1976 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

If one minus the reciprocal of $(1-x)$ equals the reciprocal of $(1-x),$ then $x$ equals

$-2$

$-1$

$\frac12$

$2$

$3$

Concepts:rational equation algebraic manipulation

Difficulty rating: 1260

Small Hint:

Multiply the equation by $1-x,$ which cannot be zero

Big Hint:

After clearing the denominator, solve $1-x-1=1$

Solution:

The equation is $1-\frac1{1-x}=\frac1{1-x}.$ Multiplying by $1-x\ne0$ gives $1-x-1=1,$ so $x=-1.$

Therefore, the correct answer is B.

2.

For how many real numbers $x$ is $\sqrt{-(x+1)^2}$ a real number?

none

one

two

a finite number greater than two

infinitely many

Concepts:radical perfect square inequality

Difficulty rating: 1180

Small Hint:

A real square root requires a nonnegative radicand

Big Hint:

The quantity $-(x+1)^2$ can be nonnegative only when the square vanishes

Solution:

Since $(x+1)^2\ge0,$ its negative is at most $0.$ It is nonnegative only when $(x+1)^2=0,$ which occurs for the single value $x=-1.$

Therefore, the correct answer is B.

3.

The sum of the distances from one vertex of a square with sides of length two to the midpoints of each of the sides of the square is

$2\sqrt5$

$2+\sqrt3$

$2+2\sqrt3$

$2+\sqrt5$

$2+2\sqrt5$

Concepts:square (geometry)midpoint distance formula

Difficulty rating: 1490

Small Hint:

Place the chosen vertex at $(0,0)$ and the other vertices at $(2,0),$ $(2,2),$ and $(0,2)$

Big Hint:

Two midpoints are one unit away, while each opposite-side midpoint forms legs $2$ and $1$

Solution:

The two midpoints on sides meeting the chosen vertex are each at distance $1.$ The other two midpoints have coordinate differences $2$ and $1,$ so each is at distance $\sqrt{2^2+1^2}=\sqrt5.$ The sum is $2+2\sqrt5.$

Therefore, the correct answer is E.

4.

Let a geometric progression with $n$ terms have first term one, common ratio $r$ and sum $s,$ where $r$ and $s$ are not zero. The sum of the geometric progression formed by replacing each term of the original progression by its reciprocal is

$\frac1s$

$\frac1{r^ns}$

$\frac{s}{r^{n-1}}$

$\frac{r^n}{s}$

$\frac{r^{n-1}}s$

Concepts:geometric sequence factoring algebraic manipulation

Difficulty rating: 1620

Small Hint:

Write the reciprocal sum as $1+r^{-1}+\cdots+r^{-(n-1)}$

Big Hint:

Factor $r^{-(n-1)}$ after reversing the order of the powers

Solution:

Let $S'$ be the reciprocal sum. Multiplying it by $r^{n-1}$ reverses the original terms: $\begin{aligned} r^{n-1}S' &=r^{n-1}+r^{n-2}+\cdots+1\\ &=s. \end{aligned}$ Hence $S'=\frac{s}{r^{n-1}}.$ This also holds when $r=1.$

Therefore, the correct answer is C.

5.

How many integers greater than ten and less than one hundred, written in base ten notation, are increased by nine when their digits are reversed?

$0$

$1$

$8$

$9$

$10$

Concepts:digits place value systematic listing

Difficulty rating: 1250

Small Hint:

Represent the tens and units digits by $t$ and $u$

Big Hint:

The increase is $10u+t-(10t+u)=9(u-t)$

Solution:

The condition is $9(u-t)=9,$ so $u=t+1.$ The possibilities are $12,$ $23,$ $34,$ $45,$ $56,$ $67,$ $78,$ and $89,$ giving $8$ integers.

Therefore, the correct answer is C.

6.

If $c$ is a real number and the negative of one of the solutions of $x^2-3x+c=0$ is a solution of $x^2+3x-c=0,$ then the solutions of $x^2-3x+c=0$ are

$1,$ $2$

$-1,$ $-2$

$0,$ $3$

$0,$ $-3$

$\frac32,$ $\frac32$

Concepts:quadratic substitution zero product property

Difficulty rating: 1210

Small Hint:

Call the first root $r$ and substitute $-r$ into the second equation

Big Hint:

The two resulting equations differ only in the sign of $c$

Solution:

For such a root $r,$ $\begin{aligned} r^2-3r+c&=0,\\ (-r)^2+3(-r)-c&=0. \end{aligned}$ Subtracting gives $2c=0.$ Thus the first equation is $x(x-3)=0,$ with roots $0$ and $3.$

Therefore, the correct answer is C.

7.

If $x$ is a real number, then the quantity $(1-|x|)(1+x)$ is positive if and only if

$|x|\lt1$

$x\lt1$

$|x|\gt1$

$x\lt-1$

$x\lt-1$ or $-1\lt x\lt1$

Concepts:absolute value inequality casework

Difficulty rating: 1570

Small Hint:

A product is positive when its two factors have the same sign

Big Hint:

Analyze the intervals cut by $-1$ and $1$

Solution:

Both factors are positive exactly when $|x|\lt1,$ giving $-1\lt x\lt1.$ Both are negative when $|x|\gt1$ and $x\lt-1,$ which reduces to $x\lt-1.$ The endpoints make the product zero.

Therefore, the correct answer is E.

8.

A point in the plane, both of whose rectangular coordinates are integers with absolute value less than or equal to four, is chosen at random, with all such points having an equal probability of being chosen. What is the probability that the distance from the point to the origin is at most two units?

$\frac{13}{81}$

$\frac{15}{81}$

$\frac{13}{64}$

$\frac{\pi}{16}$

the square of a rational number

Concepts:lattice point basic probability circle systematic listing

Difficulty rating: 1710

Small Hint:

There are $9$ choices for each coordinate

Big Hint:

Count integer pairs satisfying $x^2+y^2\le4$ by considering $x=0,$ $x=\pm1,$ and $x=\pm2$

Solution:

There are $9^2=81$ possible points. Within distance $2$ are the origin; the four points $(\pm1,0),$ $(0,\pm1);$ the four points $(\pm1,\pm1);$ and the four points $(\pm2,0),$ $(0,\pm2).$ Thus $13$ points qualify, for probability $\frac{13}{81}.$

Therefore, the correct answer is A.

9.

In triangle $ABC,$ $D$ is the midpoint of $AB;$ $E$ is the midpoint of $DB;$ and $F$ is the midpoint of $BC.$ If the area of $\triangle ABC$ is $96,$ then the area of $\triangle AEF$ is

$16$

$24$

$32$

$36$

$48$

Concepts:triangle area midpoint area ratio

Difficulty rating: 1530

Small Hint:

Compare the base $AE$ with $AB$ and the height from $F$ with the height from $C$

Big Hint:

The two ratios are $\frac34$ and $\frac12$

Solution:

Since $D$ is the midpoint of $AB$ and $E$ is the midpoint of $DB,$ $AE=\frac34AB.$ Since $F$ is the midpoint of $BC,$ its distance from line $AB$ is half that of $C.$ Therefore $\begin{aligned} [AEF]&=\frac34\cdot\frac12[ABC]\\ &=\frac38(96)=36. \end{aligned}$

Therefore, the correct answer is D.

10.

If $m,$ $n,$ $p$ and $q$ are real numbers and $f(x)=mx+n$ and $g(x)=px+q,$ then the equation $f(g(x))=g(f(x))$ has a solution

for all choices of $m,$ $n,$ $p$ and $q$

if and only if $m=p$ and $n=q$

if and only if $mq-np=0$

if and only if $n(1-p)-q(1-m)=0$

if and only if $(1-n)(1-p)$ ${}-(1-q)(1-m)=0$

Concepts:function linear equation algebraic manipulation

Difficulty rating: 1670

Small Hint:

Expand both compositions before trying to solve for $x$

Big Hint:

The coefficients of $x$ are automatically equal, leaving a condition on the constant terms

Solution:

We have $\begin{aligned} f(g(x))&=mpx+mq+n,\\ g(f(x))&=pmx+pn+q. \end{aligned}$ The $x$ -terms cancel for every $x,$ so a solution exists exactly when $mq+n=pn+q.$ Rearranging gives $n(1-p)-q(1-m)=0.$

Therefore, the correct answer is D.

11.

Which of the following statements is (are) equivalent to the statement “If the pink elephant on planet alpha has purple eyes, then the wild pig on planet beta does not have a long nose”?

$\mathrm{I}.$ “If the wild pig on planet beta has a long nose, then the pink elephant on planet alpha has purple eyes.”

$\mathrm{II}.$ “If the pink elephant on planet alpha does not have purple eyes, then the wild pig on planet beta does not have a long nose.”

$\mathrm{III}.$ “If the wild pig on planet beta has a long nose, then the pink elephant on planet alpha does not have purple eyes.”

$\mathrm{IV}.$ “The pink elephant on planet alpha does not have purple eyes, or the wild pig on planet beta does not have a long nose.”

The word “or” is used here in the inclusive sense (as is customary in mathematical writing).

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

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

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

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

$\mathrm{III}$ only

Concepts:logical deduction

Difficulty rating: 1740

Small Hint:

Write the original implication as $P\mathbin{\Rightarrow}\neg Q$

Big Hint:

Use both its contrapositive and the equivalent disjunction $\neg P\mathbin{\lor}\neg Q$

Solution:

Let $P$ mean the elephant has purple eyes and $Q$ mean the pig has a long nose. The original statement $P\Rightarrow\neg Q$ is equivalent to its contrapositive $Q\Rightarrow\neg P,$ which is $\mathrm{III},$ and to $\neg P\lor\neg Q,$ which is $\mathrm{IV}.$ Statements $\mathrm{I}$ and $\mathrm{II}$ are converses and need not follow.

Therefore, the correct answer is B.

12.

A supermarket has $128$ crates of apples. Each crate contains at least $120$ apples and at most $144$ apples. What is the largest integer $n$ such that there must be at least $n$ crates containing the same number of apples?

$4$

$5$

$6$

$24$

$25$

Concepts:pigeonhole principle basic counting

Difficulty rating: 1380

Small Hint:

Count the possible apple totals from $120$ through $144$ , inclusive

Big Hint:

If every total occurred at most five times, there would be at most $25\cdot5$ crates

Solution:

There are $144-120+1=25$ possible apple counts. Since $128>25\cdot5,$ some count occurs at least $6$ times. This is sharp: distribute $128$ crates so that three counts occur $6$ times and the other twenty-two occur $5$ times. Thus the largest guaranteed $n$ is $6.$

Therefore, the correct answer is C.

13.

If $x$ cows give $x+1$ cans of milk in $x+2$ days, how many days will it take $x+3$ cows to give $x+5$ cans of milk?

$\frac{x(x+2)(x+5)}{(x+1)(x+3)}$

$\frac{x(x+1)(x+5)}{(x+2)(x+3)}$

$\frac{(x+1)(x+3)(x+5)}{x(x+2)}$

$\frac{(x+1)(x+3)}{x(x+2)(x+5)}$

none of these

Concepts:rate ratio and proportion algebraic manipulation

Difficulty rating: 1560

Small Hint:

First find the number of cans produced per cow-day

Big Hint:

Set $\frac{x+5}{(x+3)T}=\frac{x+1}{x(x+2)}$

Solution:

The production rate is $\frac{x+1}{x(x+2)}$ cans per cow-day. If the new time is $T,$ then $\frac{x+5}{(x+3)T}=\frac{x+1}{x(x+2)}.$ Hence $T=\frac{x(x+2)(x+5)}{(x+1)(x+3)}.$

Therefore, the correct answer is A.

14.

The measures of the interior angles of a convex polygon are in arithmetic progression. If the smallest angle is $100^\circ$ and the largest angle is $140^\circ,$ then the number of sides the polygon has is

$6$

$8$

$10$

$11$

$12$

Concepts:angle sum arithmetic sequence linear equation

Difficulty rating: 1510

Small Hint:

The average of an arithmetic progression is the average of its first and last terms

Big Hint:

Equate $120n$ with the interior-angle sum $180(n-2)$

Solution:

The average angle is $\frac{100+140}{2}=120^\circ,$ so the angle sum is $120n.$ A convex $n$ -gon has angle sum $180(n-2),$ hence $120n=180(n-2)$ and $n=6.$

Therefore, the correct answer is A.

15.

If $r$ is the remainder when each of the numbers $1059,$ $1417$ and $2312$ is divided by $d,$ where $d$ is an integer greater than one, then $d-r$ equals

$1$

$15$

$179$

$d-15$

$d-1$

Concepts:modular arithmetic greatest common divisor divisibility

Difficulty rating: 1800

Small Hint:

A common remainder means $d$ divides every pairwise difference

Big Hint:

Use $1417-1059=358=2\cdot179$ and $2312-1417=895=5\cdot179$

Solution:

The divisor $d$ divides both $358=2\cdot179$ and $895=5\cdot179.$ Their greatest common divisor is $179,$ a prime, so $d=179.$ Since $1059=5\cdot179+164,$ we have $r=164$ and $d-r=15.$

Therefore, the correct answer is B.

16.

In triangles $ABC$ and $DEF,$ lengths $AC,$ $BC,$ $DF$ and $EF$ are all equal. Length $AB$ is twice the length of the altitude of $\triangle DEF$ from $F$ to $DE.$ Which of the following statements is (are) true?

$\mathrm{I}.$ $\angle ACB$ and $\angle DFE$ must be complementary.

$\mathrm{II}.$ $\angle ACB$ and $\angle DFE$ must be supplementary.

$\mathrm{III}.$ The area of $\triangle ABC$ must equal the area of $\triangle DEF.$

$\mathrm{IV}.$ The area of $\triangle ABC$ must equal twice the area of $\triangle DEF.$

$\mathrm{II}$ only

$\mathrm{III}$ only

$\mathrm{IV}$ only

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

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

Concepts:isosceles triangle altitude congruence (geometry)area

Difficulty rating: 2130

Small Hint:

Drop altitudes from $C$ and $F$ ; each altitude bisects the base of its isosceles triangle

Big Hint:

Compare a half of $ABC$ with a half of $DEF$ as right triangles

Solution:

Let the equal sides have length $s,$ let $AB=2a,$ and let the altitude from $F$ be $a.$ In the isosceles triangles, the altitude from $C$ bisects $AB,$ while the altitude from $F$ bisects $DE.$ A half of $ABC$ has hypotenuse $s$ and leg $a;$ a half of $DEF$ has the same hypotenuse and altitude leg $a.$ The two right triangles are congruent with their legs interchanged. Thus their relevant acute angles are complementary, so the full vertex angles $\angle ACB$ and $\angle DFE$ sum to $180^\circ.$ The corresponding half-areas are equal as well, hence the full triangle areas are equal. Statements $\mathrm{II}$ and $\mathrm{III}$ hold.

Therefore, the correct answer is E.

17.

If $\theta$ is an acute angle and $\sin2\theta=a,$ then $\sin\theta+\cos\theta$ equals

$\sqrt{a+1}$

$(\sqrt2-1)a+1$

$\sqrt{a+1}-\sqrt{a^2-a}$

$\sqrt{a+1}+\sqrt{a^2-a}$

$\sqrt{a+1}+a^2-a$

Concepts:trigonometric identity trigonometry radical

Difficulty rating: 1530

Small Hint:

Square the expression $\sin\theta+\cos\theta$

Big Hint:

Use $2\sin\theta\cos\theta=\sin2\theta$ and the fact that the desired sum is positive

Solution:

We have $\begin{aligned} (\sin\theta+\cos\theta)^2 &=\sin^2\theta+\cos^2\theta\\ &\quad+2\sin\theta\cos\theta\\ &=1+a. \end{aligned}$ Both sine and cosine are positive for acute $\theta,$ so the sum is $\sqrt{a+1}.$

Therefore, the correct answer is A.

18.

In the adjoining figure, $AB$ is tangent at $A$ to the circle with center $O;$ point $D$ is interior to the circle; and $DB$ intersects the circle at $C.$ If $BC=DC=3,$ $OD=2$ and $AB=6,$ then the radius of the circle is

$3+\sqrt3$

$\frac{15}{\pi}$

$\frac92$

$2\sqrt6$

$\sqrt{22}$

Concepts:circle tangent line power of a point

Difficulty rating: 2130

Small Hint:

Extend $BD$ through $D$ to meet the circle again at $E$

Big Hint:

Use $BC\cdot BE=AB^2$ first, then use the power of interior point $D$

Solution:

Extend $BD$ through $D$ to meet the circle at $E.$ Since $BC=3$ and $BD=6,$ the tangent-secant theorem gives $3\cdot BE=AB^2=36,$ so $BE=12$ and $DE=6.$ The power of $D$ gives $\begin{aligned} DC\cdot DE&=(R-OD)(R+OD)\\ &=R^2-OD^2, \end{aligned}$ hence $3\cdot6=(R-2)(R+2)=R^2-4.$ Thus $R^2=22$ and $R=\sqrt{22}.$

Therefore, the correct answer is E.

19.

A polynomial $p(x)$ has remainder three when divided by $x-1$ and remainder five when divided by $x-3.$ The remainder when $p(x)$ is divided by $(x-1)(x-3)$ is

$x-2$

$x+2$

$2$

$8$

$15$

Concepts:polynomial linear equation substitution

Difficulty rating: 1620

Small Hint:

The desired remainder has degree less than two, so write it as $ax+b$

Big Hint:

Use the remainder theorem at $x=1$ and $x=3$

Solution:

Let the remainder be $r(x)=ax+b.$ Since $p(1)=3$ and $p(3)=5,$ we have $a+b=3,$ and $3a+b=5.$ Thus $a=1,$ and $b=2,$ so the remainder is $x+2.$

Therefore, the correct answer is B.

20.

Let $a,$ $b$ and $x$ be positive real numbers distinct from one. Then $\begin{aligned} 4(\log_a x)^2&+3(\log_b x)^2\\ &=8(\log_a x)(\log_b x) \end{aligned}$

for all values of $a,$ $b$ and $x$

if and only if $a=b^2$

if and only if $b=a^2$

if and only if $x=ab$

none of these

Concepts:logarithm quadratic factoring

Difficulty rating: 2110

Small Hint:

Move all terms to one side and factor the quadratic in the two logarithms

Big Hint:

Translate each factor equation into a relation between the bases using change of base

Solution:

Factoring gives $\begin{aligned} &\bigl(2\log_a x-\log_b x\bigr)\\ &\quad\cdot \bigl(2\log_a x-3\log_b x\bigr)=0. \end{aligned}$ Since $x\ne1,$ change of base shows that the first factor vanishes when $a=b^2,$ while the second vanishes when $a^3=b^2.$ Thus the equation holds under either of two conditions, and no listed “if and only if” statement describes their union.

Therefore, the correct answer is E.

21.

What is the smallest positive odd integer $n$ such that the product $2^{\frac{1}{7}}2^{\frac{3}{7}}\cdots2^{\frac{2n+1}{7}}$ is greater than $1000?$ (In the product the denominators of the exponents are all sevens, and the numerators are the successive odd integers from $1$ to $2n+1.$ )

$7$

$9$

$11$

$17$

$19$

Concepts:exponent sum of first n odd numbers inequality

Difficulty rating: 1870

Small Hint:

Combine the powers by summing the odd numerators

Big Hint:

The odd-number sum $1+3+\cdots+(2n+1)$ equals $(n+1)^2;$ compare with $2^{10}=1024$

Solution:

The product is $2^{\frac{1+3+\cdots+(2n+1)}{7}} =2^{\frac{(n+1)^2}{7}}.$ For $n=7,$ the exponent $\frac{64}{7}$ is less than $\frac{19}{2},$ so the product is less than $2^{\frac{19}{2}}=512\sqrt2\lt768\lt1000.$ All smaller positive odd $n$ also fail. For $n=9,$ the exponent is $\frac{100}{7}\gt10,$ so the product exceeds $2^{10}=1024.$ Thus the first allowable odd $n$ is $9.$

Therefore, the correct answer is B.

22.

Given an equilateral triangle with side of length $s,$ consider the locus of all points $P$ in the plane of the triangle such that the sum of the squares of the distances from $P$ to the vertices of the triangle is a fixed number $a.$ This locus

is a circle if $a\gt s^2$

contains only three points if $a=2s^2$ and is a circle if $a\gt2s^2$

is a circle with positive radius only if $s^2\lt a\lt2s^2$

contains only a finite number of points for any value of $a$

is none of these

Concepts:centroid circle distance formula

Difficulty rating: 2110

Small Hint:

Let $G$ be the centroid and express the sum using $PG^2$

Big Hint:

For an equilateral triangle, the sum equals $3PG^2+s^2$

Solution:

For vertices $A,$ $B,$ and $C$ with centroid $G,$ put $T=GA^2+GB^2+GC^2.$ The standard centroid identity gives $PA^2+PB^2+PC^2=3PG^2+T.$ In an equilateral triangle, each $GA=\frac{s}{\sqrt3},$ so $T=s^2.$ Hence $3PG^2=a-s^2.$ The locus is empty for $a\lt s^2,$ one point for $a=s^2,$ and a circle of positive radius for $a\gt s^2.$

Therefore, the correct answer is A.

23.

For integers $k$ and $n$ such that $1\le k\lt n,$ let $\binom nk=\frac{n!}{k!(n-k)!}.$ Then $\left(\frac{n-2k-1}{k+1}\right)\binom nk$ is an integer

for all $k$ and $n$

for all even values of $k$ and $n,$ but not for all $k$ and $n$

for all odd values of $k$ and $n,$ but not for all $k$ and $n$

if $k=1$ or $n-1,$ but not for all odd values of $k$ and $n$

if $n$ is divisible by $k,$ but not for all even values of $k$ and $n$

Concepts:combinations algebraic manipulation

Difficulty rating: 2080

Small Hint:

Split $n-2k-1$ into $(n-k)-(k+1)$

Big Hint:

Try rewriting the expression as a difference of two adjacent binomial coefficients

Solution:

Using $\binom n{k+1}=\frac{n-k}{k+1}\binom nk,$ and writing $\frac{n-2k-1}{k+1}=\frac{n-k}{k+1}-1,$ the given expression becomes $\binom n{k+1}-\binom nk.$ This is an integer for every permitted $k$ and $n.$

Therefore, the correct answer is A.

24.

In the adjoining figure, circle $K$ has diameter $AB;$ circle $L$ is tangent to circle $K$ and to $AB$ at the center of circle $K;$ and circle $M$ is tangent to circle $K,$ to circle $L$ and to $AB.$ The ratio of the area of circle $K$ to the area of circle $M$ is

$12$

$14$

$16$

$18$

not an integer

Concepts:tangent circles Pythagorean Theorem area ratio

Difficulty rating: 2200

Small Hint:

Scale the radius of $K$ to $2,$ so circle $L$ has radius $1$

Big Hint:

If $M$ has radius $t$ and center $(x,t),$ use its tangencies to both larger circles

Solution:

Take $K$ to have center $(0,0)$ and radius $2,$ with $AB$ the $x$ -axis. Then $L$ has center $(0,1)$ and radius $1.$ If $M$ has center $(x,t)$ and radius $t,$ internal tangency to $K$ and external tangency to $L$ give $\begin{aligned} x^2+t^2&=(2-t)^2,\\ x^2+(t-1)^2&=(1+t)^2. \end{aligned}$ These simplify to $x^2=4-4t$ and $x^2=4t,$ so $t=\frac12.$ The radius ratio is $2:\frac{1}{2}=4:1,$ hence the area ratio is $16.$

Therefore, the correct answer is C.

25.

For a sequence $u_1,$ $u_2,$ $\ldots,$ define $\Delta^1(u_n)=u_{n+1}-u_n$ and, for all integers $k\gt1,$ $\Delta^k(u_n)=\Delta^1(\Delta^{k-1}(u_n)).$ If $u_n=n^3+n,$ then $\Delta^k(u_n)=0$ for all $n$

if $k=1$

if $k=2,$ but not if $k=1$

if $k=3,$ but not if $k=2$

if $k=4,$ but not if $k=3$

for no value of $k$

Concepts:polynomial recursion pattern recognition

Difficulty rating: 1830

Small Hint:

Each forward difference lowers the degree of a polynomial sequence by one

Big Hint:

Compute through the constant third difference to see when zero first appears

Solution:

Directly, $\begin{aligned} \Delta u_n&=3n^2+3n+2,\\ \Delta^2u_n&=6n+6,\\ \Delta^3u_n&=6,\\ \Delta^4u_n&=0. \end{aligned}$ Thus the fourth differences vanish for all $n,$ but the third differences do not.

Therefore, the correct answer is D.

26.

In the adjoining figure, every point of circle $O'$ is exterior to circle $O.$ Let $P$ and $Q$ be the points of intersection of an internal common tangent with the two external common tangents. Then the length of $PQ$ is

the average of the lengths of the internal and external common tangents

equal to the length of an external common tangent if and only if circles $O$ and $O'$ have equal radii

always equal to the length of an external common tangent

greater than the length of an external common tangent

the geometric mean of the lengths of the internal and external common tangents

Concepts:tangent circles tangent line homothety

Difficulty rating: 2200

Small Hint:

Mark the tangency point on each of the three tangent lines

Big Hint:

From either $P$ or $Q,$ tangent segments to the same circle have equal lengths

Solution:

Let the internal tangent touch $O$ and $O'$ at $R$ and $S.$ Let the external tangent through $P$ touch them at $X$ and $Y,$ and let the one through $Q$ touch them at $V$ and $W.$ Equal tangent segments from a point give $\begin{gathered} PR=PX,\\ PS=PY,\\ QR=QV,\\ QS=QW. \end{gathered}$ Along the internal tangent, $PR+QR=PQ$ and $PS+QS=PQ.$ Adding, $\begin{aligned} PR+QR+PS+QS=2PQ. \end{aligned}$ Along the external tangents, $PX+PY=XY$ and $QV+QW=VW.$ Therefore $2PQ=XY+VW.$ The two external common tangent segments have equal length, so $XY=VW$ and hence $PQ=XY=VW.$

Therefore, the correct answer is C.

27.

If $\begin{aligned} N&=\frac{\sqrt{\sqrt5+2}+\sqrt{\sqrt5-2}} {\sqrt{\sqrt5+1}}\\ &\quad-\sqrt{3-2\sqrt2}, \end{aligned}$ then $N$ equals

$1$

$2\sqrt2-1$

$\frac{\sqrt5}{2}$

$\sqrt{\frac52}$

none of these

Concepts:radical rationalizing denominator algebraic manipulation

Difficulty rating: 2080

Small Hint:

Square the numerator of the large fraction before simplifying it

Big Hint:

Recognize $3-2\sqrt2$ as $(\sqrt2-1)^2$

Solution:

Let $T$ be the first fraction. Its numerator squared is $\begin{aligned} &(\sqrt5+2)+(\sqrt5-2)\\ &\quad+2\sqrt{(\sqrt5+2)(\sqrt5-2)}\\ &=2\sqrt5+2. \end{aligned}$ Thus $T^2=\frac{2\sqrt5+2}{\sqrt5+1}=2,$ and $T=\sqrt2.$ Also $\sqrt{3-2\sqrt2}=\sqrt2-1.$ Hence $N=\sqrt2-(\sqrt2-1)=1.$

Therefore, the correct answer is A.

28.

Lines $L_1,$ $L_2,$ $\ldots,$ $L_{100}$ are distinct. All lines $L_{4n},$ $n$ a positive integer, are parallel to each other. All lines $L_{4n-3},$ $n$ a positive integer, pass through a given point $A.$ The maximum number of points of intersection of pairs of lines from the complete set $\{L_1,L_2,\ldots,L_{100}\}$ is

$4350$

$4351$

$4900$

$4901$

$9851$

Concepts:counting intersections combinations double counting

Difficulty rating: 2320

Small Hint:

Start with $\binom{100}{2}$ intersections in general position

Big Hint:

Remove the pairs among the $25$ parallel lines and collapse the pairs among the $25$ concurrent lines to one point

Solution:

There are $25$ indices divisible by $4$ and $25$ congruent to $1\pmod4.$ Starting from $\binom{100}{2}=4950,$ the parallel group contributes no intersections, removing $\binom{25}{2}=300.$ The concurrent group’s $300$ pairs all give one point rather than $300$ distinct points, removing another $299.$ All remaining intersections can be chosen distinct, so the maximum is $4950-300-299=4351.$

Therefore, the correct answer is B.

29.

Ann and Barbara were comparing their ages and found that Barbara is as old as Ann was when Barbara was as old as Ann had been when Barbara was half as old as Ann is. If the sum of their present ages is $44$ years, then Ann’s age is

$22$

$24$

$25$

$26$

$28$

Concepts:ages linear equation work backwards

Difficulty rating: 1950

Small Hint:

Let Ann’s and Barbara’s present ages be $x$ and $y,$ with $x+y=44$

Big Hint:

Translate each “when” by subtracting the same elapsed time from both ages

Solution:

Let Ann’s and Barbara’s present ages be $x$ and $y,$ and let their constant age difference be $d=x-y.$ At the first referenced time Ann was $y,$ so Barbara was $y-d.$ At the earlier referenced time Ann was that age $y-d,$ so Barbara was $y-2d.$ The final clause says this last age was $\frac{x}{2}.$ Hence $y-2(x-y)=\frac{x}{2},$ which gives $6y=5x.$ Together with $x+y=44,$ this yields $x=24$ and $y=20.$

Therefore, the correct answer is B.

30.

How many distinct ordered triples $(x,y,z)$ satisfy the equations $\begin{aligned} x+2y+4z&=12,\\ xy+4yz+2xz&=22,\\ xyz&=6? \end{aligned}$

none

$1$

$2$

$4$

$6$

Concepts:symmetry (algebra)Vieta’s Formulas permutations

Difficulty rating: 2320

Small Hint:

Rescale the variables by setting $x=2u,$ $y=v,$ and $z=\frac{w}{2}$

Big Hint:

The new variables have elementary symmetric sums $6,$ $11,$ and $6$

Solution:

Set $x=2u,$ $y=v,$ and $z=\frac{w}{2}.$ Dividing the first two equations by $2$ gives $\begin{aligned} u+v+w&=6,\\ uv+vw+uw&=11, \end{aligned}$ while $uvw=6.$ Thus $u,$ $v,$ and $w$ are the three roots of $p(t)=t^3-6t^2+11t-6.$ This polynomial factors as $p(t)=(t-1)(t-2)(t-3),$ so they are $1,$ $2,$ and $3$ in any order. Their $3!=6$ permutations produce $6$ distinct ordered triples $(x,y,z).$

Therefore, the correct answer is E.