1963 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

Which one of the following points is not on the graph of $y=\dfrac{x}{x+1}?$

$(0,0)$

$\left(-\dfrac12,-1\right)$

$\left(\dfrac12,\dfrac13\right)$

$(-1,1)$

$(-2,2)$

Concepts:function substitution

Difficulty rating: 1320

Small Hint:

Check the domain before substituting coordinates

Big Hint:

The denominator vanishes at one of the listed $x$ -coordinates

Solution:

The expression $\frac{x}{x+1}$ is undefined when $x=-1.$ Therefore no point whose first coordinate is $-1$ can lie on its graph.

Thus, the correct answer is D.

2.

Let $n=x-y^{x-y}.$ Find $n$ when $x=2$ and $y=-2.$

$-14$

$0$

$1$

$18$

$256$

Concepts:substitution exponent order of operations

Difficulty rating: 1180

Small Hint:

First compute the exponent $x-y$

Big Hint:

Evaluate the power before subtracting it from $x$

Solution:

Here $x-y=2-(-2)=4,$ so $n=2-(-2)^4=2-16=-14.$

Therefore, the correct answer is A.

3.

If the reciprocal of $x+1$ is $x-1,$ then $x$ equals:

$0$

$1$

$-1$

$\pm1$

none of these

Concepts:rational equation quadratic

Difficulty rating: 1320

Small Hint:

Translate “the reciprocal of $x+1$ ” into an equation

Big Hint:

Multiplying by $x+1$ leads to a difference of squares

Solution:

The equation is $\frac1{x+1}=x-1.$ Hence $1=x^2-1,$ so $x=\pm\sqrt2.$ Neither value is listed.

Thus, the correct answer is E.

4.

For what value(s) of $k$ does the pair of equations $y=x^2$ and $y=3x+k$ have two identical solutions?

$\dfrac49$

$-\dfrac49$

$\dfrac94$

$-\dfrac94$

$\pm\dfrac94$

Concepts:parabola tangent line quadratic

Difficulty rating: 1280

Small Hint:

Set the two expressions for $y$ equal

Big Hint:

A repeated intersection makes the resulting quadratic’s discriminant zero

Solution:

Intersections satisfy $x^2-3x-k=0.$ Two identical solutions require $(-3)^2-4(1)(-k)=9+4k=0,$ giving $k=-\frac{9}{4}.$

Therefore, the correct answer is D.

5.

If $x$ and $\log_{10}x$ are real numbers and $\log_{10}x\lt0,$ then:

$x\lt0$

$-1\lt x\lt1$

$0\lt x\leq1$

$-1\lt x\lt0$

$0\lt x\lt1$

Concepts:logarithm inequality

Difficulty rating: 940

Small Hint:

A real logarithm requires a positive argument

Big Hint:

Compare $x$ with $10^0$

Solution:

The logarithm is real only for $x\gt0.$ Since base $10$ is greater than $1,$ $\log_{10}x\lt0=\log_{10}1$ implies $x\lt1.$

Thus, the correct answer is E.

6.

Triangle $BAD$ is right-angled at $B.$ On $AD$ there is a point $C$ for which $AC=CD$ and $AB=BC.$ The magnitude of angle $DAB,$ in degrees, is:

$67\dfrac12$

$60$

$45$

$30$

$22\dfrac12$

Concepts:right triangle midpoint equilateral triangle

Difficulty rating: 1500

Small Hint:

A midpoint of a right triangle’s hypotenuse is equidistant from all three vertices

Big Hint:

Use $AB=BC=AC$

Solution:

Since $C$ is the midpoint of hypotenuse $AD,$ $AC=BC=CD.$ Given $AB=BC,$ triangle $ABC$ is equilateral. Thus $\angle DAB=\angle CAB=60^\circ.$

Therefore, the correct answer is B.

7.

Given the four equations:

$\begin{aligned} (1)\quad&3y-2x=12,\\ (2)\quad&-2x-3y=10,\\ (3)\quad&3y+2x=12,\\ (4)\quad&2y+3x=10. \end{aligned}$

The pair representing perpendicular lines is:

$(1)$ and $(4)$

$(1)$ and $(3)$

$(1)$ and $(2)$

$(2)$ and $(4)$

$(2)$ and $(3)$

Concepts:linear equation slope

Difficulty rating: 1080

Small Hint:

Rewrite each equation in slope-intercept form

Big Hint:

Perpendicular nonvertical lines have slopes whose product is $-1$

Solution:

Line $(1)$ has slope $\frac{2}{3},$ and line $(4)$ has slope $-\frac{3}{2}.$ Their product is $-1,$ so these two lines are perpendicular.

Thus, the correct answer is A.

8.

The smallest positive integer $x$ for which $1260x=N^3,$ where $N$ is an integer, is:

$1050$

$1260$

$1260^2$

$7350$

$44{,}100$

Concepts:prime factorization perfect power

Difficulty rating: 1680

Small Hint:

Factor $1260$ into primes

Big Hint:

Raise every prime exponent to the next multiple of $3$

Solution:

Since $1260=2^2\cdot3^2\cdot5\cdot7,$ the least multiplier making every exponent divisible by $3$ is $2\cdot3\cdot5^2\cdot7^2=7350.$

Thus, the correct answer is D.

9.

In the expansion of $\left(a-\dfrac1{\sqrt a}\right)^7$ the coefficient of $a^{-\frac{1}{2}}$ is:

$-7$

$7$

$-21$

$21$

$35$

Concepts:binomial theorem exponent

Difficulty rating: 1850

Small Hint:

The term choosing the second summand $r$ times has exponent $7-\frac{3r}{2}$

Big Hint:

Set that exponent equal to $-\frac{1}{2}$ before finding the binomial coefficient and sign

Solution:

The term using $-a^{-\frac{1}{2}}$ exactly $r$ times has exponent $7-r-\frac{r}{2}=7-\frac{3r}{2}.$ Setting this to $-\frac{1}{2}$ gives $r=5.$ Its coefficient is $\binom75(-1)^5=-21.$

Therefore, the correct answer is C.

10.

Point $P$ is taken interior to a square with side-length $a$ and such that it is equally distant from two consecutive vertices and from the side opposite these vertices. If $d$ represents the common distance, then $d$ equals:

$\dfrac{3a}{5}$

$\dfrac{5a}{8}$

$\dfrac{3a}{8}$

$\dfrac{a\sqrt2}{2}$

$\dfrac a2$

Concepts:square (geometry)perpendicular bisector distance formula

Difficulty rating: 1730

Small Hint:

The equal distances to two consecutive vertices put $P$ on their perpendicular bisector

Big Hint:

If the distance to the opposite side is $d,$ use a right triangle with legs $\frac{a}{2}$ and $a-d$

Solution:

Place the opposite side at height $0$ and the two vertices at height $a.$ Then $P=(\frac{a}{2},d).$ Equality of the vertex distance and $d$ gives $\left(\frac a2\right)^2+(a-d)^2=d^2,$ so $\frac{5a^2}{4}=2ad$ and $d=\frac{5a}{8}.$

Thus, the correct answer is B.

11.

The arithmetic mean of a set of $50$ numbers is $38.$ If two numbers of the set, namely $45$ and $55,$ are discarded, the arithmetic mean of the remaining set of numbers is:

$38.5$

$37.5$

$37$

$36.5$

$36$

Concepts:mean whole number operations

Difficulty rating: 1030

Small Hint:

Recover the original sum from the mean

Big Hint:

Subtract $45+55$ and divide by $48$

Solution:

The original sum is $50\cdot38=1900.$ After removing $45+55=100,$ the remaining sum is $1800,$ so the new mean is $\frac{1800}{48}=37.5.$

Thus, the correct answer is B.

12.

Three vertices of parallelogram $PQRS$ are $P(-3,-2),$ $Q(1,-5),$ $R(9,1)$ with $P$ and $R$ diagonally opposite. The sum of the coordinates of vertex $S$ is:

$13$

$12$

$11$

$10$

$9$

Concepts:parallelogram coordinate geometry midpoint

Difficulty rating: 1290

Small Hint:

The diagonals of a parallelogram bisect each other

Big Hint:

Use the vector relation $P+R=Q+S$

Solution:

Since $P+R=Q+S,$ $\begin{aligned} S&=P+R-Q\\ &=(-3,-2)+(9,1)-(1,-5)\\ &=(5,4). \end{aligned}$ The coordinate sum is $5+4=9.$

Therefore, the correct answer is E.

13.

If $2^a+2^b=3^c+3^d,$ the number of integers $a,$ $b,$ $c,$ $d$ which can possibly be negative is, at most:

$4$

$3$

$2$

$1$

$0$

Concepts:exponent divisibility parity

Difficulty rating: 2160

Small Hint:

A reduced denominator on the left can contain only $2,$ while one on the right can contain only $3$

Big Hint:

Conclude both sides must be integers, then examine when a sum of two negative powers can be integral

Solution:

In lowest terms, the left side can have only a power of $2$ in its denominator, while the right side can have only a power of $3.$ Equality therefore forces both sides to be integers. If either $c$ or $d$ were negative, $3^c+3^d$ would retain a factor $3$ in its denominator. Thus $c,d\geq0.$

Likewise, negative $a$ or $b$ can give an integer only in the exceptional case $a=b=-1,$ whose sum is $1.$ But $3^c+3^d\geq2.$ Hence $a,b\geq0$ as well, so none can be negative.

Thus, the correct answer is E.

14.

Given the equations $x^2+kx+6=0$ and $x^2-kx+6=0.$ If, when the roots of the equations are suitably listed, each root of the second equation is $5$ more than the corresponding root of the first equation, then $k$ equals:

$5$

$-5$

$7$

$-7$

none of these

Concepts:quadratic Vieta’s Formulas

Difficulty rating: 1500

Small Hint:

Compare the sums of the two pairs of roots

Big Hint:

Adding $5$ to each root increases their sum by $10$

Solution:

The first pair of roots has sum $-k,$ while the second has sum $k.$ Since both roots increase by $5,$ $k=-k+10,$ so $k=5.$

Therefore, the correct answer is A.

15.

A circle is inscribed in an equilateral triangle, and a square is inscribed in the circle. The ratio of the area of the triangle to the area of the square is:

$\sqrt3:1$

$\sqrt3:\sqrt2$

$3\sqrt3:2$

$3:\sqrt2$

$3:2\sqrt2$

Concepts:equilateral triangle incircle, incenter, and inradius square (geometry)area ratio

Difficulty rating: 1570

Small Hint:

Express both areas using the circle’s radius $r$

Big Hint:

The triangle’s side is $2\sqrt3r$ and the square’s diagonal is $2r$

Solution:

If the circle has radius $r,$ the equilateral triangle has side $2\sqrt3r$ and area $3\sqrt3r^2.$ The square has side $r\sqrt2$ and area $2r^2.$ The ratio is $3\sqrt3:2.$

Thus, the correct answer is C.

16.

Three numbers $a,$ $b,$ $c,$ none zero, form an arithmetic progression. Increasing $a$ by $1$ or increasing $c$ by $2$ results in a geometric progression. Then $b$ equals:

$16$

$14$

$12$

$10$

$8$

Concepts:arithmetic sequence geometric sequence system of equations

Difficulty rating: 1920

Small Hint:

Translate the three progression conditions into equations involving $b^2$ and $a+c$

Big Hint:

Compare $(a+1)c=b^2=a(c+2)$ first

Solution:

The conditions give $\begin{gathered} 2b=a+c,\\ b^2=(a+1)c,\qquad b^2=a(c+2). \end{gathered}$ Comparing the last two yields $c=2a.$ Thus $b=\frac{3a}{2},$ and $\frac{9a^2}{4}=2a(a+1).$ Since $a\ne0,$ $a=8,$ so $b=12.$

Therefore, the correct answer is C.

17.

The expression

$\frac{\dfrac{a}{a+y}+\dfrac{y}{a-y}} {\dfrac{y}{a+y}-\dfrac{a}{a-y}},$

$a$ real, $a\ne0,$ has the value $-1$ for:

all but two real values of $y$

only two real values of $y$

all real values of $y$

only one real value of $y$

no real values of $y$

Concepts:fraction algebraic manipulation

Difficulty rating: 1570

Small Hint:

Combine the two fractions in the numerator and denominator separately

Big Hint:

Keep track of the excluded values $y=\pm a$

Solution:

For $y\ne\pm a,$ the numerator simplifies to $\frac{a^2+y^2}{a^2-y^2}$ and the denominator to its negative. Their quotient is therefore $-1.$ The two values $y=a$ and $y=-a$ are undefined.

Thus, the correct answer is A.

18.

Chord $EF$ is the perpendicular bisector of chord $BC,$ intersecting it in $M.$ Between $B$ and $M$ point $U$ is taken, and $EU$ extended meets the circle in $A.$ Then, for any selection of $U,$ as described, triangle $EUM$ is similar to triangle:

$EFA$

$EFC$

$ABM$

$ABU$

$FMC$

Concepts:chord circle similarity

Difficulty rating: 1990

Small Hint:

Because the perpendicular bisector of a chord passes through the center, $EF$ is a diameter

Big Hint:

Compare the right angles and the shared angle at $E$

Solution:

Since $EF$ is a diameter, $\angle EAF=90^\circ.$ Also $EF\perp BC,$ so $\angle EMU=90^\circ.$ Because $E,U,A$ are collinear, the two triangles share the same acute angle at $E.$ Hence $\triangle EUM\sim\triangle EFA.$

Therefore, the correct answer is A.

19.

In counting $n$ colored balls, some red and some black, it was found that $49$ of the first $50$ counted were red. Thereafter, $7$ out of every $8$ counted were red. If, in all, $90\%$ or more of the balls counted were red, the maximum value of $n$ is:

$225$

$210$

$200$

$180$

$175$

Concepts:percentage inequality rate

Difficulty rating: 1640

Small Hint:

Write the number of red balls as $49+\frac78(n-50)$

Big Hint:

Require that quantity to be at least $0.9n$

Solution:

The condition is $49+\frac78(n-50)\geq\frac9{10}n.$ Simplifying gives $\frac{21}{4}\geq \frac{n}{40},$ so $n\leq210.$ The maximum is $210.$

Thus, the correct answer is B.

20.

Two men at points $R$ and $S,$ $76$ miles apart, set out at the same time to walk towards each other. The man at $R$ walks uniformly at $4\dfrac12$ miles per hour; the man at $S$ walks at $3\dfrac14$ miles per hour for the first hour, at $3\dfrac34$ miles per hour for the second hour, and so on, in arithmetic progression. If the men meet $x$ miles nearer $R$ than $S$ in an integral number of hours, then $x$ is:

$10$

$8$

$6$

$4$

$2$

Concepts:arithmetic sequence summation distance rate and time

Difficulty rating: 1920

Small Hint:

Let the integral meeting time be $h$ hours and sum the second man’s hourly distances

Big Hint:

Their combined distance equation simplifies to $h^2+30h-304=0$

Solution:

In $h$ hours the first man walks $\frac{9h}{2}.$ The second walks an arithmetic-series total $\begin{aligned} &\frac h2\left(2\cdot\frac{13}{4} +(h-1)\frac12\right)\\ &\qquad=3h+\frac{h^2}{4}. \end{aligned}$ Their distances sum to $76,$ giving $h^2+30h-304=0,$ whose positive root is $h=8.$ They walk $36$ and $40$ miles, so the meeting point is $4$ miles nearer $R$ than $S.$

Thus, the correct answer is D.

21.

The expression $x^2-y^2-z^2+2yz+x+y-z$ has:

no linear factor with integer coefficients and integer exponents

the factor $-x+y+z$

the factor $x-y-z+1$

the factor $x+y-z+1$

the factor $x-y+z+1$

Concepts:difference of squares factoring substitution

Difficulty rating: 1520

Small Hint:

Group $y-z$ as a single expression

Big Hint:

Rewrite the polynomial as $x^2-(y-z)^2+x+(y-z)$

Solution:

Let $u=y-z.$ The two groups factor as $\begin{gathered} x^2-u^2=(x+u)(x-u),\\ x+u=(x+u)\cdot1. \end{gathered}$ Thus the whole expression is $(x+u)(x-u+1),$ and one factor is $x-y+z+1.$

Therefore, the correct answer is E.

22.

Acute-angled triangle $ABC$ is inscribed in a circle with center at $O;$ $\overset{\frown}{AB}=120^\circ$ and $\overset{\frown}{BC}=72^\circ.$ A point $E$ is taken in minor arc $AC$ such that $OE$ is perpendicular to $AC.$ Then the ratio of the magnitudes of angles $OBE$ and $BAC$ is:

$\dfrac5{18}$

$\dfrac29$

$\dfrac14$

$\dfrac13$

$\dfrac49$

Concepts:arc inscribed angle isosceles triangle

Difficulty rating: 2030

Small Hint:

Find minor arc $AC,$ then use $OE\perp AC$ to locate $E$ at its midpoint

Big Hint:

Compute the central angle $BOE$ and the inscribed angle $BAC$

Solution:

Minor arc $AC$ has measure $360^\circ-120^\circ-72^\circ=168^\circ.$ Since $OE\perp AC,$ $E$ bisects that arc, so $\angle BOE=72^\circ+84^\circ=156^\circ.$ Isosceles triangle $OBE$ gives $\angle OBE=12^\circ,$ while $\angle BAC=\frac{72^\circ}{2}=36^\circ.$ Their ratio is $\frac{1}{3}.$

Thus, the correct answer is D.

23.

$A$ gives $B$ as many cents as $B$ has and $C$ as many cents as $C$ has. Similarly, $B$ then gives $A$ and $C$ as many cents as each then has. $C$ , similarly, then gives $A$ and $B$ as many cents as each then has. If each finally has $16$ cents, with how many cents does $A$ start?

$24$

$26$

$28$

$30$

$32$

Concepts:money work backwards

Difficulty rating: 1780

Small Hint:

Work backward from the final holdings

Big Hint:

Immediately before someone gives, each recipient must have half of the amount held just after that gift

Solution:

Reverse the transactions. Before $C$ gives, the holdings are $(8,8,32).$ Before $B$ gives, they are $(4,28,16).$ Before $A$ gives, $B$ and $C$ must have had $14$ and $8,$ while $A$ had $4+14+8=26.$

Therefore, the correct answer is B.

24.

Consider equations of the form $x^2+bx+c=0.$ How many such equations have real roots and have coefficients $b$ and $c$ selected from the set of integers $\{1,2,3,4,5,6\}?$

$20$

$19$

$18$

$17$

$16$

Concepts:quadratic basic counting inequality

Difficulty rating: 1570

Small Hint:

Real roots require $b^2\geq4c$

Big Hint:

For each $c=1,\ldots,6,$ count the allowed values of $b$

Solution:

For $c$ equal to $1,$ $2,$ $3,$ $4,$ $5,$ and $6,$ the condition $b^2\geq4c$ permits respectively $5,$ $4,$ $3,$ $3,$ $2,$ and $2$ values of $b$ in the given set. Their sum is $5+4+3+3+2+2=19.$

Thus, the correct answer is B.

25.

Point $F$ is taken in side $AD$ of square $ABCD.$ At $C$ a perpendicular is drawn to $CF,$ meeting $AB$ extended at $E.$ The area of $ABCD$ is $256$ square inches and the area of triangle $CEF$ is $200$ square inches. Then the number of inches in $BE$ is:

$12$

$14$

$15$

$16$

$20$

Concepts:square (geometry)congruence (geometry)triangle area Pythagorean Theorem

Difficulty rating: 1900

Small Hint:

Show that right triangles $CDF$ and $CBE$ are congruent

Big Hint:

Then $CF=CE,$ so use the area of right triangle $CEF$

Solution:

The complementary acute angles and the equal square sides give $\triangle CDF\cong\triangle CBE,$ hence $CF=CE.$ Since $CF\perp CE,$ $\begin{aligned} 200=[CEF] &=\frac12 CE\cdot CF\\ &=\frac12CE^2, \end{aligned}$ so $CE=20.$ The square side is $16,$ and right triangle $CBE$ gives $BE=\sqrt{20^2-16^2}=12.$

Therefore, the correct answer is A.

26.

Form I. Consider the statements

$\begin{aligned} (1)\quad&p\land\neg q\land r,\\ (2)\quad&\neg p\land\neg q\land r,\\ (3)\quad&p\land\neg q\land\neg r,\\ (4)\quad&\neg p\land q\land r, \end{aligned}$

where $p,$ $q,$ $r$ are propositions. How many of these imply the truth of $(p\to q)\to r?$

Form II. Consider the statements $(1)$ $p$ and $r$ are true and $q$ is false, $(2)$ $r$ is true and $p$ and $q$ are false, $(3)$ $p$ is true and $q$ and $r$ are false, $(4)$ $q$ and $r$ are true and $p$ is false. How many of these imply the truth of the statement “ $r$ is implied by the statement that $p$ implies $q$ ”?

$0$

$1$

$2$

$3$

$4$

Concepts:logical deduction

Difficulty rating: 1880

Small Hint:

An implication is false only when its antecedent is true and its consequent is false

Big Hint:

Evaluate $p\to q$ first in each of the four assignments

Solution:

In cases $(1)$ and $(3),$ $p\to q$ is false, so the outer implication is true. In cases $(2)$ and $(4),$ $p\to q$ is true, but $r$ is also true, so the outer implication is again true. All four statements imply it.

Thus, the correct answer is E.

27.

Six straight lines are drawn in a plane with no two parallel and no three concurrent. The number of regions into which they divide the plane is:

$16$

$20$

$22$

$24$

$26$

Concepts:counting regions counting intersections summation

Difficulty rating: 1520

Small Hint:

The $k$ th line is cut into $k$ pieces by the previous lines

Big Hint:

Start with one region and add $1+2+\cdots+6$

Solution:

Successive lines create $1,2,\ldots,6$ new regions. Thus the total is $1+\sum_{k=1}^{6}k=1+21=22.$

Therefore, the correct answer is C.

28.

Given the equation $3x^2-4x+k=0$ with real roots. The value of $k$ for which the product of the roots of the equation is a maximum is:

$\dfrac{16}{9}$

$\dfrac{16}{3}$

$\dfrac49$

$\dfrac43$

$-\dfrac43$

Concepts:quadratic Vieta’s Formulas optimization

Difficulty rating: 1210

Small Hint:

The product of the roots is $\frac{k}{3}$

Big Hint:

Use the discriminant condition to find the largest allowed $k$

Solution:

Real roots require $16-12k\geq0,$ so $k\leq\frac{4}{3}.$ The root product is $\frac{k}{3},$ which increases with $k.$ It is therefore largest at $k=\frac{4}{3}.$

Thus, the correct answer is D.

29.

A particle projected vertically upward reaches, at the end of $t$ seconds, an elevation of $s$ feet where $s=160t-16t^2.$ The highest elevation is:

$800$

$640$

$400$

$320$

$160$

Concepts:quadratic optimization

Difficulty rating: 1320

Small Hint:

The height is a downward-opening quadratic

Big Hint:

Find its vertex time using $-\frac{b}{2a}$

Solution:

The vertex occurs at $t=-\frac{160}{2\cdot-16}=5.$ Then $s(5)=160(5)-16(25)=400.$

Thus, the correct answer is C.

30.

Let

$F=\log\frac{1+x}{1-x}.$

Form a new function $G$ by replacing each $x$ in $F$ by

$\frac{3x+x^3}{1+3x^2},$

and simplify. The simplified expression $G$ is equal to:

$-F$

$F$

$3F$

$F^3$

$F^3-F$

Concepts:logarithm factoring algebraic manipulation

Difficulty rating: 2020

Small Hint:

Call the substituted fraction $u$ and simplify $\frac{1+u}{1-u}$

Big Hint:

Its numerator and denominator factor as cubes

Solution:

For $u=\frac{3x+x^3}{1+3x^2},$ $\begin{aligned} \frac{1+u}{1-u} &=\frac{1+3x+3x^2+x^3} {1-3x+3x^2-x^3}\\ &=\left(\frac{1+x}{1-x}\right)^3. \end{aligned}$ Therefore $G=\log(\frac{1+x}{1-x})^3,$ so $G=3F.$

Thus, the correct answer is C.

31.

The number of solutions in positive integers of $2x+3y=763$ is:

$255$

$254$

$128$

$127$

$0$

Concepts:Diophantine Equation parity counting integers in a range

Difficulty rating: 1710

Small Hint:

Solve for $x=\frac{763-3y}{2}$

Big Hint:

Positive integral $x$ requires $y$ to be an odd positive integer below $\frac{763}{3}$

Solution:

We need $y$ odd so that $763-3y$ is even, and positivity gives $1\leq y\leq253.$ The odd values $1,$ $3,$ $\ldots,$ and $253$ number $\frac{253+1}{2}=127.$

Therefore, the correct answer is D.

32.

The dimensions of a rectangle $R$ are $a$ and $b,$ $a\lt b.$ It is required to obtain a rectangle with dimensions $x$ and $y,$ $x\lt a,$ $y\lt a,$ so that its perimeter is one-third that of $R,$ and its area is one-third that of $R.$ The number of such (different) rectangles is:

$0$

$1$

$2$

$4$

infinitely many

Concepts:rectangle inequality algebraic manipulation

Difficulty rating: 1990

Small Hint:

Translate the conditions into $3(x+y)=a+b$ and $3xy=ab$

Big Hint:

Divide the sum equation by the product equation and compare reciprocals using $x\lt a,$ $y\lt a,$ and $a\lt b$

Solution:

The conditions give $3(x+y)=a+b$ and $3xy=ab.$ Dividing yields $\frac1x+\frac1y=\frac1a+\frac1b.$ But $x\lt a$ and $y\lt a$ make the left side greater than $\frac{2}{a},$ while $a\lt b$ makes the right side less than $\frac{2}{a}.$ This is impossible.

Thus, the correct answer is A.

33.

Given the line $y=\dfrac34x+6$ and a line $L$ parallel to the given line and $4$ units from it. A possible equation for $L$ is:

$y=\dfrac34x+1$

$y=\dfrac34x$

$y=\dfrac34x-\dfrac23$

$y=\dfrac34x-1$

$y=\dfrac34x+2$

Concepts:parallel lines distance formula slope

Difficulty rating: 1750

Small Hint:

Write parallel lines as $3x-4y+c=0$

Big Hint:

The distance equals the absolute difference of constants divided by $5$

Solution:

The given line is $3x-4y+24=0.$ A parallel line $y=\frac{3x}{4}+b$ is $3x-4y+4b=0,$ so the distance is $\frac{|24-4b|}{5}.$ Setting this equal to $4$ gives $|6-b|=5,$ hence $b=1$ or $11.$ Choice A is possible.

Thus, the correct answer is A.

34.

In triangle $ABC,$ side $a=\sqrt3,$ side $b=\sqrt3,$ and side $c\gt3.$ Let $x$ be the largest number such that the magnitude, in degrees, of the angle opposite side $c$ exceeds $x.$ Then $x$ equals:

$150$

$120$

$105$

$90$

$60$

Concepts:law of cosines inequality bounding to limit cases

Difficulty rating: 1920

Small Hint:

Use the law of cosines for the angle $C$ opposite $c$

Big Hint:

Compare $c\gt3$ with the limiting case $c=3$

Solution:

The law of cosines gives $\cos C=\frac{3+3-c^2}{2\sqrt3\sqrt3}=\frac{6-c^2}{6}.$ Since $c\gt3,$ $\cos C\lt-\frac{1}{2},$ so $C\gt120^\circ.$ Values can approach $120^\circ$ as $c$ approaches $3$ from above, so the largest guaranteed bound is $120.$

Therefore, the correct answer is B.

35.

The lengths of the sides of a triangle are integers, and its area is also an integer. One side is $21$ and the perimeter is $48.$ The shortest side is:

$8$

$10$

$12$

$14$

$16$

Concepts:Heron’s Formula casework triangle inequality

Difficulty rating: 2100

Small Hint:

Write the other sides as $x$ and $27-x$ , with semiperimeter $24$

Big Hint:

Heron’s formula makes the squared area $72(24-x)(x-3)$

Solution:

Let the other sides be $x$ and $27-x,$ with $x\leq27-x.$ The semiperimeter is $24,$ so Heron’s formula gives $\begin{aligned} K^2 &=24\cdot3(24-x)(x-3)\\ &=72(24-x)(x-3). \end{aligned}$ The triangle inequalities give $4\leq x\leq13.$ Checking these integers, the expression is not a square for $x=4,\ldots,9,$ while at $x=10$ it is $72\cdot14\cdot7=7056=84^2.$ Thus the shortest side is $10.$

Therefore, the correct answer is B.

36.

A person starting with $64$ cents and making $6$ bets, wins three times and loses three times, the wins and losses occurring in random order. The chance for a win is equal to the chance for a loss. If each wager is for half the money remaining at the time of the bet, then the final result is:

a loss of $27$ ¢

a gain of $27$ ¢

a loss of $37$ ¢

neither a gain nor a loss

a gain or a loss depending upon the order in which the wins and losses occur

Concepts:exponent process simulation

Difficulty rating: 1470

Small Hint:

A win multiplies the current amount by $\frac{3}{2}$ , while a loss multiplies it by $\frac{1}{2}$

Big Hint:

Multiplication makes the order irrelevant

Solution:

After three wins and three losses, the amount, in cents, is $\begin{aligned} 64\left(\frac32\right)^3 \left(\frac12\right)^3 &=64\cdot\frac{27}{64}\\ &=27. \end{aligned}$ The loss is $64-27=37$ cents, regardless of order.

Thus, the correct answer is C.

37.

Given points $P_1,$ $P_2,$ $\ldots,$ $P_7$ on a straight line, in the order stated (not necessarily evenly spaced). Let $P$ be an arbitrarily selected point on the line and let $s$ be the sum of the undirected lengths

$PP_1,\ PP_2,\ \ldots,\ PP_7.$

Then $s$ is smallest if and only if the point $P$ is:

midway between $P_1$ and $P_7$

midway between $P_2$ and $P_6$

midway between $P_3$ and $P_5$

at $P_4$

at $P_1$

Concepts:optimization pairing and grouping

Difficulty rating: 1640

Small Hint:

Pair the distances to $P_1,P_7$ , then $P_2,P_6$ , then $P_3,P_5$

Big Hint:

Each paired sum is minimized throughout its intervening segment; the unpaired middle point selects one location

Solution:

For any pair $P_i,P_{8-i},$ the sum $PP_i+PP_{8-i}$ is minimized when $P$ lies between them. The three such intervals all contain $P_4.$ The remaining term $PP_4$ is uniquely minimized at $P=P_4.$

Thus, the correct answer is D.

38.

Point $F$ is taken on the extension of side $AD$ of parallelogram $ABCD.$ $BF$ intersects diagonal $AC$ at $E$ and side $DC$ at $G.$ If $EF=32$ and $GF=24,$ then $BE$ equals:

$4$

$8$

$10$

$12$

$16$

Concepts:parallelogram vector ratio and proportion

Difficulty rating: 2150

Small Hint:

Parameterize points on $FB$ by their fraction of the distance from $F$ to $B$

Big Hint:

If $F=tD$ in affine coordinates based at $A,$ the parameters of $G$ and $E$ are $\frac{t-1}{t}$ and $\frac{t}{t+1}$

Solution:

Use affine coordinates $A=0,$ $B=u,$ $D=v,$ $C=u+v,$ and $F=tv.$ A point on $FB$ is $su+t(1-s)v.$ Intersecting $DC$ gives $s_G=\frac{t-1}{t},$ while intersecting $AC$ gives $s_E=\frac{t}{t+1}.$ Therefore $\begin{aligned} \frac{GF}{EF} &=\frac{s_G}{s_E} =\frac{t^2-1}{t^2}\\ &=\frac{24}{32}=\frac34, \end{aligned}$ so $t=2$ and $s_E=\frac{2}{3}.$ Thus $EF=(\frac{2}{3})FB=32,$ making $FB=48$ and $BE=16.$

Therefore, the correct answer is E.

39.

In triangle $ABC$ lines $CE$ and $AD$ are drawn so that $\dfrac{CD}{DB}=\dfrac31$ and $\dfrac{AE}{EB}=\dfrac32.$

Let $r=\dfrac{CP}{PE},$ where $P$ is the intersection point of $CE$ and $AD.$ Then $r$ equals:

$3$

$\dfrac32$

$4$

$5$

$\dfrac52$

Concepts:mass points ratio and proportion

Difficulty rating: 2030

Small Hint:

Assign endpoint masses inversely proportional to the given side ratios

Big Hint:

Choose masses $m_A=2,$ $m_B=3,$ and $m_C=1$ , then find the mass at $E$

Solution:

The ratio $AE:EB=3:2$ is represented by masses $m_A=2$ and $m_B=3.$ The ratio $CD:DB=3:1$ then gives $m_C=1.$ Point $E$ has mass $m_A+m_B=5,$ so along cevian $CE,$ $\frac{CP}{PE}=\frac{m_E}{m_C}=\frac51=5.$

Thus, the correct answer is D.

40.

If $x$ is a number satisfying the equation $\sqrt[3]{x+9}-\sqrt[3]{x-9}=3,$ then $x^2$ is between:

$55$ and $65$

$65$ and $75$

$75$ and $85$

$85$ and $95$

$95$ and $105$

Concepts:radical sum and difference of cubes symmetry (algebra)

Difficulty rating: 2290

Small Hint:

Set $u=\sqrt[3]{x+9}$ and $v=\sqrt[3]{x-9}$

Big Hint:

Use both $u-v=3$ and $u^3-v^3=18$ to find $uv$

Solution:

Let $u=\sqrt[3]{x+9}$ and $v=\sqrt[3]{x-9}.$ Then $u-v=3$ and $\begin{aligned} 18&=u^3-v^3\\ &=(u-v)(u^2+uv+v^2), \end{aligned}$ so $u^2+uv+v^2=6.$ Comparing with $(u-v)^2=9$ gives $uv=-1.$ Hence $(u+v)^2=5.$ Cubing $2u=3\pm\sqrt5$ gives $x=\pm4\sqrt5,$ and either way $x^2=80,$ between $75$ and $85.$

Therefore, the correct answer is C.