1988 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

$\sqrt8+\sqrt{18}=$

$\sqrt{20}$

$2(\sqrt2+\sqrt3)$

$7$

$5\sqrt2$

$2\sqrt{13}$

Concepts:simplifying radicals like terms

Difficulty rating: 890

Small Hint:

Extract the largest square factor from each radicand

Big Hint:

After simplifying, the two radicals are like terms

Solution:

We have $\sqrt8=2\sqrt2$ and $\sqrt{18}=3\sqrt2.$ Their sum is $5\sqrt2.$

Thus the correct answer is D.

2.

Triangles $ABC$ and $XYZ$ are similar, with $A$ corresponding to $X$ and $B$ to $Y.$ If $AB=3,$ $BC=4,$ and $XY=5,$ then $YZ$ is

$3\frac34$

$6$

$6\frac14$

$6\frac23$

$8$

Difficulty rating: 920

Small Hint:

Match $AB$ with $XY$ and $BC$ with $YZ$

Big Hint:

Use the scale factor from the first triangle to the second

Solution:

The scale factor is $\frac{XY}{AB}=\frac{5}{3}.$ Therefore $YZ=(\frac{5}{3})BC=(\frac{5}{3})(4)$ and $YZ=\frac{20}{3}=6\frac23.$

Thus the correct answer is D.

3.

Four rectangular paper strips of length $10$ and width $1$ are put flat on a table and overlap perpendicularly as shown. How much area of the table is covered?

$36$

$40$

$44$

$98$

$100$

Concepts:area inclusion-exclusion rectangles

Difficulty rating: 960

Small Hint:

First add the areas of the four strips

Big Hint:

Each perpendicular crossing is a unit square counted twice

Solution:

The strips have total area $4(10)=40.$ There are four $1$ -by- $1$ overlaps, each counted twice, so the covered area is $40-4=36.$

Thus the correct answer is A.

4.

The slope of the line $\frac{x}{3}+\frac{y}{2}=1$ is

$-\frac32$

$-\frac23$

$\frac13$

$\frac23$

$\frac32$

Concepts:slope-intercept form linear equations

Difficulty rating: 1120

Small Hint:

Solve the equation for $y$

Big Hint:

In $y=mx+b,$ the coefficient $m$ is the slope

Solution:

Multiplying by $2$ after isolating $\frac{y}{2}$ gives $y=2-\frac23x.$ The slope is $-\frac{2}{3}.$

Thus the correct answer is B.

5.

If $b$ and $c$ are constants and $(x+2)(x+b)=x^2+cx+6,$ then $c$ is

$-5$

$-3$

$-1$

$3$

$5$

Concepts:polynomial identities coefficient comparison

Difficulty rating: 1030

Small Hint:

Expand the product on the left

Big Hint:

Compare the constant term before comparing the coefficient of $x$

Solution:

Expansion gives $x^2+(b+2)x+2b.$ Thus $2b=6,$ so $b=3,$ and $c=b+2=5.$

Thus the correct answer is E.

6.

A figure is an equiangular parallelogram if and only if it is a

rectangle

regular polygon

rhombus

square

trapezoid

Concepts:quadrilaterals parallelograms angle sum

Difficulty rating: 920

Small Hint:

The four equal interior angles must add to $360^\circ$

Big Hint:

No condition here forces adjacent sides to have equal lengths

Solution:

Four equal angles summing to $360^\circ$ are all right angles. A parallelogram with four right angles is exactly a rectangle.

Thus the correct answer is A.

7.

Estimate the time it takes to send $60$ blocks of data over a communications channel if each block consists of $512$ “chunks” and the channel can transmit $120$ chunks per second.

$0.04$ seconds

$0.4$ seconds

$4$ seconds

$4$ minutes

$4$ hours

Concepts:unit rates estimation unit conversion

Difficulty rating: 1030

Small Hint:

Compute the total number of chunks first

Big Hint:

Convert the resulting number of seconds to minutes

Solution:

The transmission takes $\frac{60\cdot512}{120}=256$ seconds, which is a little over $4$ minutes.

Thus the correct answer is D.

8.

If $\frac ba=2$ and $\frac cb=3,$ what is the ratio of $a+b$ to $b+c?$

$\frac13$

$\frac38$

$\frac35$

$\frac23$

$\frac34$

Concepts:ratios substitution proportions

Difficulty rating: 1130

Small Hint:

Write both $b$ and $c$ as multiples of $a$

Big Hint:

Substitute those expressions into $\frac{a+b}{b+c}$

Solution:

We have $b=2a$ and $c=3b=6a.$ Therefore $\frac{a+b}{b+c}=\frac{3a}{8a}=\frac38.$

Thus the correct answer is B.

9.

An $8'\times10'$ table sits in the corner of a square room, as in Figure $1$ below. The owners desire to move the table to the position shown in Figure $2.$ The side of the room is $S$ feet. What is the smallest integer value of $S$ for which the table can be moved as desired without tilting it or taking it apart?

$11$

$12$

$13$

$14$

$15$

Concepts:rectangle diagonal rotation geometric bounds

Difficulty rating: 1540

Small Hint:

During a quarter-turn, a diagonal becomes perpendicular to a pair of opposite walls

Big Hint:

Compare the room side with the table’s diagonal

Solution:

The table diagonal is $\sqrt{8^2+10^2}=\sqrt{164},$ which is between $12$ and $13.$ During the turn this diagonal must fit across the room, so $S\ge\sqrt{164}.$ Conversely, a rectangle can rotate inside a circle whose diameter is its diagonal, so that bound is sufficient. The least integer $S$ is $13.$

Thus the correct answer is C.

10.

In an experiment, a scientific constant $C$ is determined to be $2.43865$ with an error of at most $\pm0.00312.$ The experimenter wishes to announce a value for $C$ in which every digit is significant. That is, whatever $C$ is, the announced value must be the correct result when $C$ is rounded to that number of digits. The most accurate value the experimenter can announce for $C$ is

$2$

$2.4$

$2.43$

$2.44$

$2.439$

Concepts:error bounds rounding significant digits

Difficulty rating: 1410

Small Hint:

Find the smallest and largest possible values of $C$

Big Hint:

Round both endpoints to increasing numbers of significant digits

Solution:

The possible interval is $[2.43553,2.44177].$ Every number in it rounds to $2.44$ at three significant digits, but the endpoints round differently at four significant digits. Hence $2.44$ is the most accurate guaranteed announcement.

Thus the correct answer is D.

11.

On each horizontal line in the figure below, the five large dots indicate the populations of cities $A,$ $B,$ $C,$ $D,$ and $E$ in the year indicated. Which city had the greatest percentage increase in population from $1970$ to $1980?$

$A$

$B$

$C$

$D$

$E$

Concepts:percent change graph interpretation data comparison

Difficulty rating: 1290

Small Hint:

Read each city’s two population values from the scale

Big Hint:

Compare increase divided by the $1970$ population, not just absolute increase

Solution:

The percentage increases for $A,B,C,D,E$ are respectively $\frac{10}{40},\frac{20}{50},\frac{30}{70},\frac{30}{100},$ and $\frac{40}{120}.$ These are $25\%,40\%,42\frac67\%,30\%,$ and $33\frac13\%,$ so city $C$ has the greatest percentage increase.

Thus the correct answer is C.

12.

Each integer $1$ through $9$ is written on a separate slip of paper and all nine slips are put into a hat. Jack picks one of these slips at random and puts it back. Then Jill picks a slip at random. Which digit is most likely to be the units digit of the sum of Jack’s integer and Jill’s integer?

$0$

$1$

$8$

$9$

each digit is equally likely

Concepts:probability modular arithmetic counting ordered pairs

Difficulty rating: 1290

Small Hint:

There are $81$ equally likely ordered pairs

Big Hint:

Count pairs whose sum has each residue modulo $10$

Solution:

For a units digit of $0,$ the ordered pairs are $(1,9),(2,8),\ldots,(9,1),$ giving $9$ pairs. Each other units digit occurs $8$ times among the $81$ ordered pairs. Therefore $0$ is most likely.

Thus the correct answer is A.

13.

If $\sin x=3\cos x,$ then what is $\sin x\cos x?$

$\frac16$

$\frac15$

$\frac29$

$\frac14$

$\frac3{10}$

Concepts:trigonometric identities algebraic substitution

Difficulty rating: 1330

Small Hint:

Square the given relation and use $\sin^2x+\cos^2x=1$

Big Hint:

The given relation also determines the sign of the product

Solution:

Squaring gives $\sin^2x=9\cos^2x.$ Hence $10\cos^2x=1.$ Multiplying $\sin x=3\cos x$ by $\cos x$ gives $\sin x\cos x=3\cos^2x=\frac{3}{10}.$

Thus the correct answer is E.

14.

For any real number $a$ and positive integer $k,$ define $\binom ak= \frac{\begin{gathered} a(a-1)(a-2)\cdots\\ {}\cdot(a-(k-1)) \end{gathered}} {\begin{gathered} k(k-1)(k-2)\cdots\\ {}\cdot(2)(1) \end{gathered}}.$ What is $\binom{-\frac12}{100}\div\binom{\frac12}{100}\ ?$

$-199$

$-197$

$-1$

$197$

$199$

Concepts:generalized binomial coefficients products cancellation

Difficulty rating: 2090

Small Hint:

Cancel the identical denominator $100!$

Big Hint:

Write the remaining factors as consecutive odd integers and look for telescoping cancellation

Solution:

After canceling $100!,$ the numerator factors have magnitudes $1,3,\ldots,199,$ while the denominator factors have magnitudes $1,1,3,\ldots,197.$ All common odd factors cancel, leaving magnitude $199.$ The numerator has $100$ negative factors and the denominator has $99,$ so the quotient is $-199.$

Thus the correct answer is A.

15.

If $a$ and $b$ are integers such that $x^2-x-1$ is a factor of $ax^3+bx^2+1,$ then $b$ is

$-2$

$-1$

$0$

$1$

$2$

Concepts:polynomial divisibility modular reduction coefficient comparison

Difficulty rating: 1760

Small Hint:

Reduce powers using $x^2\equiv x+1$

Big Hint:

A polynomial of degree below $2$ divisible by the quadratic must be zero

Solution:

Modulo $x^2-x-1,$ we have $x^2\equiv x+1$ and $x^3\equiv2x+1.$ Thus the polynomial is congruent to $(2a+b)x+(a+b+1).$ Both coefficients must vanish, so $2a+b=0$ and $a+b=-1.$ Hence $a=1$ and $b=-2.$

Thus the correct answer is A.

16.

$ABC$ and $A'B'C'$ are equilateral triangles with parallel sides and the same center, as in the figure. The distance between side $BC$ and side $B'C'$ is $\frac16$ the altitude of $\triangle ABC.$ The ratio of the area of $\triangle A'B'C'$ to the area of $\triangle ABC$ is

$\frac1{36}$

$\frac16$

$\frac14$

$\frac{\sqrt3}{4}$

$\frac{9+8\sqrt3}{36}$

Concepts:equilateral triangles centroid area scale factor

Difficulty rating: 1660

Small Hint:

A centroid lies one-third of the altitude above the base

Big Hint:

Relate the two altitudes using the distance between their parallel bases

Solution:

Let the outer and inner altitudes be $h$ and $h'.$ Their common center is $\frac{h}{3}$ above $BC$ and $\frac{h'}{3}$ above $B'C'.$ Hence $\frac{h-h'}{3}=\frac{h}{6},$ so $h'=\frac{h}{2}.$ Areas scale as the square of corresponding lengths, giving $(\frac{h'}{h})^2=\frac{1}{4}.$

Thus the correct answer is C.

17.

If $\lvert x\rvert+x+y=10$ and $x+\lvert y\rvert-y=12,$ find $x+y.$

$-2$

$2$

$\frac{18}{5}$

$\frac{22}{3}$

$22$

Concepts:absolute value systems of equations case analysis

Difficulty rating: 1570

Small Hint:

Use each equation to rule out one sign possibility

Big Hint:

After determining the signs of $x$ and $y,$ solve a linear system

Solution:

If $x\le0,$ the first equation gives $y=10,$ contradicting the second; hence $x>0.$ If $y\ge0,$ the second gives $x=12,$ contradicting the first; hence $y<0.$ The equations become $2x+y=10$ and $x-2y=12.$ Thus $x=\frac{32}{5},\ y=-\frac{14}{5},$ and $x+y=\frac{18}{5}.$

Thus the correct answer is C.

18.

At the end of a professional bowling tournament, the top $5$ bowlers have a play-off. First # $5$ bowls # $4.$ The loser receives $5$ th prize and the winner bowls # $3$ in another game. The loser of this game receives $4$ th prize and the winner bowls # $2.$ The loser of this game receives $3$ rd prize and the winner bowls # $1.$ The winner of this game gets $1$ st prize and the loser gets $2$ nd prize. In how many orders can bowlers # $1$ through # $5$ receive the prizes?

$10$

$16$

$24$

$120$

none of these

Concepts:counting binary choices tournament outcomes

Difficulty rating: 1500

Small Hint:

Exactly four games are played

Big Hint:

Each sequence of winners determines a different prize order

Solution:

Each of the four games has two possible winners. A sequence of four outcomes uniquely determines the five prize positions, so there are $2^4=16$ orders.

Thus the correct answer is B.

19.

Simplify $\frac{\begin{gathered} bx(a^2x^2+2a^2y^2+b^2y^2)\\ {}+ay(a^2x^2+2b^2x^2+b^2y^2) \end{gathered}}{bx+ay}.$

$a^2x^2+b^2y^2$

$(ax+by)^2$

$(ax+by)(bx+ay)$

$2(a^2x^2+b^2y^2)$

$(bx+ay)^2$

Concepts:factoring polynomial identities algebraic simplification

Difficulty rating: 1660

Small Hint:

Separate the $2a^2y^2$ and $2b^2x^2$ terms

Big Hint:

Try to factor the numerator first by $bx+ay$

Solution:

Regrouping the numerator gives $\begin{aligned} &(bx+ay)(a^2x^2+b^2y^2)\\ &\quad+2abxy(ay+bx)\\ &=(bx+ay)(ax+by)^2. \end{aligned}$ Dividing by $bx+ay$ yields $(ax+by)^2.$

Thus the correct answer is B.

20.

In one of the adjoining figures a square of side $2$ is dissected into four pieces so that $E$ and $F$ are the midpoints of opposite sides and $AG$ is perpendicular to $BF.$ These four pieces can then be reassembled into a rectangle as shown in the second figure. The ratio of height to base, $\frac{XY}{YZ},$ in this rectangle is

$4$

$1+2\sqrt3$

$2\sqrt5$

$\frac{8+4\sqrt3}{3}$

$5$

Concepts:dissection area preservation Pythagorean theorem

Difficulty rating: 2010

Small Hint:

Find $BF$ and $DE$ from the side length and midpoint conditions

Big Hint:

Use the unchanged area to determine the rectangle’s base

Solution:

Both $BF$ and $DE$ have horizontal and vertical changes $1$ and $2,$ so each has length $\sqrt5.$ Thus $XY=BF+DE=2\sqrt5.$ The rectangle has area $4,$ so $YZ=\frac{4}{2\sqrt5}=\frac{2}{\sqrt5}.$ Therefore $\frac{XY}{YZ}=5.$

Thus the correct answer is E.

21.

The complex number $z$ satisfies $z+\lvert z\rvert=2+8i.$ What is $\lvert z\rvert^2?$ Note: if $z=a+bi,$ then $\lvert z\rvert=\sqrt{a^2+b^2}.$

$68$

$100$

$169$

$208$

$289$

Concepts:complex numbers modulus equations with radicals

Difficulty rating: 1850

Small Hint:

Write $z=a+bi$ and compare imaginary parts

Big Hint:

Use the real part to relate $a$ to $\sqrt{a^2+b^2}$

Solution:

Writing $z=a+bi$ gives $b=8$ and $a+\sqrt{a^2+64}=2.$ Thus $\sqrt{a^2+64}=2-a.$ Squaring yields $a=-15.$ Therefore $\lvert z\rvert^2=a^2+b^2=225+64=289.$

Thus the correct answer is E.

22.

For how many integers $x$ does a triangle with side lengths $10,$ $24,$ and $x$ have all its angles acute?

$4$

$5$

$6$

$7$

more than $7$

Concepts:acute triangles Pythagorean inequalities integer counting

Difficulty rating: 1760

Small Hint:

Apply the strict Pythagorean inequality to the largest side in each possible ordering

Big Hint:

The angle opposite $24$ and the angle opposite $x$ give the restrictive bounds

Solution:

Acuteness requires $x^2<10^2+24^2=676$ and $24^2<10^2+x^2,$ so $x<26$ and $x^2>476.$ For integer $x,$ this gives $22\le x\le25.$ All four values also satisfy the triangle inequality, so there are $4$ values.

Thus the correct answer is A.

23.

The six edges of tetrahedron $ABCD$ measure $7,$ $13,$ $18,$ $27,$ $36,$ and $41$ units. If the length of edge $AB$ is $41,$ then the length of edge $CD$ is

$7$

$13$

$18$

$27$

$36$

Concepts:tetrahedron edges triangle inequality case analysis

Difficulty rating: 2160

Small Hint:

The edge of length $7$ belongs to two triangular faces

Big Hint:

In either face containing that edge, the other two edge lengths must differ by less than $7$

Solution:

For a face containing the edge $7,$ its other two sides must differ by less than $7.$ Among the remaining lengths, the only disjoint pairs with this property are $(13,18)$ and $(36,41),$ so edge $27$ is opposite edge $7.$ With $AB=41,$ the two possible placements of these pairs can be checked by the triangle inequalities; the placement making the edge opposite $AB$ equal to $18$ fails, while the valid placement has $CD=13.$

Thus the correct answer is B.

24.

An isosceles trapezoid is circumscribed around a circle. The longer base of the trapezoid is $16,$ and one of the base angles is $\arcsin(0.8).$ Find the area of the trapezoid.

$72$

$75$

$80$

$90$

not uniquely determined

Concepts:tangential quadrilateral isosceles trapezoid trigonometry area

Difficulty rating: 2160

Small Hint:

For a circumscribed quadrilateral, the sums of opposite side lengths are equal

Big Hint:

Use $\sin\alpha=\frac{4}{5}$ and $\cos\alpha=\frac{3}{5}$ to relate the leg, height, and difference of the bases

Solution:

Let the shorter base be $b$ and each leg be $\ell.$ Tangency gives $16+b=2\ell.$ If the base angle is $\alpha,$ then $\sin\alpha=\frac{4}{5},\ \cos\alpha=\frac{3}{5},$ and $16-b=2\ell\cos\alpha=(\frac{6}{5})\ell.$ Solving gives $b=4,\ \ell=10,$ and height $\ell\sin\alpha=8.$ The area is $\frac12(16+4)(8)=80.$

Thus the correct answer is C.

25.

$X,$ $Y,$ and $Z$ are pairwise disjoint sets of people. The average ages of people in the sets $X,$ $Y,$ $Z,$ $X\cup Y,$ $X\cup Z,$ and $Y\cup Z$ are given in the table below.

Set	$X$	$Y$	$Z$	$X\cup Y$	$X\cup Z$	$Y\cup Z$
Average age of people in the set	$37$	$23$	$41$	$29$	$39.5$	$33$

Find the average age of the people in the set

X\cup Y\cup Z.

$33$

$33.5$

$33.66$

$33.833$

$34$

Concepts:weighted averages systems of equations set cardinality

Difficulty rating: 2110

Small Hint:

Let $x,y,z$ be the sizes of the three sets

Big Hint:

Each union average gives a linear relation among $x,y,z$

Solution:

Let the set sizes be $x,y,z.$ The three union averages give $\begin{aligned} 37x+23y&=29(x+y),\\ 37x+41z&=39.5(x+z), \end{aligned}$ and $23y+41z=33(y+z).$ Thus $y=\frac{4x}{3}$ and $z=\frac{5x}{3}.$ The total average is $\frac{\begin{gathered} 37x+23(\frac{4x}{3})\\ {}+41(\frac{5x}{3}) \end{gathered}} {x+\frac{4x}{3}+\frac{5x}{3}}=34.$

Thus the correct answer is E.

26.

Suppose that $p$ and $q$ are positive numbers for which $\begin{aligned} \log_9(p)&=\log_{12}(q)\\ &=\log_{16}(p+q). \end{aligned}$ What is the value of $\frac{q}{p}?$

$\frac43$

$\frac12(1+\sqrt3)$

$\frac85$

$\frac12(1+\sqrt5)$

$\frac{16}{9}$

Concepts:logarithms exponential equations quadratic equations

Difficulty rating: 2360

Small Hint:

Call the common logarithm value $t$

Big Hint:

If $r=\frac{q}{p},$ compare $(\frac{16}{9})^t$ with $(\frac{12}{9})^t$

Solution:

Let the common value be $t.$ Then $p=9^t,\ q=12^t,$ and $p+q=16^t.$ Put $r=\frac{q}{p}=(\frac{4}{3})^t.$ Dividing the last equation by $p$ gives $\begin{aligned} 1+r&=(\frac{16}{9})^t\\ &=((\frac{4}{3})^t)^2=r^2. \end{aligned}$ Since $r>0,$ $r=\frac{1+\sqrt5}{2}.$

Thus the correct answer is D.

27.

In the figure, $AB\perp BC,$ $BC\perp CD,$ and $BC$ is tangent to the circle with center $O$ and diameter $AD.$ In which one of the following cases is the area of $ABCD$ an integer?

$AB=3,$ $CD=1$

$AB=5,$ $CD=2$

$AB=7,$ $CD=3$

$AB=9,$ $CD=4$

$AB=11,$ $CD=5$

Concepts:circle tangency power of a point trapezoid area

Difficulty rating: 2280

Small Hint:

Let the tangent point be $M$ and use the diameter to form a rectangle inside the trapezoid

Big Hint:

Power of point $B$ relates half of $BC$ to $AB$ and $CD$

Solution:

The tangent point is the midpoint of $BC,$ and the tangent-secant relation gives $(\frac{BC}{2})^2=AB\cdot CD.$ Thus $BC=2\sqrt{AB\cdot CD},$ and the trapezoid area is $\begin{aligned} \text{Area} &=\frac{AB+CD}{2}BC\\ &=(AB+CD)\sqrt{AB\cdot CD}. \end{aligned}$ Only $AB=9,\ CD=4$ makes the product under the radical a square; the area is $13\cdot6=78.$

Thus the correct answer is D.

28.

An unfair coin has probability $p$ of coming up heads on a single toss. Let $w$ be the probability that, in $5$ independent tosses of this coin, heads come up exactly $3$ times. If $w=\frac{144}{625},$ then

$p$ must be $\frac{2}{5}$

$p$ must be $\frac{3}{5}$

$p$ must be greater than $\frac{3}{5}$

$p$ is not uniquely determined

there is no value of $p$ for which $w=\frac{144}{625}$

Concepts:binomial probability continuity nonunique solutions

Difficulty rating: 2360

Small Hint:

Write $w$ as a function of $p$ using the binomial coefficient

Big Hint:

Check one simple value, then compare the function at $p=\frac{3}{5}$ and $p=1$

Solution:

Here $w(p)=10p^3(1-p)^2.$ At $p=\frac{2}{5},$ $w=10\left(\frac25\right)^3\left(\frac35\right)^2 =\frac{144}{625}.$ Also $w(\frac{3}{5})=\frac{216}{625}>\frac{144}{625},$ while $w(1)=0.$ By continuity there is another solution between $\frac{3}{5}$ and $1.$ Hence $p$ is not unique.

Thus the correct answer is D.

29.

You plot weight $(y)$ against height $(x)$ for three of your friends and obtain the points $(x_1,y_1),$ $(x_2,y_2),$ $(x_3,y_3).$ If $\begin{aligned} x_1&\lt x_2\lt x_3,\\ x_3-x_2&=x_2-x_1, \end{aligned}$ which of the following is necessarily the slope of the line which best fits the data? “Best fits” means that the sum of the squares of the vertical distances from the data points to the line is smaller than for any other line.

$\frac{y_3-y_1}{x_3-x_1}$

$\frac{(y_2-y_1)-(y_3-y_2)}{x_3-x_1}$

$\frac{2y_3-y_1-y_2}{2x_3-x_1-x_2}$

$\frac{y_2-y_1}{x_2-x_1}+\frac{y_3-y_2}{x_3-x_2}$

none of these

Concepts:least squares slope symmetric coordinates

Difficulty rating: 2480

Small Hint:

Translate and scale the $x$ -coordinates to $-1,0,1$

Big Hint:

For a least-squares line through symmetric $x$ -values, compute the slope from the covariance numerator

Solution:

Translate and scale so the $x$ -coordinates are $-d,0,d.$ Their mean is $0,$ so the least-squares slope is $\begin{aligned} \frac{(-d)y_1+0y_2+dy_3}{d^2+0+d^2} &=\frac{y_3-y_1}{2d}\\ &=\frac{y_3-y_1}{x_3-x_1}. \end{aligned}$

Thus the correct answer is A.

30.

Let $f(x)=4x-x^2.$ Given $x_0,$ consider the sequence defined by $x_n=f(x_{n-1})$ for all $n\ge1.$ For how many real numbers $x_0$ will the sequence $x_0,$ $x_1,$ $x_2,$ $\ldots$ take on only a finite number of different values?

$0$

$1$ or $2$

$3,$ $4,$ $5$ or $6$

more than $6$ but finitely many

infinitely many

Concepts:function iteration preimages induction finite orbits

Difficulty rating: 2930

Small Hint:

Start with $0,$ then find numbers mapping successively to $0$

Big Hint:

For every $a\le4,$ solve $4x-x^2=a$ and choose a new real preimage

Solution:

The starting values $0,4,2$ give finite orbits $0$ , $4\to0$ , and $2\to4\to0.$ More generally, if $a_n$ begins a finite chain ending at $0,$ solve $4a_{n+1}-a_{n+1}^2=a_n.$ Its real solutions are $a_{n+1}=2\pm\sqrt{4-a_n}.$ Choosing a preimage not already in the chain extends it by one new value. Repeating produces infinitely many distinct starting values with finite orbits.

Thus the correct answer is E.