1973 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

A chord which is the perpendicular bisector of a radius of length $12$ in a circle has length

$3\sqrt3$

$27$

$6\sqrt3$

$12\sqrt3$

none of these

Concepts:circle chord Pythagorean Theorem

Difficulty rating: 1330

Small Hint:

The chord meets the radius halfway from the center and at a right angle

Big Hint:

Find half the chord with the Pythagorean theorem, then double it

Solution:

Let $M$ be the midpoint of the chord and also the midpoint of the radius. The distance from the center $O$ to $M$ is $6,$ while the circle’s radius is $12.$ If $A$ is one endpoint of the chord, then $AM=\sqrt{12^2-6^2}=6\sqrt3.$ The full chord has length $2AM=12\sqrt3.$

Therefore, the correct answer is D.

2.

One thousand unit cubes are fastened together to form a large cube with edge length $10$ units; this is painted and then separated into the original cubes. The number of these unit cubes which have at least one face painted is

$600$

$520$

$488$

$480$

$400$

Concepts:cube geometry complementary counting

Difficulty rating: 1230

Small Hint:

Count the cubes with no painted face instead

Big Hint:

Removing the outer layer leaves an $8\times8\times8$ cube

Solution:

The unpainted cubes are precisely the interior $8\times8\times8$ cube. Therefore the number having at least one painted face is $10^3-8^3=1000-512=488.$

Therefore, the correct answer is C.

3.

The stronger Goldbach conjecture states that any even integer greater than $7$ can be written as the sum of two different prime numbers. For such representations of the even number $126,$ the largest possible difference between the two primes is

Note: The regular Goldbach conjecture states that any even integer greater than $3$ is expressible as a sum of two primes. Neither this conjecture nor the stronger version has been settled.

$112$

$100$

$92$

$88$

$80$

Concepts:prime optimization systematic listing

Difficulty rating: 1360

Small Hint:

For a fixed sum, maximize the difference by minimizing the smaller prime

Big Hint:

Test the primes $3,$ $5,$ $7,$ $11,$ $13,$ $\ldots$ until the complement to $126$ is prime

Solution:

For primes $p\lt q$ with $p+q=126,$ the difference $q-p=126-2p$ is largest when $p$ is as small as possible. The complements of $3,$ $5,$ $7,$ and $11$ are $123,$ $121,$ $119,$ and $115,$ none prime. For $p=13,$ the complement is $113,$ which is prime. Thus the largest difference is $113-13=100.$

Therefore, the correct answer is B.

4.

Two congruent $30^\circ$ - $60^\circ$ - $90^\circ$ triangles are placed so that they overlap partly and their hypotenuses coincide. If the hypotenuse of each triangle is $12,$ the area common to both triangles is

$6\sqrt3$

$8\sqrt3$

$9\sqrt3$

$12\sqrt3$

$24$

Concepts:special right triangle coordinate geometry triangle area

Difficulty rating: 1670

Small Hint:

Place the common hypotenuse on the $x$ -axis and reverse which endpoint has the $30^\circ$ angle

Big Hint:

The upper boundary of the overlap consists of two lines making $30^\circ$ angles with the hypotenuse

Solution:

Put the common hypotenuse from $A=(0,0)$ to $B=(12,0).$ The two triangles have their third vertices at $(9,3\sqrt3)$ and $(3,3\sqrt3).$ Their common region is the triangle with base $AB$ and apex where $y=\frac{x}{\sqrt3}$ and $y=\frac{12-x}{\sqrt3}$ meet. This occurs at $x=6,$ with height $2\sqrt3.$ Hence the common area is $\frac12(12)(2\sqrt3)=12\sqrt3.$

Therefore, the correct answer is D.

5.

Of the following five statements, $\mathrm{I}$ to $\mathrm{V},$ about the binary operation of averaging (arithmetic mean),

$\mathrm{I}.$ Averaging is associative

$\mathrm{II}.$ Averaging is commutative

$\mathrm{III}.$ Averaging distributes over addition

$\mathrm{IV}.$ Addition distributes over averaging

$\mathrm{V}.$ Averaging has an identity element

those which are always true are

All

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

$\mathrm{II}$ and $\mathrm{III}$ only

$\mathrm{II}$ and $\mathrm{IV}$ only

$\mathrm{II}$ and $\mathrm{V}$ only

Concepts:custom operation distributive property counterexample

Difficulty rating: 1750

Small Hint:

Write the averaging operation as $a*b=\frac{a+b}{2}$

Big Hint:

Expand each proposed identity; one counterexample is enough to reject a property

Solution:

Let $a*b=\frac{a+b}{2}.$ Commutativity is immediate. Also $\begin{aligned} a+(b*c) &=\frac{2a+b+c}{2}\\ &=(a+b)*(a+c). \end{aligned}$ so addition distributes over averaging.

Associativity fails because $\begin{aligned} (a*b)*c&=\frac{a+b+2c}{4},\\ a*(b*c)&=\frac{2a+b+c}{4}. \end{aligned}$ Averaging does not distribute over addition since generally $a*(b+c)\ne(a*b)+(a*c).$ Finally, an identity $e$ would require $\frac{e+a}{2}=a$ for every $a,$ or $e=a,$ which is impossible for a fixed $e.$ Thus only $\mathrm{II}$ and $\mathrm{IV}$ always hold.

Therefore, the correct answer is D.

6.

If $554$ is the base $b$ representation of the square of the number whose base $b$ representation is $24,$ then $b,$ when written in base $10,$ equals

$6$

$8$

$12$

$14$

$16$

Concepts:number base quadratic place value

Difficulty rating: 1670

Small Hint:

Translate $24_b$ and $554_b$ into base-ten polynomial expressions

Big Hint:

After solving the resulting quadratic, enforce that a base containing digit $5$ must exceed $5$

Solution:

In base ten, $\begin{aligned} 24_b&=2b+4,\\ 554_b&=5b^2+5b+4. \end{aligned}$ Therefore $\begin{aligned} (2b+4)^2&=5b^2+5b+4,\\ b^2-11b-12&=0,\\ (b-12)(b+1)&=0. \end{aligned}$ A base is positive and must exceed $5,$ so $b=12.$

Therefore, the correct answer is C.

7.

The sum of all the integers between $50$ and $350$ which end in $1$ is

$5880$

$5539$

$5208$

$4877$

$4566$

Concepts:arithmetic sequence summation

Difficulty rating: 1180

Small Hint:

The integers form an arithmetic sequence from $51$ through $341$

Big Hint:

Use the common difference to count the terms, then pair the first and last

Solution:

The sequence is $51,61,\ldots,341.$ It has $\frac{341-51}{10}+1=30$ terms. Its sum is therefore $\begin{aligned} \frac{30(51+341)}2 &=15\cdot392\\ &=5880. \end{aligned}$

Therefore, the correct answer is A.

8.

If $1$ pint of paint is needed to paint a statue $6$ ft. high, then the number of pints it will take to paint (to the same thickness) $540$ statues similar to the original but only $1$ ft. high is

$90$

$72$

$45$

$30$

$15$

Difficulty rating: 1250

Small Hint:

Paint required scales with surface area, not volume

Big Hint:

Reducing every length by a factor of $6$ reduces the paint per statue by the square of that factor

Solution:

A $1$ -ft. statue has linear scale $\frac{1}{6}$ relative to the original, so its surface area and paint requirement are $\frac{1}{36}$ as large. The $540$ small statues therefore require $540\cdot\frac1{36}=15$ pints.

Therefore, the correct answer is E.

9.

In $\triangle ABC$ with right angle at $C,$ altitude $CH$ and median $CM$ trisect the right angle. If the area of $\triangle CHM$ is $K,$ then the area of $\triangle ABC$ is

$6K$

$4\sqrt3K$

$3\sqrt3K$

$3K$

$4K$

Concepts:right triangle median (geometry)congruence (geometry)area ratio

Difficulty rating: 1870

Small Hint:

Compare the right triangles $CHM$ and $CHB$ using the trisection angles

Big Hint:

After locating $M$ and $H$ on $AB,$ compare $HM$ with the whole base $AB$

Solution:

Since $CH\perp AB,$ both $\triangle CHM$ and $\triangle CHB$ are right at $H.$ The trisection gives equal acute angles at $C,$ and the triangles share side $CH,$ so they are congruent. Hence $HM=HB.$

Because $M$ is the midpoint of $AB,$ $AM=MB.$ Along the hypotenuse the order is $A,M,H,B,$ and $MB=MH+HB=2MH.$ Thus $AB=2MB=4MH.$ Triangles $ABC$ and $CHM$ have bases on the same line and the same altitude from $C,$ so their areas are in the ratio $AB:HM=4:1.$ Therefore the area of $\triangle ABC$ is $4K.$

Therefore, the correct answer is E.

10.

If $n$ is a real number, then the simultaneous system $\begin{cases} nx+y=1,\\ ny+z=1,\\ x+nz=1 \end{cases}$ has no solution if and only if $n$ is equal to

$-1$

$0$

$1$

$0$ or $1$

$\frac12$

Concepts:system of equations linear equation substitution

Difficulty rating: 1690

Small Hint:

Add all three equations and inspect the coefficient of $x+y+z$

Big Hint:

For every remaining value of $n,$ try the symmetric choice $x=y=z$

Solution:

Adding the equations gives $(n+1)(x+y+z)=3.$ When $n=-1,$ this becomes $0=3,$ so no solution exists. If $n\ne-1,$ the symmetric assignment $x=y=z=\frac1{n+1}$ satisfies all three equations. Hence the system has no solution exactly when $n=-1.$

Therefore, the correct answer is A.

11.

A circle with a circumscribed and an inscribed square centered at the origin $O$ of a rectangular coordinate system with positive $x$ - and $y$ -axes $OX$ and $OY$ is shown in each figure $\mathrm{I}$ to $\mathrm{IV}$ below.

The inequalities $\begin{aligned} |x|+|y| &\le \sqrt{2(x^2+y^2)}\\ &\le 2\operatorname{Max}(|x|,|y|). \end{aligned}$ are represented geometrically by the figure numbered

Geometric representation: An inequality $f(x,y)\le g(x,y)$ for all $x,$ $y$ is represented by a figure showing, for a typical real number $a,$ the containment $\begin{aligned} &\{(x,y):g(x,y)\le a\}\\ &\quad\subset \{(x,y):f(x,y)\le a\}. \end{aligned}$

$\mathrm{I}$

$\mathrm{II}$

$\mathrm{III}$

$\mathrm{IV}$

none of these

Concepts:inequality coordinate geometry circle square (geometry)

Difficulty rating: 2190

Small Hint:

Identify the shapes of the sublevel sets for $|x|+|y|,$ $\sqrt{2(x^2+y^2)},$ and $2\operatorname{Max}(|x|,|y|)$

Big Hint:

The inequality reverses the order of containment of the corresponding sublevel sets

Solution:

For a fixed positive $a,$ the set $2\operatorname{Max}(|x|,|y|)\le a$ is an axis-aligned square. The set $\sqrt{2(x^2+y^2)}\le a$ is its circumscribed circle, and $|x|+|y|\le a$ is the diamond circumscribed about that circle. Thus the required nesting is an inner axis-aligned square, then a circle, then an outer diamond, which is figure $\mathrm{II}.$

Therefore, the correct answer is B.

12.

The average (arithmetic mean) age of a group consisting of doctors and lawyers is $40.$ If the doctors average $35$ and the lawyers $50$ years old, then the ratio of the number of doctors to the number of lawyers is

$3:2$

$3:1$

$2:3$

$2:1$

$1:2$

Concepts:weighted mean ratio and proportion linear equation

Difficulty rating: 1290

Small Hint:

Let $d$ and $\ell$ be the numbers of doctors and lawyers

Big Hint:

Equate the total age $35d+50\ell$ to $40(d+\ell)$

Solution:

If there are $d$ doctors and $\ell$ lawyers, then $35d+50\ell=40(d+\ell).$ Thus $5d=10\ell,$ so $d:\ell=2:1.$

Therefore, the correct answer is D.

13.

The fraction $\frac{2(\sqrt2+\sqrt6)}{3\sqrt{2+\sqrt3}}$ is equal to

$\dfrac{2\sqrt2}{3}$

$1$

$\dfrac{2\sqrt3}{3}$

$\dfrac43$

$\dfrac{16}{9}$

Concepts:radical algebraic manipulation

Difficulty rating: 1530

Small Hint:

All quantities are positive, so compare the square of the fraction with the squares of the choices

Big Hint:

Use $(\sqrt2+\sqrt6)^2=8+4\sqrt3$

Solution:

The fraction is positive, and its square is $\begin{aligned} &\frac{4(\sqrt2+\sqrt6)^2} {9(2+\sqrt3)}\\ &\quad=\frac{4(8+4\sqrt3)} {9(2+\sqrt3)}\\ &\quad=\frac{16(2+\sqrt3)} {9(2+\sqrt3)}\\ &\quad=\frac{16}{9}. \end{aligned}$ Therefore the original fraction is $\frac{4}{3}.$

Therefore, the correct answer is D.

14.

Each valve $A,$ $B,$ and $C,$ when open, releases water into a tank at its own constant rate. With all three valves open, the tank fills in $1$ hour, with only valves $A$ and $C$ open it takes $1.5$ hours, and with only valves $B$ and $C$ open it takes $2$ hours. The number of hours required with only valves $A$ and $B$ open is

$1.1$

$1.15$

$1.2$

$1.25$

$1.75$

Concepts:rate system of equations rational equation

Difficulty rating: 1560

Small Hint:

Let $a,$ $b,$ $c$ be the fractions of the tank filled per hour by the three valves

Big Hint:

Combine $a+b+c=1,$ $a+c=\frac{2}{3},$ and $b+c=\frac{1}{2}$ to find $a+b$

Solution:

Let $a,$ $b,$ $c$ be the hourly rates in tankfuls. Then $\begin{aligned} a+b+c&=1,\\ a+c&=\frac23,\\ b+c&=\frac12. \end{aligned}$ Twice the first equation minus the other two gives $a+b=2-\frac23-\frac12=\frac56.$ Thus valves $A$ and $B$ fill the tank in $\frac1{a+b}=\frac65=1.2$ hours.

Therefore, the correct answer is C.

15.

A sector with acute central angle $\theta$ is cut from a circle of radius $6.$ The radius of the circle circumscribed about the sector is

$3\cos\theta$

$3\sec\theta$

$3\cos\frac12\theta$

$3\sec\frac12\theta$

$3$

Concepts:sector circumcircle, circumcenter, and circumradius law of sines

Difficulty rating: 2100

Small Hint:

Treat the sector’s two radii and chord as an isosceles triangle

Big Hint:

Express the chord using $\frac{\theta}{2},$ then apply the extended sine rule

Solution:

The two radii and the sector’s chord form an isosceles triangle with equal sides $6$ and vertex angle $\theta.$ Its base has length $12\sin(\frac{\theta}{2}).$ If $R$ is the triangle’s circumradius, the extended sine rule gives $2R=\frac{12\sin(\frac{\theta}{2})}{\sin\theta}.$ Since $\sin\theta=2\sin(\frac{\theta}{2})\cos(\frac{\theta}{2}),$ $R=3\sec\frac{\theta}{2}.$

Therefore, the correct answer is D.

16.

If the sum of all the angles except one of a convex polygon is $2190^\circ,$ then the number of sides of the polygon must be

$13$

$15$

$17$

$19$

$21$

Concepts:angle sum inequality bounding to limit cases

Difficulty rating: 1600

Small Hint:

Let the omitted interior angle be $x^\circ$ and the number of sides be $n$

Big Hint:

Use $0\lt x\lt180$ to trap the integer $n-2$ in an interval of length $1$

Solution:

If the omitted angle is $x^\circ,$ then convexity gives $0\lt x\lt180.$ Hence $180(n-2)=2190+x,$ so $\frac{2190}{180} \lt n-2 \lt\frac{2370}{180}.$ This is $12\frac16\lt n-2\lt13\frac16.$ The only possible integer is $n-2=13,$ so $n=15.$

Therefore, the correct answer is B.

17.

If $\theta$ is an acute angle and $\sin\frac12\theta=\sqrt{\frac{x-1}{2x}},$ then $\tan\theta$ equals

$x$

$\dfrac1x$

$\dfrac{\sqrt{x-1}}{x+1}$

$\dfrac{\sqrt{x^2-1}}x$

$\sqrt{x^2-1}$

Concepts:trigonometric identity radical algebraic manipulation

Difficulty rating: 1930

Small Hint:

Use $\cos\theta=1-2\sin^2(\frac{\theta}{2})$

Big Hint:

Once $\cos\theta$ is known, apply $\tan^2\theta=\sec^2\theta-1$ and use that $\theta$ is acute

Solution:

The half-angle identity gives $\begin{aligned} \cos\theta &=1-2\sin^2\frac{\theta}{2}\\ &=1-\frac{x-1}{x}\\ &=\frac1x. \end{aligned}$ Therefore $\tan^2\theta=\sec^2\theta-1=x^2-1.$ Because $\theta$ is acute, its tangent is positive, so $\tan\theta=\sqrt{x^2-1}.$

Therefore, the correct answer is E.

18.

If $p\ge5$ is a prime number, then $24$ divides $p^2-1$ without remainder

never

sometimes only

always

only if $p=5$

none of these

Concepts:divisibility prime modular arithmetic

Difficulty rating: 1610

Small Hint:

Factor $p^2-1=(p-1)(p+1)$

Big Hint:

Among the three consecutive integers $p-1,$ $p,$ $p+1,$ locate factors of $3$ and $8$

Solution:

A prime $p\ge5$ is odd, so $p-1$ and $p+1$ are consecutive even integers. One is divisible by $4,$ making their product divisible by $8.$ Among the three consecutive integers $p-1,$ $p,$ $p+1,$ one is divisible by $3.$ It cannot be $p,$ since $p\ge5$ is prime, so $3$ also divides $(p-1)(p+1).$ Because $3$ and $8$ are relatively prime, $(p-1)(p+1)=p^2-1$ is always divisible by $24.$

Therefore, the correct answer is C.

19.

Define $n_a!$ for positive $n$ and $a$ to be $\begin{aligned} n_a!={}&n(n-a)(n-2a)\\ &\cdot(n-3a)\cdots(n-ka), \end{aligned}$ where $k$ is the greatest integer for which $n\gt ka.$ Then the quotient $\frac{72_8!}{18_2!}$ is equal to

$4^5$

$4^6$

$4^8$

$4^9$

$4^{12}$

Concepts:custom operation factorial exponent

Difficulty rating: 1530

Small Hint:

Write both generalized factorials as nine explicit factors

Big Hint:

Factor $8$ from every numerator term and $2$ from every denominator term

Solution:

The two products are $\begin{aligned} 72_8!&=72\cdot64\cdots8 =8^9(9!),\\ 18_2!&=18\cdot16\cdots2 =2^9(9!). \end{aligned}$ Their quotient is therefore $\frac{8^9(9!)}{2^9(9!)}=4^9.$

Therefore, the correct answer is D.

20.

A cowboy is $4$ miles south of a stream which flows due east. He is also $8$ miles west and $7$ miles north of his cabin. He wishes to water his horse at the stream and return home. The shortest distance (in miles) he can travel and accomplish this is

$4+\sqrt{185}$

$16$

$17$

$18$

$\sqrt{32}+\sqrt{137}$

Concepts:reflection (geometry)distance formula optimization

Difficulty rating: 1850

Small Hint:

Reflect the cowboy’s starting point across the straight stream

Big Hint:

After reflection, the two-leg trip through the stream becomes one straight segment to the cabin

Solution:

Take the stream as the $x$ -axis and put the cowboy at $C=(0,-4).$ His cabin is then $H=(8,-11).$ Reflect $C$ across the stream to $D=(0,4).$ For any point $S$ on the stream, $CS=DS,$ so minimizing $CS+SH$ is the same as minimizing $DS+SH.$ This occurs when $D,$ $S,$ $H$ are collinear. The minimum distance is $DH=\sqrt{8^2+15^2}=\sqrt{289}=17.$

Therefore, the correct answer is C.

21.

The number of sets of two or more consecutive positive integers whose sum is $100$ is

$1$

$2$

$3$

$4$

$5$

Concepts:arithmetic sequence factor parity systematic listing

Difficulty rating: 1970

Small Hint:

If there are $k$ terms beginning with $a,$ double the sum to obtain $200=k(2a+k-1)$

Big Hint:

Check divisors $k\lt\sqrt{200}$ and enforce that $a$ is a positive integer

Solution:

For $k\ge2$ consecutive positive integers starting at $a,$ $200=k(2a+k-1).$ Positivity gives $k\lt\sqrt{200},$ so the possible divisors $k$ of $200$ are $2,4,5,8,10.$ From $2a=\frac{200}{k}-k+1,$ only $k=5$ and $k=8$ give positive even right-hand sides. They yield the sets $18,19,20,21,22$ and $9,10,\ldots,16.$ Hence there are $2$ sets.

Therefore, the correct answer is B.

22.

The set of all real solutions of the inequality $|x-1|+|x+2|\lt5$ is

$\{x:-3\lt x\lt2\}$

$\{x:-1\lt x\lt2\}$

$\{x:-2\lt x\lt1\}$

$\left\{x:-\dfrac32\lt x\lt\dfrac72\right\}$

$\varnothing$ (empty)

Concepts:absolute value inequality

Difficulty rating: 1470

Small Hint:

Interpret the two absolute values as the distances from $x$ to $1$ and to $-2$

Big Hint:

Between $-2$ and $1$ the sum is constant; outside that interval it increases by twice the distance from the nearer endpoint

Solution:

For $-2\le x\le1,$ the sum of the distances from $x$ to $-2$ and $1$ is $3.$ If $x$ lies a distance $u$ outside this interval, the sum is $3+2u.$ Thus $3+2u\lt5 \quad\Longleftrightarrow\quad u\lt1.$ Extending the interval $[-2,1]$ by $1$ at each end gives $-3\lt x\lt2.$

Therefore, the correct answer is A.

23.

There are two cards; one is red on both sides and the other is red on one side and blue on the other. The cards have the same probability $(\frac{1}{2})$ of being chosen, and one is chosen and placed on the table. If the upper side of the card on the table is red, then the probability that the under-side is also red is

$\dfrac14$

$\dfrac13$

$\dfrac12$

$\dfrac23$

$\dfrac34$

Concepts:conditional probability basic probability

Difficulty rating: 1670

Small Hint:

Condition on the individual card sides that could be showing red

Big Hint:

There are three equally likely visible red faces; determine the color behind each

Solution:

Among outcomes having a red upper side, either of the two red faces of the red-red card or the red face of the red-blue card can be uppermost. These three visible red faces are equally likely. The underside is red in the first two cases and blue in the third, so the conditional probability is $\frac23.$

Therefore, the correct answer is D.

24.

The check for a luncheon of $3$ sandwiches, $7$ cups of coffee and one piece of pie came to $\$3.15.$ The check for a luncheon consisting of $4$ sandwiches, $10$ cups of coffee and one piece of pie came to $\$4.20$ at the same place. The cost of a luncheon consisting of one sandwich, one cup of coffee and one piece of pie at the same place will come to

$\$1.70$

$\$1.65$

$\$1.20$

$\$1.05$

$\$0.95$

Concepts:system of equations money algebraic manipulation

Difficulty rating: 1580

Small Hint:

Let $s,$ $c,$ $p$ be the three item prices; only $s+c+p$ is required

Big Hint:

A suitable linear combination of $3s+7c+p=3.15$ and $4s+10c+p=4.20$ isolates the desired sum

Solution:

Let $s,$ $c,$ $p$ be the prices of a sandwich, coffee, and pie. The checks give $\begin{aligned} 3s+7c+p&=3.15,\\ 4s+10c+p&=4.20. \end{aligned}$ Three times the first equation minus twice the second gives $\begin{aligned} s+c+p &=3(3.15)-2(4.20)\\ &=1.05. \end{aligned}$ The requested luncheon costs $\$1.05.$

Therefore, the correct answer is D.

25.

A circular grass plot $12$ feet in diameter is cut by a straight gravel path $3$ feet wide, one edge of which passes through the center of the plot. The number of square feet in the remaining grass area is

$36\pi-34$

$30\pi-15$

$36\pi-33$

$35\pi-9\sqrt3$

$30\pi-9\sqrt3$

Concepts:circle area sector area decomposition special right triangle

Difficulty rating: 2230

Small Hint:

Bisect the path by the perpendicular diameter of the circular plot

Big Hint:

Each half of the path is a $30^\circ$ sector together with a $30^\circ$ - $60^\circ$ - $90^\circ$ triangle

Solution:

The plot has radius $6.$ At the path’s other edge, the perpendicular distance from the center is $3,$ so the radius to an intersection point makes a $30^\circ$ angle with the edge through the center. Half of the path consists of a $30^\circ$ sector of radius $6$ and a right triangle with legs $3$ and $3\sqrt3.$ Its area is $\begin{aligned} &\frac{30}{360}\pi(6^2) +\frac12(3)(3\sqrt3)\\ &\qquad=3\pi+\frac{9\sqrt3}{2}. \end{aligned}$ Thus the whole path has area $6\pi+9\sqrt3,$ and the remaining grass area is $36\pi-(6\pi+9\sqrt3) =30\pi-9\sqrt3.$

Therefore, the correct answer is E.

26.

The number of terms in an A.P. (Arithmetic Progression) is even. The sums of the odd- and even-numbered terms are $24$ and $30,$ respectively. If the last term exceeds the first by $10.5,$ the number of terms in the A.P. is

$20$

$18$

$12$

$10$

$8$

Concepts:arithmetic sequence system of equations pairing and grouping

Difficulty rating: 1850

Small Hint:

Write the number of terms as $2n$ and the common difference as $d$

Big Hint:

Pair each odd-numbered term with the following even-numbered term to get $nd=6$

Solution:

Let the progression have $2n$ terms and common difference $d.$ Pairing each odd-numbered term with its successor shows that $nd=30-24=6.$ The difference between the last and first terms is $(2n-1)d=10.5.$ Since $2nd=12,$ subtraction gives $d=1.5.$ Hence $n=\frac{6}{1.5}=4,$ and the progression has $2n=8$ terms.

Therefore, the correct answer is E.

27.

Cars $A$ and $B$ travel the same distance. Car $A$ travels half that distance at $u$ miles per hour and half at $v$ miles per hour. Car $B$ travels half the time at $u$ miles per hour and half at $v$ miles per hour. The average speed of Car $A$ is $x$ miles per hour and that of Car $B$ is $y$ miles per hour. Then we always have

$x\le y$

$x\ge y$

$x=y$

$x\lt y$

$x\gt y$

Concepts:harmonic mean inequality algebraic manipulation

Difficulty rating: 1800

Small Hint:

Car $A$ ’s average is the harmonic mean of $u$ and $v,$ while Car $B$ ’s is their arithmetic mean

Big Hint:

Subtract the two means and factor the numerator as a square

Solution:

Car $A$ ’s equal-distance average and Car $B$ ’s equal-time average are $x=\frac{2uv}{u+v}, \qquad y=\frac{u+v}{2}.$ Their difference is $\begin{aligned} y-x &=\frac{(u+v)^2-4uv}{2(u+v)}\\ &=\frac{(u-v)^2}{2(u+v)}\\ &\ge0. \end{aligned}$ Thus $x\le y,$ with equality possible when $u=v.$

Therefore, the correct answer is A.

28.

If $a,$ $b,$ and $c$ are in geometric progression (G.P.) with $1\lt a\lt b\lt c$ and $n\gt1$ is an integer, then $\log_a n,$ $\log_b n,$ $\log_c n$ form a sequence

which is a G.P.

which is an arithmetic progression (A.P.)

in which the reciprocals of the terms form an A.P.

in which the second and third terms are the $n$ th powers of the first and second respectively

none of these

Concepts:logarithm geometric sequence arithmetic sequence

Difficulty rating: 2060

Small Hint:

Take reciprocals and use $\dfrac1{\log_a n}=\log_n a$

Big Hint:

Apply logarithms to the geometric-progression relation $b^2=ac$

Solution:

By change of base, $\begin{aligned} \frac1{\log_a n}&=\log_n a,\\ \frac1{\log_b n}&=\log_n b,\\ \frac1{\log_c n}&=\log_n c. \end{aligned}$ Since $a,$ $b,$ $c$ are in geometric progression, $b^2=ac.$ Taking logarithms to base $n$ gives $2\log_n b=\log_n a+\log_n c.$ Hence the reciprocals of the three given terms form an arithmetic progression.

Therefore, the correct answer is C.

29.

Two boys start moving from the same point $A$ on a circular track but in opposite directions. Their speeds are $5$ ft. per sec. and $9$ ft. per sec. If they start at the same time and finish when they first meet at the point $A$ again, then the number of times they meet, excluding the start and finish, is

$13$

$25$

$44$

infinity

none of these

Concepts:relative speed greatest common divisor counting intersections

Difficulty rating: 1830

Small Hint:

Because $5$ and $9$ are relatively prime, determine when both boys first complete whole numbers of laps

Big Hint:

Before that finish time, meetings occur whenever their combined distance is another whole lap

Solution:

Let the track length be $L.$ Since $\gcd(5,9)=1,$ the first positive time when both boys are back at $A$ is $t=L$ : they have completed $5$ and $9$ laps. Their relative speed is $5+9=14,$ so before time $L$ they meet at $t=\frac{kL}{14}, \qquad k=1,2,\ldots,13.$ Thus there are $13$ meetings excluding the start and finish.

Therefore, the correct answer is A.

30.

Let $[t]$ denote the greatest integer not exceeding $t,$ where $t\ge0,$ and $\begin{aligned} S=\{(x,y):{}&(x-T)^2+y^2\\ &\le T^2\},\\ T&=t-[t]. \end{aligned}$ Then we have

the point $(0,0)$ does not belong to $S$ for any $t$

$0\le\operatorname{Area} S\le\pi$ for all $t$

$S$ is contained in the first quadrant for all $t\ge5$

the center of $S$ for any $t$ is on the line $y=x$

none of the other statements is true

Concepts:floor and ceiling functions circle area coordinate geometry

Difficulty rating: 1720

Small Hint:

Recognize $T=t-[t]$ as the fractional part of $t$

Big Hint:

Interpret the equation for $S$ as a disk and identify its center and radius

Solution:

The fractional part satisfies $0\le T\lt1.$ The set $S$ is the closed disk centered at $(T,0)$ with radius $T.$ Its area is $\pi T^2,$ so $0\le\operatorname{Area}S\lt\pi,$ which in particular gives $0\le\operatorname{Area}S\le\pi.$ The origin lies on every such disk, the disk extends below the $x$ -axis when $T\gt0,$ and its center is generally not on $y=x.$

Therefore, the correct answer is B.

31.

In the following equation, each of the letters represents uniquely a different digit in base ten: $(YE)\cdot(ME)=TTT.$ The sum $E+M+T+Y$ equals

$19$

$20$

$21$

$22$

$24$

Concepts:cryptarithm divisibility digits units digit

Difficulty rating: 2190

Small Hint:

Use $TTT=111T=3\cdot37\cdot T$

Big Hint:

The prime $37$ must divide one of the two-digit factors; test its two-digit multiples ending in the common digit $E$

Solution:

Since $TTT=111T=3\cdot37\cdot T,$ the prime $37$ divides one of $YE$ and $ME.$ A two-digit multiple of $37$ is $37$ or $74.$ The value $74$ is impossible: the other two-digit factor ending in $4$ is at least $14,$ and $74\cdot14\gt999.$ Hence one factor is $37,$ so $E=7.$

The units digit of the product is the units digit of $7^2,$ so $T=9.$ Thus the product is $999,$ and the other factor is $\frac{999}{37}=27.$ The four digits are $2,$ $3,$ $7,$ $9,$ whose sum is $21.$

Therefore, the correct answer is C.

32.

The volume of a pyramid whose base is an equilateral triangle of side length $6$ and whose other edges are each of length $\sqrt{15}$ is

$9$

$\dfrac92$

$\dfrac{27}{2}$

$\dfrac{9\sqrt3}{2}$

none of these

Concepts:pyramid equilateral triangle volume Pythagorean Theorem

Difficulty rating: 2040

Small Hint:

The altitude from the apex meets the base at the equilateral triangle’s circumcenter

Big Hint:

Use the base circumradius $2\sqrt3$ and a lateral edge $\sqrt{15}$ to find the height

Solution:

The base area is $\frac{\sqrt3}{4}(6^2)=9\sqrt3.$ Because the apex is equally distant from all three base vertices, its perpendicular projection is the base circumcenter. The circumradius of the equilateral base is $\frac{6}{\sqrt3}=2\sqrt3.$ If $h$ is the pyramid’s height, then $h^2+(2\sqrt3)^2=(\sqrt{15})^2,$ so $h=\sqrt3.$ The volume is $\frac13(9\sqrt3)(\sqrt3)=9.$

Therefore, the correct answer is A.

33.

When one ounce of water is added to a mixture of acid and water, the new mixture is $20\%$ acid. When one ounce of acid is added to the new mixture, the result is $33\frac13\%$ acid. The percentage of acid in the original mixture is

$22\%$

$24\%$

$25\%$

$30\%$

$33\frac13\%$

Concepts:mixture percentage system of equations

Difficulty rating: 1830

Small Hint:

Let $x$ and $y$ be the original ounces of water and acid

Big Hint:

Write one concentration equation after adding water and another after subsequently adding acid

Solution:

Let the original mixture contain $x$ ounces of water and $y$ ounces of acid. The two additions give $\begin{aligned} \frac{y}{x+y+1}&=\frac15,\\ \frac{y+1}{x+y+2}&=\frac13. \end{aligned}$ These simplify to $x+1=4y, \qquad x=2y+1.$ Hence $y=1$ and $x=3.$ The original acid percentage was $100\cdot\frac{y}{x+y} =100\cdot\frac14=25\%.$

Therefore, the correct answer is C.

34.

A plane flew straight against a wind between two towns in $84$ minutes and returned with that wind in $9$ minutes less than it would take in still air. The number of minutes (two answers) for the return trip was

$54$ or $18$

$60$ or $15$

$63$ or $12$

$72$ or $36$

$75$ or $20$

Concepts:distance rate and time quadratic system of equations

Difficulty rating: 2410

Small Hint:

Let $x$ be the return time; the still-air time is then $x+9$

Big Hint:

The plane’s still-air speed is the average of its against-wind and with-wind ground speeds

Solution:

Let the distance be $d$ and the return time be $x$ minutes. The against-wind and with-wind speeds are $\frac{d}{84}$ and $\frac{d}{x}.$ Their average is the still-air speed $\frac{d}{x+9}.$ Therefore $\frac{2}{x+9}=\frac1{84}+\frac1x.$ Clearing denominators gives $\begin{gathered} 168x=(x+9)(x+84),\\ x^2-75x+756=0,\\ (x-63)(x-12)=0. \end{gathered}$ Both positive values are consistent with the stated conditions, so the two return times are $63$ and $12$ minutes.

Therefore, the correct answer is C.

35.

In the unit circle shown in the figure, chords $PQ$ and $MN$ are parallel to the unit radius $OR$ of the circle with center at $O.$ Chords $MP,$ $PQ,$ and $NR$ are each $s$ units long and chord $MN$ is $d$ units long.

Of the three equations $\begin{array}{rl} \mathrm{I}.&d-s=1,\\ \mathrm{II}.&ds=1,\\ \mathrm{III}.&d^2-s^2=\sqrt5 \end{array}$ those which are necessarily true are

$\mathrm{I}$ only

$\mathrm{II}$ only

$\mathrm{III}$ only

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

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

Concepts:chord circle trigonometric identity algebraic manipulation

Difficulty rating: 2520

Small Hint:

Use symmetry across the vertical diameter to see that the upper semicircle is divided into five equal chords

Big Hint:

Write $s=2\sin18^\circ$ and $d=2\sin54^\circ,$ then relate each to the square of the other

Solution:

Let $K$ be the left endpoint of the horizontal diameter. Reflection across the vertical diameter shows that $KM=NR=s$ and $QN=MP=s.$ Together with the given equal chords, the upper semicircle is split into five equal arcs. Each subtends $36^\circ$ at $O.$ Thus $\begin{aligned} s&=2\sin18^\circ,\\ d&=2\sin54^\circ\\ &=2\cos36^\circ. \end{aligned}$

Using the double-angle identities, $\begin{aligned} d&=2(1-2\sin^218^\circ)\\ &=2-s^2,\\ s&=2\cos72^\circ\\ &=4\cos^236^\circ-2\\ &=d^2-2. \end{aligned}$ Adding these equations gives $d+s=d^2-s^2=(d-s)(d+s).$ Since $d+s\gt0,$ it follows that $d-s=1.$ Substituting $d=s+1$ into $d=2-s^2$ yields $s^2+s=1, \qquad s=\frac{\sqrt5-1}{2}.$ Therefore $ds=s(s+1)=1$ and $\begin{aligned} d^2-s^2 &=(d-s)(d+s)\\ &=2s+1\\ &=\sqrt5. \end{aligned}$ All three equations are necessarily true.

Therefore, the correct answer is E.