2002 AMC 12B — Problems & Online Timed Exam

Timed

1:15:00

1.

The arithmetic mean of the nine numbers in the set $\{9,$ $99,$ $999,$ $9999,$ $\ldots,$ $999999999\}$ is a $9$ -digit number $M,$ all of whose digits are distinct. The number $M$ does not contain the digit

$0$

$2$

$4$

$6$

$8$

Answer: A

Concepts:place value digits

Difficulty rating: 950

Small Hint:

The mean is $\dfrac{1}{9}(9+99+\cdots+999{,}999{,}999)$

Big Hint:

Dividing each term by $9$ turns the sum into $1+11+111+\cdots$ ; add these and see which digit never appears

Solution:

Each of the nine numbers is $10^k-1,$ so their sum is $9+99+\cdots+999{,}999{,}999.$ Dividing by $9,$ $\begin{aligned} M &= 1+11+111+\cdots \\ &\quad {}+111{,}111{,}111 \\ &= 123{,}456{,}789. \end{aligned}$ Its digits are $1$ through $9,$ so the missing digit is $0.$

Thus, the correct answer is A.

2.

What is the value of $\begin{aligned} &(3x-2)(4x+1) \\ &\quad {}-(3x-2)4x+1 \end{aligned}$ when $x=4?$

$0$

$1$

$10$

$11$

$12$

Answer: D

Concepts:algebraic manipulation factoring

Difficulty rating: 980

Small Hint:

Group the two terms containing $3x-2$ using the distributive law

Big Hint:

$(3x-2)(4x+1)-(3x-2)4x$ $=(3x-2)(4x+1-4x)$

Solution:

Factor $3x-2$ out of the first two terms: $\begin{gathered} (3x-2)(4x+1) \\ {}-(3x-2)4x+1 \\ {}=(3x-2)(4x+1-4x)+1 \\ {}=3x-1. \end{gathered}$ At $x=4$ this equals $3\cdot4-1=11.$

Thus, the correct answer is D.

3.

For how many positive integers $n$ is $n^2-3n+2$ a prime number?

none

one

two

more than two, but finitely many

infinitely many

Answer: B

Concepts:factoring prime

Difficulty rating: 1120

Small Hint:

Factor $n^2-3n+2$

Big Hint:

$(n-1)(n-2)$ is a product of two integers; for it to be prime one factor must equal $1$

Solution:

Factor $n^2-3n+2=(n-1)(n-2).$ For $n\ge4$ both factors exceed $1,$ so the value is composite. Checking $n=1,2,3$ gives $0,$ $0,$ and $2.$ Only $n=3$ yields a prime, so there is exactly one such $n.$

Thus, the correct answer is B.

4.

Let $n$ be a positive integer such that $\dfrac12+\dfrac13+\dfrac17+\dfrac1n$ is an integer. Which of the following statements is not true:

$2$ divides $n$

$3$ divides $n$

$6$ divides $n$

$7$ divides $n$

$n\gt84$

Answer: E

Concepts:fraction divisibility bounding to limit cases

Difficulty rating: 1270

Small Hint:

$\dfrac12+\dfrac13+\dfrac17=\dfrac{41}{42}$

Big Hint:

The whole sum lies strictly between $0$ and $2,$ so it must equal $1$ ; solve for $n$

Solution:

Since $\dfrac12+\dfrac13+\dfrac17=\dfrac{41}{42},$ the sum $\dfrac{41}{42}+\dfrac1n$ lies strictly between $0$ and $2,$ so it must equal $1.$ Then $\dfrac1n=\dfrac1{42},$ giving $n=42.$

Now $2,$ $3,$ $6,$ and $7$ all divide $42,$ but $n\gt84$ is false. The untrue statement is $n\gt84.$

Thus, the correct answer is E.

5.

Let $v,$ $w,$ $x,$ $y,$ and $z$ be the degree measures of the five angles of a pentagon. Suppose $v\lt w\lt x\lt y\lt z$ and $v,$ $w,$ $x,$ $y,$ $z$ form an arithmetic sequence. Find the value of $x.$

$72$

$84$

$90$

$108$

$120$

Answer: D

Concepts:angle sum arithmetic sequence

Difficulty rating: 1080

Small Hint:

The interior angles of a pentagon sum to $540^\circ$

Big Hint:

Write the five terms as $x-2d,$ $x-d,$ $x,$ $x+d,$ $x+2d$ ; their sum is $5x$

Solution:

The five interior angles sum to $540^\circ.$ Writing the sequence as $x-2d,$ $x-d,$ $x,$ $x+d,$ $x+2d,$ the sum is $5x=540,$ so $x=108.$

Thus, the correct answer is D.

6.

Suppose that $a$ and $b$ are nonzero real numbers, and that the equation $x^2+ax+b=0$ has solutions $a$ and $b.$ Then the pair $(a,b)$ is

$(-2,1)$

$(-1,2)$

$(1,-2)$

$(2,-1)$

$(4,4)$

Answer: C

Concepts:Vieta’s Formulas quadratic

Difficulty rating: 1190

Small Hint:

If $a$ and $b$ are the roots, then $x^2+ax+b=(x-a)(x-b)$

Big Hint:

Match coefficients: $a+b=-a$ and $ab=b,$ using $b\neq0$

Solution:

Since $a$ and $b$ are the roots, $\begin{gathered} x^2+ax+b \\ {}=(x-a)(x-b) \\ {}=x^2-(a+b)x+ab. \end{gathered}$ Matching coefficients gives $a+b=-a$ and $ab=b.$

As $b\neq0,$ the second equation gives $a=1,$ and then $a+b=-a$ gives $b=-2.$ So $(a,b)=(1,-2).$

Thus, the correct answer is C.

7.

The product of three consecutive positive integers is $8$ times their sum. What is the sum of their squares?

$50$

$77$

$110$

$149$

$194$

Answer: B

Concepts:quadratic algebraic manipulation

Difficulty rating: 1190

Small Hint:

Call the integers $n-1,$ $n,$ $n+1$ ; their product is $n^3-n$ and their sum is $3n$

Big Hint:

$n^2-1=24,$ so find $n$ and then compute $(n-1)^2+n^2+(n+1)^2$

Solution:

Let the integers be $n-1,$ $n,$ $n+1.$ Then $(n-1)n(n+1)=8\cdot3n,$ so $n^2-1=24$ and $n=5.$

The three integers $4,$ $5,$ $6$ have squares summing to $16+25+36=77.$

Thus, the correct answer is B.

8.

Suppose July of year $N$ has five Mondays. Which of the following must occur five times in August of year $N?$ (Note: Both months have $31$ days.)

Monday

Tuesday

Wednesday

Thursday

Friday

Answer: D

Concepts:date and time modular arithmetic casework

Difficulty rating: 1370

Small Hint:

With $31$ days, the weekday of July $1$ is the one that repeats five times, so Monday is July $1,$ $2,$ or $3$

Big Hint:

August $1$ falls three weekdays after July $1$ ; test the three cases and find the August weekday common to all

Solution:

Since July has $31=4\cdot7+3$ days, the weekday of July $1$ occurs five times, so Monday falls on July $1,$ $2,$ or $3.$ The days that occur five times in August are those of August $1,$ $2,$ $3,$ and August $1$ is three weekdays after July $1.$

Testing the three cases, the August weekdays occurring five times are Thursday through Saturday, Wednesday through Friday, and Tuesday through Thursday. Thursday appears in every case.

Thus, the correct answer is D.

9.

If $a,$ $b,$ $c,$ $d$ are positive real numbers such that $a,$ $b,$ $c,$ $d$ form an increasing arithmetic sequence and $a,$ $b,$ $d$ form a geometric sequence, then $\dfrac{a}{d}$ is

$\dfrac{1}{12}$

$\dfrac{1}{6}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

Answer: C

Concepts:arithmetic sequence geometric sequence

Difficulty rating: 1330

Small Hint:

Write $b=a+r,$ $c=a+2r,$ $d=a+3r$

Big Hint:

The geometric condition $b^2=ad$ gives $r^2=ar,$ so $r=a$

Solution:

Let $b=a+r,$ $c=a+2r,$ $d=a+3r.$ The geometric condition $b^2=ad$ gives $(a+r)^2=a(a+3r),$ i.e. $r^2=ar,$ so $r=a.$

Then $d=a+3a=4a$ and $\dfrac{a}{d}=\dfrac14.$

Thus, the correct answer is C.

10.

How many different integers can be expressed as the sum of three distinct members of the set $\{1,4,7,10,13,16,19\}?$

$13$

$16$

$24$

$30$

$35$

Answer: A

Concepts:modular arithmetic counting integers in a range

Difficulty rating: 1270

Small Hint:

Every element is one more than a multiple of $3,$ so each three-element sum is a multiple of $3$

Big Hint:

Count the multiples of $3$ from the smallest sum $1+4+7$ to the largest sum $13+16+19$

Solution:

Every element is one more than a multiple of $3,$ so any sum of three of them is a multiple of $3.$ The smallest sum is $1+4+7=12$ and the largest is $13+16+19=48,$ and every multiple of $3$ between them is attainable.

There are $13$ multiples of $3$ from $12$ to $48.$

Thus, the correct answer is A.

11.

The positive integers $A,$ $B,$ $A-B,$ and $A+B$ are all prime numbers. The sum of these four primes is

even

divisible by $3$

divisible by $5$

divisible by $7$

prime

Answer: E

Concepts:parity prime

Difficulty rating: 1430

Small Hint:

$A-B$ and $A+B$ have the same parity, and both are prime, so both are odd

Big Hint:

Then $B=2,$ making $A-2,$ $A,$ $A+2$ three primes in arithmetic progression

Solution:

$A-B$ and $A+B$ have the same parity; being prime, both are odd, so $A$ and $B$ have opposite parity. If $A$ were even, then the prime $A$ would equal $2,$ but the positive prime $B$ would satisfy $B\ge2$ and make $A-B\le0.$ Hence $A$ is odd and the even prime $B$ is $2.$

Then $A-2,$ $A,$ $A+2$ are three primes. One is divisible by $3,$ so that one must equal $3;$ the triple is $3,$ $5,$ $7.$ Their sum together with $2$ is $2+3+5+7=17,$ a prime.

Thus, the correct answer is E.

12.

For how many integers $n$ is $\dfrac{n}{20-n}$ the square of an integer?

$1$

$2$

$3$

$4$

$10$

Answer: D

Concepts:perfect square divisibility

Difficulty rating: 1490

Small Hint:

Set $\dfrac{n}{20-n}=k^2$ and solve for $n$ in terms of $k$

Big Hint:

$n=\dfrac{20k^2}{k^2+1}$ ; since $\gcd(k^2,k^2+1)=1,$ $k^2+1$ must divide $20$

Solution:

Set $\dfrac{n}{20-n}=k^2.$ Solving, $n=\dfrac{20k^2}{k^2+1}.$ Since $k^2$ and $k^2+1$ are coprime, $k^2+1$ must divide $20,$ which happens only for $|k|=0,1,2,3.$

The signs of $k$ do not change $n;$ these give $n=0,$ $10,$ $16,$ and $18,$ which is four values.

Thus, the correct answer is D.

13.

The sum of $18$ consecutive positive integers is a perfect square. The smallest possible value of this sum is

$169$

$225$

$289$

$361$

$441$

Answer: B

Concepts:perfect square arithmetic sequence

Difficulty rating: 1430

Small Hint:

The sum of $n,$ $n+1,\ldots,$ $n+17$ equals $9(2n+17)$

Big Hint:

Since $9$ is already a square, $2n+17$ must be an odd perfect square; take the smallest valid one

Solution:

The sum of $n,$ $n+1,\ldots,$ $n+17$ is $18n+\dfrac{17\cdot18}{2}=9(2n+17).$ Since $9$ is a perfect square, $2n+17$ must be one too.

The smallest positive integer $n$ making $2n+17$ a perfect square is $n=4,$ giving $2n+17=25$ and a sum of $9\cdot25=225.$

Thus, the correct answer is B.

14.

Four distinct circles are drawn in a plane. What is the maximum number of points where at least two of the circles intersect?

$8$

$9$

$10$

$12$

$16$

Answer: D

Concepts:counting intersections counting pairs

Difficulty rating: 1220

Small Hint:

Two distinct circles meet in at most $2$ points

Big Hint:

There are $\binom{4}{2}$ pairs of circles; multiply by $2$

Solution:

Each pair of circles meets in at most $2$ points, and there are $\binom{4}{2}=6$ pairs, giving at most $6\cdot2=12$ intersection points.

The bound is attainable by taking four circles in general position so that every pair crosses twice and no three pass through the same point.

Thus, the correct answer is D.

15.

How many four-digit numbers $N$ have the property that the three-digit number obtained by removing the leftmost digit is one ninth of $N?$

$4$

$5$

$6$

$7$

$8$

Answer: D

Concepts:place value digits

Difficulty rating: 1510

Small Hint:

Write $N=1000a+x,$ where $a$ is the leading digit and $x$ the last three digits

Big Hint:

$N=9x$ gives $1000a=8x,$ so $x=125a$ ; count the digits $a$ keeping $x$ three digits

Solution:

Let $a$ be the leading digit and $x$ the three-digit number after removing it, so $N=1000a+x.$ The condition $N=9x$ gives $1000a=8x,$ i.e. $x=125a.$

For $a=1,\ldots,7$ this makes $x$ a three-digit number, while $a=8$ gives $x=1000.$ So there are $7$ such numbers.

Thus, the correct answer is D.

16.

Juan rolls a fair regular octahedral die marked with the numbers $1$ through $8.$ Then Amal rolls a fair six-sided die. What is the probability that the product of the two rolls is a multiple of $3?$

$\dfrac{1}{12}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{7}{12}$

$\dfrac{2}{3}$

Answer: C

Concepts:complementary probability dice (probability)

Difficulty rating: 1430

Small Hint:

The product is a multiple of $3$ exactly when at least one roll is $3$ or $6$

Big Hint:

Compute $1$ minus the probability that neither die shows a $3$ or $6$

Solution:

The product is a multiple of $3$ if and only if at least one die shows $3$ or $6.$ The octahedral die avoids $3,6$ with probability $\dfrac68=\dfrac34,$ and the six-sided die avoids them with probability $\dfrac46=\dfrac23.$

So neither shows a multiple of $3$ with probability $\dfrac34\cdot\dfrac23=\dfrac12,$ and the answer is $1-\dfrac12=\dfrac12.$

Thus, the correct answer is C.

17.

Andy’s lawn has twice as much area as Beth’s lawn and three times as much area as Carlos’ lawn. Carlos’ lawn mower cuts half as fast as Beth’s mower and one third as fast as Andy’s mower. If they all start to mow their lawns at the same time, who will finish first?

Andy

Beth

Carlos

Andy and Carlos tie for first.

All three tie.

Answer: B

Concepts:rate ratio and proportion

Difficulty rating: 1370

Small Hint:

Let Andy’s area be $A$ ; then the areas are $A,$ $\dfrac A2,$ $\dfrac A3$ and the mowing rates are $3R,$ $2R,$ $R$

Big Hint:

Time equals area divided by rate; compare $\dfrac{A}{3R},$ $\dfrac{A}{4R},$ $\dfrac{A}{3R}$

Solution:

Let Andy’s lawn have area $A;$ then Beth’s is $\dfrac A2$ and Carlos’ is $\dfrac A3.$ With Carlos’ rate $R,$ Beth mows at $2R$ and Andy at $3R.$

Their times are $\dfrac{A}{3R},$ $\dfrac{\frac{A}{2}}{2R}=\dfrac{A}{4R},$ and $\dfrac{\frac{A}{3}}{R}=\dfrac{A}{3R}$ respectively. Beth’s time $\dfrac{A}{4R}$ is the smallest, so Beth finishes first.

Thus, the correct answer is B.

18.

A point $P$ is randomly selected from the rectangular region with vertices $(0,0),$ $(2,0),$ $(2,1),$ $(0,1).$ What is the probability that $P$ is closer to the origin than it is to the point $(3,1)?$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

$\dfrac{3}{4}$

$\dfrac{4}{5}$

$1$

Answer: C

Concepts:geometric probability perpendicular bisector trapezoid

Difficulty rating: 1610

Small Hint:

$P$ is closer to the origin on one side of the perpendicular bisector of the segment from $(0,0)$ to $(3,1)$

Big Hint:

That bisector is the line $3x+y=5$ ; find the area of the rectangle on the origin side and divide by the total area

Solution:

The points closer to $(0,0)$ than to $(3,1)$ lie on the origin side of the perpendicular bisector of that segment, the line $3x+y=5.$

Within the rectangle, this region is a trapezoid whose parallel sides have lengths $\dfrac53$ (at $y=0$ ) and $\dfrac43$ (at $y=1$ ), so its area is $\dfrac12\left(\dfrac53+\dfrac43\right)=\dfrac32.$ The rectangle has area $2,$ so the probability is $\dfrac{\frac{3}{2}}{2}=\dfrac34.$

Thus, the correct answer is C.

19.

If $a,$ $b,$ and $c$ are positive real numbers such that $a(b+c)=152,$ $b(c+a)=162,$ and $c(a+b)=170,$ then $abc$ is

$672$

$688$

$704$

$720$

$750$

Answer: D

Concepts:system of equations symmetry (algebra)

Difficulty rating: 1540

Small Hint:

Add all three equations to get $ab+bc+ca$

Big Hint:

Subtract each original equation from that total to find $ab,$ $bc,$ $ca,$ then multiply and take a square root

Solution:

Adding the three equations gives $2(ab+bc+ca)=484,$ so $ab+bc+ca=242.$ Subtracting each original equation from this yields $bc=90,$ $ca=80,$ and $ab=72.$

Multiplying, $(abc)^2=90\cdot80\cdot72=720^2,$ and since $abc\gt0,$ we get $abc=720.$

Thus, the correct answer is D.

20.

Let $\triangle XOY$ be a right-angled triangle with $m\angle XOY=90^\circ.$ Let $M$ and $N$ be the midpoints of legs $OX$ and $OY,$ respectively. Given that $XN=19$ and $YM=22,$ find $XY.$

$24$

$26$

$28$

$30$

$32$

Answer: B

Concepts:right triangle median (geometry)Pythagorean Theorem

Difficulty rating: 1660

Small Hint:

Let $OM=a,$ $ON=b$ ; then $19^2=(2a)^2+b^2$ and $22^2=a^2+(2b)^2$

Big Hint:

Add the equations to get $5(a^2+b^2)$ ; note that $XY=2\sqrt{a^2+b^2}$

Solution:

Let $OM=a$ and $ON=b.$ Since $M,N$ are midpoints, $XN^2=(2a)^2+b^2=361$ and $YM^2=a^2+(2b)^2=484.$

Adding, $5(a^2+b^2)=845,$ so $a^2+b^2=169.$ Then $MN=\sqrt{a^2+b^2}=13,$ and since $MN$ is the midsegment joining $M$ and $N,$ we have $XY=2\,MN=26.$

Thus, the correct answer is B.

21.

For all positive integers $n$ less than $2002,$ let $a_n=\begin{cases} 11, & \text{if } n \text{ is divisible by } 13 \text{ and } 14;\\ 13, & \text{if } n \text{ is divisible by } 14 \text{ and } 11;\\ 14, & \text{if } n \text{ is divisible by } 11 \text{ and } 13;\\ 0, & \text{otherwise}. \end{cases}$ Calculate $\displaystyle\sum_{n=1}^{2001} a_n.$

$448$

$486$

$1560$

$2001$

$2002$

Answer: A

Concepts:divisibility least common multiple multiple

Difficulty rating: 1650

Small Hint:

Because $11,$ $13,$ $14$ are pairwise coprime, “divisible by two of them” means divisible by their product

Big Hint:

Count multiples of $182,$ $154,$ $143$ below $2002$ ; no number under $2002$ is divisible by all three

Solution:

Since $2002=11\cdot13\cdot14$ with $11,$ $13,$ $14$ pairwise coprime, $a_n=11$ when $n$ is divisible by $182,$ $a_n=13$ when $n$ is divisible by $154,$ and $a_n=14$ when $n$ is divisible by $143$ (and no $n\lt2002$ is divisible by all three).

For $n\le2001$ there are $10$ multiples of $182,$ $12$ of $154,$ and $13$ of $143.$ So the sum is $\begin{gathered} 11\cdot10+13\cdot12+14\cdot13 \\ {}=110+156+182 \\ {}=448. \end{gathered}$

Thus, the correct answer is A.

22.

For all integers $n$ greater than $1,$ define $a_n=\dfrac{1}{\log_n 2002}.$ Let $b=a_2+a_3+a_4+a_5$ and $c=a_{10}+a_{11}+a_{12}+a_{13}+a_{14}.$ Then $b-c$ equals

$-2$

$-1$

$\dfrac{1}{2002}$

$\dfrac{1}{1001}$

$\dfrac{1}{2}$

Answer: B

Concepts:logarithm

Difficulty rating: 1630

Small Hint:

$\dfrac{1}{\log_n 2002}=\log_{2002} n$

Big Hint:

Then $b-c$ $=\log_{2002}\dfrac{2\cdot3\cdot4\cdot5}{10\cdot11\cdot12\cdot13\cdot14}$ ; simplify the fraction

Solution:

By change of base, $a_n=\dfrac{1}{\log_n 2002}=\log_{2002} n.$ So $\begin{gathered} b-c \\ {}=\log_{2002}\frac{2\cdot3\cdot4\cdot5}{10\cdot11\cdot12\cdot13\cdot14}. \end{gathered}$

The fraction equals $\dfrac{120}{240240}=\dfrac{1}{2002},$ so $b-c=\log_{2002}\dfrac{1}{2002}=-1.$

Thus, the correct answer is B.

23.

In $\triangle ABC,$ we have $AB=1$ and $AC=2.$ Side $BC$ and the median from $A$ to $BC$ have the same length. What is $BC?$

$\dfrac{1+\sqrt{2}}{2}$

$\dfrac{1+\sqrt{3}}{2}$

$\sqrt{2}$

$\dfrac{3}{2}$

$\sqrt{3}$

Answer: C

Concepts:median (geometry)law of cosines

Difficulty rating: 1820

Small Hint:

Let $M$ be the midpoint of $BC,$ $AM=2a,$ and $\theta=\angle AMB$

Big Hint:

Apply the Law of Cosines to $\triangle ABM$ and $\triangle AMC$ (angles $\theta$ and $180^\circ-\theta$ ) and add

Solution:

Let $M$ be the midpoint of $BC,$ set $AM=2a,$ and let $\theta=\angle AMB,$ so $\angle AMC=180^\circ-\theta.$ With $BM=CM=a,$ so that $BC=2a,$ the Law of Cosines in $\triangle ABM$ and $\triangle AMC$ gives $a^2+4a^2-4a^2\cos\theta=1,$ $a^2+4a^2+4a^2\cos\theta=4.$

Adding, $10a^2=5,$ so $a=\dfrac{\sqrt2}{2}$ and $BC=2a=\sqrt2.$

Thus, the correct answer is C.

24.

A convex quadrilateral $ABCD$ with area $2002$ contains a point $P$ in its interior such that $PA=24,$ $PB=32,$ $PC=28,$ and $PD=45.$ Find the perimeter of $ABCD.$

$4\sqrt{2002}$

$2\sqrt{8465}$

$2\left(48+\sqrt{2002}\right)$

$2\sqrt{8633}$

$4\left(36+\sqrt{113}\right)$

Answer: E

Concepts:area diagonal Pythagorean Theorem bounding to limit cases

Difficulty rating: 2150

Small Hint:

For a quadrilateral with diagonals $d_1,d_2,$ the area is at most $\tfrac12 d_1 d_2,$ with equality when the diagonals are perpendicular

Big Hint:

Here $\tfrac12(PA+PC)(PB+PD)$ $=2002$ forces perpendicular diagonals meeting at $P$ ; find each side with the Pythagorean theorem

Solution:

For any quadrilateral, the area is at most $\tfrac12\,d_1 d_2$ where $d_1,d_2$ are the diagonals, with equality exactly when they are perpendicular. Here $\begin{gathered} 2002=\text{Area} \\ {}\le \tfrac12\,AC\cdot BD \\ {}\le \tfrac12(PA+PC)(PB+PD) \\ {}= \tfrac12\cdot52\cdot77 \\ {}= 2002. \end{gathered}$

Equality forces the diagonals to be perpendicular and to intersect at $P.$ Then $\begin{aligned} AB &= \sqrt{24^2+32^2}=40, \\ BC &= \sqrt{28^2+32^2}=4\sqrt{113}, \end{aligned}$ $\begin{aligned} CD &= \sqrt{28^2+45^2}=53, \\ DA &= \sqrt{45^2+24^2}=51. \end{aligned}$

The perimeter is $144+4\sqrt{113}=4\left(36+\sqrt{113}\right).$

Thus, the correct answer is E.

25.

Let $f(x)=x^2+6x+1,$ and let $R$ denote the set of points $(x,y)$ in the coordinate plane such that $f(x)+f(y)\le0$ and $f(x)-f(y)\le0.$ The area of $R$ is closest to

$21$

$22$

$23$

$24$

$25$

Answer: E

Concepts:completing the square circle area

Difficulty rating: 2260

Small Hint:

Complete the square: $f(x)+f(y)$ $=(x+3)^2+(y+3)^2-16$

Big Hint:

The first condition is a disk of radius $4$ ; the second, $(x-y)(x+y+6)\le0,$ keeps half of it

Solution:

Completing the square, $\begin{gathered} f(x)+f(y) \\ {}=(x+3)^2+(y+3)^2-16, \end{gathered}$ so the first condition is the disk of radius $4$ centered at $(-3,-3).$

Also $\begin{gathered} f(x)-f(y) \\ {}=(x-y)(x+y+6), \end{gathered}$ so the second condition $(x-y)(x+y+6)\le0$ describes two half-planes bounded by the perpendicular lines through $(-3,-3)$ of slopes $1$ and $-1.$ These cut the disk into two equal halves.

Thus $R$ has area $\tfrac12\pi\cdot4^2=8\pi\approx25.13,$ which is closest to $25.$

Thus, the correct answer is E.