1973 AMC 12 Problems

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

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

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.

Count the cubes with no painted face instead

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

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.

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

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

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.

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

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

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.

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

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

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.

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

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

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.

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

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

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.

Paint required scales with surface area, not volume

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

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.

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

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

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.

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

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

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.

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

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

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.

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

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

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.

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

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

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.

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

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

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.

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

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

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.

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

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

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.

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

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

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.

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

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

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.

Write both generalized factorials as nine explicit factors

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

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.

Reflect the cowboy’s starting point across the straight stream

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

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.

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

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

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.

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

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

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.

Condition on the individual card sides that could be showing red

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

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.

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

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

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.

Bisect the path by the perpendicular diameter of the circular plot

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

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.

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

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

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.

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

Subtract the two means and factor the numerator as a square

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 $u=v.$