1994 AMC 12 — Problems & Online Timed Exam

Timed

1:15:00

1.

$4^4\cdot9^4\cdot4^9\cdot9^9=$

$13^{13}$

$13^{36}$

$36^{13}$

$36^{36}$

$1296^{26}$

Answer: C

Concepts:laws of exponents factoring

Difficulty rating: 960

Small Hint:

Group the powers of $4$ and the powers of $9$

Big Hint:

Both bases have total exponent $4+9$

Solution:

Combining like bases and then pairing them, $\begin{aligned} 4^4\cdot4^9\cdot9^4\cdot9^9 &=4^{13}9^{13}\\ &=(4\cdot9)^{13}\\ &=36^{13}. \end{aligned}$ Thus the correct answer is C.

2.

A large rectangle is partitioned into four rectangles by two segments parallel to its sides. The areas of three of the resulting rectangles are shown. What is the area of the fourth rectangle?

$6$	$14$
?	$35$

$10$

$15$

$20$

$21$

$25$

Answer: B

Concepts:rectangle areas proportionality

Difficulty rating: 960

Small Hint:

Represent the two column widths and two row heights by variables

Big Hint:

Products of diagonally opposite areas are equal

Solution:

Let the column widths be $u,v$ and the row heights be $s,t.$ The displayed areas give $us=6,$ $vs=14,$ and $vt=35.$ Therefore the missing area is $ut=\frac{(us)(vt)}{vs}=\frac{6\cdot35}{14}=15.$ Thus the correct answer is B.

3.

How many of the following are equal to $x^x+x^x$ for all $x\gt0?$ $\begin{aligned} \mathrm{I:}\quad&2x^x\\ \mathrm{II:}\quad&x^{2x}\\ \mathrm{III:}\quad&(2x)^x\\ \mathrm{IV:}\quad&(2x)^{2x} \end{aligned}$

$0$

$1$

$2$

$3$

$4$

Answer: B

Concepts:exponential expressions identities

Difficulty rating: 1260

Small Hint:

First combine the two identical terms

Big Hint:

Test whether changing the base or exponent preserves $2x^x$ for every $x$

Solution:

The given sum is $2x^x,$ so expression I is always equal to it. Expression II fails at $x=1,$ where its value is $1$ instead of $2.$ Expression III fails at $x=2,$ where its value is $16$ instead of $8.$ Expression IV fails at $x=1,$ where its value is $4$ instead of $2.$ Hence only one expression works. Thus the correct answer is B.

4.

In the $xy$ -plane, the segment with endpoints $(-5,0)$ and $(25,0)$ is the diameter of a circle. If the point $(x,15)$ is on the circle, then $x=$

$10$

$12.5$

$15$

$17.5$

$20$

Answer: A

Concepts:circle equation diameter

Difficulty rating: 1070

Small Hint:

Find the center and radius from the diameter endpoints

Big Hint:

The point’s vertical distance from the center already equals the radius

Solution:

The center is $(10,0)$ and the radius is $15.$ Thus $(x-10)^2+15^2=15^2,$ so $x=10.$ Thus the correct answer is A.

5.

Pat intended to multiply a number by $6$ but instead divided by $6.$ Pat then meant to add $14$ but instead subtracted $14.$ After these mistakes, the result was $16.$ If the correct operations had been used, the value produced would have been

less than $400$

between $400$ and $600$

between $600$ and $800$

between $800$ and $1000$

greater than $1000$

Answer: E

Concepts:reverse operations linear equations

Difficulty rating: 890

Small Hint:

Undo the mistaken subtraction and division to recover the starting number

Big Hint:

Apply $6x+14$ to that starting number

Solution:

If the starting number is $x,$ the mistaken calculation gives $\frac{x}{6}-14=16,$ so $x=180.$ The intended calculation would produce $6(180)+14=1094,$ which is greater than $1000.$ Thus the correct answer is E.

6.

In the sequence $\ldots,a,b,c,d,0,1,1,2,3,5,8,\ldots$ each term is the sum of the two terms to its left. Find $a.$

$-3$

$-1$

$0$

$1$

$3$

Answer: A

Concepts:Fibonacci recurrence working backward

Difficulty rating: 1150

Small Hint:

Work backward by subtracting the preceding known term

Big Hint:

Determine $d,c,b$ in that order before finding $a$

Solution:

Working backward, $d+0=1$ gives $d=1,$ then $c+d=0$ gives $c=-1,$ and $b+c=d$ gives $b=2.$ Finally $a+b=c,$ so $a=-1-2=-3.$ Thus the correct answer is A.

7.

Squares $ABCD$ and $EFGH$ are congruent, $AB=10,$ and $G$ is the center of square $ABCD.$ The area of the region in the plane covered by these squares is

$75$

$100$

$125$

$150$

$175$

Answer: E

Concepts:overlapping squares area diagonals

Difficulty rating: 1460

Small Hint:

The overlap is triangle $GAB$

Big Hint:

The distance from the center $G$ to side $\overline{AB}$ is $5$

Solution:

Each square has area $100.$ Their overlap is $\triangle GAB,$ whose base is $AB=10$ and whose altitude from the center $G$ to $\overline{AB}$ is $5.$ Its area is $\frac12(10)(5)=25.$ Hence the union has area $100+100-25=175.$ Thus the correct answer is E.

8.

In the polygon shown, each side is perpendicular to its adjacent sides, and all $28$ of the sides are congruent. The perimeter of the polygon is $56.$ The area of the region bounded by the polygon is

$84$

$96$

$100$

$112$

$196$

Answer: C

Concepts:orthogonal polygon area decomposition

Difficulty rating: 1180

Small Hint:

Each side has length $\frac{56}{28}$

Big Hint:

In units of one side length, split the shape into horizontal strips of widths $1,3,5,7,5,3,1$

Solution:

Each side has length $2.$ Measured in side-length units, the seven horizontal strips have widths $1,3,5,7,5,3,1,$ totaling $25$ unit squares. Each such square has area $2^2=4,$ so the polygon’s area is $25\cdot4=100.$ Thus the correct answer is C.

9.

If $\angle A$ is four times $\angle B,$ and the complement of $\angle B$ is four times the complement of $\angle A,$ then $\angle B=$

$10^\circ$

$12^\circ$

$15^\circ$

$18^\circ$

$22.5^\circ$

Answer: D

Concepts:complementary angles linear equations

Difficulty rating: 1260

Small Hint:

Let $\angle B=x$ and write $\angle A=4x$

Big Hint:

Translate the second condition as $90-x=4(90-4x)$

Solution:

Let $\angle B=x,$ so $\angle A=4x.$ The complement condition gives $90-x=4(90-4x).$ Hence $15x=270$ and $x=18^\circ.$ Thus the correct answer is D.

10.

For distinct real numbers $x$ and $y,$ let $M(x,y)$ be the larger of $x$ and $y$ and let $m(x,y)$ be the smaller of $x$ and $y.$ If $a\lt b\lt c\lt d\lt e,$ then $\begin{aligned} &M\bigl(M(a,m(b,c)),\\ &\qquad m(d,m(a,e))\bigr)= \end{aligned}$

$a$

$b$

$c$

$d$

$e$

Answer: B

Concepts:nested minimum and maximum inequalities

Difficulty rating: 1590

Small Hint:

Evaluate the innermost $m$ expressions first

Big Hint:

The two arguments of the outermost $M$ reduce to $b$ and $a$

Solution:

From the ordering, $m(b,c)=b,$ so $M(a,m(b,c))=M(a,b)=b.$ Also $m(a,e)=a,$ hence $m(d,m(a,e))=m(d,a)=a.$ The outer expression is therefore $M(b,a)=b.$ Thus the correct answer is B.

11.

Three cubes of volume $1,$ $8$ and $27$ are glued together at their faces. The smallest possible surface area of the resulting configuration is

$36$

$56$

$70$

$72$

$74$

Answer: D

Concepts:surface area cubes maximizing contact area

Difficulty rating: 1900

Small Hint:

The cube side lengths are $1,2,$ and $3$

Big Hint:

Minimize exposed area by maximizing the total areas of glued face portions

Solution:

Before gluing, the total surface area is $6(1^2+2^2+3^2)=84.$ The side- $2$ cube can share area $4$ with the side- $3$ cube, while the side- $1$ cube is placed at their common edge so that it shares area $1$ with each larger cube. Thus the total contact area is $6.$ Each glued area removes two exposed copies, giving $84-2(6)=72.$ No pair can share more than the smaller face, so this is maximal contact and minimal surface area. Thus the correct answer is D.

12.

If $i^2=-1,$ then $\left(i-i^{-1}\right)^{-1}=$

$0$

$-2i$

$2i$

$-\frac{i}{2}$

$\frac{i}{2}$

Answer: D

Concepts:complex numbers reciprocals

Difficulty rating: 1570

Small Hint:

Use $i^{-1}=-i$

Big Hint:

After simplifying the parentheses, rationalize $\frac{1}{2i}$

Solution:

Because $i^{-1}=-i,$ $\begin{aligned} \left(i-i^{-1}\right)^{-1} &=(i+i)^{-1}\\ &=\frac1{2i}\\ &=-\frac i2. \end{aligned}$ Thus the correct answer is D.

13.

In triangle $ABC,$ $AB=AC.$ If there is a point $P$ strictly between $A$ and $B$ such that $AP=PC=CB,$ then $\angle A=$

$30^\circ$

$36^\circ$

$48^\circ$

$60^\circ$

$72^\circ$

Answer: B

Concepts:isosceles triangles angle chasing

Difficulty rating: 1960

Small Hint:

Let $\angle A=\theta$ and use $AP=PC$ in $\triangle APC$

Big Hint:

Use $PC=CB$ to relate $\angle PCB$ to a base angle of $\triangle ABC$

Solution:

Let $\angle A=\theta.$ Since $AP=PC,$ triangle $APC$ has $\angle PAC=\angle PCA=\theta.$ Since $AB=AC,$ each base angle of $\triangle ABC$ is $\frac{180^\circ-\theta}{2}.$ Also $PC=CB,$ so the base angles of $\triangle PCB$ at $P$ and $B$ are equal; the one at $B$ is $\frac{180^\circ-\theta}{2}.$ Hence $\angle PCB=\theta.$ At $C,$ $2\theta=\frac{180^\circ-\theta}{2},$ giving $5\theta=180^\circ$ and $\theta=36^\circ.$ Thus the correct answer is B.

14.

Find the sum of the arithmetic series $20+20\frac15+20\frac25+\cdots+40.$

$3000$

$3030$

$3150$

$4100$

$6000$

Answer: B

Concepts:arithmetic series mixed numbers

Difficulty rating: 1420

Small Hint:

The common difference is $\frac{1}{5}$

Big Hint:

Find the number of terms, then multiply by the average of the endpoints

Solution:

The number of terms is $\frac{40-20}{\frac{1}{5}}+1=101.$ Their average is $\frac{20+40}{2}=30,$ so the sum is $101\cdot30=3030.$ Thus the correct answer is B.

15.

For how many $n$ in $\{1,2,3,\ldots,100\}$ is the tens digit of $n^2$ odd?

$10$

$20$

$30$

$40$

$50$

Answer: B

Concepts:squares modulo 100 decimal digits

Difficulty rating: 1750

Small Hint:

The parity of the tens digit of $n^2$ depends only on the units digit of $n$

Big Hint:

Check the squares of the ten possible units digits

Solution:

Write $n=10a+b.$ Modulo $100,$ $n^2\equiv20ab+b^2.$ The term $20ab$ changes the tens digit by an even amount, so only the tens digit of $b^2$ matters. Among $b=0,1,\ldots,9,$ it is odd only for $b=4$ and $b=6.$ Each units digit occurs $10$ times from $1$ through $100,$ giving $2\cdot10=20.$ Thus the correct answer is B.

16.

Some marbles in a bag are red and the rest are blue. If one red marble is removed, then one-seventh of the remaining marbles are red. If two blue marbles are removed instead of one red, then one-fifth of the remaining marbles are red. How many marbles were in the bag originally?

$8$

$22$

$36$

$57$

$71$

Answer: B

Concepts:systems of equations proportions

Difficulty rating: 1510

Small Hint:

Let $R$ be the red count and $T$ the total count

Big Hint:

Translate the two experiments into $\frac{R-1}{T-1}=\frac{1}{7}$ and $\frac{R}{T-2}=\frac{1}{5}$

Solution:

The two conditions give $\frac{R-1}{T-1}=\frac17$ and $\frac{R}{T-2}=\frac15.$ Thus $T=7R-6$ and $T=5R+2.$ Hence $R=4$ and $T=22.$ Thus the correct answer is B.

17.

An $8$ by $2\sqrt2$ rectangle has the same center as a circle of radius $2.$ The area of the region common to both the rectangle and the circle is

$2\pi$

$2\pi+2$

$4\pi-4$

$2\pi+4$

$4\pi-2$

Answer: D

Concepts:circle segments area symmetry

Difficulty rating: 1960

Small Hint:

The rectangle removes two congruent caps from the circle

Big Hint:

The cap chord is at distance $\sqrt2$ from the center, so its half central angle is $45^\circ$

Solution:

The rectangle is wider than the circle, so the common region is the circle with the top and bottom caps beyond distance $\sqrt2$ from the center removed. One cap has area $\begin{aligned} 2^2\left(\frac{\pi}{4}\right) &-\sqrt2\sqrt{2^2-(\sqrt2)^2}\\ &=\pi-2. \end{aligned}$ Therefore the common area is $4\pi-2(\pi-2)=2\pi+4.$ Thus the correct answer is D.

18.

Triangle $ABC$ is inscribed in a circle, and $\angle B=\angle C=4\angle A.$ If $B$ and $C$ are adjacent vertices of a regular polygon of $n$ sides inscribed in this circle, then $n=$

$5$

$7$

$9$

$15$

$18$

Answer: C

Concepts:inscribed angles regular polygons

Difficulty rating: 1800

Small Hint:

Use the triangle angle sum to find $\angle A$

Big Hint:

The central angle subtending $\overline{BC}$ is twice $\angle A$

Solution:

The angle sum gives $9\angle A=180^\circ,$ so $\angle A=20^\circ.$ The central angle subtending chord $\overline{BC}$ is therefore $40^\circ.$ Adjacent vertices of a regular $n$ -gon subtend $\frac{360^\circ}{n},$ so $\frac{360^\circ}{n}=40^\circ,$ and $n=9.$ Thus the correct answer is C.

19.

Label one disk “ $1$ ,” two disks “ $2$ ,” three disks “ $3$ ,” $\ldots,$ fifty disks “ $50$ .” Put these $1+2+3+\cdots+50=1275$ labeled disks in a box. Disks are then drawn from the box at random without replacement. The minimum number of disks that must be drawn to guarantee drawing at least ten disks with the same label is

$10$

$51$

$415$

$451$

$501$

Answer: C

Concepts:pigeonhole principle extremal counting

Difficulty rating: 1750

Small Hint:

Count the most disks that can be drawn while taking at most nine of each label

Big Hint:

All disks labeled $1$ through $9$ may be drawn, but only nine of each label $10$ through $50$

Solution:

To avoid ten equal labels, one may draw all $1+2+\cdots+9=45$ disks with labels below $10,$ and at most $9$ from each of the $41$ labels $10$ through $50.$ Thus $45+9(41)=414$ disks can be drawn without forcing ten alike, and the next draw guarantees them. The minimum is $415.$ Thus the correct answer is C.

20.

Suppose $x,$ $y,$ $z$ is a geometric sequence with common ratio $r$ and $x\ne y.$ If $x,$ $2y,$ $3z$ is an arithmetic sequence, then $r$ is

$\frac14$

$\frac13$

$\frac12$

$2$

$4$

Answer: B

Concepts:geometric sequences arithmetic sequences quadratic equations

Difficulty rating: 1570

Small Hint:

Write $y=xr$ and $z=xr^2$

Big Hint:

The arithmetic-sequence condition is $4y=x+3z$

Solution:

Because $y=xr$ and $z=xr^2,$ the arithmetic-sequence condition gives $4xr=x+3xr^2.$ Here $x\ne0,$ so $3r^2-4r+1=0,$ yielding $r=1$ or $r=\frac{1}{3}.$ The condition $x\ne y$ excludes $r=1,$ leaving $r=\frac{1}{3}.$ Thus the correct answer is B.

21.

Find the number of counterexamples to the statement:

“If $N$ is an odd positive integer the sum of whose digits is $4$ and none of whose digits is $0,$ then $N$ is prime.”

$0$

$1$

$2$

$3$

$4$

Answer: C

Concepts:digit sums primality enumeration

Difficulty rating: 1900

Small Hint:

A number satisfying the digit conditions has at most four digits

Big Hint:

List the odd possibilities by composing $4$ into positive digits

Solution:

The odd possibilities are $13,31,121,211,$ and $1111.$ The numbers $13,31,$ and $211$ are prime, while $121=11^2$ and $1111=11\cdot101.$ Thus there are $2$ counterexamples. Thus the correct answer is C.

22.

Nine chairs in a row are to be occupied by six students and Professors Alpha, Beta and Gamma. These three professors arrive before the six students and decide to choose their chairs so that each professor will be between two students. In how many ways can Professors Alpha, Beta and Gamma choose their chairs?

$12$

$36$

$60$

$84$

$630$

Answer: C

Concepts:seating arrangements nonconsecutive selections

Difficulty rating: 1900

Small Hint:

Professor chairs cannot be endpoints or adjacent to one another

Big Hint:

First choose three nonconsecutive positions from chairs $2$ through $8$ , then assign the professors

Solution:

The professors must occupy three nonconsecutive positions among chairs $2,3,\ldots,8.$ The number of such $3$ -subsets of $7$ consecutive positions is $\binom{7-3+1}{3}=\binom53=10.$ The three distinct professors can be assigned to the selected chairs in $3!=6$ ways, for $10\cdot6=60.$ Thus the correct answer is C.

23.

In the $xy$ -plane, consider the L-shaped region bounded by horizontal and vertical segments with vertices at $(0,0),$ $(0,3),$ $(3,3),$ $(3,1),$ $(5,1)$ and $(5,0).$ The slope of the line through the origin that divides the area of this region exactly in half is

$\frac27$

$\frac13$

$\frac23$

$\frac34$

$\frac79$

Answer: E

Concepts:coordinate geometry area bisectors slope

Difficulty rating: 1960

Small Hint:

The L-shaped region has total area $11$

Big Hint:

For the desired slope $m,$ the area below the line is the $2$ -unit extension plus a triangle of base $3$

Solution:

The region has area $3\cdot3+2\cdot1=11,$ so each half has area $\frac{11}{2}.$ If the line is $y=mx,$ the part below it consists of the entire $2$ -unit rectangle from $x=3$ to $x=5,$ together with a triangle of base $3$ and height $3m.$ Therefore $2+\frac12(3)(3m)=\frac{11}{2}.$ This gives $m=\frac{7}{9},$ which indeed lies in the assumed range. Thus the correct answer is E.

24.

A sample consisting of five observations has an arithmetic mean of $10$ and a median of $12.$ The smallest value that the range (largest observation minus smallest) can assume for such a sample is

$2$

$3$

$5$

$7$

$10$

Answer: C

Concepts:mean median range extremal construction

Difficulty rating: 1960

Small Hint:

Order the observations as $a\le b\le12\le d\le e$

Big Hint:

Use the total sum $50$ , then test whether a range below $5$ is possible

Solution:

The sample $7,7,12,12,12$ has sum $50,$ median $12,$ and range $5,$ so range $5$ is possible. Suppose the range were at most $4.$ If the largest value were at most $12,$ then the top three values would all be $12,$ leaving the bottom two with sum $14;$ yet each would be at least $8,$ a contradiction. If the largest value exceeded $12,$ then the smallest would exceed $8,$ making the total exceed $8+8+12+12+12=52.$ Thus no smaller range works. The correct answer is C.

25.

If $x$ and $y$ are non-zero real numbers such that

$|x|+y=3$

and

$|x|y+x^3=0,$ then the integer nearest to $x-y$ is

$-3$

$-1$

$2$

$3$

$5$

Answer: A

Concepts:absolute value systems of equations

Difficulty rating: 2080

Small Hint:

Consider the signs of $x$ separately

Big Hint:

For $x\lt0,$ put $t=-x$ and combine $y=t^2$ with $t+y=3$

Solution:

If $x\gt0,$ the second equation gives $y=-x^2,$ while the first gives $y=3-x;$ these would require $x^2-x+3=0,$ which has no real root. Hence $x\lt0.$ Put $t=-x\gt0.$ Then the equations become $t+y=3$ and $ty-t^3=0,$ so $y=t^2.$ Therefore $x-y=-t-t^2=-3,$ already an integer. Thus the correct answer is A.

26.

A regular polygon of $m$ sides is exactly enclosed (no overlaps, no gaps) by $m$ regular polygons of $n$ sides each. (Shown here for $m=4,$ $n=8.$ ) If $m=10,$ what is the value of $n?$

$5$

$6$

$14$

$20$

$26$

Answer: A

Concepts:regular polygons interior angles angle sums

Difficulty rating: 1900

Small Hint:

At each vertex of the inner polygon, one $m$ -gon angle and two $n$ -gon angles fill $360^\circ$

Big Hint:

Use the regular $k$ -gon interior angle $180^\circ(1-\frac{2}{k})$

Solution:

At a vertex of the inner polygon, its interior angle and two angles from the surrounding polygons total $360^\circ.$ Thus $\begin{gathered} 180^\circ\left(1-\frac2m\right) \\ {}+2\cdot180^\circ\left(1-\frac2n\right)\\ {}=360^\circ. \end{gathered}$ This simplifies to $\frac{1}{m}+\frac{2}{n}=\frac{1}{2}.$ With $m=10,$ we get $\frac{2}{n}=\frac{2}{5},$ so $n=5.$ Thus the correct answer is A.

27.

A bag of popping corn contains $\frac{2}{3}$ white kernels and $\frac{1}{3}$ yellow kernels. Only $\frac{1}{2}$ of the white kernels will pop, whereas $\frac{2}{3}$ of the yellow ones will pop. A kernel is selected at random from the bag, and pops when placed in the popper. What is the probability that the kernel selected was white?

$\frac12$

$\frac59$

$\frac47$

$\frac35$

$\frac23$

Answer: D

Concepts:conditional probability Bayes’ theorem

Difficulty rating: 1750

Small Hint:

Compute the probabilities of selecting-and-popping for each color

Big Hint:

Condition the white-and-popped probability on the total probability of popping

Solution:

The probabilities of selecting a kernel that pops are $\begin{gathered} P(W\text{ and pop})=\frac23\cdot\frac12=\frac13, \\ P(Y\text{ and pop})=\frac13\cdot\frac23=\frac29. \end{gathered}$ Hence $P(\text{pop})=\frac{5}{9},$ and $P(W\mid\text{pop})=\frac{\frac{1}{3}}{\frac{5}{9}}=\frac35.$ Thus the correct answer is D.

28.

In the $xy$ -plane, how many lines whose $x$ -intercept is a positive prime number and whose $y$ -intercept is a positive integer pass through the point $(4,3)?$

$0$

$1$

$2$

$3$

$4$

Answer: C

Concepts:intercept form primes divisibility

Difficulty rating: 1960

Small Hint:

Let the intercepts be $p$ and $q$ , and use $\frac{4}{p}+\frac{3}{q}=1$

Big Hint:

Rearrange to $q=3+\frac{12}{p-4}$

Solution:

Let the $x$ -intercept be the prime $p$ and the $y$ -intercept be the positive integer $q.$ Intercept form gives $\begin{gathered} \frac4p+\frac3q=1, \\ q=\frac{3p}{p-4}=3+\frac{12}{p-4}. \end{gathered}$ Thus $p-4$ is a positive divisor of $12.$ Testing $1,2,3,4,6,12$ gives prime $p$ only for $p=5$ and $p=7.$ Therefore there are $2$ lines. Thus the correct answer is C.

29.

Points $A,$ $B$ and $C$ on a circle of radius $r$ are situated so that $AB=AC,$ $AB\gt r,$ and the length of minor arc $\overset{\frown}{BC}$ is $r.$ If angles are measured in radians, then $\frac{AB}{BC}=$

$\frac12\csc\frac14$

$2\cos\frac12$

$4\sin\frac12$

$\csc\frac12$

$2\sec\frac12$

Answer: A

Concepts:arc length chord length trigonometric identities

Difficulty rating: 2280

Small Hint:

The minor arc $\overset{\frown}{BC}$ subtends a central angle of $1$ radian

Big Hint:

Use the chord formula $2r\sin(\frac{\theta}{2})$ for both $BC$ and $AB$

Solution:

The central angle subtending minor arc $\overset{\frown}{BC}$ is $\frac{r}{r}=1,$ so $BC=2r\sin\frac12.$ Since $AB=AC$ and $AB\gt r,$ point $A$ is the midpoint of the major arc $BC.$ The minor central angle from $A$ to $B$ is $\pi-\frac12,$ so $AB=2r$ $\sin\left(\frac{\pi-\frac{1}{2}}{2}\right)=2r\cos\frac14.$ Therefore $\frac{AB}{BC}=\frac{\cos(\frac{1}{4})}{\sin(\frac{1}{2})}$ $=\frac1{2\sin(\frac{1}{4})}=\frac12\csc\frac14.$ Thus the correct answer is A.

30.

When $n$ standard $6$ -sided dice are rolled, the probability of obtaining a sum of $1994$ is greater than zero and is the same as the probability of obtaining a sum of $S.$ The smallest possible value of $S$ is

$333$

$335$

$337$

$339$

$341$

Answer: C

Concepts:dice sums symmetry extremal values

Difficulty rating: 2150

Small Hint:

A sum of $1994$ is possible only when $6n\ge1994$

Big Hint:

Replacing every die value $d$ by $7-d$ pairs sums $T$ and $7n-T$

Solution:

The smallest feasible number of dice is $n=\left\lceil\frac{1994}{6}\right\rceil=333.$ Replacing every die result $d$ by $7-d$ is a bijection between outcomes of sum $T$ and outcomes of sum $7n-T.$ Thus for $n=333,$ the sum paired with $1994$ is $S=7(333)-1994=337.$ Dice-sum counts increase strictly from the minimum sum up to the mean, so for any feasible $n$ no sum below $337$ can match the count at $1994.$ Thus the smallest value is $337,$ and the correct answer is C.