2011 AMC 12B — Problems & Online Timed Exam

Timed

1:15:00

1.

What is

$\dfrac{2+4+6}{1+3+5} - \dfrac{1+3+5}{2+4+6}?$

$-1$

$\dfrac{5}{36}$

$\dfrac{7}{12}$

$\dfrac{147}{60}$

$\dfrac{43}{3}$

Answer: C

Concepts:fraction

Difficulty rating: 770

Small Hint:

Add the terms in each numerator and denominator first

Big Hint:

The expression becomes $\dfrac{12}{9}-\dfrac{9}{12}$

Solution:

The sums are $2+4+6=12$ and $1+3+5=9,$ so the expression equals $\dfrac{12}{9}-\dfrac{9}{12}=\dfrac{4}{3}-\dfrac{3}{4}.$

Over a common denominator this is $\dfrac{16}{12}-\dfrac{9}{12}=\dfrac{7}{12}.$

Thus, the correct answer is C.

2.

Josanna’s test scores to date are $90,$ $80,$ $70,$ $60,$ and $85.$ Her goal is to raise her test average at least $3$ points with her next test. What is the minimum test score she would need to accomplish this goal?

$80$

$82$

$85$

$90$

$95$

Answer: E

Concepts:mean

Difficulty rating: 880

Small Hint:

Find the current average of the five scores

Big Hint:

The six scores must total $6$ times the target average

Solution:

The five scores sum to $90+80+70+60+85=385,$ giving an average of $77.$ The goal is a new average of at least $80.$

Six tests averaging $80$ must total $6\cdot80=480,$ so the sixth score must be at least $480-385=95.$

Thus, the correct answer is E.

3.

LeRoy and Bernardo went on a week-long trip together and agreed to share the costs equally. Over the week, each of them paid for various joint expenses such as gasoline and car rental. At the end of the trip it turned out that LeRoy had paid $A$ dollars and Bernardo had paid $B$ dollars, where $A \lt B.$ How many dollars must LeRoy give to Bernardo so that they share the costs equally?

$\dfrac{A+B}{2}$

$\dfrac{A-B}{2}$

$\dfrac{B-A}{2}$

$B-A$

$A+B$

Answer: C

Concepts:algebraic manipulation

Difficulty rating: 990

Small Hint:

Each person should end up paying $\dfrac{A+B}{2}$

Big Hint:

LeRoy must cover the gap between what he paid and his fair share

Solution:

The total cost is $A+B,$ so each person’s fair share is $\dfrac{A+B}{2}.$

LeRoy paid $A,$ which is less than his share, so he must give Bernardo $\dfrac{A+B}{2}-A=\dfrac{B-A}{2}.$

Thus, the correct answer is C.

4.

In multiplying two positive integers $a$ and $b,$ Ron reversed the digits of the two-digit number $a.$ His erroneous product was $161.$ What is the correct value of the product of $a$ and $b?$

$116$

$161$

$204$

$214$

$224$

Answer: E

Concepts:prime factorization digits

Difficulty rating: 1040

Small Hint:

Factor $161$ into primes

Big Hint:

The reversed $a$ is the two-digit factor; reverse it back before multiplying

Solution:

Since $161=7\cdot23,$ the only two-digit factor is $23.$ This must be the reversed value of $a,$ so the true value of $a$ is $32,$ and $b=7.$

The correct product is $32\cdot7=224.$

Thus, the correct answer is E.

5.

Let $N$ be the second smallest positive integer that is divisible by every positive integer less than $7.$ What is the sum of the digits of $N?$

$3$

$4$

$5$

$6$

$9$

Answer: A

Concepts:least common multiple digits

Difficulty rating: 990

Small Hint:

Find the least common multiple of $1$ through $6$

Big Hint:

The valid numbers are the multiples of that value; take the second one

Solution:

A number divisible by every integer from $1$ to $6$ must be a multiple of $\operatorname{lcm}(1,2,3,4,5,6)=60.$

The second smallest positive multiple of $60$ is $120,$ whose digit sum is $1+2+0=3.$

Thus, the correct answer is A.

6.

Two tangents to a circle are drawn from a point $A.$ The points of contact $B$ and $C$ divide the circle into arcs with lengths in the ratio $2:3.$ What is the degree measure of $\angle BAC?$

$24$

$30$

$36$

$48$

$60$

Answer: C

Concepts:tangent line arc angle chasing

Difficulty rating: 1240

Small Hint:

The two arcs split $360^\circ$ in the ratio $2:3$

Big Hint:

A tangent meets the radius at its point of contact at $90^\circ$ ; use quadrilateral $ABOC$

Solution:

Let $O$ be the center. The arcs measure $2x$ and $3x$ with $2x+3x=360^\circ,$ so $x=72^\circ$ and the minor arc $BC$ gives central angle $\angle BOC=144^\circ.$

The radii to $B$ and $C$ are perpendicular to the tangents, so $\angle ABO=\angle ACO=90^\circ.$ In quadrilateral $ABOC,$ $\begin{gathered} \angle BAC=360^\circ-144^\circ-90^\circ \\ {}-90^\circ=36^\circ. \end{gathered}$

Thus, the correct answer is C.

7.

Let $x$ and $y$ be two-digit positive integers with mean $60.$ What is the maximum value of the ratio $\dfrac{x}{y}?$

$3$

$\dfrac{33}{7}$

$\dfrac{39}{7}$

$9$

$\dfrac{99}{10}$

Answer: B

Concepts:optimization bounding to limit cases

Difficulty rating: 1200

Small Hint:

The mean condition gives $x+y=120$

Big Hint:

To maximize $\dfrac{x}{y},$ make $x$ as large and $y$ as small as allowed

Solution:

Since $\dfrac{x+y}{2}=60,$ we have $x+y=120.$ To maximize $\dfrac{x}{y}$ we make $y$ small.

Because $x\le99,$ it follows that $y=120-x\ge21.$ Taking $x=99$ and $y=21$ gives the maximum $\dfrac{99}{21}=\dfrac{33}{7}.$

Thus, the correct answer is B.

8.

Keiko walks once around a track at exactly the same constant speed every day. The sides of the track are straight, and the ends are semicircles. The track has width $6$ meters, and it takes her $36$ seconds longer to walk around the outside edge of the track than around the inside edge. What is Keiko’s speed in meters per second?

$\dfrac{\pi}{3}$

$\dfrac{2\pi}{3}$

$\pi$

$\dfrac{4\pi}{3}$

$\dfrac{5\pi}{3}$

Answer: A

Concepts:circumference distance rate and time

Difficulty rating: 1330

Small Hint:

The straight portions have the same length on both edges, so only the curved ends differ

Big Hint:

The two semicircular ends form full circles whose radii differ by $6$

Solution:

The straight sides are the same length for both paths, so the difference in length comes only from the two semicircular ends. If the inner radius is $r,$ those ends combine into a full circle, and the extra length is $2\pi(r+6)-2\pi r=12\pi.$

If her speed is $x$ meters per second, then the extra time gives $36x=12\pi,$ so $x=\dfrac{\pi}{3}.$

Thus, the correct answer is A.

9.

Two real numbers are selected independently at random from the interval $[-20, 10].$ What is the probability that the product of those numbers is greater than zero?

$\dfrac{1}{9}$

$\dfrac{1}{3}$

$\dfrac{4}{9}$

$\dfrac{5}{9}$

$\dfrac{2}{3}$

Answer: D

Concepts:geometric probability independent events

Difficulty rating: 1390

Small Hint:

The product is positive exactly when both numbers share the same sign

Big Hint:

Each number is negative with probability $\dfrac{20}{30}$ and positive with probability $\dfrac{10}{30}$

Solution:

The interval has length $30,$ with $20$ of it negative and $10$ of it positive. So each number is positive with probability $\dfrac13$ and negative with probability $\dfrac23.$

The product is positive when both are positive or both are negative: $\left(\dfrac13\right)^2+\left(\dfrac23\right)^2=\dfrac19+\dfrac49=\dfrac59.$

Thus, the correct answer is D.

10.

Rectangle $ABCD$ has $AB=6$ and $BC=3.$ Point $M$ is chosen on side $AB$ so that $\angle AMD=\angle CMD.$ What is the degree measure of $\angle AMD?$

$15$

$30$

$45$

$60$

$75$

Answer: E

Concepts:parallel lines isosceles triangle special right triangle

Difficulty rating: 1450

Small Hint:

Since $AB\parallel CD,$ the angle $\angle CDM$ equals $\angle AMD$

Big Hint:

That makes $\triangle CMD$ isosceles with $CM=CD=6,$ so $\triangle MBC$ is $30$ - $60$ - $90$

Solution:

Because $AB\parallel CD,$ we have $\angle CDM=\angle AMD.$ Combined with $\angle AMD=\angle CMD,$ this gives $\angle CDM=\angle CMD,$ so $\triangle CMD$ is isosceles with $CM=CD=6.$

Then $\triangle MBC$ is right-angled at $B$ with hypotenuse $CM=6$ and leg $BC=3,$ so it is a $30$ - $60$ - $90^\circ$ triangle with $\angle BMC=30^\circ.$

Finally, $\angle AMD+\angle CMD$ $+\angle BMC=180^\circ,$ so $2\angle AMD+30^\circ=180^\circ,$ giving $\angle AMD=75^\circ.$

Thus, the correct answer is E.

11.

A frog located at $(x, y),$ with both $x$ and $y$ integers, makes successive jumps of length $5$ and always lands on points with integer coordinates. Suppose that the frog starts at $(0, 0)$ and ends at $(1, 0).$ What is the smallest possible number of jumps the frog makes?

$2$

$3$

$4$

$5$

$6$

Answer: B

Concepts:lattice point distance formula

Difficulty rating: 1510

Small Hint:

One jump is impossible since the distance is $1,$ not $5$

Big Hint:

A midpoint of two jumps would lie on $x=\tfrac12,$ which has no lattice points; try three jumps

Solution:

One jump cannot work, since $(0,0)$ and $(1,0)$ are only $1$ apart. Two jumps also fail: the intermediate point would be at distance $5$ from both, forcing it onto the perpendicular bisector $x=\dfrac12,$ which contains no lattice points.

Three jumps suffice, for example $(0,0)\to(3,4)\to(6,0)\to(1,0),$ where each step has length $5.$

Thus, the correct answer is B.

12.

A dart board is a regular octagon divided into regions as shown. Suppose that a dart thrown at the board is equally likely to land anywhere on the board. What is the probability that the dart lands within the center square?

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

$\dfrac{1}{4}$

$\dfrac{2-\sqrt{2}}{2}$

$\dfrac{\sqrt{2}}{4}$

$2-\sqrt{2}$

Answer: A

Concepts:geometric probability area decomposition regular polygon

Difficulty rating: 1480

Small Hint:

Set the octagon’s edge to $1$ and split it into the center square, four rectangles, and four corner triangles

Big Hint:

The corner triangles are right isosceles with legs $\dfrac{\sqrt2}{2}$

Solution:

Assume the octagon has edge length $1.$ The four corner triangles are right isosceles with legs $\dfrac{\sqrt2}{2}$ and area $\dfrac14$ each. The four rectangles are $1$ by $\dfrac{\sqrt2}{2}$ with area $\dfrac{\sqrt2}{2}$ each, and the center square has area $1.$

The total area is $4\cdot\dfrac14+4\cdot\dfrac{\sqrt2}{2}+1=2+2\sqrt2.$ The probability of hitting the center square is $\dfrac{1}{2+2\sqrt2}=\dfrac{\sqrt2-1}{2}.$

Thus, the correct answer is A.

13.

Brian writes down four integers $w \gt x \gt y \gt z$ whose sum is $44.$ The pairwise positive differences of these numbers are $1,$ $3,$ $4,$ $5,$ $6,$ and $9.$ What is the sum of the possible values for $w?$

$16$

$31$

$48$

$62$

$93$

Answer: B

Concepts:system of equations casework

Difficulty rating: 1610

Small Hint:

The largest difference is $w-z=9,$ and the smallest positive difference must be $x-y=1$

Big Hint:

The second largest difference $6$ is either $w-y$ or $x-z$ ; handle each case

Solution:

The largest difference is $w-z=9.$ For either interior number $n,$ we have $9=(w-n)+(n-z).$ The only pairs among the listed differences that sum to $9$ are $3+6$ and $4+5,$ so the remaining difference must be $x-y=1.$

The second largest difference $6$ is either $w-y$ or $x-z.$ If $w-y=6,$ the numbers are $\{w,w-5,w-6,w-9\},$ so $4w-20=44$ and $w=16.$ If $x-z=6,$ the numbers are $\{w,w-3,w-4,w-9\},$ so $4w-16=44$ and $w=15.$

The possible values are $16$ and $15,$ which sum to $31.$

Thus, the correct answer is B.

14.

A segment through the focus $F$ of a parabola with vertex $V$ is perpendicular to $\overline{FV}$ and intersects the parabola in points $A$ and $B.$ What is $\cos(\angle AVB)?$

$-\dfrac{3\sqrt{5}}{7}$

$-\dfrac{2\sqrt{5}}{5}$

$-\dfrac{4}{5}$

$-\dfrac{3}{5}$

$-\dfrac{1}{2}$

Answer: D

Concepts:parabola trigonometric identity Pythagorean Theorem

Difficulty rating: 1710

Small Hint:

Let $p=FV.$ Use the focus-directrix property to find $FB$ and then $VB$

Big Hint:

Find $\cos(\angle FVB),$ then $\angle AVB=2\angle FVB$ via the double angle formula

Solution:

Let $p=FV$ and let the directrix be $\ell.$ Projecting $F$ and $B$ onto $\ell,$ the focus-directrix property gives $FB=2p$ (the distance from $B$ to $\ell$ ), and by the Pythagorean Theorem $\begin{aligned} VB&=\sqrt{FV^2+FB^2} \\ &=\sqrt{p^2+4p^2}=\sqrt5\,p. \end{aligned}$

Then $\cos(\angle FVB)=\dfrac{FV}{VB}$ $=\dfrac{p}{\sqrt5\,p}$ $=\dfrac{1}{\sqrt5}.$ Since $\angle AVB=2\angle FVB,$ $\begin{aligned} \cos(\angle AVB) &=2\cos^2(\angle FVB) \\ &\quad {}-1 \\ &=2\cdot\dfrac15-1 \\ &=-\dfrac35. \end{aligned}$

Thus, the correct answer is D.

15.

How many positive two-digit integers are factors of $2^{24}-1?$

$4$

$8$

$10$

$12$

$14$

Answer: D

Concepts:difference of squares prime factorization factor

Difficulty rating: 1740

Small Hint:

Repeatedly apply the difference of squares to factor $2^{24}-1$

Big Hint:

The prime factorization is $3^2\cdot5\cdot7\cdot13\cdot17\cdot241;$ list two-digit products

Solution:

Factoring, $\begin{aligned} 2^{24}-1 &=(2^{12}-1)(2^{12}+1) \\ &=(2^6-1)(2^6+1) \\ &\quad {}\cdot(2^4+1)(2^8-2^4+1), \end{aligned}$ which equals $63\cdot65\cdot17\cdot241$ $=3^2\cdot5\cdot7\cdot13\cdot17\cdot241.$

Since $241$ is a three-digit prime, the two-digit factors come from $3^2\cdot5\cdot7\cdot13\cdot17.$ They are $\begin{gathered} 13,15,17,21,35,39,45,51, \\ 63,65,85,91, \end{gathered}$ for a total of $12.$

Thus, the correct answer is D.

16.

Rhombus $ABCD$ has side length $2$ and $\angle B=120^\circ.$ Region $R$ consists of all points inside the rhombus that are closer to vertex $B$ than any of the other three vertices. What is the area of $R?$

$\dfrac{\sqrt{3}}{3}$

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

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

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

$2$

Answer: C

Concepts:perpendicular bisector rhombus area

Difficulty rating: 1850

Small Hint:

The boundary of $R$ lies along the perpendicular bisectors of the sides through $B$

Big Hint:

$R$ is a pentagon made of four congruent $30$ - $60$ - $90$ triangles, each with a leg of length $1$

Solution:

Let $E$ and $H$ be the midpoints of $AB$ and $BC.$ The perpendicular bisector of $AB$ through $E$ meets diagonal $AC$ at $F,$ and the perpendicular bisector of $BC$ through $H$ meets $AC$ at $G.$ The region $R$ is the pentagon $BEFGH.$

Triangle $AFE$ is a $30$ - $60$ - $90^\circ$ triangle with $AE=1,$ so its area is $\dfrac12\cdot1\cdot\dfrac{1}{\sqrt3}=\dfrac{\sqrt3}{6}.$ Triangles $BFE$ and $BGH$ are congruent to it, and $\triangle FBG$ is equilateral, splitting into two more copies.

Hence $R$ consists of four congruent triangles, giving area $4\cdot\dfrac{\sqrt3}{6}=\dfrac{2\sqrt3}{3}.$

Thus, the correct answer is C.

17.

Let $f(x)=10^{10x},$ $g(x)=\log_{10}\!\left(\dfrac{x}{10}\right),$ $h_1(x)=g(f(x)),$ and $h_n(x)=h_1(h_{n-1}(x))$ for integers $n\ge2.$ What is the sum of the digits of $h_{2011}(1)?$

$16{,}081$

$16{,}089$

$18{,}089$

$18{,}098$

$18{,}099$

Answer: B

Concepts:logarithm recursion digits

Difficulty rating: 1980

Small Hint:

Simplify $h_1(x)=\log_{10}\!\left(\dfrac{10^{10x}}{10}\right)$ to a linear function

Big Hint:

Iterating gives $h_n(1)$ as a number whose digits are all $8$ ’s ending in a $9$

Solution:

First, $\begin{aligned} h_1(x) &=\log_{10}\!\left(\dfrac{10^{10x}}{10}\right) \\ &=\log_{10}\!\left(10^{10x-1}\right) \\ &=10x-1. \end{aligned}$

Iterating, $h_n(x)=10^n x$ $-(1+10+\cdots+10^{n-1}).$ Therefore $h_n(1)$ is an $n$ -digit integer whose units digit is $9$ and all of whose other digits are $8.$

For $n=2011,$ the digit sum is $8\cdot2010+9=16{,}089.$

Thus, the correct answer is B.

18.

A pyramid has a square base with sides of length $1$ and has lateral faces that are equilateral triangles. A cube is placed within the pyramid so that one face is on the base of the pyramid and its opposite face has all its edges on the lateral faces of the pyramid. What is the volume of this cube?

$5\sqrt{2}-7$

$7-4\sqrt{3}$

$\dfrac{2\sqrt{2}}{27}$

$\dfrac{\sqrt{2}}{9}$

$\dfrac{\sqrt{3}}{9}$

Answer: A

Concepts:pyramid cube geometry 3D geometry

Difficulty rating: 2030

Small Hint:

Slice the pyramid through the plane containing a diagonal of the base and the apex

Big Hint:

That cross-section is an isosceles right triangle with hypotenuse $\sqrt2;$ the cube meets it in a rectangle of height $x$ and width $\sqrt2\,x$

Solution:

Let the apex be $A$ and the base be square $BCDE.$ Then $AB=AD=1$ and $BD=\sqrt2,$ so $\triangle BAD$ is an isosceles right triangle.

Let the cube have edge length $x.$ Its intersection with the plane of $\triangle BAD$ is a rectangle of height $x$ and width $\sqrt2\,x,$ whose top corners lie on $AB$ and $AD.$ Because the legs $AB$ and $AD$ meet the base at $45^\circ,$ each portion of $BD$ outside the rectangle has length $x,$ so $\sqrt2=BD=\sqrt2\,x+2x,$ which reduces to $x=\dfrac{\sqrt2}{2+\sqrt2}=\sqrt2-1.$

The volume is $(\sqrt2-1)^3=5\sqrt2-7.$

Thus, the correct answer is A.

19.

A lattice point in an $xy$ -coordinate system is any point $(x, y)$ where both $x$ and $y$ are integers. The graph of $y=mx+2$ passes through no lattice point with $0 \lt x \le 100$ for all $m$ such that $\dfrac{1}{2} \lt m \lt a.$ What is the maximum possible value of $a?$

$\dfrac{51}{101}$

$\dfrac{50}{99}$

$\dfrac{51}{100}$

$\dfrac{52}{101}$

$\dfrac{13}{25}$

Answer: B

Concepts:lattice point slope

Difficulty rating: 2090

Small Hint:

For each $x,$ find the smallest slope $\gt\dfrac12$ from $(0,2)$ that reaches a lattice point at that $x$

Big Hint:

Compare the minimum slopes for even $x$ and odd $x;$ the tighter bound wins

Solution:

For $0\lt x\le100,$ the nearest lattice point above the line $y=\tfrac12x+2$ is $\left(x,\tfrac12x+3\right)$ if $x$ is even and $\left(x,\tfrac12x+\tfrac52\right)$ if $x$ is odd.

The slope from $(0,2)$ to that point is $\dfrac12+\dfrac1x$ for even $x$ and $\dfrac12+\dfrac{1}{2x}$ for odd $x.$ The minimum such slope is $\dfrac{51}{100}$ for even $x$ and $\dfrac{50}{99}$ for odd $x.$

Since $\dfrac{50}{99}\lt\dfrac{51}{100},$ the line avoids all these lattice points exactly when $\dfrac12\lt m\lt\dfrac{50}{99},$ so the maximum is $a=\dfrac{50}{99}.$

Thus, the correct answer is B.

20.

Triangle $ABC$ has $AB=13,$ $BC=14,$ and $AC=15.$ The points $D,$ $E,$ and $F$ are the midpoints of $AB,$ $BC,$ and $AC$ respectively. Let $X\ne E$ be the intersection of the circumcircles of $\triangle BDE$ and $\triangle CEF.$ What is $XA+XB+XC?$

$24$

$14\sqrt{3}$

$\dfrac{195}{8}$

$\dfrac{129\sqrt{7}}{14}$

$\dfrac{69\sqrt{2}}{4}$

Answer: C

Concepts:circumcircle, circumcenter, and circumradius inscribed angle Heron’s Formula

Difficulty rating: 2220

Small Hint:

Use the parallel midsegments and the Inscribed Angle Theorem to show $X$ is equidistant from $A,$ $B,$ $C$

Big Hint:

$X$ is the circumcenter, so $XA+XB+XC=3R$ where $R=\dfrac{abc}{4\cdot[ABC]}$

Solution:

Since $DE\parallel AC$ and $EF\parallel AB,$ we get $\angle BDE=\angle BAC=\angle EFC.$ By the Inscribed Angle Theorem, $\angle BXE=\angle BDE$ and $\angle EXC=\angle EFC,$ so $\angle BXE=\angle EXC.$ With $BE=EC,$ this forces $XB=XC.$

Also $\angle BXC=2\angle BAC.$ The chord formula in $\triangle BXC$ gives $BC=2XB\sin(\angle BAC),$ while the same formula in $\triangle ABC$ gives $BC=2R\sin(\angle BAC).$ Thus $XB=XC=R.$ The point at distance $R$ from both $B$ and $C$ on this side of $BC$ is the circumcenter of $\triangle ABC,$ so $XA=XB=XC=R.$

The area of the $13$ - $14$ - $15$ triangle is $84$ by Heron’s formula, so $R=\dfrac{13\cdot14\cdot15}{4\cdot84}=\dfrac{65}{8},$ and $XA+XB+XC=3R=\dfrac{195}{8}.$

Thus, the correct answer is C.

21.

The arithmetic mean of two distinct positive integers $x$ and $y$ is a two-digit integer. The geometric mean of $x$ and $y$ is obtained by reversing the digits of the arithmetic mean. What is $|x-y|?$

$24$

$48$

$54$

$66$

$70$

Answer: D

Concepts:difference of squares perfect square digits

Difficulty rating: 2180

Small Hint:

Write the arithmetic mean as $10a+b$ and the geometric mean as $10b+a$

Big Hint:

Compute $(x-y)^2=(x+y)^2-4xy;$ it factors as $396(a+b)(a-b)$

Solution:

Let the arithmetic mean be $10a+b$ and the geometric mean be $10b+a.$ Then $x+y=2(10a+b)$ and $xy=(10b+a)^2.$

Therefore $\begin{aligned} (x-y)^2 &=(x+y)^2-4xy \\ &=396(a^2-b^2) \\ &=11\cdot6^2 \\ &\quad {}\cdot(a+b)(a-b). \end{aligned}$ Since the arithmetic mean exceeds the geometric mean for distinct positive numbers, $a>b.$ Put $u=a-b$ and $v=a+b.$ Then $1\le u\le9$ and $1\le v\le17.$ For $11uv$ to be a square, $uv$ must contain an odd power of $11.$ Therefore $v=11,$ because $u<11$ and $v<22.$ Now $11uv=121u$ is a square, so $u$ is a square. Also $u$ and $v$ have the same parity, leaving $u=1$ or $9.$ The latter gives $a=10,$ not a digit, so $u=1$ and $(a,b)=(6,5).$

Then $(x-y)^2=11\cdot6^2\cdot11=66^2,$ so $|x-y|=66.$ (Indeed $\{x,y\}=\{32,98\}.$ )

Thus, the correct answer is D.

22.

Let $T_1$ be a triangle with sides $2011,$ $2012,$ and $2013.$ For $n\ge1,$ if $T_n=\triangle ABC$ and $D,$ $E,$ and $F$ are the points of tangency of the incircle of $\triangle ABC$ to the sides $AB,$ $BC,$ and $AC,$ respectively, then $T_{n+1}$ is a triangle with side lengths $AD,$ $BE,$ and $CF,$ if it exists. What is the perimeter of the last triangle in the sequence $(T_n)?$

$\dfrac{1509}{8}$

$\dfrac{1509}{32}$

$\dfrac{1509}{64}$

$\dfrac{1509}{128}$

$\dfrac{1509}{256}$

Answer: D

Concepts:incircle, incenter, and inradius recursion triangle inequality

Difficulty rating: 2350

Small Hint:

The tangent lengths satisfy $AD=\tfrac12(b+c-a),$ and similarly for $BE$ and $CF$

Big Hint:

Each triangle keeps the form $(y-1,y,y+1)$ with $y$ halving; it fails the triangle inequality once $y\le2$

Solution:

For a triangle with sides $a,b,c,$ the tangent lengths are $AD=\tfrac12(b+c-a),$ $BE=\tfrac12(a+c-b),$ and $CF=\tfrac12(a+b-c).$ If $T_n$ has sides $(y-1,y,y+1),$ then $T_{n+1}$ has sides $\left(\tfrac{y}{2}-1,\tfrac{y}{2},\tfrac{y}{2}+1\right).$

Starting from $T_1$ with middle side $2012,$ the middle side halves each step and the perimeter of $T_{n+1}$ is $\tfrac12$ the perimeter of $T_n.$ A triangle of this form exists only while its middle side exceeds $2.$

The middle side of $T_n$ is $\dfrac{2012}{2^{n-1}}.$ This first drops to $2$ or below at $n=11,$ so the last valid triangle is $T_{10},$ whose middle side is $\dfrac{2012}{2^9}$ and whose perimeter is $3\cdot\dfrac{2012}{2^9}=\dfrac{6036}{512}=\dfrac{1509}{128}.$

Thus, the correct answer is D.

23.

A bug travels in the coordinate plane, moving only along the lines that are parallel to the $x$ -axis or $y$ -axis. Let $A=(-3, 2)$ and $B=(3, -2).$ Consider all possible paths of the bug from $A$ to $B$ of length at most $20.$ How many points with integer coordinates lie on at least one of these paths?

$161$

$185$

$195$

$227$

$255$

Answer: C

Concepts:lattice point casework symmetry

Difficulty rating: 2390

Small Hint:

A point $X$ lies on such a path iff its taxicab distances satisfy $d(A,X)+d(X,B)\le20$

Big Hint:

This is $|x-3|+|x+3|$ $+|y-2|+|y+2|\le20;$ count by symmetry across the axes

Solution:

A lattice point $X=(x,y)$ lies on some path exactly when $\begin{aligned} d&=|x-3|+|x+3| \\ &\quad {}+|y-2|+|y+2|\le20. \end{aligned}$ This expression is unchanged when $x\to-x$ or $y\to-y,$ so we count points with $x\ge0,$ $y\ge0,$ multiply by $4,$ and correct for the axes.

If $0\le x\le3$ and $0\le y\le2,$ all $4\cdot3=12$ points work. If $0\le x\le3$ and $y\ge3,$ then $y\le7,$ giving $4\cdot5=20$ points. If $x\ge4$ and $0\le y\le2,$ then $x\le8,$ giving $5\cdot3=15$ points. Finally, for $x\ge4$ and $y\ge3,$ the condition is $x+y\le10,$ giving $4+3+2+1=10$ points. Thus there are $57$ in the first quadrant, including $15$ on the nonnegative axes. By symmetry the total is $4\cdot57-2\cdot15-3=195.$

Thus, the correct answer is C.

24.

Let $P(z)=z^8+(4\sqrt{3}+6)z^4$ $-(4\sqrt{3}+7).$ What is the minimum perimeter among all the $8$ -sided polygons in the complex plane whose vertices are precisely the zeros of $P(z)?$

$4\sqrt{3}+4$

$8\sqrt{2}$

$3\sqrt{2}+3\sqrt{6}$

$4\sqrt{2}+4\sqrt{3}$

$4\sqrt{3}+6$

Answer: B

Concepts:complex number roots of unity factoring

Difficulty rating: 2520

Small Hint:

Treat $P$ as a quadratic in $z^4$ and factor as $(z^4-1)\big(z^4+(4\sqrt3+7)\big)$

Big Hint:

Since $4\sqrt3+7=(\sqrt3+2)^2,$ the eight roots lie on two squares; the shortest edges have length $\sqrt2$

Solution:

Factoring in $z^4,$ $\begin{aligned} P(z) &=(z^4-1) \\ &\quad {}\cdot\big(z^4+(4\sqrt3+7)\big). \end{aligned}$ The first factor gives the roots $1,-1,i,-i.$ Since $4\sqrt3+7=(\sqrt3+2)^2$ and $2(\sqrt3+2)=(\sqrt3+1)^2,$ writing $w=\tfrac12(\sqrt3+1)$ the other four roots are $w(\pm1\pm i).$

The eight roots are symmetric about the origin with $4$ -fold symmetry, and every segment joining two of them has length at least $\sqrt2.$ Thus any such polygon has perimeter at least $8\sqrt2,$ and the polygon with vertices $1,$ $w(1+i),$ $i,$ $w(-1+i),$ $-1,$ $w(-1-i),$ $-i,$ $w(1-i)$ achieves it.

Thus, the correct answer is B.

25.

For every $m$ and $k$ integers with $k$ odd, denote by $\left[\dfrac{m}{k}\right]$ the integer closest to $\dfrac{m}{k}.$ For every odd integer $k,$ let $P(k)$ be the probability that $\left[\dfrac{n}{k}\right]+\left[\dfrac{100-n}{k}\right]=\left[\dfrac{100}{k}\right]$ for an integer $n$ randomly chosen from the interval $1\le n\le99!.$ What is the minimum possible value of $P(k)$ over the odd integers $k$ in the interval $1\le k\le99?$

$\dfrac{1}{2}$

$\dfrac{50}{99}$

$\dfrac{44}{87}$

$\dfrac{34}{67}$

$\dfrac{7}{13}$

Answer: D

Concepts:modular arithmetic floor and ceiling functions basic probability

Difficulty rating: 2650

Small Hint:

Whether $n$ works depends only on $n\bmod k,$ and each residue class is equally likely

Big Hint:

Write $100=qk+r$ with $|r|\le\tfrac{k-1}{2};$ then $P(k)=1-\dfrac{|r|}{k}$

Solution:

Because $\left[\dfrac{n+mk}{k}\right]=\left[\dfrac{n}{k}\right]+m,$ whether $n$ satisfies the identity depends only on $n\bmod k.$ Since $99!$ is divisible by $k$ for $1\le k\le99,$ every residue class is equally likely.

Write $100=qk+r$ and $n=q_1k+r_1,$ choosing both remainders in $[-\frac{(k-1)}{2},\frac{k-1}{2}].$ If $r\ge0,$ no carry occurs precisely when $r-\frac{k-1}{2}\le r_1\le\frac{k-1}{2};$ this gives $k-r$ residue classes. The case $r<0$ similarly gives $k+r$ classes. Hence in both cases $P(k)=1-\dfrac{|r|}{k}.$

To minimize $P(k)$ we maximize $|r|/k.$ If $r=\frac{k-1}{2},$ then $201=k(2q+1),$ and the largest possible $k\le99$ is the divisor $67$ of $201.$ If $r=-\frac{(k-1)}{2},$ then $199=k(2q-1);$ because $199$ is prime, only $k=1$ is possible. In every remaining case $|r|\le\frac{k-3}{2},$ so $\begin{aligned} P(k)&\ge\dfrac12+\dfrac{3}{2k} \\ &\ge\dfrac12+\dfrac{3}{198} \\ &>\dfrac{34}{67}. \end{aligned}$ For $k=67,$ $P(67)=\dfrac12+\dfrac{1}{2\cdot67}=\dfrac{34}{67}.$

Thus, the correct answer is D.