1972 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

The lengths in inches of the three sides of each of four triangles $\mathrm{I},$ $\mathrm{II},$ $\mathrm{III},$ and $\mathrm{IV}$ are as follows: $\begin{aligned} \mathrm{I}\quad&3,4,5\\ \mathrm{II}\quad&4,7\frac12,8\frac12\\ \mathrm{III}\quad&7,24,25\\ \mathrm{IV}\quad&3\frac12,4\frac12,5\frac12 \end{aligned}$ Of these four given triangles, the only right triangles are:

$\mathrm{I}$ and $\mathrm{II}$

$\mathrm{I}$ and $\mathrm{III}$

$\mathrm{I}$ and $\mathrm{IV}$

$\mathrm{I},$ $\mathrm{II},$ and $\mathrm{III}$

$\mathrm{I},$ $\mathrm{II},$ and $\mathrm{IV}$

Concepts:Pythagorean Theorem right triangle systematic listing

Difficulty rating: 1180

Small Hint:

Compare the square of each longest side with the sum of the squares of the other two sides

Big Hint:

Clear the halves in triangles $\mathrm{II}$ and $\mathrm{IV}$ before comparing

Solution:

For triangles $\mathrm{I},$ $\mathrm{II},$ and $\mathrm{III},$ the relevant equalities are $\begin{aligned} 3^2+4^2&=5^2,\\ 4^2+\left(\frac{15}{2}\right)^2 &=\left(\frac{17}{2}\right)^2,\\ 7^2+24^2&=25^2. \end{aligned}$ For $\mathrm{IV},$ $\begin{aligned} \left(\frac72\right)^2+\left(\frac92\right)^2 &=\frac{130}{4},\\ \left(\frac{11}{2}\right)^2&=\frac{121}{4}. \end{aligned}$ These values are unequal. Thus precisely $\mathrm{I},$ $\mathrm{II},$ and $\mathrm{III}$ are right triangles.

Therefore, the correct answer is D.

2.

If a dealer could get his goods for $8\%$ less while keeping his selling price fixed, his profit, based on cost, would be increased to $(x+10)\%$ from his present profit of $x\%,$ which is:

$12\%$

$15\%$

$30\%$

$50\%$

$75\%$

Concepts:percentage money linear equation

Difficulty rating: 1380

Small Hint:

Let the present cost be $C$ and express the same selling price in two ways

Big Hint:

The reduced cost is $0.92C$ and its profit rate is $(x+10)\%$

Solution:

The fixed selling price gives $\begin{aligned} 1+0.01x&=0.92\\ &\quad+0.0092(x+10). \end{aligned}$ Simplifying yields $1+0.01x=1.012+0.0092x,$ so $0.0008x=0.012$ and $x=15.$

Therefore, the correct answer is B.

3.

If $x=\dfrac{1-i\sqrt3}{2},$ where $i=\sqrt{-1},$ then $\dfrac{1}{x^2-x}$ is equal to:

$-2$

$-1$

$1+i\sqrt3$

$1$

$2$

Concepts:complex number quadratic algebraic manipulation

Difficulty rating: 1690

Small Hint:

Compute $x^2-x$ before taking the reciprocal

Big Hint:

The number $x$ satisfies $x^2-x+1=0$

Solution:

Directly, $\begin{aligned} x^2-x &=\left(\frac{1-i\sqrt3}{2}\right)^2\\ &\quad-\frac{1-i\sqrt3}{2}\\ &=-1. \end{aligned}$ Its reciprocal is therefore also $-1.$

Therefore, the correct answer is B.

4.

The number of solutions to $\{1,2\}\subseteq X\subseteq\{1,2,3,4,5\},$ where $X$ is a subset of $\{1,2,3,4,5\},$ is:

$2$

$4$

$6$

$8$

None of these

Concepts:subsets multiplication principle

Difficulty rating: 1310

Small Hint:

The elements $1$ and $2$ are forced

Big Hint:

Each of $3,4,5$ may independently be included or omitted

Solution:

Every valid $X$ contains $1,2,$ while each of $3,4,5$ may be chosen independently. Hence there are $2^3=8$ possible sets.

Therefore, the correct answer is D.

5.

From among $2^{\frac{1}{2}},$ $3^{\frac{1}{3}},$ $8^{\frac{1}{8}},$ $9^{\frac{1}{9}},$ those which have the greatest and the next to the greatest values, in that order, are:

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

$3^{\frac{1}{3}},$ $8^{\frac{1}{8}}$

$3^{\frac{1}{3}},$ $9^{\frac{1}{9}}$

$8^{\frac{1}{8}},$ $9^{\frac{1}{9}}$

None of these

Concepts:exponent inequality algebraic manipulation

Difficulty rating: 1740

Small Hint:

Compare two roots by raising both positive numbers to a common power

Big Hint:

First compare $3^{\frac{1}{3}}$ with $2^{\frac{1}{2}}$ , then compare $2^{\frac{1}{2}}$ with each remaining number

Solution:

Raising to convenient common powers gives $(3^{\frac{1}{3}})^6=9>8=(2^{\frac{1}{2}})^6,$ $(2^{\frac{1}{2}})^8=16>8=(8^{\frac{1}{8}})^8,$ and $(2^{\frac{1}{2}})^{18}=512>81=(9^{\frac{1}{9}})^{18}.$ Thus $3^{\frac{1}{3}}$ is greatest and $2^{\frac{1}{2}}$ is next.

Therefore, the correct answer is A.

6.

If $3^{2x}+9=10(3^x),$ then the value of $x^2+1$ is:

$1$ only

$5$ only

$1$ or $5$

$2$

$10$

Concepts:exponent quadratic substitution factoring

Difficulty rating: 1620

Small Hint:

Set $u=3^x$

Big Hint:

Factor $u^2-10u+9$ and convert each positive root back to $x$

Solution:

Let $u=3^x.$ Then $\begin{aligned} u^2-10u+9&=(u-1)(u-9)\\ &=0. \end{aligned}$ Thus $x=0$ or $x=2,$ so $x^2+1$ is respectively $1$ or $5.$

Therefore, the correct answer is C.

7.

If $yz:zx:xy=1:2:3,$ then $\dfrac{x}{yz}:\dfrac{y}{zx}$ is equal to:

$3{:}2$

$1{:}2$

$1{:}4$

$2{:}1$

$4{:}1$

Concepts:ratio and proportion fraction algebraic manipulation

Difficulty rating: 1670

Small Hint:

Simplify the quotient of the two fractions in the requested ratio

Big Hint:

From $yz:zx=1:2$ , obtain $y:x=1:2$

Solution:

The requested ratio has quotient $\frac{\frac{x}{yz}}{\frac{y}{zx}}=\frac{x^2}{y^2}.$ Also $yz:zx=y:x=1:2,$ so $x:y=2:1.$ Therefore $x^2:y^2=4:1.$

Therefore, the correct answer is E.

8.

If $\lvert x-\log y\rvert=x+\log y,$ where $x$ and $\log y$ are real, then:

$x=0$

$y=1$

$x=0$ and $y=1$

$x(y-1)=0$

None of these

Concepts:absolute value logarithm casework

Difficulty rating: 1740

Small Hint:

Separate the cases $x-\log y\ge0$ and $x-\log y\lt0$

Big Hint:

One case forces $\log y=0$ , while the other forces $x=0$

Solution:

If $x-\log y\ge0,$ the equation gives $\log y=0,$ hence $y=1.$ If $x-\log y\lt0,$ it gives $x=0.$ Therefore every solution satisfies either $x=0$ or $y=1,$ which is expressed by $x(y-1)=0.$

Therefore, the correct answer is D.

9.

Ann and Sue bought identical boxes of stationery. Ann used hers to write $1$ -sheet letters and Sue used hers to write $3$ -sheet letters. Ann used all the envelopes and had $50$ sheets of paper left, while Sue used all of the sheets of paper and had $50$ envelopes left. The number of sheets of paper in each box was:

$150$

$125$

$120$

$100$

$80$

Concepts:system of equations linear equation

Difficulty rating: 1290

Small Hint:

Let $S$ and $E$ be the numbers of sheets and envelopes in a box

Big Hint:

Ann gives $S-E=50$ , while Sue gives $E-\frac{S}{3}=50$

Solution:

Let $S$ and $E$ denote sheets and envelopes per box. The two accounts give $S-E=50$ and $E-\frac S3=50.$ Adding yields $\frac{2S}{3}=100,$ so $S=150.$

Therefore, the correct answer is A.

10.

For $x$ real, the inequality $1\le\lvert x-2\rvert\le7$ is equivalent to:

$x\le1$ or $x\ge3$

$1\le x\le3$

$-5\le x\le9$

$-5\le x\le1$ or $3\le x\le9$

$-6\le x\le1$ or $3\le x\le10$

Concepts:absolute value inequality casework

Difficulty rating: 1400

Small Hint:

The upper bound places $x$ within $7$ units of $2$

Big Hint:

The lower bound removes the open interval of points less than $1$ unit from $2$

Solution:

The condition $\lvert x-2\rvert\le7$ gives $-5\le x\le9.$ The condition $\lvert x-2\rvert\ge1$ gives $x\le1$ or $x\ge3.$ Their intersection is $[-5,1]\cup[3,9].$

Therefore, the correct answer is D.

11.

The value(s) of $y$ for which the following pair of equations $\begin{aligned} x^2+y^2-16&=0,\\ x^2-3y+12&=0 \end{aligned}$ may have a real common solution, are:

$4$ only

$-7,$ $4$

$0,$ $4$

no $y$

all $y$

Concepts:system of equations quadratic substitution

Difficulty rating: 1740

Small Hint:

Eliminate $x^2$ between the equations

Big Hint:

After finding the two candidate $y$ -values, check whether the corresponding $x^2$ is nonnegative

Solution:

From the second equation, $x^2=3y-12.$ Substitution into the first gives $\begin{aligned} y^2+3y-28&=(y-4)(y+7)\\ &=0. \end{aligned}$ If $y=4,$ then $x=0.$ If $y=-7,$ then $x^2=-33,$ so no real $x$ exists. Thus only $y=4$ is possible.

Therefore, the correct answer is A.

12.

The number of cubic feet in the volume of a cube is the same as the number of square inches in its surface area. The length of the edge expressed as a number of feet is:

$6$

$864$

$1728$

$6\times1728$

$2304$

Concepts:cube geometry surface area volume unit conversion

Difficulty rating: 1790

Small Hint:

If the edge is $f$ feet, its length in inches is $12f$

Big Hint:

Equate the numerical values $f^3$ and $6(12f)^2$

Solution:

Let the edge length be $f$ feet, or $12f$ inches. The stated numerical equality is $f^3=6(12f)^2.$ Since $f\gt0,$ division by $f^2$ gives $f=6\cdot144=864.$

Therefore, the correct answer is B.

13.

Inside square $ABCD$ with sides of length $12$ inches, segment $AE$ is drawn, where $E$ is the point on $DC$ which is $5$ inches from $D.$ The perpendicular bisector of $AE$ is drawn and intersects $AE,$ $AD,$ and $BC$ at points $M,$ $P,$ and $Q,$ respectively. The ratio of segment $PM$ to $MQ$ is:

$5{:}12$

$5{:}13$

$5{:}19$

$1{:}4$

$5{:}21$

Concepts:square (geometry)midpoint perpendicular bisector similarity

Difficulty rating: 1950

Small Hint:

Draw through $M$ a line parallel to $AB$

Big Hint:

The horizontal distances from $M$ to $AD$ and $BC$ are $\frac{5}{2}$ and $\frac{19}{2}$

Solution:

Because $M$ is the midpoint of $AE,$ its horizontal distance from $AD$ is $\frac{5}{2}.$ Its distance from $BC$ is $12-\frac52=\frac{19}{2}.$ The two right triangles cut from line $PQ$ by the horizontal through $M$ are similar, so their hypotenuses are in the ratio of these horizontal legs: $PM:MQ=\frac52:\frac{19}{2}=5:19.$

Therefore, the correct answer is C.

14.

A triangle has angles of $30^\circ$ and $45^\circ.$ If the side opposite the $45^\circ$ angle has length $8,$ then the side opposite the $30^\circ$ angle has length:

$4$

$4\sqrt2$

$4\sqrt3$

$4\sqrt6$

$6$

Concepts:law of sines trigonometry

Difficulty rating: 1530

Small Hint:

Relate the two sides with the Law of Sines

Big Hint:

Use $\sin30^\circ=\frac{1}{2}$ and $\sin45^\circ=\frac{\sqrt2}{2}$

Solution:

If $s$ is the side opposite $30^\circ,$ the Law of Sines gives $\frac{s}{\sin30^\circ}=\frac8{\sin45^\circ}.$ Therefore $s=8\frac{\frac{1}{2}}{\frac{\sqrt2}{2}}=4\sqrt2.$

Therefore, the correct answer is B.

15.

A contractor estimated that one of his two bricklayers would take $9$ hours to build a certain wall and the other $10$ hours. However, he knew from experience that when they worked together, their combined output fell by $10$ bricks per hour. Being in a hurry, he put both men on the job and found that it took exactly $5$ hours to build the wall. The number of bricks in the wall was:

$500$

$550$

$900$

$950$

$960$

Concepts:rate linear equation

Difficulty rating: 1790

Small Hint:

Let $N$ be the number of bricks in the wall

Big Hint:

Their actual combined hourly rate is $\frac{N}{9}+\frac{N}{10}-10$

Solution:

If the wall contains $N$ bricks, the individual rates are $\frac{N}{9}$ and $\frac{N}{10}.$ Thus $5\left(\frac N9+\frac N{10}-10\right)=N.$ Multiplying by $18$ and simplifying gives $19N-900=18N,$ hence $N=900.$

Therefore, the correct answer is C.

16.

There are two positive numbers that may be inserted between $3$ and $9$ such that the first three are in geometric progression while the last three are in arithmetic progression. The sum of those two positive numbers is:

$13\frac12$

$11\frac14$

$10\frac12$

$10$

$9\frac12$

Concepts:geometric sequence arithmetic sequence system of equations quadratic

Difficulty rating: 1930

Small Hint:

Call the inserted numbers $x$ and $y,$ in that order

Big Hint:

Use $x^2=3y$ and $2y=x+9$

Solution:

The progression conditions give $x^2=3y$ and $2y=x+9.$ Eliminating $y$ yields $\begin{aligned} 2x^2-3x-27&=(2x-9)(x+3)\\ &=0. \end{aligned}$ Positivity gives $x=\frac{9}{2},$ and then $y=\frac{27}{4}.$ Their sum is $\frac92+\frac{27}{4}=\frac{45}{4}=11\frac14.$

Therefore, the correct answer is B.

17.

A piece of string is cut in two at a point selected at random. The probability that the longer piece is at least $x$ times as large as the shorter piece is:

$\dfrac12$

$\dfrac2x$

$\dfrac1{x+1}$

$\dfrac1x$

$\dfrac2{x+1}$

Concepts:geometric probability ratio and proportion inequality

Difficulty rating: 1740

Small Hint:

Normalize the string length to $1$ and let the cut be $t$ units from one end

Big Hint:

Near either endpoint, solve $1-t\ge xt$

Solution:

Take the string to have length $1.$ Near one endpoint, the condition is $1-t\ge xt,$ so $t\le\frac{1}{x+1}.$ The same interval occurs at the other endpoint. Their total length, hence the probability, is $\frac2{x+1}.$

Therefore, the correct answer is E.

18.

Let $ABCD$ be a trapezoid with the measure of base $AB$ twice that of base $DC,$ and let $E$ be the point of intersection of the diagonals. If the measure of diagonal $AC$ is $11,$ then that of segment $EC$ is equal to:

$3\frac23$

$3\frac34$

$4$

$3\frac12$

$3$

Concepts:trapezoid similarity ratio and proportion

Difficulty rating: 1500

Small Hint:

Triangles $ABE$ and $CDE$ are similar

Big Hint:

The base ratio $AB:DC=2:1$ also equals $AE:EC$

Solution:

Since $AB\parallel DC,$ triangles $ABE$ and $CDE$ are similar. Hence $AE:EC=AB:DC=2:1.$ Thus $AC=AE+EC=3EC=11,$ so $EC=\frac{11}{3}=3\frac23.$

Therefore, the correct answer is A.

19.

The sum of the first $n$ terms of the sequence $\begin{gathered} 1,\ (1+2),\ (1+2+2^2),\\ \ldots,\ (1+2+2^2+\cdots+2^{n-1}) \end{gathered}$ in terms of $n$ is:

$2^n$

$2^n-n$

$2^{n+1}-n$

$2^{n+1}-n-2$

$n\cdot2^n$

Concepts:geometric sequence summation algebraic manipulation

Difficulty rating: 1790

Small Hint:

The $k$ -th parenthesized sum is $2^k-1$

Big Hint:

Sum $2^1-1,$ $2^2-1,$ $\ldots,$ and $2^n-1$

Solution:

The $k$ -th term is the geometric sum $2^k-1.$ Therefore the desired total is $\begin{aligned} \sum_{k=1}^n(2^k-1) &=(2^{n+1}-2)-n\\ &=2^{n+1}-n-2. \end{aligned}$

Therefore, the correct answer is D.

20.

If $\tan x=\dfrac{2ab}{a^2-b^2},$ where $a\gt b\gt0$ and $0^\circ\lt x\lt90^\circ,$ then $\sin x$ is equal to:

$\dfrac ab$

$\dfrac ba$

$\dfrac{\sqrt{a^2-b^2}}{2a}$

$\dfrac{\sqrt{a^2-b^2}}{2ab}$

$\dfrac{2ab}{a^2+b^2}$

Concepts:trigonometric identity right triangle Pythagorean Theorem algebraic manipulation

Difficulty rating: 1850

Small Hint:

Model the tangent as the ratio of legs $2ab$ and $a^2-b^2$

Big Hint:

The corresponding hypotenuse simplifies because $(2ab)^2+(a^2-b^2)^2=(a^2+b^2)^2$

Solution:

Use a right triangle with opposite leg $2ab$ and adjacent leg $a^2-b^2.$ Its hypotenuse is $\begin{aligned} &\sqrt{(2ab)^2+(a^2-b^2)^2}\\ &\qquad=\sqrt{(a^2+b^2)^2}\\ &\qquad=a^2+b^2. \end{aligned}$ Hence $\sin x=\frac{2ab}{a^2+b^2}.$

Therefore, the correct answer is E.

21.

If the sum of the measures in degrees of angles $A,$ $B,$ $C,$ $D,$ $E,$ and $F$ in the figure is $90n,$ then $n$ is equal to:

$2$

$3$

$4$

$5$

$6$

Concepts:angle sum angle chasing

Difficulty rating: 1910

Small Hint:

Name the two intersections of $AD$ with $BF$ and $CE$

Big Hint:

Combine the angle sums of the central quadrilateral and the two triangles attached to it

Solution:

Let $P=AD\cap BF$ and $Q=AD\cap CE.$ In quadrilateral $EFPQ,$ $E+F+\angle P+\angle Q=360^\circ.$ The triangles $BPD$ and $AQC$ give $B+D=\angle P$ and $A+C=\angle Q,$ using the supplementary angles at $P$ and $Q.$ Adding yields $\begin{gathered} A+B+C+D+E+F\\ =360^\circ=90^\circ\cdot4. \end{gathered}$ Thus $n=4.$

Therefore, the correct answer is C.

22.

If $a\pm bi$ $(b\ne0)$ are imaginary roots of the equation $x^3+qx+r=0,$ where $a,$ $b,$ $q,$ and $r$ are real numbers, then $q$ in terms of $a$ and $b$ is:

$a^2+b^2$

$2a^2-b^2$

$b^2-a^2$

$b^2-2a^2$

$b^2-3a^2$

Concepts:complex number polynomial Vieta’s Formulas

Difficulty rating: 2080

Small Hint:

The conjugate $a-bi$ is another root

Big Hint:

Because the $x^2$ coefficient is zero, determine the third root and then use the sum of pairwise products

Solution:

The conjugate root is $a-bi.$ Since the sum of the three roots is $0,$ the third root is $-2a.$ By Vieta’s formulas, $\begin{aligned} q&=(a+bi)(a-bi)\\ &\quad-2a(a+bi)\\ &\quad-2a(a-bi)\\ &=a^2+b^2-4a^2\\ &=b^2-3a^2. \end{aligned}$

Therefore, the correct answer is E.

23.

The radius of the smallest circle containing the symmetric figure composed of $3$ unit squares shown is:

$\sqrt2$

$\sqrt{1.25}$

$1.25$

$\dfrac{5\sqrt{17}}{16}$

None of these

Concepts:circle coordinate geometry symmetry optimization

Difficulty rating: 2380

Small Hint:

By symmetry, place the circle’s center on the vertical axis of the figure

Big Hint:

Equate its distances to a lower outer corner and an upper outer corner

Solution:

Put the midpoint of the bottom edge at $O=(0,0).$ A lower outer corner is $A=(1,0),$ and an upper outer corner is $B=(\frac{1}{2},2).$ By symmetry the center of the smallest enclosing circle is $P=(0,k).$ At the optimum both types of outer corner lie on the circle, so $1+k^2=\left(\frac12\right)^2+(2-k)^2.$ This gives $k=\frac{13}{16}.$ Hence $\begin{aligned} r^2&=1+\left(\frac{13}{16}\right)^2 =\frac{425}{256},\\ r&=\frac{5\sqrt{17}}{16}. \end{aligned}$

Therefore, the correct answer is D.

24.

A man walked a certain distance at a constant rate. If he had gone $\frac12$ mile per hour faster, he would have walked the distance in four-fifths of the time; if he had gone $\frac12$ mile per hour slower, he would have been $2\frac12$ hours longer on the road. The distance in miles he walked was:

$13\frac12$

$15$

$17\frac12$

$20$

$25$

Concepts:distance rate and time system of equations

Difficulty rating: 1930

Small Hint:

Let the actual speed and time be $v$ and $t$

Big Hint:

First use $vt=(v+\frac{1}{2})(\frac{4t}{5})$ to determine $v$

Solution:

Let the actual speed and time be $v$ and $t.$ The faster case gives $vt=\left(v+\frac12\right)\frac{4t}{5},$ so $v=2.$ The slower case then gives $2t=\frac32\left(t+\frac52\right),$ whence $t=\frac{15}{2}.$ The distance was $vt=2\cdot\frac{15}{2}=15.$

Therefore, the correct answer is B.

25.

Inscribed in a circle is a quadrilateral having sides of lengths $25,$ $39,$ $52,$ and $60$ taken consecutively. The diameter of this circle has length:

$62$

$63$

$65$

$66$

$69$

Concepts:cyclic quadrilateral Pythagorean Triple right triangle

Difficulty rating: 2140

Small Hint:

Notice two scaled Pythagorean triples among the four side lengths

Big Hint:

The $25,60$ pair and the $39,52$ pair share the same possible hypotenuse

Solution:

The consecutive sides may be labeled $\begin{gathered} AB=25,\\ BC=39,\\ CD=52,\\ DA=60. \end{gathered}$ Since $25^2+60^2=65^2=39^2+52^2,$ triangles $ABD$ and $BCD$ are right triangles with common hypotenuse $BD=65.$ Their union is the cyclic quadrilateral, and a right triangle’s hypotenuse is a diameter of its circumcircle. Thus the diameter is $65.$

Therefore, the correct answer is C.

26.

In the circle shown, $M$ is the midpoint of arc $\overset{\frown}{CAB},$ and segment $MP$ is perpendicular to chord $AB$ at $P.$ If the measure of chord $AC$ is $x$ and that of segment $AP$ is $x+1,$ then segment $PB$ has measure equal to:

$3x+2$

$3x+1$

$2x+3$

$2x+2$

$2x+1$

Concepts:arc chord rectangle symmetry

Difficulty rating: 2230

Small Hint:

Copy arc $\overset{\frown}{CA}$ to an arc $\overset{\frown}{MN}$ ending at a new point $N$ between $M$ and $B$

Big Hint:

Drop $NQ\perp AB$ ; use equal arcs to identify a rectangle and equal horizontal end segments

Solution:

Choose $N$ on arc $\overset{\frown}{MB}$ so that arcs $\overset{\frown}{MN}$ and $\overset{\frown}{CA}$ are equal, and drop $NQ\perp AB.$ Then $MN=AC=x.$ The remaining equal arcs $\overset{\frown}{AM}$ and $\overset{\frown}{NB}$ give equal corresponding chord projections, so $QB=AP=x+1.$ Also $MNPQ$ is a rectangle, hence $PQ=MN=x.$ Therefore $\begin{aligned} PB&=PQ+QB\\ &=x+(x+1)\\ &=2x+1. \end{aligned}$

Therefore, the correct answer is E.

27.

If the area of $\triangle ABC$ is $64$ square units and the geometric mean (mean proportional) between sides $AB$ and $AC$ is $12$ inches, then $\sin A$ is equal to:

$\dfrac{\sqrt3}{2}$

$\dfrac35$

$\dfrac45$

$\dfrac89$

$\dfrac{15}{17}$

Concepts:triangle area trigonometry algebraic manipulation

Difficulty rating: 1470

Small Hint:

The geometric-mean condition determines the product $AB\cdot AC$

Big Hint:

Use $[\triangle ABC]=\frac12(AB)(AC)\sin A$

Solution:

The geometric-mean condition says $AB\cdot AC=12^2=144.$ Therefore $\begin{aligned} 64&=\frac12(AB)(AC)\sin A\\ &=72\sin A, \end{aligned}$ so $\sin A=\frac{8}{9}.$

Therefore, the correct answer is D.

28.

A circular disc with diameter $D$ is placed on an $8\times8$ checkerboard with width $D$ so that the centers coincide. The number of checkerboard squares which are completely covered by the disc is:

$48$

$44$

$40$

$36$

$32$

Concepts:circle distance formula basic counting

Difficulty rating: 2080

Small Hint:

No square touching the outside border can be completely covered

Big Hint:

Among the $6\times6$ interior squares, test the four corner squares separately

Solution:

The $28$ border squares are not fully covered. Consider the remaining $6\times6$ interior grid and measure distances in square side lengths, so the disc has radius $4.$ The four outer corners of this interior grid are at distance $\sqrt{3^2+3^2}=3\sqrt2\gt4$ from the center, so those four corner squares are not fully covered. Every other interior square lies within the disc: its farthest possible corner is at distance at most $\sqrt{3^2+2^2}=\sqrt{13}\lt4.$ Hence $36-4=32$ squares are completely covered.

Therefore, the correct answer is E.

29.

If $f(x)=\log\left(\dfrac{1+x}{1-x}\right)$ for $-1\lt x\lt1,$ then $f\left(\dfrac{3x+x^3}{1+3x^2}\right)$ in terms of $f(x)$ is:

$-f(x)$

$2f(x)$

$3f(x)$

$[f(x)]^2$

$[f(x)]^3-f(x)$

Concepts:logarithm function factoring substitution

Difficulty rating: 1860

Small Hint:

Substitute the rational expression into $\frac{1+u}{1-u}$

Big Hint:

Its numerator and denominator factor as $(1+x)^3$ and $(1-x)^3$

Solution:

Let $u=\frac{3x+x^3}{1+3x^2}.$ Then $\begin{aligned} \frac{1+u}{1-u} &=\frac{1+3x^2+3x+x^3} {1+3x^2-3x-x^3}\\ &=\left(\frac{1+x}{1-x}\right)^3. \end{aligned}$ Consequently $f(u)=3\log\left(\frac{1+x}{1-x}\right)=3f(x).$

Therefore, the correct answer is C.

30.

A rectangular piece of paper $6$ inches wide is folded as in the diagram so that one corner touches the opposite side. The length in inches of the crease $L$ in terms of angle $\theta$ is:

$3\sec^2\theta\csc\theta$

$6\sin\theta\sec\theta$

$3\sec\theta\csc\theta$

$6\sec\theta\csc^2\theta$

None of these

Concepts:paper folding trigonometry right triangle

Difficulty rating: 2270

Small Hint:

Let $h$ be the height of the rectangular sheet and use the equal lengths created by the fold

Big Hint:

The diagram gives $\frac{6}{h}=\sin(2\theta)$ and $\frac{L}{h}=\sec\theta$

Solution:

Let $h$ be the sheet’s height. The congruent right triangles created by reflecting the folded corner across the crease give $\frac6h=\sin(2\theta)=2\sin\theta\cos\theta,$ so $h=\frac{3}{\sin\theta\cos\theta}.$ They also give $\frac{L}{h}=\sec\theta.$ Hence $\begin{aligned} L&=h\sec\theta\\ &=\frac{3\sec\theta} {\sin\theta\cos\theta}\\ &=3\sec^2\theta\csc\theta. \end{aligned}$

Therefore, the correct answer is A.

31.

When the number $2^{1000}$ is divided by $13,$ the remainder in the division is:

$1$

$2$

$3$

$7$

$11$

Concepts:modular arithmetic modular exponentiation

Difficulty rating: 1740

Small Hint:

Use $2^6\equiv-1\pmod{13}$

Big Hint:

Write $1000=6\cdot166+4$

Solution:

Since $2^6=64\equiv-1\pmod{13},$ $\begin{aligned} 2^{1000}&=2^{6\cdot166+4}\\ &\equiv(-1)^{166}2^4\\ &\equiv16\\ &\equiv3\pmod{13}. \end{aligned}$

Therefore, the correct answer is C.

32.

Chords $AB$ and $CD$ in the circle shown intersect at $E$ and are perpendicular to each other. If segments $AE,$ $EB,$ and $ED$ have measures $2,$ $6,$ and $3$ respectively, then the length of the diameter of the circle is:

$4\sqrt5$

$\sqrt{65}$

$2\sqrt{17}$

$3\sqrt7$

$6\sqrt2$

Concepts:power of a point coordinate geometry distance formula circle

Difficulty rating: 1890

Small Hint:

First use $AE\cdot EB=CE\cdot ED$

Big Hint:

Place $E$ at the origin; the center is at the intersection of the two chord perpendicular bisectors

Solution:

The intersecting-chords theorem gives $2\cdot6=CE\cdot3,$ so $CE=4.$ Put $E=(0,0),$ $A=(-2,0),$ $B=(6,0),$ $C=(0,4),$ and $D=(0,-3).$ The perpendicular bisectors of $AB$ and $CD$ meet at $O=\left(2,\frac12\right).$ Thus $r^2=OA^2=4^2+\left(\frac12\right)^2=\frac{65}{4},$ and the diameter is $2r=\sqrt{65}.$

Therefore, the correct answer is B.

33.

The minimum value of the quotient of a (base ten) number of three different nonzero digits divided by the sum of its digits is:

$9.7$

$10.1$

$10.5$

$10.9$

$20.5$

Concepts:digits optimization inequality

Difficulty rating: 2140

Small Hint:

Let the hundreds, tens, and units digits be $H,$ $T,$ and $U,$ and consider which position should contain the largest digit

Big Hint:

At a minimum $U=9$ ; then maximize $T$ and minimize $H$ subject to distinct nonzero digits

Solution:

Let $Q$ denote the quotient. Then $Q=1+\frac{99H+9T}{H+T+U}.$ Interchanging $U$ with a larger digit in either earlier position decreases the quotient, so the units digit must be the largest. Increasing that units digit lowers a quotient greater than $1,$ so $U=9.$ Then $\frac{T+11H}{T+H+9} =1+\frac{10H-9}{T+H+9}.$ This is minimized by taking the largest available $T=8$ and then the smallest $H=1.$ The number is $189,$ and $\frac{189}{1+8+9}=\frac{189}{18}=10.5.$

Therefore, the correct answer is C.

34.

Three times Dick’s age plus Tom’s age equals twice Harry’s age. Double the cube of Harry’s age is equal to three times the cube of Dick’s age added to the cube of Tom’s age. Their respective ages are relatively prime to each other. The sum of the squares of their ages is:

$42$

$46$

$122$

$290$

$326$

Concepts:ages sum and difference of cubes Diophantine Equation greatest common divisor

Difficulty rating: 2270

Small Hint:

Let the ages be $D,$ $T,$ and $H,$ and rewrite the linear equation as $2(H-D)=D+T$

Big Hint:

Factor both differences of cubes after rewriting the cubic equation

Solution:

The equations are $\begin{aligned} 3D+T&=2H,\\ 2H^3&=3D^3+T^3. \end{aligned}$ Rewrite them as $2(H-D)=D+T$ and $\begin{gathered} 2(H-D)(H^2+HD+D^2)\\ =(D+T)\\ \qquad\cdot(D^2-DT+T^2). \end{gathered}$ Canceling the equal positive factors gives $\begin{aligned} H^2+HD+DT-T^2&=0,\\ (H+T)(H+D-T)&=0, \end{aligned}$ so $T=H+D.$ The linear equation then gives $H=4D,$ and pairwise relative primality forces $(D,H,T)=(1,4,5).$ The requested sum is $1^2+4^2+5^2=42.$

Therefore, the correct answer is A.

35.

Equilateral triangle $ABP$ with side $AB$ of length $2$ inches is placed inside square $AXYZ$ with side of length $4$ inches so that $B$ is on side $AX.$ The triangle is rotated clockwise about $B,$ then $P,$ and so on along the sides of the square until $P,$ $A,$ and $B$ all return to their original positions. The length of the path in inches traversed by vertex $P$ is equal to:

$\dfrac{20\pi}{3}$

$\dfrac{32\pi}{3}$

$12\pi$

$\dfrac{40\pi}{3}$

$15\pi$

Concepts:equilateral triangle transformation arc process simulation

Difficulty rating: 2450

Small Hint:

Track the orientation after one eight-pivot circuit of the square

Big Hint:

Three circuits are needed; classify the pivots according to whether $P$ is fixed or moves through a $120^\circ$ or $30^\circ$ arc

Solution:

One circuit of the square uses $8$ pivots and changes the triangle’s orientation by $\frac{2}{3}$ of a full turn. Therefore $3$ circuits, or $24$ pivots, are required to restore the entire triangle. In $8$ of those pivots the rotation is about $P,$ so $P$ does not move. In the other $16,$ eight arcs subtend $120^\circ$ and eight subtend $30^\circ,$ all with radius $2.$ Hence the total path length is $\begin{aligned} &8\left(\frac13\cdot4\pi\right) +8\left(\frac1{12}\cdot4\pi\right)\\ &\qquad=\frac{32\pi}{3}+\frac{8\pi}{3}\\ &\qquad=\frac{40\pi}{3}. \end{aligned}$

Therefore, the correct answer is D.