2018 AMC 12B Solutions | LIVE by Po-Shen Loh

1.

Kate bakes a $20$ -inch by $18$ -inch pan of cornbread. The cornbread is cut into pieces that measure $2$ inches by $2$ inches. How many pieces of cornbread does the pan contain?

$90$

$100$

$180$

$200$

$360$

Concepts:area

Difficulty rating: 840

Small Hint:

Divide the total area of the pan by the area of one piece

Big Hint:

The pan has area $20\times18$ and each piece has area $2\times2$

Solution:

The pan has area $20\cdot18=360$ square inches, and each piece has area $2\cdot2=4$ square inches.

The number of pieces is $\dfrac{360}{4}=90.$

Thus, the correct answer is A.

2.

Sam drove $96$ miles in $90$ minutes. His average speed during the first $30$ minutes was $60$ mph (miles per hour), and his average speed during the second $30$ minutes was $65$ mph. What was his average speed, in mph, during the last $30$ minutes?

$64$

$65$

$66$

$67$

$68$

Concepts:distance rate and time

Difficulty rating: 1080

Small Hint:

Find the distance covered in each of the first two $30$ -minute segments

Big Hint:

The remaining distance is covered in $\tfrac12$ hour; divide to get the speed

Solution:

In the first $30$ minutes Sam covered $60\cdot\tfrac12=30$ miles, and in the second he covered $65\cdot\tfrac12=32.5$ miles.

The last $30$ minutes covered $96-30-32.5=33.5$ miles, so the speed was $\dfrac{33.5}{\frac{1}{2}}=67\text{ mph}.$

Thus, the correct answer is D.

3.

A line with slope $2$ intersects a line with slope $6$ at the point $(40, 30).$ What is the distance between the $x$ -intercepts of these two lines?

$5$

$10$

$20$

$25$

$50$

Concepts:linear equation slope coordinate geometry

Difficulty rating: 1240

Small Hint:

Write each line in point-slope form through $(40,30)$

Big Hint:

Set $y=0$ in each equation to find its $x$ -intercept

Solution:

The line of slope $2$ is $y-30=2(x-40);$ setting $y=0$ gives $x=25.$ The line of slope $6$ is $y-30=6(x-40);$ setting $y=0$ gives $x=35.$

The distance between the intercepts is $|35-25|=10.$

Thus, the correct answer is B.

4.

A circle has a chord of length $10,$ and the distance from the center of the circle to the chord is $5.$ What is the area of the circle?

$25\pi$

$50\pi$

$75\pi$

$100\pi$

$125\pi$

Concepts:chord Pythagorean Theorem circle area

Difficulty rating: 1310

Small Hint:

The radius, half the chord, and the distance to the chord form a right triangle

Big Hint:

With legs $5$ and $5,$ the radius satisfies $r^2=5^2+5^2$

Solution:

Dropping a perpendicular from the center to the chord bisects it, forming a right triangle with legs $5$ (half the chord) and $5$ (the distance), and hypotenuse $r.$

Then $r^2=5^2+5^2=50,$ so the area is $\pi r^2=50\pi.$

Thus, the correct answer is B.

5.

How many subsets of $\{2, 3, 4, 5, 6, 7, 8, 9\}$ contain at least one prime number?

$128$

$192$

$224$

$240$

$256$

Concepts:complementary counting subsets

Difficulty rating: 1390

Small Hint:

Count all subsets, then subtract those containing no prime number

Big Hint:

The non-primes are $\{4,6,8,9\},$ so subsets with no prime number are subsets of that set

Solution:

The set has $8$ elements, giving $2^8=256$ subsets. The subsets with no prime use only the four non-primes $\{4,6,8,9\},$ and there are $2^4=16$ of these.

So the number containing at least one prime is $256-16=240.$

Thus, the correct answer is D.

6.

Suppose $S$ cans of soda can be purchased from a vending machine for $Q$ quarters. Which of the following expressions describes the number of cans of soda that can be purchased for $D$ dollars, where $1$ dollar is worth $4$ quarters?

$\dfrac{4DQ}{S}$

$\dfrac{4DS}{Q}$

$\dfrac{4Q}{DS}$

$\dfrac{DQ}{4S}$

$\dfrac{DS}{4Q}$

Concepts:ratio and proportion unit conversion

Difficulty rating: 1430

Small Hint:

One can costs $\tfrac{Q}{S}$ quarters

Big Hint:

Convert $D$ dollars to $4D$ quarters, then divide by the price per can

Solution:

One can costs $\tfrac{Q}{S}$ quarters, which is $\tfrac{Q}{4S}$ dollars. The number of cans that $D$ dollars can buy is

$\dfrac{D}{\tfrac{Q}{4S}}=\dfrac{4DS}{Q}.$

Thus, the correct answer is B.

7.

What is the value of $\begin{gathered} \log_3 7\cdot\log_5 9\cdot\log_7 11 \\ {}\cdot\log_9 13\cdots\log_{21} 25\cdot\log_{23} 27? \end{gathered}$

$3$

$3\log_7 23$

$6$

$9$

$10$

Concepts:logarithm telescoping

Difficulty rating: 1580

Small Hint:

Apply the change-of-base formula $\log_a b=\dfrac{\log b}{\log a}$

Big Hint:

Separate into two telescoping chains: bases $3,7,11,\dots$ and bases $5,9,13,\dots$

Solution:

The factors split into two telescoping chains. The odd-position factors form $\begin{gathered} \log_3 7\cdot\log_7 11 \\ {}\cdot\log_{11} 15\cdots\log_{23} 27 \\ =\log_3 27=3, \end{gathered}$ and the even-position factors form $\begin{gathered} \log_5 9\cdot\log_9 13\cdots\log_{21} 25 \\ =\log_5 25=2. \end{gathered}$

The product is $3\cdot2=6.$

Thus, the correct answer is C.

8.

Line segment $\overline{AB}$ is a diameter of a circle with $AB=24.$ Point $C,$ not equal to $A$ or $B,$ lies on the circle. As point $C$ moves around the circle, the centroid (center of mass) of $\triangle ABC$ traces out a closed curve missing two points. To the nearest positive integer, what is the area of the region bounded by this curve?

$25$

$38$

$50$

$63$

$75$

Concepts:centroid homothety circle area

Difficulty rating: 1600

Small Hint:

Let $O$ be the center; the centroid lies one-third of the way from $O$ toward $C$

Big Hint:

As $C$ moves on the circle of radius $12,$ the centroid traces a circle of radius $\tfrac13\cdot12$

Solution:

Let $O$ be the center of the circle. The centroid of $\triangle ABC$ is the average of $A,$ $B,$ and $C;$ since $O$ is the midpoint of $\overline{AB},$ the centroid lies one-third of the way from $O$ to $C.$

As $C$ traces the circle of radius $12,$ the centroid traces a circle of radius $\tfrac13\cdot12=4.$ Its area is $16\pi\approx50.$

Thus, the correct answer is C.

9.

What is $\sum_{i=1}^{100}\sum_{j=1}^{100}(i+j)?$

$100{,}100$

$500{,}500$

$505{,}000$

$1{,}001{,}000$

$1{,}010{,}000$

Concepts:summation arithmetic sequence

Difficulty rating: 1620

Small Hint:

Split the sum into $\sum\sum i+\sum\sum j$

Big Hint:

The sum of the first $100$ positive integers is $5050$

Solution:

Splitting the sum, $\begin{gathered} \sum_{i=1}^{100}\sum_{j=1}^{100}(i+j) \\ =\sum_{i=1}^{100}\sum_{j=1}^{100}i \\ {}+\sum_{i=1}^{100}\sum_{j=1}^{100}j \\ =100\sum_{i=1}^{100}i \\ {}+100\sum_{j=1}^{100}j. \end{gathered}$

Since $\sum_{k=1}^{100}k=5050,$ this equals $100\cdot5050$ $+100\cdot5050$ $=1{,}010{,}000.$

Thus, the correct answer is E.

10.

A list of $2018$ positive integers has a unique mode, which occurs exactly $10$ times. What is the least number of distinct values that can occur in the list?

$202$

$223$

$224$

$225$

$234$

Concepts:mode pigeonhole principle

Difficulty rating: 1700

Small Hint:

The $2018-10=2008$ non-mode entries can each appear at most $9$ times

Big Hint:

Find the fewest distinct non-mode values needed, then add $1$ for the mode

Solution:

The mode uses $10$ of the entries, leaving $2008.$ Because the mode is unique, every other value appears at most $9$ times, so at least $\left\lceil\tfrac{2008}{9}\right\rceil=224$ distinct non-mode values are needed.

Adding the mode gives $224+1=225.$ This is achievable: use $9$ copies each of $1$ through $223,$ ten copies of $224,$ and one copy of $225.$

Thus, the correct answer is D.

11.

A closed box with a square base is to be wrapped with a square sheet of wrapping paper. The box is centered on the wrapping paper with the vertices of the base lying on the midlines of the square sheet of paper, as shown in the figure on the left. The four corners of the wrapping paper are to be folded up over the sides and brought together to meet at the center of the top of the box, point $A$ in the figure on the right. The box has base length $w$ and height $h.$ What is the area of the sheet of wrapping paper?

$2(w+h)^2$

$\dfrac{(w+h)^2}{2}$

$2w^2+4wh$

$2w^2$

$w^2h$

Concepts:special right triangle area

Difficulty rating: 1760

Small Hint:

The distance from a corner of the sheet to its center is $\tfrac{w}{2}+h+\tfrac{w}{2}$

Big Hint:

A side of the square sheet is $\sqrt2$ times that corner-to-center distance

Solution:

Following a fold from a corner of the paper to the center of the box top, the distance from a corner of the sheet to its center is $\dfrac{w}{2}+h+\dfrac{w}{2}=w+h.$

That segment is a leg of a $45$ - $45$ - $90$ triangle whose hypotenuse is a full side of the square sheet, so the side length is $\sqrt2\,(w+h).$

The area of the sheet is $\left(\sqrt2\,(w+h)\right)^2=2(w+h)^2.$

Thus, the correct answer is A.

12.

Side $\overline{AB}$ of $\triangle ABC$ has length $10.$ The bisector of angle $A$ meets $\overline{BC}$ at $D,$ and $CD=3.$ The set of all possible values of $AC$ is an open interval $(m, n).$ What is $m+n?$

$16$

$17$

$18$

$19$

$20$

Concepts:angle bisector theorem triangle inequality

Difficulty rating: 1820

Small Hint:

By the angle bisector theorem, $\dfrac{AC}{CD}=\dfrac{AB}{BD}$

Big Hint:

Let $AC=q,$ so $BD=\dfrac{30}{q};$ then apply the three triangle inequalities

Solution:

Let $q=AC$ and $r=BD.$ The angle bisector theorem gives $\tfrac{q}{3}=\tfrac{10}{r},$ so $r=\tfrac{30}{q}.$

Applying the triangle inequalities to sides $q,$ $10,$ and $3+r$ and substituting $r=\tfrac{30}{q}$ yields $(q-15)(q+2)\lt0$ and $(q-3)(q+10)\gt0$ (the third inequality holds automatically). Together these force $3\lt q\lt15.$

So $(m,n)=(3,15)$ and $m+n=18.$

Thus, the correct answer is C.

13.

Square $ABCD$ has side length $30.$ Point $P$ lies inside the square so that $AP=12$ and $BP=26.$ The centroids of $\triangle ABP,$ $\triangle BCP,$ $\triangle CDP,$ and $\triangle DAP$ are the vertices of a convex quadrilateral. What is the area of that quadrilateral?

$100\sqrt{2}$

$100\sqrt{3}$

$200$

$200\sqrt{2}$

$200\sqrt{3}$

Concepts:centroid coordinate geometry area

Difficulty rating: 1810

Small Hint:

Place the square in coordinates and write $P=(3x,3y)$

Big Hint:

Each centroid averages three vertices; the four centroids form a square whose diagonals have length $20$

Solution:

Place $A=(0,30),$ $B=(0,0),$ $C=(30,0),$ $D=(30,30),$ and $P=(3x,3y).$ Averaging the vertices, the four centroids are $\begin{gathered} (x,\,y+10),\ \\ (x+10,\,y),\ \\ (x+20,\,y+10),\ \\ (x+10,\,y+20). \end{gathered}$

These form a square whose diagonals, one horizontal and one vertical, each have length $20.$ Its area is $\tfrac12\cdot20\cdot20=200,$ independent of where $P$ lies.

Thus, the correct answer is C.

14.

Joey and Chloe and their daughter Zoe all have the same birthday. Joey is $1$ year older than Chloe, and Zoe is exactly $1$ year old today. Today is the first of the $9$ birthdays on which Chloe’s age will be an integral multiple of Zoe’s age. What will be the sum of the two digits of Joey’s age the next time his age is a multiple of Zoe’s age?

$7$

$8$

$9$

$10$

$11$

Concepts:factor counting divisibility

Difficulty rating: 1870

Small Hint:

Let Chloe be $n$ today; $y$ years from now her age is a multiple of Zoe’s exactly when $1+y$ divides $n-1$

Big Hint:

So $n-1$ must have exactly $9$ divisors; find the only two-digit such value

Solution:

Let Chloe be $n$ today, so she is $n-1$ years older than Zoe. In $y$ years Chloe’s age $n+y$ is a multiple of Zoe’s age $1+y$ exactly when $1+y$ divides $n-1.$ Having $9$ such birthdays means $n-1$ has exactly $9$ divisors.

A number with exactly $9$ divisors has the form $p^2q^2$ for distinct primes $p,q,$ or $p^8.$ Because Joey’s age at the requested future birthday has two digits, $n-1\lt99;$ the only possibility is $2^2\cdot3^2=36.$ So Chloe is $37$ and Joey is $38.$

Joey’s age $38+y$ is a multiple of $1+y$ exactly when $1+y$ divides $37.$ The next time is $y=36,$ making Joey $74,$ with digit sum $7+4=11.$

Thus, the correct answer is E.

15.

How many $3$ -digit positive odd multiples of $3$ do not include the digit $3?$

$96$

$97$

$98$

$102$

$120$

Concepts:digits divisibility multiplication principle

Difficulty rating: 1930

Small Hint:

Choose the hundreds digit (not $0$ or $3$ ) and the units digit (odd, not $3$ ) first

Big Hint:

For each such pair, the allowed tens digits split evenly into three residue classes mod $3$

Solution:

Write the number as $\overline{abc}.$ The hundreds digit $a$ has $8$ choices ( $1,2,4,5,6,7,8,9$ ), and the units digit $c$ has $4$ choices ( $1,5,7,9$ ).

The tens digit $b$ may be any of $\{0,1,2,4,5,6,7,8,9\}.$ These split into three residue classes mod $3$ of equal size $\{0,6,9\},\{1,4,7\},\{2,5,8\},$ so exactly $3$ choices of $b$ make $a+b+c$ divisible by $3.$

The count is $8\cdot4\cdot3=96.$

Thus, the correct answer is A.

16.

The solutions to the equation $(z+6)^8=81$ are connected in the complex plane to form a convex regular polygon, three of whose vertices are labeled $A,$ $B,$ and $C.$ What is the least possible area of $\triangle ABC?$

$\dfrac{1}{6}\sqrt{6}$

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

$2\sqrt{3}-2\sqrt{2}$

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

$\sqrt{3}-1$

Concepts:roots of unity regular polygon triangle area

Difficulty rating: 1990

Small Hint:

Shifting by $6$ does not change the shape; the roots of $z^8=81$ form a regular octagon

Big Hint:

The circumradius is $81^{\frac{1}{8}}=\sqrt3,$ and the smallest triangle uses three consecutive vertices

Solution:

Translating by $6,$ the solutions of $z^8=81$ are eight points on a circle of radius $81^{\frac{1}{8}}=\sqrt3,$ forming a regular octagon. The minimum-area triangle uses three consecutive vertices.

Take $A=\left(\tfrac12\sqrt6,\tfrac12\sqrt6\right),$ $B=(\sqrt3,0),$ and $C=\left(\tfrac12\sqrt6,-\tfrac12\sqrt6\right).$ Then $AC=\sqrt6$ and the height is $\sqrt3-\tfrac12\sqrt6,$ so the area is $\begin{gathered} \tfrac12\cdot\sqrt6\left(\sqrt3-\tfrac12\sqrt6\right) \\ =\tfrac{3}{2}\sqrt2-\tfrac{3}{2}. \end{gathered}$

Thus, the correct answer is B.

17.

Let $p$ and $q$ be positive integers such that $\dfrac{5}{9}\lt\dfrac{p}{q}\lt\dfrac{4}{7}$ and $q$ is as small as possible. What is $q-p?$

$7$

$11$

$13$

$17$

$19$

Concepts:fraction inequality bounding to limit cases

Difficulty rating: 2090

Small Hint:

Since $9p\gt5q$ and $4q\gt7p$ with integer sides, $9p-5q\ge1$ and $4q-7p\ge1$

Big Hint:

Add these after writing $\tfrac{4}{7}-\tfrac{5}{9}=\tfrac{1}{63}$ to bound $q$

Solution:

From $\tfrac59\lt\tfrac pq$ we get $9p-5q\ge1,$ and from $\tfrac pq\lt\tfrac47$ we get $4q-7p\ge1.$ Now $\begin{gathered} \dfrac{1}{63}=\dfrac47-\dfrac59 \\ =\dfrac{4q-7p}{7q}+\dfrac{9p-5q}{9q} \\ \ge\dfrac{1}{7q}+\dfrac{1}{9q} \\ =\dfrac{16}{63q}. \end{gathered}$

Hence $q\ge16.$ With $q=16,$ the fraction $\tfrac{9}{16}$ lies strictly between $\tfrac59$ and $\tfrac47,$ so $p=9$ and $q-p=16-9=7.$

Thus, the correct answer is A.

18.

A function $f$ is defined recursively by $f(1)=f(2)=1$ and $f(n)=f(n-1)-f(n-2)+n$ for all integers $n\ge3.$ What is $f(2018)?$

$2016$

$2017$

$2018$

$2019$

$2020$

Concepts:recursion telescoping

Difficulty rating: 2150

Small Hint:

Compute several terms and look for a pattern in $f(n)-n$

Big Hint:

Expanding the recursion four times shows $f(n)=f(n-6)+6$

Solution:

Repeatedly substituting the recursion into itself gives $f(n)=f(n-6)+6.$ So $f$ increases by $6$ every time $n$ increases by $6.$

Since $2018=2+6\cdot336,$ we have $f(2018)=f(2)+6\cdot336$ $=1+2016=2017.$

Thus, the correct answer is B.

19.

Mary chose an even $4$ -digit number $n.$ She wrote down all the divisors of $n$ in increasing order from left to right: $1,$ $2,$ $\ldots,$ $\tfrac{n}{2},$ $n.$ At some moment Mary wrote $323$ as a divisor of $n.$ What is the smallest possible value of the next divisor written to the right of $323?$

$324$

$330$

$340$

$361$

$646$

Concepts:divisibility greatest common divisor prime factorization

Difficulty rating: 2170

Small Hint:

Factor $323=17\cdot19$

Big Hint:

If the next divisor $d$ were coprime to $323,$ then $n\ge323d\gt9999;$ so $d$ shares a prime with $323$

Solution:

Let $d$ be the next divisor after $323.$ If $\gcd(d,323)=1,$ then $323d$ divides $n,$ forcing $n\ge323d\gt323^2\gt9999,$ impossible for a $4$ -digit number. So $d$ shares a prime factor with $323=17\cdot19.$

Then $d-323\ge\gcd(d,323)\ge17,$ so $d\ge340.$ Indeed $d=340=17\cdot20$ occurs for $n=17\cdot19\cdot20=6460,$ which is even and $4$ -digit.

Thus, the correct answer is C.

20.

Let $ABCDEF$ be a regular hexagon with side length $1.$ Denote by $X,$ $Y,$ and $Z$ the midpoints of sides $AB,$ $CD,$ and $EF,$ respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ?$

$\dfrac{3}{8}\sqrt{3}$

$\dfrac{7}{16}\sqrt{3}$

$\dfrac{15}{32}\sqrt{3}$

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

$\dfrac{9}{16}\sqrt{3}$

Concepts:regular polygon area ratio equilateral triangle

Difficulty rating: 2270

Small Hint:

Put the regular hexagon on the unit circle and write coordinates for $A,C,E$ and the three midpoints

Big Hint:

Intersect the side lines of the two equilateral triangles to find the six vertices, then use shoelace

Solution:

Place the regular hexagon on the unit circle with $A=(1,0),$ $C=(-\tfrac12,\tfrac{\sqrt3}{2}),$ and $E=(-\tfrac12,-\tfrac{\sqrt3}{2}).$ The three specified midpoints are $X=(\tfrac34,\tfrac{\sqrt3}{4}),$ $Y=(-\tfrac34,\tfrac{\sqrt3}{4}),$ and $Z=(0,-\tfrac{\sqrt3}{2}).$

Intersecting the side lines of $\triangle ACE$ and $\triangle XYZ$ gives the six vertices of their common interior, in cyclic order: $\begin{gathered} (-\tfrac12,0),\quad (-\tfrac18,-\tfrac{3\sqrt3}{8}),\\ (\tfrac14,-\tfrac{\sqrt3}{4}),\quad (\tfrac58,\tfrac{\sqrt3}{8}),\\ (\tfrac14,\tfrac{\sqrt3}{4}),\quad (-\tfrac12,\tfrac{\sqrt3}{4}). \end{gathered}$ The shoelace formula applied to these vertices gives area $\dfrac{15\sqrt3}{32}.$

Thus, the correct answer is C.

21.

In $\triangle ABC$ with side lengths $AB=13,$ $AC=12,$ and $BC=5,$ let $O$ and $I$ denote the circumcenter and incenter, respectively. A circle with center $M$ is tangent to the legs $AC$ and $BC$ and to the circumcircle of $\triangle ABC.$ What is the area of $\triangle MOI?$

$\dfrac{5}{2}$

$\dfrac{11}{4}$

$3$

$\dfrac{13}{4}$

$\dfrac{7}{2}$

Concepts:coordinate geometry circumcircle, circumcenter, and circumradius incircle, incenter, and inradius tangent circles

Difficulty rating: 2360

Small Hint:

The triangle is right-angled at $C;$ put $C$ at the origin with the legs on the axes

Big Hint:

$O$ is the midpoint of the hypotenuse and $I=(2,2);$ the circle at $M=(\rho,\rho)$ is tangent to the circumcircle

Solution:

Since $5^2+12^2=13^2,$ the triangle is right-angled at $C.$ Set $C=(0,0),$ $A=(12,0),$ and $B=(0,5).$ Then $O$ is the midpoint of $\overline{AB},$ namely $O=\left(6,\tfrac52\right),$ with circumradius $\tfrac{13}{2}.$ The inradius is $\tfrac{\text{area}}{s}=\tfrac{30}{15}=2,$ so $I=(2,2).$

Because $M$ ’s circle is tangent to both legs, $M=(\rho,\rho).$ Internal tangency to the circumcircle gives $MO=\tfrac{13}{2}-\rho.$ Setting this equal to $\sqrt{(\rho-6)^2+\left(\rho-\tfrac52\right)^2}$ and solving gives $\rho=4,$ so $M=(4,4).$

The shoelace formula on $M=(4,4),$ $O=\left(6,\tfrac52\right),$ $I=(2,2)$ gives area $\tfrac72.$

Thus, the correct answer is E.

22.

Consider polynomials $P(x)$ of degree at most $3,$ each of whose coefficients is an element of $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}.$ How many such polynomials satisfy $P(-1)=-9?$

$110$

$143$

$165$

$220$

$286$

Concepts:stars and bars substitution

Difficulty rating: 2330

Small Hint:

Write $P(x)=ax^3+bx^2+cx+d;$ then $P(-1)=-a+b-c+d=-9$

Big Hint:

Substitute $a'=9-a$ and $c'=9-c$ to turn it into $a'+b+c'+d=9$ with each variable in $[0,9]$

Solution:

Write $P(x)=ax^3+bx^2+cx+d$ with each of $a,b,c,d$ in $\{0,\ldots,9\}.$ The condition is $-a+b-c+d=-9.$

Let $a'=9-a$ and $c'=9-c,$ both in $[0,9].$ Then $a'+b+c'+d=9.$ By stars and bars the number of nonnegative solutions is $\binom{9+3}{3}=\binom{12}{3}=220,$ and each automatically satisfies the upper bounds since the sum is $9.$

Thus, the correct answer is D.

23.

Ajay is standing at point $A$ near Pontianak, Indonesia, $0^\circ$ latitude and $110^\circ$ E longitude. Billy is standing at point $B$ near Big Baldy Mountain, Idaho, USA, $45^\circ$ N latitude and $115^\circ$ W longitude. Assume that Earth is a perfect sphere with center $C.$ What is the degree measure of $\angle ACB?$

$105$

$112\tfrac{1}{2}$

$120$

$135$

$150$

Concepts:3D geometry sphere vector

Difficulty rating: 2400

Small Hint:

The longitude difference is $360^\circ-(110^\circ+115^\circ)=135^\circ$

Big Hint:

Put both points on a unit sphere and take the dot product of their position vectors

Solution:

The longitudes differ by $360^\circ-(110^\circ+115^\circ)=135^\circ,$ and $B$ is at latitude $45^\circ$ N. Place $A=(1,0,0)$ on the unit sphere.

Then $B=\tiny\left(\cos45^\circ\cos135^\circ,\ \cos45^\circ\sin135^\circ,\ \sin45^\circ\right)$ $=\left(-\tfrac12,\tfrac12,\tfrac{\sqrt2}{2}\right).$ The dot product is $A\cdot B=-\tfrac12,$ so $\cos\angle ACB=-\tfrac12$ and $\angle ACB=120^\circ.$

Thus, the correct answer is C.

24.

Let $\lfloor x\rfloor$ denote the greatest integer less than or equal to $x.$ How many real numbers $x$ satisfy the equation $x^2+10{,}000\lfloor x\rfloor=10{,}000x?$

$197$

$198$

$199$

$200$

$201$

Concepts:floor and ceiling functions counting intersections

Difficulty rating: 2500

Small Hint:

Rewrite the equation as $x^2=10{,}000\,(x-\lfloor x\rfloor)$ $=10{,}000\{x\}$

Big Hint:

Then $0\le\dfrac{x^2}{10{,}000}\lt1,$ so $-100\lt x\lt100;$ count one solution per unit interval

Solution:

Let $\{x\}=x-\lfloor x\rfloor.$ The equation becomes $x^2=10{,}000\{x\},$ so $\tfrac{x^2}{10{,}000}=\{x\}.$ Since $0\le\{x\}\lt1,$ we need $0\le x^2\lt10{,}000,$ i.e. $-100\lt x\lt100.$

On each interval $[k,k+1),$ write $x=k+t$ with $0\le t\lt1.$ The equation becomes $(k+t)^2-10{,}000t=0.$ For $-100\le k\le98,$ the left side is strictly decreasing; at $t=0$ it is $k^2\ge0,$ while as $t$ approaches $1$ it approaches $(k+1)^2-10{,}000\lt0.$ Thus each of these intervals contains exactly one solution. There are $98-(-100)+1=199$ such intervals.

Thus, the correct answer is C.

25.

Circles $\omega_1,$ $\omega_2,$ and $\omega_3$ each have radius $4$ and are placed in the plane so that each circle is externally tangent to the other two. Points $P_1,$ $P_2,$ and $P_3$ lie on $\omega_1,$ $\omega_2,$ and $\omega_3,$ respectively, so that $P_1P_2=P_2P_3=P_3P_1$ and line $P_iP_{i+1}$ is tangent to $\omega_i$ for each $i=1,$ $2,$ $3,$ where $P_4=P_1.$ See the figure below. The area of $\triangle P_1P_2P_3$ can be written in the form $\sqrt{a}+\sqrt{b},$ where $a$ and $b$ are positive integers. What is $a+b?$

$546$

$548$

$550$

$552$

$554$

Concepts:tangent circles special right triangle law of cosines equilateral triangle

Difficulty rating: 2840

Small Hint:

Since $P_iP_{i+1}$ is tangent to $\omega_i$ at $P_i,$ the radius $O_iP_i$ is perpendicular to $P_iP_{i+1}$

Big Hint:

Let $K=O_1P_1\cap O_2P_2;$ with $d=P_1K,$ apply the Law of Cosines in $\triangle O_1KO_2$

Solution:

Let $O_i$ be the center of $\omega_i,$ and let $K$ be the intersection of lines $O_1P_1$ and $O_2P_2.$ Because $\angle P_1P_2P_3=60^\circ,$ triangle $P_2KP_1$ is a $30$ - $60$ - $90^\circ$ triangle. With $d=P_1K,$ we get $P_2K=2d$ and $P_1P_2=\sqrt3\,d.$

The Law of Cosines in $\triangle O_1KO_2$ (with $O_1O_2=8$ ) gives $\begin{gathered} 8^2=(d+4)^2+(2d-4)^2 \\ {}-2(d+4)(2d-4)\cos60^\circ, \end{gathered}$ which simplifies to $3d^2-12d-16=0,$ so $d=2+\tfrac23\sqrt{21}.$

Then $P_1P_2=\sqrt3\,d=2\sqrt3+2\sqrt7,$ and the area is $\begin{gathered} \dfrac{\sqrt3}{4}\left(2\sqrt3+2\sqrt7\right)^2 \\ =10\sqrt3+6\sqrt7 \\ =\sqrt{300}+\sqrt{252}. \end{gathered}$

So $a+b=300+252=552.$

Thus, the correct answer is D.