The complete set of x -values satisfying the inequality (x^2-4)/(x^2-1)\gt0 is the set of all x such that:

(A) x\gt2 or x\lt-2 or -1\lt x\lt1

(E) x is any real number except 1 or -1

If the graph of x^2+y^2=m is tangent to that of x+y=√2m, then:

(E) m may be any nonnegative real number

1969 AMC 12 — Problems & Online Timed Exam

Timed

1:15:00

1.

When $x$ is added to both the numerator and the denominator of the fraction $\frac{a}{b},$ $a\ne b,$ $b\ne0,$ the value of the fraction is changed to $\frac{c}{d}.$ Then $x$ equals:

$\dfrac1{c-d}$

$\dfrac{ad-bc}{c-d}$

$\dfrac{ad-bc}{c+d}$

$\dfrac{bc-ad}{c-d}$

$\dfrac{bc-ad}{c+d}$

Answer: B

Concepts:rational equation algebraic manipulation fraction

Difficulty rating: 1340

Small Hint:

Translate the change into $\frac{a+x}{b+x}=\frac{c}{d}$

Big Hint:

Cross-multiply and collect the terms containing $x$

Solution:

From $\frac{a+x}{b+x}=\frac cd$ we obtain $ad+dx=bc+cx.$ Hence $x(d-c)=bc-ad,$ so $x=\frac{bc-ad}{d-c}=\frac{ad-bc}{c-d}.$

Therefore, the correct answer is B.

2.

If an item is sold for $x$ dollars, there is a loss of $15\%$ based on the cost. If, however, the same item is sold for $y$ dollars, there is a profit of $15\%$ based on the cost. The ratio $y:x$ is:

$23:17$

$17y:23$

$23x:17$

dependent upon the cost

none of these

Answer: A

Concepts:percentage ratio and proportion

Difficulty rating: 1180

Small Hint:

Let the cost be $C$

Big Hint:

Write $x=0.85C$ and $y=1.15C,$ then form the ratio

Solution:

If the cost is $C,$ then $x=0.85C$ and $y=1.15C.$ Therefore $\begin{aligned} y:x&=1.15C:0.85C\\ &=115:85=23:17. \end{aligned}$

Therefore, the correct answer is A.

3.

If $N,$ written in base $2,$ is $11000,$ the integer immediately preceding $N,$ written in base $2,$ is:

$10001$

$10010$

$10011$

$10110$

$10111$

Answer: E

Concepts:number base place value whole number operations

Difficulty rating: 1180

Small Hint:

Subtract $1$ from the binary number

Big Hint:

Borrow through the three trailing zeros

Solution:

Subtracting $1$ from $11000_2$ requires borrowing through the three trailing zeros. Thus $11000_2-1_2=10111_2.$

Therefore, the correct answer is E.

4.

Let a binary operation $*$ on ordered pairs of integers be defined by $(a,b)*(c,d)=(a-c,b+d).$ Then, if $(3,2)*(0,0)$ and $(x,y)*(3,2)$ represent identical pairs, $x$ equals:

$-3$

$0$

$2$

$3$

$6$

Answer: E

Concepts:custom operation system of equations

Difficulty rating: 1160

Small Hint:

Evaluate each ordered-pair operation directly

Big Hint:

Equate the first coordinates $3$ and $x-3$

Solution:

The first pair is $(3-0,2+0)=(3,2).$ The second is $(x-3,y+2).$ Equality of the first coordinates gives $x-3=3,$ so $x=6.$

Therefore, the correct answer is E.

5.

If a number $N,$ $N\ne0,$ diminished by four times its reciprocal, equals a given real constant $R,$ then, for this given $R,$ the sum of all such possible values of $N$ is:

$\dfrac1R$

$R$

$4$

$\dfrac14$

$-R$

Answer: B

Concepts:quadratic Vieta’s Formulas rational equation

Difficulty rating: 1400

Small Hint:

Rewrite $N-\frac{4}{N}=R$ as a quadratic in $N$

Big Hint:

Use the sum of the roots of that quadratic

Solution:

Multiplying $N-\frac{4}{N}=R$ by $N$ gives $N^2-RN-4=0.$ The sum of its two roots is $R$ by Vieta’s formulas. Both roots are nonzero because their product is $-4.$

Therefore, the correct answer is B.

6.

The area of the ring between two concentric circles is $12\dfrac12\pi$ square inches. The length of a chord of the larger circle tangent to the smaller circle, in inches, is:

$\dfrac5{\sqrt2}$

$5$

$5\sqrt2$

$10$

$10\sqrt2$

Answer: C

Concepts:annulus chord Pythagorean Theorem

Difficulty rating: 1670

Small Hint:

If the radii are $R$ and $r,$ the area gives $R^2-r^2=\frac{25}{2}$

Big Hint:

Half the tangent chord, the smaller radius, and the larger radius form a right triangle

Solution:

Let the radii be $R$ and $r.$ The ring area gives $\pi(R^2-r^2)=\frac{25}{2}\pi.$ The radius to the tangent point bisects the chord. If its length is $L,$ then $\left(\frac L2\right)^2+r^2=R^2.$ Thus $L=2\sqrt{R^2-r^2},$ so $L=2\sqrt{\frac{25}{2}}=5\sqrt2.$

Therefore, the correct answer is C.

7.

If the points $(1,y_1)$ and $(-1,y_2)$ lie on the graph of $y=ax^2+bx+c,$ and $y_1-y_2=-6,$ then $b$ equals:

$-3$

$0$

$3$

$\sqrt{ac}$

$\dfrac{a+c}{2}$

Answer: A

Concepts:quadratic substitution symmetry (algebra)

Difficulty rating: 1240

Small Hint:

Substitute $x=1$ and $x=-1$

Big Hint:

Subtract the two resulting values; the $a$ and $c$ terms cancel

Solution:

We have $y_1=a+b+c$ and $y_2=a-b+c.$ Therefore $y_1-y_2=2b=-6,$ so $b=-3.$

Therefore, the correct answer is A.

8.

Triangle $ABC$ is inscribed in a circle. The measures of the non-overlapping minor arcs $AB,$ $BC,$ and $CA$ are, respectively, $x+75^\circ,$ $2x+25^\circ,$ $3x-22^\circ.$ Then one interior angle of the triangle, in degrees, is:

$57\dfrac12$

$59$

$60$

$61$

$122$

Answer: D

Concepts:arc inscribed angle angle sum

Difficulty rating: 1500

Small Hint:

The three non-overlapping arcs sum to $360^\circ$

Big Hint:

Each inscribed angle is half the measure of its intercepted arc

Solution:

The arc sum gives $\begin{aligned} 6x+78&=360,\\ x&=47. \end{aligned}$ The arcs then measure $122^\circ,$ $119^\circ,$ and $119^\circ.$ The angle intercepting the $122^\circ$ arc measures $61^\circ.$

Therefore, the correct answer is D.

9.

The arithmetic mean (ordinary average) of the fifty-two successive positive integers beginning with $2$ is:

$27$

$27\dfrac14$

$27\dfrac12$

$28$

$28\dfrac12$

Answer: C

Concepts:arithmetic sequence mean

Difficulty rating: 920

Small Hint:

Find the fifty-second integer in the list

Big Hint:

The mean of an arithmetic sequence is the average of its first and last terms

Solution:

The integers run from $2$ through $2+51=53.$ Their mean is the average of the first and last: $\frac{2+53}{2}=\frac{55}{2}=27\frac12.$

Therefore, the correct answer is C.

10.

The number of points equidistant from a circle and two parallel tangents to the circle is:

$0$

$2$

$3$

$4$

infinite

Answer: C

Concepts:circle tangent line symmetry

Difficulty rating: 1960

Small Hint:

Points equidistant from the two parallel tangents lie on their midway parallel line

Big Hint:

On that line, compare distance to either tangent with distance to the circle

Solution:

Let the circle have center $O$ and radius $r.$ A point equidistant from the two parallel tangents must lie on their midline, which passes through $O.$ If its distance from $O$ is $t,$ its distance from either tangent is $r,$ while its distance from the circle is $\lvert t-r\rvert.$ Thus $\lvert t-r\rvert=r,$ giving $t=0$ or $t=2r.$ There is one point with $t=0$ and two with $t=2r,$ for a total of $3.$

Therefore, the correct answer is C.

11.

Given points $P(-1,-2)$ and $Q(4,2)$ in the $xy$ -plane, point $R(1,m)$ is taken so that $PR+RQ$ is a minimum. Then $m$ equals:

$-\dfrac35$

$-\dfrac25$

$-\dfrac15$

$\dfrac15$

either $-\dfrac15$ or $\dfrac15$

Answer: B

Concepts:coordinate geometry slope triangle inequality

Difficulty rating: 1640

Small Hint:

The triangle inequality is sharp when $P,$ $R,$ and $Q$ are collinear

Big Hint:

Find the point on line $PQ$ whose $x$ -coordinate is $1$

Solution:

By the triangle inequality, $PR+RQ\ge PQ,$ with equality when $R$ lies on segment $PQ.$ The slope of $PQ$ is $\frac{2-(-2)}{4-(-1)}=\frac{4}{5}.$ Moving from $x=-1$ to $x=1$ raises the $y$ -coordinate by $(\frac{4}{5})(2)=\frac{8}{5},$ so $m=-2+\frac85=-\frac25.$

Therefore, the correct answer is B.

12.

Let $F=\dfrac{6x^2+16x+3m}{6}$ be the square of an expression which is linear in $x.$ Then $m$ has a particular value between:

$3$ and $4$

$4$ and $5$

$5$ and $6$

$-4$ and $-3$

$-6$ and $-5$

Answer: A

Concepts:completing the square perfect square quadratic

Difficulty rating: 1580

Small Hint:

Rewrite $F=x^2+\frac83x+\frac m2$

Big Hint:

Match the constant term to the square of half the linear coefficient

Solution:

We have $F=x^2+\frac83x+\frac m2.$ For this to be a square, it must equal $(x+\frac{4}{3})^2,$ whose constant term is $\frac{16}{9}.$ Hence $\frac{m}{2}=\frac{16}{9},$ so $m=\frac{32}{9},$ which lies between $3$ and $4.$

Therefore, the correct answer is A.

13.

A circle with radius $r$ is contained within the region bounded by a circle with radius $R.$ The area bounded by the larger circle is $\frac{a}{b}$ times the area of the region outside the smaller circle and inside the larger circle. Then $R:r$ equals:

$\sqrt a:\sqrt b$

$\sqrt a:\sqrt{a-b}$

$\sqrt b:\sqrt{a-b}$

$a:\sqrt{a-b}$

$b:\sqrt{a-b}$

Answer: B

Concepts:annulus area ratio ratio and proportion

Difficulty rating: 1690

Small Hint:

Write $\pi R^2=(\frac{a}{b})\pi(R^2-r^2)$

Big Hint:

Solve for the ratio $\frac{R^2}{r^2}$ before taking square roots

Solution:

The area condition is $\pi R^2=\frac ab\pi(R^2-r^2).$ Thus $bR^2=aR^2-ar^2,$ so $ar^2=(a-b)R^2.$ Therefore $\frac Rr=\sqrt{\frac a{a-b}},$ and $R:r=\sqrt a:\sqrt{a-b}.$

Therefore, the correct answer is B.

14.

The complete set of $x$ -values satisfying the inequality $\dfrac{x^2-4}{x^2-1}\gt0$ is the set of all $x$ such that:

$x\gt2$ or $x\lt-2$ or $-1\lt x\lt1$

$x\gt2$ or $x\lt-2$

$x\gt1$ or $x\lt-2$

$x\gt1$ or $x\lt-1$

$x$ is any real number except $1$ or $-1$

Answer: A

Concepts:factoring inequality casework

Difficulty rating: 1780

Small Hint:

The sign can change only at $-2,$ $-1,$ $1,$ and $2$

Big Hint:

Test one point in each interval and exclude zeros and undefined points

Solution:

Factor the expression as $\frac{(x-2)(x+2)}{(x-1)(x+1)}.$ A sign chart at $-2,$ $-1,$ $1,$ and $2$ shows it is positive on $(-\infty,-2)\cup(-1,1)\cup(2,\infty).$ The endpoints $\pm2$ give zero and $\pm1$ are undefined.

Therefore, the correct answer is A.

15.

In a circle with center at $O$ and radius $r,$ chord $AB$ is drawn with length equal to $r$ units. From $O$ a perpendicular to $AB$ meets $AB$ at $M.$ From $M$ a perpendicular to $OA$ meets $OA$ at $D.$ In terms of $r,$ the area of triangle $MDA,$ in appropriate square units, is:

$\dfrac{3r^2}{16}$

$\dfrac{\pi r^2}{16}$

$\dfrac{\pi r^2\sqrt2}{8}$

$\dfrac{r^2\sqrt3}{32}$

$\dfrac{r^2\sqrt6}{48}$

Answer: D

Concepts:chord equilateral triangle special right triangle triangle area

Difficulty rating: 1720

Small Hint:

Because $OA=OB=AB=r,$ triangle $OAB$ is equilateral

Big Hint:

Use the two $30$ - $60$ - $90$ triangles to find $AD$ and $MD$

Solution:

Triangle $OAB$ is equilateral. Since $OM\perp AB,$ $M$ is the midpoint of $AB,$ so $AM=\frac{r}{2}$ and $\angle OAM=60^\circ.$ In right triangle $AMD,$ the hypotenuse is $AM=\frac{r}{2},$ giving $AD=\frac r4,\qquad MD=\frac{r\sqrt3}{4}.$ Hence $\begin{aligned} [MDA]&=\frac12(AD)(MD)\\ &=\frac{r^2\sqrt3}{32}. \end{aligned}$

Therefore, the correct answer is D.

16.

When $(a-b)^n,$ $n\ge2,$ $ab\ne0,$ is expanded by the binomial theorem, it is found that, when $a=kb,$ where $k$ is a positive integer, the sum of the second and third terms is zero. Then $n$ equals:

$\dfrac12k(k-1)$

$\dfrac12k(k+1)$

$2k-1$

$2k$

$2k+1$

Answer: E

Concepts:binomial theorem algebraic manipulation

Difficulty rating: 2010

Small Hint:

The second and third terms are $-na^{n-1}b$ and $\binom n2a^{n-2}b^2$

Big Hint:

Substitute $a=kb$ and cancel the common nonzero factor

Solution:

The sum of the second and third terms is $-na^{n-1}b+\frac{n(n-1)}2a^{n-2}b^2.$ Substituting $a=kb$ and dividing by the nonzero quantity $na^{n-2}b$ gives $-k+\frac{n-1}{2}=0.$ Therefore $n=2k+1.$

Therefore, the correct answer is E.

17.

The equation $2^{2x}-8\cdot2^x+12=0$ is satisfied by:

$\log 3$

$\dfrac12\log 6$

$1+\log\dfrac34$

$1+\dfrac{\log3}{\log2}$

none of these

Answer: D

Concepts:exponent logarithm quadratic substitution

Difficulty rating: 1880

Small Hint:

Let $t=2^x$ and solve a quadratic in $t$

Big Hint:

One root is $t=6$ ; take logarithms base $2$

Solution:

Set $t=2^x.$ Then $\begin{aligned} t^2-8t+12&=0,\\ (t-2)(t-6)&=0. \end{aligned}$ The root $t=6$ gives $\begin{aligned} x&=\log_2 6\\ &=1+\log_2 3\\ &=1+\frac{\log3}{\log2}. \end{aligned}$ (The equation is also satisfied by $x=1,$ which is not a listed choice.)

Therefore, the correct answer is D.

18.

The number of points common to the graphs of

$(x-y+2)(3x+y-4)=0$

and

$(x+y-2)(2x-5y+7)=0$

is:

$2$

$4$

$6$

$16$

infinite

Answer: B

Concepts:coordinate geometry system of equations counting intersections

Difficulty rating: 1640

Small Hint:

Each factored equation represents a pair of lines

Big Hint:

Intersect each line from the first pair with each line from the second pair and check distinctness

Solution:

The first graph is the pair $x-y+2=0,$ and $3x+y-4=0,$ and the second is $x+y-2=0,$ and $2x-5y+7=0.$ Each line in the first pair meets each line in the second pair. Solving the four pairings gives four distinct points: $\begin{gathered} (0,2),\quad(-1,1),\\ (1,1),\quad\left(\frac{13}{17},\frac{29}{17}\right). \end{gathered}$ Thus there are $4$ common points.

Therefore, the correct answer is B.

19.

The number of distinct ordered pairs $(x,y),$ where $x$ and $y$ have positive integral values satisfying the equation $x^4y^4-10x^2y^2+9=0,$ is:

$0$

$3$

$4$

$12$

infinite

Answer: B

Concepts:Diophantine Equation factoring counting pairs

Difficulty rating: 1640

Small Hint:

Treat $x^2y^2$ as one variable

Big Hint:

After factoring, translate the possible values into conditions on the positive integer product $xy$

Solution:

Let $u=x^2y^2.$ Then $\begin{aligned} u^2-10u+9&=0,\\ (u-1)(u-9)&=0. \end{aligned}$ Since $x$ and $y$ are positive integers, this means $xy=1$ or $xy=3.$ The pairs are $(1,1),$ $(1,3),$ and $(3,1),$ for a total of $3.$

Therefore, the correct answer is B.

20.

Let $P$ equal the product of $3{,}659{,}893{,}456{,}789{,}325{,}678$ and $342{,}973{,}489{,}379{,}256.$ The number of digits in $P$ is:

$36$

$35$

$34$

$33$

$32$

Answer: C

Concepts:digits estimation

Difficulty rating: 1460

Small Hint:

Bound the factors between $3.6\cdot10^{18}$ and $3.7\cdot10^{18},$ and between $3.4\cdot10^{14}$ and $3.5\cdot10^{14}$

Big Hint:

Show the product lies between $10^{33}$ and $10^{34}$

Solution:

The factors satisfy $\begin{aligned} 3.6\cdot10^{18}&\lt A\lt3.7\cdot10^{18},\\ 3.4\cdot10^{14}&\lt B\lt3.5\cdot10^{14}. \end{aligned}$ Hence $12.24\cdot10^{32}\lt AB\lt12.95\cdot10^{32},$ so $10^{33}\lt P\lt10^{34}.$ Therefore $P$ has $34$ digits.

Therefore, the correct answer is C.

21.

If the graph of $x^2+y^2=m$ is tangent to that of $x+y=\sqrt{2m},$ then:

$m$ must equal $\dfrac12$

$m$ must equal $\dfrac1{\sqrt2}$

$m$ must equal $\sqrt2$

$m$ must equal $2$

$m$ may be any nonnegative real number

Answer: E

Concepts:circle distance formula tangent line

Difficulty rating: 1720

Small Hint:

The circle has center the origin and radius $\sqrt m$

Big Hint:

Find the distance from the origin to $x+y-\sqrt{2m}=0$

Solution:

The circle has radius $\sqrt m.$ The distance from the origin to the line is $\frac{\lvert-\sqrt{2m}\rvert}{\sqrt{1^2+1^2}}=\sqrt m.$ Thus the line is tangent for every $m\ge0$ (with the $m=0$ case degenerate at the origin).

Therefore, the correct answer is E.

22.

Let $K$ be the measure of the area bounded by the $x$ -axis, the line $x=8,$ and the curve defined by

$\begin{aligned} f=\{(x,y)\mid{}&\\ y&=x\\ &\text{when }0\le x\le5,\\ y&=2x-5\\ &\text{when }5\le x\le8\}. \end{aligned}$

Then $K$ is:

$21.5$

$36.4$

$36.5$

$44$

less than $44$ but arbitrarily close to it

Answer: C

Concepts:area decomposition trapezoid triangle area

Difficulty rating: 1510

Small Hint:

Split the region at $x=5$

Big Hint:

Use a triangle from $0$ to $5$ and a trapezoid from $5$ to $8$

Solution:

From $x=0$ to $5,$ the region is a triangle of area $\frac12(5)(5)=12.5.$ From $x=5$ to $8,$ the endpoint heights are $5$ and $11,$ so the trapezoid has area $\frac12(5+11)(3)=24.$ Thus $K=12.5+24=36.5.$

Therefore, the correct answer is C.

23.

For any integer $n$ greater than $1,$ the number of prime numbers greater than $n!+1$ and less than $n!+n$ is:

Here $n!=1\cdot2\cdots(n-1)n.$

$0$

$1$

$\dfrac n2$ for $n$ even, $\dfrac{n+1}{2}$ for $n$ odd

$n-1$

$n$

Answer: A

Concepts:factorial divisibility prime

Difficulty rating: 1780

Small Hint:

Every integer strictly between the endpoints has the form $n!+k$ with $2\le k\le n-1$

Big Hint:

Use the fact that $k$ divides both $n!$ and $n!+k$

Solution:

Every integer in the interval has the form $n!+k$ for some $2\le k\le n-1.$ Because $n!$ is divisible by $k,$ so is $n!+k,$ and $n!+k\gt k.$ Thus every such integer is composite. (For $n=2,$ the interval is empty.) Hence there are no primes in the interval.

Therefore, the correct answer is A.

24.

When the natural numbers $P$ and $P',$ with $P\gt P',$ are divided by the natural number $D,$ the remainders are $R$ and $R',$ respectively. When $PP'$ and $RR'$ are divided by $D,$ the remainders are $r$ and $r',$ respectively. Then:

$r\gt r'$ always

$r\lt r'$ always

$r\gt r'$ sometimes, and $r\lt r'$ sometimes

$r\gt r'$ sometimes, and $r=r'$ sometimes

$r=r'$ always

Answer: E

Concepts:modular arithmetic

Difficulty rating: 1670

Small Hint:

Write $P\equiv R\pmod D$ and $P'\equiv R'\pmod D$

Big Hint:

Multiply the two congruences

Solution:

By definition of the remainders, $\begin{aligned} P&\equiv R\pmod D,\\ P'&\equiv R'\pmod D. \end{aligned}$ Multiplying gives $PP'\equiv RR'\pmod D.$ Since each has a unique remainder between $0$ and $D-1,$ their remainders must be equal: $r=r'.$

Therefore, the correct answer is E.

25.

If it is known that $\log_2a+\log_2b\ge6,$ then the least value that can be taken on by $a+b$ is:

$2\sqrt6$

$6$

$8\sqrt2$

$16$

none of these

Answer: D

Concepts:logarithm AM-GM Inequality optimization

Difficulty rating: 1880

Small Hint:

Combine the logarithms to bound $ab$

Big Hint:

Apply $a+b\ge2\sqrt{ab}$

Solution:

The logarithm condition gives $\log_2(ab)\ge6,$ so $ab\ge64.$ By AM-GM, $a+b\ge2\sqrt{ab}\ge2\sqrt{64}=16.$ Equality is attained at $a=b=8.$

Therefore, the correct answer is D.

26.

A parabolic arch has a height of $16$ inches and a span of $40$ inches. The height, in inches, of the arch at a point $5$ inches from the center $M,$ is:

$1$

$15$

$15\dfrac13$

$15\dfrac12$

$15\dfrac34$

Answer: B

Concepts:parabola coordinate geometry substitution

Difficulty rating: 1560

Small Hint:

Place the center at the origin and write the parabola as $y=16-ax^2$

Big Hint:

Use the endpoint $(20,0)$ to find $a$ , then substitute $x=5$

Solution:

Place $M$ at the origin with the span on the $x$ -axis. The arch has equation $y=16-ax^2.$ Since $(20,0)$ lies on it, $0=16-400a,$ so $a=\frac{1}{25}.$ At $x=5,$ $y=16-\frac{25}{25}=15.$

Therefore, the correct answer is B.

27.

A particle moves so that its speed for the second and subsequent miles varies inversely as the integral number of miles already traveled. For each subsequent mile the speed is constant. If the second mile is traversed in $2$ hours, then the time, in hours, needed to traverse the $n$ th mile is:

$\dfrac2{n-1}$

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

$\dfrac2n$

$2n$

$2(n-1)$

Answer: E

Concepts:distance rate and time rate ratio and proportion

Difficulty rating: 1500

Small Hint:

During the $n$ th mile, the number of miles already traveled is $n-1$

Big Hint:

Time for a fixed one-mile distance varies directly as $n-1$

Solution:

For the $n$ th mile, $v_n=\frac{k}{n-1}.$ Since the distance is one mile, the time is $T_n=\frac1{v_n}=\frac{n-1}{k}.$ The condition $T_2=2$ gives $\frac{1}{k}=2.$ Therefore $T_n=2(n-1).$

Therefore, the correct answer is E.

28.

Let $n$ be the number of points $P$ interior to the region bounded by a circle with radius $1,$ such that the sum of the squares of the distances from $P$ to the endpoints of a given diameter is $3.$ Then $n$ is:

$0$

$1$

$2$

$4$

infinite

Answer: E

Concepts:coordinate geometry distance formula circle

Difficulty rating: 1720

Small Hint:

Place the circle at the origin with diameter endpoints $(-1,0)$ and $(1,0)$

Big Hint:

Simplify the sum of the two squared distances

Solution:

Let $P=(x,y)$ and take the diameter endpoints as $(-1,0)$ and $(1,0).$ The condition becomes $\begin{aligned} PA^2+PB^2&=3,\\ 2x^2+2y^2+2&=3. \end{aligned}$ Thus $x^2+y^2=\frac{1}{2}.$ This is an entire circle of radius $\frac{1}{\sqrt2},$ lying inside the given unit circle. It contains infinitely many points.

Therefore, the correct answer is E.

29.

If $x=t^{\frac{1}{t-1}}$ and $y=t^{\frac{t}{t-1}},$ $t\gt0,$ $t\ne1,$ a relation between $x$ and $y$ is:

$y^x=x^{\frac{1}{y}}$

$y^{\frac{1}{x}}=x^y$

$y^x=x^y$

$x^x=y^y$

none of these

Answer: C

Concepts:exponent algebraic manipulation substitution

Difficulty rating: 2100

Small Hint:

Divide $y$ by $x$ to express $t$ in terms of $x$ and $y$

Big Hint:

Also observe that $y=x^t$ , then eliminate $t$

Solution:

Dividing the definitions gives $\frac yx=t^{\frac{t-1}{t-1}}=t.$ Also, $y=t^{\frac{t}{t-1}}=\left(t^{\frac{1}{t-1}}\right)^t=x^t.$ Substituting $t=\frac{y}{x}$ gives $y=x^{\frac{y}{x}}.$ Raising both sides to the power $x$ yields $y^x=x^y.$

Therefore, the correct answer is C.

30.

Let $P$ be a point of hypotenuse $AB$ (or its extension) of isosceles right triangle $ABC.$ Let $s=AP^2+PB^2.$ Then:

$s\lt2CP^2$ for a finite number of positions of $P$

$s\lt2CP^2$ for an infinite number of positions of $P$

$s=2CP^2$ only if $P$ is the midpoint of $AB$ or an endpoint of $AB$

$s=2CP^2$ always

$s\gt2CP^2$ if $P$ is a trisection point of $AB$

Answer: D

Concepts:coordinate geometry distance formula isosceles triangle

Difficulty rating: 1670

Small Hint:

Put the midpoint of $AB$ at the origin and $AB$ on the $x$ -axis

Big Hint:

Use $A=(-a,0),$ $B=(a,0),$ $C=(0,a),$ and $P=(p,0)$

Solution:

Take $A=(-a,0),$ $B=(a,0),$ $C=(0,a),$ and $P=(p,0).$ This describes every point on the hypotenuse line. Then $\begin{aligned} s&=(p+a)^2+(p-a)^2\\ &=2p^2+2a^2,\\ 2CP^2&=2(p^2+a^2)\\ &=2p^2+2a^2. \end{aligned}$ Thus $s=2CP^2$ for every position of $P.$

Therefore, the correct answer is D.

31.

Let $OABC$ be a unit square in the $xy$ -plane with $O(0,0),$ $A(1,0),$ $B(1,1),$ and $C(0,1).$ Let $u=x^2-y^2$ and $v=2xy$ define a transformation of the $xy$ -plane into the $uv$ -plane. The transform (or image) of the square is:

Answer: D

Concepts:coordinate geometry parabola transformation

Difficulty rating: 2120

Small Hint:

Map the four vertices and then transform each side separately

Big Hint:

The horizontal sides and vertical sides become two line segments and two parabolic arcs

Solution:

The vertices map as $\begin{aligned} O&\mapsto(0,0),\\ A&\mapsto(1,0),\\ B&\mapsto(0,2),\\ C&\mapsto(-1,0). \end{aligned}$ Side $OA$ maps to the segment from $(0,0)$ to $(1,0),$ and $CO$ maps to the segment from $(-1,0)$ to $(0,0).$ On $AB,$ $x=1,$ so $u=1-\frac{v^2}{4};$ on $BC,$ $y=1,$ so $u=\frac{v^2}{4}-1.$ These are the two upper parabolic arcs joining $(\pm1,0)$ to $(0,2).$

Therefore, the correct answer is D.

32.

Let a sequence $\{u_n\}$ be defined by $u_1=5$ and the relation $u_{n+1}-u_n=3+4(n-1),$ $n=1,$ $2,$ $3,$ $\ldots.$ If $u_n$ is expressed as a polynomial in $n,$ the algebraic sum of its coefficients is:

$3$

$4$

$5$

$6$

$11$

Answer: C

Concepts:polynomial recursion substitution

Difficulty rating: 1530

Small Hint:

The sum of a polynomial’s coefficients is its value at $n=1$

Big Hint:

The problem already gives $u_1$

Solution:

For any polynomial $p(n),$ the sum of its coefficients is $p(1).$ Here the polynomial represents $u_n,$ and the initial condition gives $u_1=5.$ Therefore the coefficient sum is $5.$

Therefore, the correct answer is C.

33.

Let $S_n$ and $T_n$ be the respective sums of the first $n$ terms of two arithmetic series. If $S_n:T_n=(7n+1):(4n+27)$ for all $n,$ the ratio of the eleventh term of the first series to the eleventh term of the second series is:

$4:3$

$3:2$

$7:4$

$78:71$

undetermined

Answer: A

Concepts:arithmetic sequence ratio and proportion

Difficulty rating: 1840

Small Hint:

In an arithmetic sequence, the middle term of the first $21$ terms equals their average

Big Hint:

Express each eleventh term as its corresponding $21$ -term sum divided by $21$

Solution:

For an arithmetic sequence, the eleventh term is the average of the first $21$ terms. Thus the respective eleventh terms are $\frac{S_{21}}{21}$ and $\frac{T_{21}}{21}.$ Their ratio is $\frac{S_{21}}{T_{21}} =\frac{7(21)+1}{4(21)+27} =\frac{148}{111} =\frac43.$

Therefore, the correct answer is A.

34.

The remainder $R$ obtained by dividing $x^{100}$ by $x^2-3x+2$ is a polynomial of degree less than $2.$ Then $R$ may be written as:

$2^{100}-1$

$2^{100}(x-1)-(x-2)$

$2^{100}(x-3)$

$x(2^{100}-1)+2(2^{99}-1)$

$2^{100}(x+1)-(x+2)$

Answer: B

Concepts:polynomial system of equations substitution

Difficulty rating: 2010

Small Hint:

Write $R(x)=ax+b$ and use $x^2-3x+2=(x-1)(x-2)$

Big Hint:

Evaluate the division identity at $x=1$ and $x=2$

Solution:

Write $R(x)=ax+b.$ Since the divisor is $(x-1)(x-2),$ evaluating the division identity at its roots gives $R(1)=1,\qquad R(2)=2^{100}.$ Therefore $\begin{aligned} R(x)&=1+(2^{100}-1)(x-1)\\ &=2^{100}(x-1)-(x-2). \end{aligned}$

Therefore, the correct answer is B.

35.

Let $L(m)$ be the $x$ -coordinate of the left endpoint of the intersection of the graphs of $y=x^2-6$ and $y=m,$ where $-6\lt m\lt6.$ Let $r=\frac{[L(-m)-L(m)]}{m}.$ Then, as $m$ is made arbitrarily close to zero, the value of $r$ is:

arbitrarily close to zero

arbitrarily close to $\dfrac1{\sqrt6}$

arbitrarily close to $\dfrac2{\sqrt6}$

arbitrarily large

undetermined

Answer: B

Concepts:calculus radical rationalizing denominator

Difficulty rating: 1970

Small Hint:

The left intersection coordinate is $L(m)=-\sqrt{6+m}$

Big Hint:

Substitute into $r$ and rationalize the numerator

Solution:

The left intersection satisfies $L(m)=-\sqrt{6+m}.$ Hence $\begin{aligned} r&=\frac{-\sqrt{6-m}+\sqrt{6+m}}{m}\\ &=\frac{2}{\sqrt{6+m}+\sqrt{6-m}}. \end{aligned}$ As $m$ approaches $0,$ this approaches $\frac{2}{2\sqrt6}=\frac{1}{\sqrt6}.$

Therefore, the correct answer is B.