1991 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

If for any three distinct numbers $a,$ $b,$ and $c$ we define $\boxed{a,b,c}=\frac{c+a}{c-b},$ then $\boxed{1,-2,-3}=$

$-2$

$-\frac25$

$-\frac14$

$\frac25$

$2$

Concepts:defined operation substitution

Difficulty rating: 840

Small Hint:

Substitute $a=1,$ $b=-2,$ and $c=-3$ into the definition

Big Hint:

Be careful with the two negative signs in the denominator

Solution:

Direct substitution gives $\begin{aligned} \boxed{1,-2,-3} &=\frac{-3+1}{-3-(-2)}\\ &=\frac{-2}{-1}=2. \end{aligned}$

Thus the correct answer is E.

2.

$|3-\pi|=$

$\frac17$

$0.14$

$3-\pi$

$3+\pi$

$\pi-3$

Concepts:absolute value comparing constants

Difficulty rating: 770

Small Hint:

Recall that $\pi\gt3$

Big Hint:

The absolute value of a negative number is its opposite

Solution:

Since $\pi\gt3,$ the quantity $3-\pi$ is negative. Therefore $|3-\pi|=-(3-\pi)=\pi-3.$

Thus the correct answer is E.

3.

$\left(4^{-1}-3^{-1}\right)^{-1}=$

$-12$

$-1$

$\frac1{12}$

$1$

$12$

Concepts:negative exponent reciprocal fraction subtraction

Difficulty rating: 840

Small Hint:

Rewrite the negative first powers as reciprocals

Big Hint:

First simplify $\frac14-\frac13$ , then take its reciprocal

Solution:

We have $4^{-1}-3^{-1}=\frac14-\frac13=-\frac1{12}.$ Raising this result to the power $-1$ takes its reciprocal, giving $-12.$

Thus the correct answer is A.

4.

Which of the following triangles cannot exist?

An acute isosceles triangle

An isosceles right triangle

An obtuse right triangle

A scalene right triangle

A scalene obtuse triangle

Concepts:triangle angle classification

Difficulty rating: 800

Small Hint:

A right triangle already contains an angle of $90^\circ$

Big Hint:

An obtuse triangle must contain an angle greater than $90^\circ$

Solution:

A right triangle has one $90^\circ$ angle, leaving only $90^\circ$ for its other two positive angles together. Neither remaining angle can therefore be obtuse. Hence a triangle cannot be both right and obtuse.

Thus the correct answer is C.

5.

In the arrow-shaped polygon shown, the angles at vertices $A,$ $C,$ $D,$ $E,$ and $F$ are right angles, $BC=FG=5,$ $CD=FE=20,$ $DE=10,$ and $AB=AG.$ The area of the polygon is closest to

$288$

$291$

$294$

$297$

$300$

Concepts:polygon area decomposition isosceles right triangle

Difficulty rating: 1360

Small Hint:

Separate the arrowhead from the rectangular shaft

Big Hint:

Use $AB=AG$ and the two right angles at $C$ and $F$ to determine the arrowhead

Solution:

The shaft $CDEF$ is a $20$ -by- $10$ rectangle, so its area is $200.$ Also $BG=BC+CF+FG$ $=5+10+5=20.$ Because $\angle BAG=90^\circ$ and $AB=AG,$ triangle $ABG$ is an isosceles right triangle with hypotenuse $20.$ Its legs are $10\sqrt2,$ so its area is $\frac12(10\sqrt2)^2=100.$ The total area is $200+100=300.$

Thus the correct answer is E.

6.

If $x\ge0,$ then $\sqrt{x\sqrt{x\sqrt{x}}}=$

$x\sqrt{x}$

$x\sqrt[4]{x}$

$\sqrt[8]{x}$

$\sqrt[8]{x^3}$

$\sqrt[8]{x^7}$

Concepts:nested radical fractional exponent

Difficulty rating: 1260

Small Hint:

Work from the innermost square root outward

Big Hint:

Rewrite each square root as a power of $\frac{1}{2}$

Solution:

Starting inside and using fractional exponents, $\begin{aligned} \sqrt{x\sqrt{x\sqrt{x}}} &=\sqrt{x\cdot x^{\frac{3}{4}}}\\ &=x^{\frac{7}{8}} =\sqrt[8]{x^7}. \end{aligned}$

Thus the correct answer is E.

7.

If $x=\frac{a}{b},$ $a\ne b,$ and $b\ne0,$ then $\dfrac{a+b}{a-b}=$

$\frac{x}{x+1}$

$\frac{x+1}{x-1}$

$1$

$x-\frac1x$

$x+\frac1x$

Concepts:algebraic substitution rational expression

Difficulty rating: 1030

Small Hint:

Divide the numerator and denominator by $b$

Big Hint:

Replace each occurrence of $\frac{a}{b}$ by $x$

Solution:

Dividing both numerator and denominator by the nonzero number $b$ gives $\frac{a+b}{a-b}=\frac{\frac{a}{b}+1}{\frac{a}{b}-1}=\frac{x+1}{x-1}.$

Thus the correct answer is B.

8.

Liquid $X$ does not mix with water. Unless obstructed, it spreads out on the surface of water to form a circular film $0.1$ cm thick. A rectangular box measuring $6$ cm by $3$ cm by $12$ cm is filled with liquid $X.$ Its contents are poured onto a large body of water. What will be the radius, in centimeters, of the resulting circular film?

$\frac{\sqrt{216}}{\pi}$

$\sqrt{\frac{216}{\pi}}$

$\sqrt{\frac{2160}{\pi}}$

$\frac{216}{\pi}$

$\frac{2160}{\pi}$

Concepts:volume conservation cylinder circle area

Difficulty rating: 1260

Small Hint:

Equate the volume in the box to the volume of the thin circular film

Big Hint:

The film volume is $\pi r^2(0.1)$

Solution:

The liquid volume is $6\cdot3\cdot12=216$ cubic centimeters. Thus $0.1\pi r^2=216,$ so $r^2=\frac{2160}{\pi}$ and $r=\sqrt{\frac{2160}{\pi}}.$

Thus the correct answer is C.

9.

From time $t=0$ to time $t=1$ a population increased by $i\%,$ and from time $t=1$ to time $t=2$ the population increased by $j\%.$ Therefore, from time $t=0$ to time $t=2$ the population increased by

$(i+j)\%$

$ij\%$

$(i+ij)\%$

$\left(i+j+\frac{ij}{100}\right)\%$

$\left(i+j+\frac{i+j}{100}\right)\%$

Concepts:percent change compound growth

Difficulty rating: 1310

Small Hint:

Represent the two increases by multiplication factors

Big Hint:

Expand $\left(1+\frac{i}{100}\right)\left(1+\frac{j}{100}\right)$

Solution:

The combined growth factor is $\begin{aligned} &\left(1+\frac{i}{100}\right) \left(1+\frac{j}{100}\right)\\ &\qquad=1+\frac{i+j}{100} +\frac{ij}{10000}. \end{aligned}$ Hence the percent increase is $i+j+\frac{ij}{100}.$

Thus the correct answer is D.

10.

Point $P$ is $9$ units from the center of a circle of radius $15.$ How many different chords of the circle contain $P$ and have integer lengths?

$11$

$12$

$13$

$14$

$29$

Concepts:circle chord distance counting

Difficulty rating: 1610

Small Hint:

Find the shortest and longest chord through $P$

Big Hint:

Except at the extreme lengths, each attainable chord length occurs in two directions

Solution:

The longest chord through $P$ is a diameter of length $30.$ The shortest is perpendicular to the radius through $P,$ and has length $2\sqrt{15^2-9^2}=24.$ As the chord rotates between these positions, its length varies continuously from $24$ to $30.$ The endpoint lengths each occur once, while each of the five interior integer lengths $25,26,27,28,29$ occurs twice. Thus the count is $1+2(5)+1=12.$

Thus the correct answer is B.

11.

Jack and Jill run $10$ kilometers. They start at the same point, run $5$ kilometers up a hill, and return to the starting point by the same route. Jack has a $10$ -minute head start and runs at the rate of $15$ km/hr uphill and $20$ km/hr downhill. Jill runs $16$ km/hr uphill and $22$ km/hr downhill. How far from the top of the hill are they when they pass going in opposite directions?

$\frac54$ km

$\frac{35}{27}$ km

$\frac{27}{20}$ km

$\frac73$ km

$\frac{28}{9}$ km

Concepts:relative motion rates distance

Difficulty rating: 1680

Small Hint:

Determine Jill’s position when Jack reaches the top

Big Hint:

After Jack turns around, use their combined closing speed

Solution:

Jack reaches the top in $\frac{5}{15}=\frac{1}{3}$ hour. Jill has then run for $\frac{1}{3}-\frac{1}{6}=\frac{1}{6}$ hour and is $\frac{16}{6}=\frac{8}{3}$ km uphill, so they are $5-\frac{8}{3}=\frac{7}{3}$ km apart. Their closing speed is $20+16=36$ km/hr, so they meet after $\frac{\frac{7}{3}}{36}=\frac{7}{108}$ hour. Jack descends $20\left(\frac7{108}\right)=\frac{35}{27}$ kilometers.

Thus the correct answer is B.

12.

The measures (in degrees) of the interior angles of a convex hexagon form an arithmetic sequence of positive integers. Let $m^\circ$ be the measure of the largest interior angle of the hexagon. The largest possible value of $m^\circ$ is

$165^\circ$

$167^\circ$

$170^\circ$

$175^\circ$

$179^\circ$

Concepts:polygon angle sum arithmetic sequence divisibility

Difficulty rating: 1630

Small Hint:

Write the angles as $a,a+d,\ldots,a+5d$ and use their sum

Big Hint:

The equation forces $d$ to be even, and convexity bounds the largest angle below $180^\circ$

Solution:

The angle sum is $720^\circ,$ so $\begin{aligned} 6a+15d&=720,\\ 2a+5d&=240. \end{aligned}$ Hence $d$ is even. The largest angle is $a+5d=120+\frac52d,$ which must be less than $180.$ Thus $d\lt24,$ and the largest allowable even value is $d=22.$ It gives $a=65$ and $m=65+5(22)=175.$

Thus the correct answer is D.

13.

Horses $X,$ $Y,$ and $Z$ are entered in a three-horse race in which ties are not possible. If the odds against $X$ winning are $3$ -to- $1$ and the odds against $Y$ winning are $2$ -to- $3,$ what are the odds against $Z$ winning? (By “odds against $H$ winning are $p$ -to- $q$ ” we mean that the probability of $H$ winning the race is $\frac{q}{p+q}.$ )

$3$ -to- $20$

$5$ -to- $6$

$8$ -to- $5$

$17$ -to- $3$

$20$ -to- $3$

Concepts:probability odds complementary events

Difficulty rating: 1520

Small Hint:

Convert each set of odds to a probability

Big Hint:

The three winning probabilities sum to $1$

Solution:

The given probabilities are $P(X)=\frac14$ and $P(Y)=\frac35.$ Therefore $P(Z)=1-\frac14-\frac35=\frac3{20}.$ The probability that $Z$ loses is $\frac{17}{20},$ so the odds against $Z$ are $17$ -to- $3.$

Thus the correct answer is D.

14.

If $x$ is the cube of a positive integer and $d$ is the number of positive integers that are divisors of $x,$ then $d$ could be

$200$

$201$

$202$

$203$

$204$

Concepts:divisor count prime factorization perfect cube

Difficulty rating: 1750

Small Hint:

In the prime factorization of a cube, every exponent is divisible by $3$

Big Hint:

Try a cube having only one prime factor

Solution:

For any prime $p,$ take $x=p^{201}=(p^{67})^3.$ This is a cube, and its positive divisors are $1,p,p^2,\ldots,p^{201},$ a total of $202.$ Thus $202$ can occur.

Thus the correct answer is C.

15.

A circular table has exactly $60$ chairs around it. There are $N$ people seated at this table in such a way that the next person to be seated must sit next to someone. The smallest possible value of $N$ is

$15$

$20$

$30$

$40$

$58$

Concepts:circular arrangement covering extremal counting

Difficulty rating: 1490

Small Hint:

No empty chair can have both neighboring chairs empty

Big Hint:

Between consecutive occupied chairs there can be at most two empty chairs

Solution:

The condition says that every empty chair has an occupied neighbor, so no three consecutive chairs may all be empty. Therefore each occupied chair can account for at most itself and the two empty chairs following it, giving $60\le3N$ and $N\ge20.$ Repeating the pattern occupied-empty-empty around the table attains $N=20.$

Thus the correct answer is B.

16.

One hundred students at Century High School participated in the AHSME last year, and their mean score was $100.$ The number of non-seniors taking the AHSME was $50\%$ more than the number of seniors, and the mean score of the seniors was $50\%$ higher than that of the non-seniors. What was the mean score of the seniors?

$100$

$112.5$

$120$

$125$

$150$

Concepts:weighted average percent linear equation

Difficulty rating: 1410

Small Hint:

First determine the numbers of seniors and non-seniors

Big Hint:

Let the non-senior mean be $x$ and form a weighted-average equation

Solution:

If there are $s$ seniors, then there are $1.5s$ non-seniors, so $2.5s=100$ and $s=40.$ Let the non-senior mean be $x;$ the senior mean is $1.5x.$ The total score equation is $60x+40(1.5x)=100(100),$ so $120x=10000,$ and the senior mean is $1.5x=125.$

Thus the correct answer is D.

17.

A positive integer $N$ is a palindrome if the integer obtained by reversing the sequence of digits of $N$ is equal to $N.$ The year $1991$ is the only year in the current century with the following two properties:

(a) It is a palindrome.

(b) It factors as a product of a $2$ -digit prime palindrome and a $3$ -digit prime palindrome.

How many years in the millennium between $1000$ and $2000$ (including the year $1991$ ) have properties (a) and (b)?

$1$

$2$

$3$

$4$

$5$

Concepts:palindrome prime factorization divisibility

Difficulty rating: 1880

Small Hint:

Every four-digit palindrome from $1000$ to $2000$ has the form $1dd1$

Big Hint:

Factor $1dd1$ as $11(91+10d)$ and test when the three-digit factor is prime

Solution:

The palindromes in the interval are $1dd1,$ where $d$ is a digit. Algebraically, $\begin{aligned} 1dd1&=1001+110d\\ &=11(91+10d). \end{aligned}$ The two-digit prime palindrome must therefore be $11,$ and $91+10d$ must be a three-digit prime palindrome. For $d=0,1,\ldots,9,$ the prime values occur for $d=1,4,6,9,$ giving $101,131,151,181.$ Thus there are $4$ years.

Thus the correct answer is D.

18.

If $S$ is the set of points $z$ in the complex plane such that $(3+4i)z$ is a real number, then $S$ is a

right triangle

circle

hyperbola

line

parabola

Concepts:complex plane real and imaginary parts locus

Difficulty rating: 1780

Small Hint:

Write $z=x+yi$ and expand the product

Big Hint:

Set the imaginary part of $(3+4i)(x+yi)$ equal to zero

Solution:

Writing $z=x+yi,$ $\begin{aligned} (3+4i)z &=(3x-4y)\\ &\quad +(4x+3y)i. \end{aligned}$ This is real exactly when $4x+3y=0,$ which is the equation of a line through the origin.

Thus the correct answer is D.

19.

Triangle $ABC$ has a right angle at $C,$ $AC=3,$ and $BC=4.$ Triangle $ABD$ has a right angle at $A$ and $AD=12.$ Points $C$ and $D$ are on opposite sides of $AB.$ The line through $D$ parallel to $AC$ meets $CB$ extended at $E.$ If $\frac{DE}{DB}=\frac mn,$ where $m$ and $n$ are relatively prime positive integers, then $m+n=$

$25$

$128$

$153$

$243$

$256$

Concepts:coordinate geometry perpendicular vectors distance

Difficulty rating: 2110

Small Hint:

Place $C=(0,0),$ $A=(0,3),$ and $B=(4,0)$

Big Hint:

Find $D$ by using a length- $12$ vector perpendicular to $\overrightarrow{AB}$

Solution:

Place $C=(0,0),$ $A=(0,3),$ and $B=(4,0).$ Since $\overrightarrow{AB}=(4,-3),$ the length- $12$ vector perpendicular to $\overrightarrow{AB}$ and directed away from $C$ is $(\frac{36}{5},\frac{48}{5}).$ Hence $D=(\frac{36}{5},\frac{63}{5}).$ Since $DE\parallel AC,$ we have $E=(\frac{36}{5},0),$ so $DE=\frac{63}{5}$ and $DB=13.$ Therefore $\frac{DE}{DB}=\frac{\frac{63}{5}}{13}=\frac{63}{65}.$ Thus $m+n=63+65=128.$

Thus the correct answer is B.

20.

The sum of all real $x$ such that $\begin{aligned} &(2^x-4)^3+(4^x-2)^3\\ &\qquad=(4^x+2^x-6)^3 \end{aligned}$ is

$\frac32$

$2$

$\frac52$

$3$

$\frac72$

Concepts:sum of cubes exponential equation factoring

Difficulty rating: 1950

Small Hint:

Set $a=2^x-4$ and $b=4^x-2$

Big Hint:

Factor $a^3+b^3-(a+b)^3$ as $-3ab(a+b)$

Solution:

Let $a=2^x-4$ and $b=4^x-2.$ The right side is $(a+b)^3,$ so $a^3+b^3=(a+b)^3$ is equivalent to $ab(a+b)=0.$ The cases give $2^x=4,$ $4^x=2,$ or $4^x+2^x=6,$ whose real solutions are $x=2,\frac12,$ and $1,$ respectively. Their sum is $2+\frac12+1=\frac72.$

Thus the correct answer is E.

21.

If $f\left(\frac{x}{x-1}\right)=\frac1x$ for all $x\notin\{0,1\},$ and $0\lt\theta\lt\frac{\pi}{2},$ then $f(\sec^2\theta)=$

$\sin^2\theta$

$\cos^2\theta$

$\tan^2\theta$

$\cot^2\theta$

$\csc^2\theta$

Concepts:function substitution trigonometric identity

Difficulty rating: 1800

Small Hint:

Choose $x$ so that $\frac{1}{x}=\sec^2\theta$

Big Hint:

Use $1-\cos^2\theta=\sin^2\theta$

Solution:

Since $\sec^2\theta=\frac1{\cos^2\theta} =\frac{\csc^2\theta}{\csc^2\theta-1},$ choose $x=\csc^2\theta$ in the definition. Then $f(\sec^2\theta)=\frac1{\csc^2\theta}=\sin^2\theta.$

Thus the correct answer is A.

22.

Two circles are externally tangent. Lines $PAB$ and $PA'B'$ are common tangents with $A$ and $A'$ on the smaller circle and $B$ and $B'$ on the larger circle. If $PA=AB=4,$ then the area of the smaller circle is

$1.44\pi$

$2\pi$

$2.56\pi$

$\sqrt8\pi$

$4\pi$

Concepts:tangent circles homothety power of a point

Difficulty rating: 2160

Small Hint:

The two circles are related by a dilation centered at $P$

Big Hint:

Use $PA=4,$ $PB=8$ to find the dilation ratio, then apply the tangent-length relation to the smaller circle

Solution:

Since $PA=4$ and $AB=4,$ we have $PB=8.$ The homothety centered at $P$ taking the smaller circle to the larger therefore has ratio $2.$ If the smaller radius is $r,$ its center is $3r$ from $P:$ the centers are on the same ray, their distances from $P$ have ratio $2,$ and their difference is $r+2r=3r.$ Applying the right triangle formed by $P,$ the small center, and $A$ , $PA^2=(3r)^2-r^2=8r^2.$ Thus $16=8r^2,$ so $r^2=2$ and the area is $2\pi.$

Thus the correct answer is B.

23.

If $ABCD$ is a $2\times2$ square, $E$ is the midpoint of $AB,$ $F$ is the midpoint of $BC,$ $AF$ and $DE$ intersect at $I,$ and $BD$ and $AF$ intersect at $H,$ then the area of quadrilateral $BEIH$ is

$\frac13$

$\frac25$

$\frac7{15}$

$\frac8{15}$

$\frac35$

Concepts:coordinate geometry line intersection polygon area

Difficulty rating: 2250

Small Hint:

Assign coordinates to the square and write equations for $AF,$ $DE,$ and $BD$

Big Hint:

Find $I$ and $H,$ then use the shoelace formula on $B,E,I,H$

Solution:

Set $B=(0,0),$ $C=(2,0),$ $D=(2,2),$ and $A=(0,2).$ Then $E=(0,1),$ $F=(1,0),$ and the line intersections are $\begin{aligned} I&=AF\cap DE =\left(\frac25,\frac65\right),\\ H&=AF\cap BD =\left(\frac23,\frac23\right). \end{aligned}$ The shoelace formula gives $\begin{aligned} [BEIH] &=\frac12\left| \frac4{15}-\frac25-\frac45 \right|\\ &=\frac7{15}. \end{aligned}$

Thus the correct answer is C.

24.

The graph, $G,$ of $y=\log_{10}x$ is rotated $90^\circ$ counter-clockwise about the origin to obtain a new graph $G'.$ Which of the following is an equation for $G'?$

$y=\log_{10}\left(\frac{x+90}{9}\right)$

$y=\log_x10$

$y=\frac1{x+1}$

$y=10^{-x}$

$y=10^x$

Concepts:graph transformation rotation logarithm exponential

Difficulty rating: 1710

Small Hint:

A $90^\circ$ counter-clockwise rotation sends $(x,y)$ to $(-y,x)$

Big Hint:

Rename the rotated coordinates and solve the logarithmic relation for the new $y$

Solution:

A point $(u,\log_{10}u)$ rotates to $(x,y)=(-\log_{10}u,u).$ Hence $x=-\log_{10}y,$ so $\log_{10}y=-x$ and $y=10^{-x}.$

Thus the correct answer is D.

25.

If $T_n=1+2+3+\cdots+n$ and $\begin{aligned} P_n&=\frac{T_2}{T_2-1} \cdot\frac{T_3}{T_3-1}\\ &\quad\cdot\frac{T_4}{T_4-1} \cdots\frac{T_n}{T_n-1} \end{aligned}$ for $n=2,$ $3,$ $4,$ $\ldots,$ then $P_{1991}$ is closest to which of the following numbers?

$2.0$

$2.3$

$2.6$

$2.9$

$3.2$

Concepts:telescoping product triangular number approximation

Difficulty rating: 1950

Small Hint:

Factor $T_k-1$ and simplify each factor $\frac{T_k}{T_k-1}$

Big Hint:

Write each factor as $\frac{k}{k-1}\cdot\frac{k+1}{k+2}$ and telescope

Solution:

Since $T_k=\frac{k(k+1)}{2}$ and $T_k-1=\frac{(k-1)(k+2)}{2},$ $\frac{T_k}{T_k-1} =\frac{k}{k-1}\cdot\frac{k+1}{k+2}.$ The product telescopes: $\begin{aligned} P_n &=\left(\prod_{k=2}^n\frac{k}{k-1}\right) \left(\prod_{k=2}^n\frac{k+1}{k+2}\right)\\ &=n\cdot\frac3{n+2}\\ &=\frac{3n}{n+2}. \end{aligned}$ For $n=1991,$ this is just under $3$ and is closest to $2.9.$

Thus the correct answer is D.

26.

An $n$ -digit positive integer is cute if its $n$ digits are an arrangement of the set $\{1,2,\ldots,n\}$ and its first $k$ digits form an integer that is divisible by $k,$ for $k=1,$ $2,$ $\ldots,$ $n.$ For example, $321$ is a cute $3$ -digit integer because $1$ divides $3,$ $2$ divides $32,$ and $3$ divides $321.$ How many cute $6$ -digit integers are there?

$0$

$1$

$2$

$3$

$4$

Concepts:divisibility permutations case analysis

Difficulty rating: 2270

Small Hint:

The second digit must be even, the third-prefix digit sum must be divisible by $3,$ and the fourth prefix must be divisible by $4$

Big Hint:

The fifth digit must be $5,$ and the full number must be even and divisible by $3$

Solution:

Divisibility by $5$ forces the fifth digit to be $5.$ Divisibility by $2$ and $6$ forces the second and sixth digits to be even. Now apply the divisibility tests successively: the first three digits must have sum divisible by $3,$ and the two-digit number formed by the third and fourth digits must be divisible by $4.$ Checking the remaining choices from $\{1,2,3,4,6\}$ leaves $123654$ and $321654.$ Each number directly satisfies all six prefix divisibility conditions, so there are $2.$

Thus the correct answer is C.

27.

If $\begin{aligned} x+\sqrt{x^2-1} &+\frac1{x-\sqrt{x^2-1}}\\ &=20, \end{aligned}$ then $\begin{aligned} x^2+\sqrt{x^4-1} &+\frac1{x^2+\sqrt{x^4-1}}\\ &= \end{aligned}$

$5.05$

$20$

$51.005$

$61.25$

$400$

Concepts:conjugate radicals reciprocal algebraic identity

Difficulty rating: 2110

Small Hint:

Rationalize $\frac1{x-\sqrt{x^2-1}}$

Big Hint:

The last two terms of the requested expression are conjugates

Solution:

Because $\frac1{x-\sqrt{x^2-1}}=x+\sqrt{x^2-1},$ the given equation implies $x+\sqrt{x^2-1}=10.$ Its reciprocal is $x-\sqrt{x^2-1}=\frac{1}{10},$ so adding yields $2x=10.1$ and $x=5.05.$ Also $\frac1{x^2+\sqrt{x^4-1}}=x^2-\sqrt{x^4-1}.$ Therefore the requested expression is $2x^2=2(5.05)^2=51.005.$

Thus the correct answer is C.

28.

Initially an urn contains $100$ black marbles and $100$ white marbles. Repeatedly, three marbles are removed from the urn and replaced from a pile outside the urn as follows:

Marbles removed	Replaced with
$3$ black	$1$ black
$2$ black, $1$ white	$1$ black, $1$ white
$1$ black, $2$ white	$2$ white
$3$ white	$1$ black, $1$ white

Which of the following sets of marbles could be the contents of the urn after repeated applications of this procedure?

$2$ black marbles

$2$ white marbles

$1$ black marble

$1$ black and $1$ white marble

$1$ white marble

Concepts:invariant constructive process parity

Difficulty rating: 2330

Small Hint:

The number of white marbles always changes by $0$ or $2$

Big Hint:

When an operation first leaves at most two marbles, inspect the possible outputs of that final operation

Solution:

The parity of the number of white marbles never changes, so choices with one white marble are impossible. Also a final operation starting with three marbles can leave $1$ black, $1$ black and $1$ white, or $2$ white, but never $2$ black. The state with $2$ white is attainable: repeatedly replace $3$ black by $1$ black until $2$ black remain; twice use the $1$ -black- $2$ -white rule, leaving $100$ white; then alternate the $3$ -white rule with the $1$ -black- $2$ -white rule, reducing the number of white marbles by $2$ per pair until $2$ remain.

Thus the correct answer is B.

29.

Equilateral triangle $ABC$ has been creased and folded so that vertex $A$ now rests at $A'$ on $BC$ as shown. If $BA'=1$ and $A'C=2,$ then the length of crease $PQ$ is

$\frac85$

$\frac7{20}\sqrt{21}$

$\frac{1+\sqrt5}{2}$

$\frac{13}{8}$

$\sqrt3$

Concepts:folding perpendicular bisector coordinate geometry distance

Difficulty rating: 2360

Small Hint:

A fold crease is the perpendicular bisector of the segment joining a point to its image

Big Hint:

Place $B=(0,0),$ $C=(3,0),$ $A=(\frac{3}{2},\frac{3\sqrt3}{2}),$ and $A'=(1,0)$

Solution:

With the coordinates in the hint, the perpendicular-bisector condition for $X=(x,y)$ is $|X-A|=|X-A'|,$ which simplifies to $x+3\sqrt3\,y=8.$ Intersecting this line with $AB,\ y=\sqrt3x,$ gives $P=(\frac{4}{5},\frac{4\sqrt3}{5}).$ Intersecting it with $AC,\ y=\sqrt3(3-x),$ gives $Q=(\frac{19}{8},\frac{5\sqrt3}{8}).$ Hence $\begin{aligned} PQ &=\sqrt{\left(\frac{63}{40}\right)^2 +\left(\frac{7\sqrt3}{40}\right)^2}\\ &=\frac7{20}\sqrt{21}. \end{aligned}$

Thus the correct answer is B.

30.

For any set $S,$ let $|S|$ denote the number of elements in $S,$ and let $n(S)$ be the number of subsets of $S,$ including the empty set and the set $S$ itself. If $A,$ $B,$ and $C$ are sets for which $\begin{gathered} n(A)+n(B)+n(C)\\ {}=n(A\cup B\cup C),\\ |A|=|B|=100, \end{gathered}$ then what is the minimum possible value of $|A\cap B\cap C|?$

$96$

$97$

$98$

$99$

$100$

Concepts:set cardinality subsets powers of two extremal counting

Difficulty rating: 2430

Small Hint:

Use $n(S)=2^{|S|}$ and determine $|C|$ and $|A\cup B\cup C|$

Big Hint:

Within the union, count how many elements each of $A,$ $B,$ and $C$ can omit

Solution:

Let $c=|C|$ and $u=|A\cup B\cup C|.$ Since $n(S)=2^{|S|},$ the equation becomes $2^{101}+2^c=2^u.$ The two summands must be equal, so $c=101$ and $u=102.$ Within the union, $A,$ $B,$ and $C$ omit $2,$ $2,$ and $1$ elements. Thus at most five distinct elements are absent from the triple intersection, giving $|A\cap B\cap C|\ge97.$ Equality is attained by making those omissions distinct.

Thus the correct answer is B.