1960 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

If $2$ is a solution (root) of $x^3+hx+10=0,$ then $h$ equals:

$10$

$9$

$2$

$-2$

$-9$

Concepts:polynomial substitution linear equation

Difficulty rating: 840

Small Hint:

Substitute $x=2$ into the equation

Big Hint:

Solve the resulting linear equation for $h$

Solution:

Substituting the given root gives $2^3+2h+10=0.$ Thus $2h=-18,$ so $h=-9.$

Therefore, the correct answer is E.

2.

It takes $5$ seconds for a clock to strike $6$ o’clock beginning at $6{:}00$ o’clock precisely. If the strikings are uniformly spaced, how long, in seconds, does it take to strike $12$ o’clock?

$9\dfrac13$

$10$

$11$

$14\dfrac23$

none of these

Concepts:clock rate fencepost counting

Difficulty rating: 960

Small Hint:

Six strikes contain only five time intervals

Big Hint:

Twelve strikes contain eleven intervals of the same length

Solution:

There are $5$ equal intervals from the first of $6$ strikes to the last, so each interval lasts $1$ second. Twelve strikes contain $11$ such intervals and therefore take $11$ seconds.

Thus, the correct answer is C.

3.

Applied to a bill for $\$10{,}000$ the difference between a discount of $40\%$ and two successive discounts of $36\%$ and $4\%,$ expressed in dollars, is:

$0$

$144$

$256$

$400$

$416$

Concepts:percentage money

Difficulty rating: 1060

Small Hint:

Successive discounts are applied one after the other, not added

Big Hint:

Compare $40\%$ of the bill with $36\%$ plus $4\%$ of the remaining $64\%$

Solution:

A $40\%$ discount is $\$4000.$ The successive discounts total $3600+0.04(6400)=3856.$ Their difference is $4000-3856=144$ dollars.

Therefore, the correct answer is B.

4.

Each of two angles of a triangle is $60^\circ$ and the included side is $4$ inches. The area of the triangle, in square inches, is:

$8\sqrt3$

$8$

$4\sqrt3$

$4$

$2\sqrt3$

Concepts:equilateral triangle triangle area

Difficulty rating: 890

Small Hint:

Determine the third angle of the triangle

Big Hint:

Use the area formula for an equilateral triangle of side $4$

Solution:

The third angle is also $60^\circ,$ so the triangle is equilateral with side $4.$ Its area is $\frac{\sqrt3}{4}(4)^2=4\sqrt3.$

Thus, the correct answer is C.

5.

The number of distinct points common to the graphs of $x^2+y^2=9$ and $y^2=9$ is:

infinitely many

four

two

one

none

Concepts:system of equations circle

Difficulty rating: 840

Small Hint:

Substitute $y^2=9$ into the circle equation

Big Hint:

After finding $x,$ count the possible values of $y$

Solution:

Substitution gives $x^2+9=9,$ so $x=0.$ Then $y^2=9$ gives $y=3$ or $y=-3.$ Hence there are two common points.

Therefore, the correct answer is C.

6.

The circumference of a circle is $100$ inches. The side of a square inscribed in this circle, expressed in inches, is:

$\dfrac{25\sqrt2}{\pi}$

$\dfrac{50\sqrt2}{\pi}$

$\dfrac{100}{\pi}$

$\dfrac{100\sqrt2}{\pi}$

$50\sqrt2$

Concepts:circumference square (geometry)diagonal

Difficulty rating: 1320

Small Hint:

Find the diameter from the circumference

Big Hint:

The diagonal of the inscribed square is the circle’s diameter

Solution:

The diameter is $\frac{100}{\pi}.$ If the square has side $s,$ its diagonal is $s\sqrt2,$ so $s\sqrt2=\frac{100}{\pi}.$ Therefore $s=\frac{50\sqrt2}{\pi}.$

Thus, the correct answer is B.

7.

Circle I passes through the center of, and is tangent to, circle II. The area of circle I is $4$ square inches. Then the area of circle II, in square inches, is:

$8$

$8\sqrt2$

$8\sqrt\pi$

$16$

$16\sqrt2$

Concepts:tangent circles circle area power scaling of length, area, and volume

Difficulty rating: 1140

Small Hint:

Draw the line through the two centers and the tangency point

Big Hint:

The radius of circle II is twice the radius of circle I

Solution:

Because circle I passes through the center of circle II, the distance between the centers equals the radius $r$ of circle I. Internal tangency also makes that distance $R-r,$ where $R$ is the radius of circle II. Thus $R=2r,$ and the area is multiplied by $2^2=4.$ It is therefore $4\cdot4=16.$

Thus, the correct answer is D.

8.

The number $2.5252525\ldots$ can be written as a fraction. When reduced to lowest terms the sum of the numerator and denominator of this fraction is:

$7$

$29$

$141$

$349$

none of these

Concepts:repeating decimal fraction greatest common divisor

Difficulty rating: 1320

Small Hint:

Write the repeating part $0.525252\ldots$ using a denominator of $99$

Big Hint:

Add the integer part and then reduce the fraction

Solution:

We have $2.525252\ldots=2+\frac{52}{99}=\frac{250}{99}.$ This fraction is already in lowest terms, and $250+99=349.$

Therefore, the correct answer is D.

9.

The fraction $\frac{a^2+b^2-c^2+2ab}{a^2+c^2-b^2+2ac}$ is (with suitable restrictions on the values of $a,$ $b,$ and $c$ ):

irreducible

reducible to $-1$

reducible to a polynomial of three terms

reducible to $\dfrac{a-b+c}{a+b-c}$

reducible to $\dfrac{a+b-c}{a-b+c}$

Concepts:factoring difference of squares algebraic manipulation

Difficulty rating: 1180

Small Hint:

Rewrite each quadratic expression as a difference of squares

Big Hint:

The numerator and denominator share a factor $a+b+c$

Solution:

Factoring gives $\begin{gathered} a^2+b^2-c^2+2ab\\ {}=(a+b-c)(a+b+c),\\ a^2+c^2-b^2+2ac\\ {}=(a-b+c)(a+b+c). \end{gathered}$ Cancelling the common factor, under the stated suitable restrictions, leaves $\frac{a+b-c}{a-b+c}.$

Thus, the correct answer is E.

10.

Given the following six statements:

$(1)$ All women are good drivers.

$(2)$ Some women are good drivers.

$(3)$ No men are good drivers.

$(4)$ All men are bad drivers.

$(5)$ At least one man is a bad driver.

$(6)$ All men are good drivers.

The statement that negates statement $(6)$ is:

$(1)$

$(2)$

$(3)$

$(4)$

$(5)$

Concepts:logical deduction

Difficulty rating: 890

Small Hint:

Negating “all” produces an existence statement

Big Hint:

The negation needs only one man who is not a good driver

Solution:

The negation of “All men are good drivers” is “At least one man is not a good driver.” In the terminology of the choices, that is statement $(5),$ “At least one man is a bad driver.”

Therefore, the correct answer is E.

11.

For a given value of $k,$ the product of the roots of $x^2-3kx+2k^2-1=0$ is $7.$ The roots may be characterized as:

integral and positive

integral and negative

rational, but not integral

irrational

imaginary

Concepts:Vieta’s Formulas quadratic radical

Difficulty rating: 1210

Small Hint:

Use the constant term to determine the possible values of $k$

Big Hint:

For either value of $k,$ examine the quadratic formula or discriminant

Solution:

By Vieta’s formulas, $2k^2-1=7,$ so $k=\pm2.$ The roots are $3\pm\sqrt2$ when $k=2,$ and $-3\pm\sqrt2$ when $k=-2.$ In either case both roots are irrational.

Thus, the correct answer is D.

12.

The locus of the centers of all circles of given radius $a,$ in the same plane, passing through a fixed point, is:

a point

a straight line

two straight lines

a circle

two circles

Concepts:circle distance formula

Difficulty rating: 800

Small Hint:

Every center must be exactly one radius from the fixed point

Big Hint:

Identify the locus of points at a fixed positive distance from one point

Solution:

A circle of radius $a$ passes through the fixed point exactly when its center is distance $a$ from that point. The locus of all such centers is a circle of radius $a.$

Therefore, the correct answer is D.

13.

The polygon(s) formed by $y=3x+2,$ $y=-3x+2,$ and $y=-2,$ is (are):

an equilateral triangle

an isosceles triangle

a right triangle

a triangle and a trapezoid

a quadrilateral

Concepts:coordinate geometry slope isosceles triangle

Difficulty rating: 1140

Small Hint:

Find the three pairwise intersections of the lines

Big Hint:

The two slanted lines are mirror images across the $y$ -axis

Solution:

The slanted lines meet at $(0,2),$ while their intersections with $y=-2$ are $(-\frac{4}{3},-2)$ and $(\frac{4}{3},-2).$ The latter two points are symmetric about the $y$ -axis, so the two slanted sides have equal length. The figure is an isosceles triangle.

Thus, the correct answer is B.

14.

If $a$ and $b$ are real numbers, the equation $3x-5+a=bx+1$ has a unique solution $x$ [the symbol $a\ne0$ means that $a$ is different from zero]:

for all $a$ and $b$

if $a\ne2b$

if $a\ne6$

if $b\ne0$

if $b\ne3$

Concepts:linear equation

Difficulty rating: 890

Small Hint:

Collect the $x$ -terms on one side

Big Hint:

A linear equation has a unique solution when the coefficient of $x$ is nonzero

Solution:

Rearranging gives $(3-b)x=6-a.$ This has a unique solution exactly when $3-b\ne0,$ or $b\ne3.$

Thus, the correct answer is E.

15.

Triangle I is equilateral with side $A,$ perimeter $P,$ area $K,$ and circumradius $R$ (radius of the circumscribed circle). Triangle II is equilateral with side $a,$ perimeter $p,$ area $k,$ and circumradius $r.$ If $A$ is different from $a,$ then:

$P:p=R:r$ only sometimes

$P:p=R:r$ always

$P:p=K:k$ only sometimes

$R:r=K:k$ always

$R:r=K:k$ only sometimes

Difficulty rating: 960

Small Hint:

All equilateral triangles are similar

Big Hint:

Perimeter and circumradius both scale linearly with side length

Solution:

For equilateral triangles, $P=3A$ and $R=\frac{A}{\sqrt3},$ with analogous formulas for $p$ and $r.$ Therefore $P:p=A:a=R:r$ for every such pair of triangles.

Thus, the correct answer is B.

16.

In the numeration system with base $5,$ counting is as follows: $1,$ $2,$ $3,$ $4,$ $10,$ $11,$ $12,$ $13,$ $14,$ $20,$ $\ldots$ The number whose description in the decimal system is $69,$ when described in the base $5$ system, is a number with:

two consecutive digits

two non-consecutive digits

three consecutive digits

three non-consecutive digits

four digits

Concepts:number base place value digits

Difficulty rating: 1140

Small Hint:

Express $69$ using powers of $5$

Big Hint:

After finding the base- $5$ digits, compare their values

Solution:

Since $69=2\cdot25+3\cdot5+4,$ its base- $5$ representation is $234_5.$ Its three digits $2,$ $3,$ and $4$ are consecutive.

Therefore, the correct answer is C.

17.

The formula $N=8\cdot10^8\cdot x^{-\frac{3}{2}}$ gives, for a certain group, the number of individuals whose income exceeds $x$ dollars. The lowest income, in dollars, of the wealthiest $800$ individuals is at least:

$10^4$

$10^6$

$10^8$

$10^{12}$

$10^{16}$

Concepts:exponent algebraic manipulation

Difficulty rating: 1210

Small Hint:

Set $N=800$ at the income cutoff

Big Hint:

After simplifying powers of $10,$ solve $x^{\frac{3}{2}}=10^6$

Solution:

At the cutoff, $800=8\cdot10^8x^{-\frac{3}{2}}.$ Hence $x^{\frac{3}{2}}=10^6,$ and raising both sides to the $\frac{2}{3}$ power gives $x=10^4.$

Thus, the correct answer is A.

18.

The pair of equations $3^{x+y}=81$ and $81^{x-y}=3$ has:

no common solution

the solution $x=2,$ $y=2$

the solution $x=2\dfrac12,$ $y=1\dfrac12$

a common solution in positive and negative integers

none of these

Concepts:exponent system of equations substitution

Difficulty rating: 1280

Small Hint:

Rewrite $81$ as $3^4$ in both equations

Big Hint:

Solve the resulting system for $x+y$ and $x-y$

Solution:

The first equation gives $x+y=4.$ The second gives $3^{4(x-y)}=3,$ so $x-y=\frac{1}{4}.$ Therefore $x=\frac{17}{8}$ and $y=\frac{15}{8}.$ This pair is not listed among the first four choices.

Thus, the correct answer is E.

19.

Consider equation I: $x+y+z=46,$ where $x,$ $y,$ and $z$ are positive integers, and equation II: $x+y+z+w=46,$ where $x,$ $y,$ $z,$ and $w$ are positive integers. Then:

$\mathrm{I}$ can be solved in consecutive integers

$\mathrm{I}$ can be solved in consecutive even integers

$\mathrm{II}$ can be solved in consecutive integers

$\mathrm{II}$ can be solved in consecutive even integers

$\mathrm{II}$ can be solved in consecutive odd integers

Concepts:arithmetic sequence divisibility linear equation

Difficulty rating: 1140

Small Hint:

Write the sums of three and four consecutive integers

Big Hint:

Check which required average is compatible with $46$

Solution:

The sum of three consecutive integers is divisible by $3,$ so it cannot be $46.$ Four consecutive integers beginning with $n$ have sum $4n+6.$ Setting $4n+6=46$ gives $n=10,$ and indeed $10+11+12+13=46.$

Therefore, the correct answer is C.

20.

The coefficient of $x^7$ in the expansion of $\left(\frac{x^2}{2}-\frac2x\right)^8$ is:

$56$

$-56$

$14$

$-14$

$0$

Concepts:binomial theorem exponent

Difficulty rating: 1570

Small Hint:

In a term using $k$ copies of $-\frac{2}{x},$ determine the exponent of $x$

Big Hint:

Solve $16-3k=7$ and then compute that binomial coefficient

Solution:

The term using $k$ factors of $-\frac{2}{x}$ has exponent $2(8-k)-k=16-3k.$ To obtain $x^7,$ we need $k=3.$ Its coefficient is $\binom83\left(\frac12\right)^5(-2)^3=-14.$

Thus, the correct answer is D.

21.

The diagonal of square I is $a+b.$ The perimeter of square II with twice the area of I is:

$(a+b)^2$

$\sqrt2(a+b)^2$

$2(a+b)$

$\sqrt8(a+b)$

$4(a+b)$

Concepts:square (geometry)diagonal power scaling of length, area, and volume

Difficulty rating: 1140

Small Hint:

Express the area of square I in terms of its diagonal

Big Hint:

Square II’s area is $(a+b)^2,$ so find its side length

Solution:

A square with diagonal $a+b$ has area $\frac{(a+b)^2}{2}.$ Square II therefore has area $(a+b)^2,$ so its side is $a+b$ and its perimeter is $4(a+b).$

Thus, the correct answer is E.

22.

The equality $(x+m)^2-(x+n)^2=(m-n)^2,$ where $m$ and $n$ are unequal non-zero constants, is satisfied by $x=am+bn$ where:

$a=0,$ $b$ has a unique non-zero value

$a=0,$ $b$ has two non-zero values

$b=0,$ $a$ has a unique non-zero value

$b=0,$ $a$ has two non-zero values

$a$ and $b$ each have a unique non-zero value

Concepts:difference of squares linear equation substitution

Difficulty rating: 1280

Small Hint:

Factor the difference of the two squares

Big Hint:

Cancel the nonzero factor $m-n$ and solve for $x$

Solution:

Factoring and using $m\ne n$ gives $\begin{gathered} (m-n)(2x+m+n)\\ {}=(m-n)^2. \end{gathered}$ Hence $2x+m+n=m-n,$ so $x=-n.$ Thus $a=0$ and $b=-1,$ a unique nonzero value.

Therefore, the correct answer is A.

23.

The radius $R$ of a cylindrical box is $8$ inches, the height $H$ is $3$ inches. The volume $V=\pi R^2H$ is to be increased by the same fixed positive amount when $R$ is increased by $x$ inches as when $H$ is increased by $x$ inches. This condition is satisfied by:

no real value of $x$

one integral value of $x$

one rational, but not integral, value of $x$

one irrational value of $x$

two real values of $x$

Concepts:cylinder volume quadratic

Difficulty rating: 1500

Small Hint:

Write the volume increase once with radius $8+x$ and once with height $3+x$

Big Hint:

Equate the increases and discard the nonpositive solution

Solution:

Increasing the radius changes the volume by $3\pi\big((8+x)^2-8^2\big),$ while increasing the height changes it by $64\pi x.$ Equating these gives $48x+3x^2=64x,$ so $x=0$ or $x=\frac{16}{3}.$ The increase must be positive, leaving one rational but nonintegral value.

Thus, the correct answer is C.

24.

If $\log_{2x}216=x,$ where $x$ is real, then $x$ is:

a non-square, non-cube integer

a non-square, non-cube, non-integral rational number

an irrational number

a perfect square

a perfect cube

Concepts:logarithm exponent function

Difficulty rating: 1280

Small Hint:

Convert the logarithmic equation to $(2x)^x=216$

Big Hint:

Use $216=6^3$ to recognize a real solution

Solution:

The equation is equivalent to $(2x)^x=216.$ Since $216=6^3,$ $x=3$ satisfies the equation. To see it is the only admissible real solution, note that the base requires $x>0$ and $2x\ne1;$ on $0\lt x\lt\frac{1}{2},$ $x\ln(2x)\lt0,$ while for $x>\frac{1}{2}$ the function $x\ln(2x)$ is strictly increasing. Thus $x=3,$ a non-square, non-cube integer.

Therefore, the correct answer is A.

25.

Let $m$ and $n$ be any two odd numbers, with $n$ less than $m.$ The largest integer which divides all possible numbers of the form $m^2-n^2$ is:

$2$

$4$

$6$

$8$

$16$

Concepts:divisibility parity difference of squares

Difficulty rating: 1280

Small Hint:

Factor $m^2-n^2$ into two even factors

Big Hint:

Among $m-n$ and $m+n,$ one is divisible by $4$ ; then test a small pair

Solution:

We have $m^2-n^2=(m-n)(m+n).$ Both factors are even, and one is divisible by $4,$ so every such difference is divisible by $8.$ Taking $m=3,n=1$ gives $m^2-n^2=8,$ proving that no larger integer always divides it.

Thus, the correct answer is D.

26.

Find the set of $x$ -values satisfying the inequality $\left|\frac{5-x}{3}\right|\lt2.$ [The symbol $|a|$ means $+a$ if $a$ is positive, $-a$ if $a$ is negative, $0$ if $a$ is zero. The notation $1\lt a\lt2$ means that $a$ can have any value between $1$ and $2,$ excluding $1$ and $2.$ ]

$1\lt x\lt11$

$-1\lt x\lt11$

$x\lt11$

x>11

$|x|\lt6$

Concepts:absolute value inequality

Difficulty rating: 960

Small Hint:

Multiply the inequality by $3$

Big Hint:

Rewrite $|5-x|\lt6$ as a compound inequality

Solution:

The inequality is equivalent to $|5-x|\lt6,$ so $-6\lt5-x\lt6.$ Subtracting $5$ and reversing signs as needed gives $-1\lt x\lt11.$

Therefore, the correct answer is B.

27.

Let $S$ be the sum of the interior angles of a polygon $P$ for which each interior angle is $7\dfrac12$ times the exterior angle at the same vertex. Then:

$S=2660^\circ$ and $P$ may be regular

$S=2660^\circ$ and $P$ is not regular

$S=2700^\circ$ and $P$ is regular

$S=2700^\circ$ and $P$ is not regular

$S=2700^\circ$ and $P$ may or may not be regular

Concepts:angle sum equiangular polygon regular polygon

Difficulty rating: 1500

Small Hint:

An interior angle and its corresponding exterior angle sum to $180^\circ$

Big Hint:

The condition fixes every angle but says nothing about the side lengths

Solution:

If an exterior angle is $e,$ then $e+\frac{15}{2}e=180^\circ,$ so $e=\frac{360^\circ}{17}.$ Since the exterior angles total $360^\circ,$ the polygon has $17$ vertices and $S=(17-2)180^\circ=2700^\circ.$ All its angles are equal, but its sides need not be equal, so it may or may not be regular.

Thus, the correct answer is E.

28.

The equation $x-\frac7{x-3}=3-\frac7{x-3}$ has:

infinitely many integral roots

no root

one integral root

two equal integral roots

two equal non-integral roots

Concepts:rational equation algebraic manipulation

Difficulty rating: 890

Small Hint:

The identical fractional terms cancel wherever the equation is defined

Big Hint:

Check whether the resulting value lies in the original domain

Solution:

For $x\ne3,$ subtracting the identical fractions from both sides leaves $x=3.$ But $x=3$ makes the original denominators zero, so it is not in the domain. The equation has no root.

Therefore, the correct answer is B.

29.

Five times $A$ ’s money added to $B$ ’s money is more than $\$51.00.$ Three times $A$ ’s money minus $B$ ’s money is $\$21.00.$ If $a$ represents $A$ ’s money in dollars and $b$ represents $B$ ’s money in dollars, then:

a>9, b>6

a>9, $b\lt6$

a>9, $b=6$

a>9, but we can put no bounds on $b$

$2a=3b$

Concepts:inequality system of equations money

Difficulty rating: 1320

Small Hint:

Use $3a-b=21$ to express $b$ in terms of $a$

Big Hint:

Substitute into $5a+b>51$

Solution:

From $3a-b=21,$ we have $b=3a-21.$ Substitution into the inequality gives $5a+3a-21>51,$ so $a>9.$ It follows that $b=3a-21>6.$

Thus, the correct answer is A.

30.

Given the line $3x+5y=15$ and a point on this line equidistant from the coordinate axes. Such a point exists in:

none of the quadrants

quadrant $\mathrm{I}$ only

quadrants $\mathrm{I},$ $\mathrm{II}$ only

quadrants $\mathrm{I},$ $\mathrm{II},$ $\mathrm{III}$ only

each of the quadrants

Concepts:coordinate geometry absolute value system of equations

Difficulty rating: 1210

Small Hint:

A point equidistant from the axes satisfies $|x|=|y|$

Big Hint:

Intersect the given line with both $y=x$ and $y=-x$

Solution:

For $y=x,$ the line gives $8x=15,$ producing a point in quadrant I. For $y=-x,$ it gives $-2x=15,$ so $x\lt0$ and $y>0,$ producing a point in quadrant II. There are no other possibilities.

Therefore, the correct answer is C.

31.

For $x^2+2x+5$ to be a factor of $x^4+px^2+q,$ the values of $p$ and $q$ must be, respectively:

$-2,$ $5$

$5,$ $25$

$10,$ $20$

$6,$ $25$

$14,$ $25$

Concepts:polynomial algebraic manipulation factor

Difficulty rating: 1690

Small Hint:

Work modulo $x^2+2x+5,$ so $x^2=-2x-5$

Big Hint:

Reduce $x^4$ to a linear expression and make both remainder coefficients zero

Solution:

Modulo $x^2+2x+5,$ we have $x^2=-2x-5.$ Then $x^3=10-x,\qquad x^4=12x+5.$ Thus $\begin{gathered} x^4+px^2+q\\ {}\equiv(12-2p)x+(5-5p+q). \end{gathered}$ Both coefficients vanish when $p=6$ and $q=25.$

Therefore, the correct answer is D.

32.

In this figure the center of the circle is $O.$ $\overline{AB}\perp\overline{BC},$ $ADOE$ is a straight line, $\overline{AP}=\overline{AD},$ and $\overline{AB}$ has a length twice the radius. Then:

$\overline{AP}^{\,2}=\overline{PB}\cdot\overline{AB}$

$\overline{AP}\cdot\overline{DO}=\overline{PB}\cdot\overline{AD}$

$\overline{AB}^{\,2}=\overline{AD}\cdot\overline{DE}$

$\overline{AB}\cdot\overline{AD}=\overline{OB}\cdot\overline{AO}$

none of these

Concepts:tangent line power of a point circle

Difficulty rating: 2000

Small Hint:

Let the radius be $r$ and write $AD$ in terms of $AO$ and $r$

Big Hint:

Use the tangent relation $AB^2=AD\cdot AE$ together with $AB=2r$

Solution:

Let the radius be $r.$ Since $AB$ is tangent at $B,$ $AB^2=AD\cdot AE.$ Write $AD=t.$ Because the secant passes through the center, $AE=t+2r,$ while $AB=2r.$ Hence $t(t+2r)=4r^2.$ Also $AP=AD=t$ and $PB=AB-AP=2r-t.$ The displayed equation rearranges to $t^2=2r(2r-t)=PB\cdot AB.$ Therefore $AP^2=PB\cdot AB.$

Thus, the correct answer is A.

33.

You are given a sequence of $58$ terms; each term has the form $P+n$ where $P$ stands for the product $2\cdot3\cdot5\cdots61$ of all prime numbers less than or equal to $61,$ and $n$ takes, successively, the values $2,$ $3,$ $4,$ $\ldots,$ $59.$ Let $N$ be the number of primes appearing in this sequence. Then $N$ is:

$0$

$16$

$17$

$57$

$58$

Concepts:prime divisibility prime factorization

Difficulty rating: 1500

Small Hint:

Every $n$ from $2$ through $59$ has a prime divisor at most $59$

Big Hint:

That same prime divisor divides both $P$ and $n$

Solution:

For each $n$ in the given range, choose a prime divisor $p$ of $n.$ Since $p\le59,$ it is one of the factors of $P.$ Therefore $p$ divides $P+n.$ Moreover $P+n>p,$ so $P+n$ is composite. No term is prime, and $N=0.$

Thus, the correct answer is A.

34.

Two swimmers, at opposite ends of a $90$ -foot pool, start to swim the length of the pool, one at the rate of $3$ feet per second, the other at $2$ feet per second. They swim back and forth for $12$ minutes. Allowing no loss of time at the turns, find the number of times they pass each other.

$24$

$21$

$20$

$19$

$18$

Concepts:relative speed reflection (geometry)least common multiple

Difficulty rating: 1870

Small Hint:

Reflect the pool each time a swimmer turns so both paths become straight

Big Hint:

The joint position pattern repeats every $180$ seconds; inspect one full cycle, including endpoint coincidences

Solution:

The faster swimmer’s position repeats every $60$ seconds, and the slower swimmer’s position repeats every $90$ seconds. Thus their joint position pattern repeats every $180$ seconds. During one such cycle, their positions coincide at $t=18,54,90,126,162$ seconds. The keyed interpretation counts the simultaneous turn at the endpoint when $t=90$ as one of these encounters. Thus there are $5$ counted encounters per cycle and $4$ cycles in $12$ minutes, giving $5\cdot4=20.$

Therefore, the correct answer is C.

35.

From point $P$ outside a circle, with a circumference of $10$ units, a tangent is drawn. Also from $P$ a secant is drawn dividing the circle into unequal arcs with lengths $m$ and $n.$ It is found that $t,$ the length of the tangent, is the mean proportional between $m$ and $n.$ If $m$ and $t$ are integers, then $t$ may have the following number of values:

zero

one

two

three

infinitely many

Concepts:ratio and proportion perfect square arc

Difficulty rating: 1870

Small Hint:

Use $m+n=10$ and $t^2=mn$

Big Hint:

Test the integer values $m=1,2,\ldots,9,$ excluding equal arcs

Solution:

The arc lengths satisfy $m+n=10,$ while the mean-proportional condition gives $t^2=mn=m(10-m).$ For integral $m$ from $1$ through $9,$ the possible products, up to symmetry, are $9,$ $16,$ $21,$ $24,$ and $25.$ The last comes from equal arcs $m=n=5$ and is excluded. The remaining squares give $t=3$ and $t=4,$ two values.

Thus, the correct answer is C.

36.

Let $s_1,$ $s_2,$ $s_3$ be the respective sums of $n,$ $2n,$ $3n$ terms of the same arithmetic progression with $a$ as the first term and $d$ as the common difference. Let $R=s_3-s_2-s_1.$ Then $R$ is dependent on:

$a$ and $d$

$d$ and $n$

$a$ and $n$

$a,$ $d,$ and $n$

neither $a$ nor $d$ nor $n$

Concepts:arithmetic sequence summation algebraic manipulation

Difficulty rating: 1670

Small Hint:

Use $S_k=\dfrac{k}{2}(2a+(k-1)d)$

Big Hint:

Substitute $k=n,2n,3n$ and collect the $a$ -terms and $d$ -terms separately

Solution:

Using the arithmetic-series formula, $s_j=\frac{jn}{2}\bigl(2a+(jn-1)d\bigr).$ In $s_3-s_2-s_1,$ the coefficient of $a$ is $3n-2n-n=0.$ Simplifying the remaining terms gives $R=2n^2d.$ Thus $R$ depends on $d$ and $n,$ but not on $a.$

Therefore, the correct answer is B.

37.

The base of a triangle is of length $b,$ and the altitude is of length $h.$ A rectangle of height $x$ is inscribed in the triangle with the base of the rectangle in the base of the triangle. The area of the rectangle is:

$\dfrac{bx}{h}(h-x)$

$\dfrac{hx}{b}(b-x)$

$\dfrac{bx}{h}(h-2x)$

$x(b-x)$

$x(h-x)$

Concepts:similarity rectangle area

Difficulty rating: 1280

Small Hint:

At height $x,$ use similarity to find the horizontal width of the triangle

Big Hint:

The available width is $b(1-\frac{x}{h})$ ; multiply it by the rectangle’s height

Solution:

The cross-section parallel to the base at height $x$ has width $b\left(1-\frac{x}{h}\right)=\frac{b(h-x)}{h}$ by similarity. Multiplying by the rectangle’s height $x$ gives $\frac{bx}{h}(h-x).$

Thus, the correct answer is A.

38.

In this diagram $\overline{AB}$ and $\overline{AC}$ are the equal sides of an isosceles triangle $ABC,$ in which is inscribed equilateral triangle $DEF.$ Designate angle $BFD$ by $a,$ angle $ADE$ by $b,$ and angle $FEC$ by $c.$ Then:

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

$b=\dfrac{a-c}{2}$

$a=\dfrac{b-c}{2}$

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

none of these

Concepts:angle chasing isosceles triangle equilateral triangle

Difficulty rating: 2000

Small Hint:

Let each base angle of $ABC$ be $\theta$

Big Hint:

Compare the directions of $DF,$ $DE,$ and $EF$ using the $60^\circ$ angles of the equilateral triangle

Solution:

Let each base angle of $ABC$ be $\theta.$ Measured from the direction $BC,$ the line $FD$ has direction $180^\circ-a,$ so the equilateral condition makes $DE$ have direction $60^\circ-a.$ At $D,$ $b=\theta-(60^\circ-a)=\theta-60^\circ+a.$ Similarly, angle chasing at $E$ gives $c=60^\circ+a-\theta.$ Adding these equations yields $b+c=2a.$

Therefore $a=\frac{b+c}{2},$ and the correct answer is D.

39.

To satisfy the equation $\frac{a+b}{a}=\frac{b}{a+b},$ $a$ and $b$ must be:

both rational

both real but not rational

both not real

one real, one not real

one real, one not real or both not real

Concepts:complex number quadratic

Difficulty rating: 1730

Small Hint:

Let $t=\frac{b}{a}$ after noting that the denominators must be nonzero

Big Hint:

The resulting quadratic $t^2+t+1=0$ has negative discriminant

Solution:

The denominators require $a\ne0$ and $a+b\ne0.$ Setting $t=\frac{b}{a}$ and cross-multiplying gives $(1+t)^2=t,$ or $t^2+t+1=0.$ Its discriminant is $-3,$ so $t$ is not real. Therefore $a$ and $b$ cannot both be real. Depending on the nonzero complex scale chosen for $a,$ one can be real and the other nonreal, or both can be nonreal.

Thus, the correct answer is E.

40.

Given right triangle $ABC$ with legs $\overline{BC}=3,$ $\overline{AC}=4.$ Find the length of the shorter angle trisector from $C$ to the hypotenuse:

$\dfrac{32\sqrt3-24}{13}$

$\dfrac{12\sqrt3-9}{13}$

$6\sqrt3-8$

$\dfrac{5\sqrt{10}}6$

$\dfrac{25}{12}$

Concepts:coordinate geometry trigonometry right triangle

Difficulty rating: 2000

Small Hint:

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

Big Hint:

The two trisector rays make angles $30^\circ$ and $60^\circ$ with $CB$ ; intersect each with $\frac{x}{3}+\frac{y}{4}=1$

Solution:

Place $C=(0,0),$ $B=(3,0),$ $A=(0,4).$ The hypotenuse has equation $\frac{x}{3}+\frac{y}{4}=1.$ The $30^\circ$ trisector ray is $y=\frac{x}{\sqrt3}.$ Substitution gives $x=\frac{12\sqrt3}{4\sqrt3+3}.$ Its length is $\frac{x}{\cos30^\circ}=\frac{2x}{\sqrt3},$ hence $\frac{24}{4\sqrt3+3} =\frac{32\sqrt3-24}{13}.$ The $60^\circ$ ray gives the longer trisector.

Thus, the correct answer is A.