the intervals on which $x$ and $2.5-x$ round to integers whose sum is $3$

Solution:

Let the first part be $x\in[0,2.5],$ so the second is $2.5-x.$ Ignoring endpoints of probability zero, the rounded values sum to $3$ exactly when $\begin{gathered} \frac12\lt x\lt1\\ \text{or}\\ \frac32\lt x\lt2. \end{gathered}$ These intervals have total length $1,$ out of a sample interval of length $2.5.$ The probability is therefore $\frac{1}{2.5}=\frac{2}{5}.$

Thus the correct answer is B.

27.

A cube of cheese $C=\{(x,y,z)\mid0\le x,y,z\le1\}$ 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.)

$5$

$6$

$7$

$8$

$9$

Answer: B

Concepts:three-dimensional geometry coordinate ordering permutations

Difficulty rating: 1940

Small Hint:

Within one piece, the relative order of $x,y,z$ cannot change

Big Hint:

Count the strict orderings of three distinct coordinates

Solution:

The three planes are precisely the boundaries where two coordinates are equal. Away from them, each piece is determined by a strict ordering of $x,y,z,$ such as $x\lt y\lt z.$ Every one of the $3!=6$ orderings occurs inside the cube, so there are $6$ pieces.

Thus the correct answer is B.

28.

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

$a$

$b$

$c$

$-a$

$-b$

Answer: D

Concepts:complex conjugates unit circle Vieta’s formulas

Difficulty rating: 2340

Small Hint:

For a complex number $r$ on the unit circle, compare $\frac{1}{r}$ with $\overline r$

Big Hint:

Real polynomial coefficients make the sum of the roots real

Solution:

If $|r|=1,$ then $\frac{1}{r}=\overline r.$ Therefore the sum of the reciprocals is the conjugate of the sum of the roots. By Vieta’s formulas, the sum of the roots is $-a,$ which is real. Its conjugate is still $-a.$

Thus the correct answer is D.

29.

Consider the sequence of numbers defined recursively by $t_1=1$ and for $n\gt1$ by $t_n=1+t_{\frac{n}{2}}$ 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

$15$

$17$

$19$

$21$

$23$

Answer: A

Concepts:recursive sequence inverse recursion rational numbers

Difficulty rating: 2440

Small Hint:

Values greater than $1$ come from even indices; values between $0$ and $1$ come from odd indices

Big Hint:

Reverse the recursion by subtracting $1$ from values above $1$ and taking reciprocals of values below $1$

Solution:

Let $N(r)$ be the index at which the value $r$ occurs. Reversing the recursion gives $N(r)=2N(r-1)$ for $r\gt1,$ and $N(r)=N(\frac{1}{r})+1$ for $0\lt r\lt1.$

Starting with $N(1)=1,$ repeated use gives $N(2)=2,$ $N(\frac{1}{2})=3,$ $N(\frac{3}{2})=6,$ and $N(\frac{2}{3})=7.$ Continuing gives $N(\frac{5}{3})=14,$ $N(\frac{8}{3})=28,$ $N(\frac{3}{8})=29,$ and $N(\frac{11}{8})=58.$

Next, $N(\frac{8}{11})=59,$ $N(\frac{19}{11})=118,$ $N(\frac{11}{19})=119,$ $N(\frac{30}{19})=238,$ $N(\frac{49}{19})=476,$ and $N(\frac{68}{19})=952.$ Finally, $N(\frac{87}{19})=1904$ and $N(\frac{19}{87})=1905.$ Thus $n=1905,$ whose digit sum is $1+9+0+5=15.$

Therefore the correct answer is A.

30.

In the figure, $\triangle ABC$ has $\angle A=45^\circ$ and $\angle B=30^\circ.$ A line $DE,$ with $D$ on $AB$ and $\angle ADE=60^\circ,$ divides $\triangle ABC$ into two pieces of equal area. (Note: the figure may not be accurate; perhaps $E$ is on $CB$ instead of $AC.$ ) The ratio $\frac{AD}{AB}$ is

$\frac1{\sqrt2}$

$\frac2{2+\sqrt2}$

$\frac1{\sqrt3}$

$\frac1{\sqrt[3]{6}}$

$\frac1{\sqrt[4]{12}}$

Answer: E

Concepts:coordinate geometry triangle area similar scaling trigonometry

Difficulty rating: 2400

Small Hint:

Scale so $AB=1$ and place $A=(0,0),\ B=(1,0)$

Big Hint:

Find the height of $C$ and the height of $E$ in terms of $d=AD$ , then equate half the total area to the area of $\triangle ADE$

Solution:

Set $AB=1,\ A=(0,0),$ and $B=(1,0).$ Since $\angle A=45^\circ,$ side $AC$ lies on $y=x.$ The line through $B$ making angle $30^\circ$ with $BA$ meets it at height $h_C=\frac1{1+\sqrt3}.$ Let $d=AD.$ The ray $DE$ has slope $-\sqrt3,$ so its intersection with $y=x$ has height $h_E=\frac{\sqrt3\,d}{1+\sqrt3}.$ Equal areas require $\frac12d h_E=\frac12\left(\frac12h_C\right).$ Therefore $\begin{aligned} \frac{\sqrt3\,d^2}{2(1+\sqrt3)} &=\frac1{4(1+\sqrt3)},\\ d^2&=\frac1{2\sqrt3} =\frac1{\sqrt{12}}. \end{aligned}$ Hence $\frac{AD}{AB}=d=\frac{1}{\sqrt[4]{12}}.$

Thus the correct answer is E.