1978 AMC 12 — Problems & Online Timed Exam

Timed

1:15:00

1.

If $1-\frac4x+\frac4{x^2}=0,$ then $\frac2x$ equals

$-1$

$1$

$2$

$-1$ or $2$

$-1$ or $-2$

Answer: B

Concepts:quadratic substitution completing the square

Difficulty rating: 1240

Small Hint:

Let $u=\frac2x$

Big Hint:

The resulting quadratic is a perfect square

Solution:

With $u=\frac{2}{x},$ the equation becomes $1-2u+u^2=0,$ or $(u-1)^2=0.$ Therefore $\frac{2}{x}=u=1.$

Therefore, the correct answer is B.

2.

If four times the reciprocal of the circumference of a circle equals the diameter of the circle, then the area of the circle is

$\frac1{\pi^2}$

$\frac1\pi$

$1$

$\pi$

$\pi^2$

Answer: C

Concepts:circle circumference area

Difficulty rating: 1280

Small Hint:

Write the circumference and diameter in terms of the radius $r$

Big Hint:

The condition directly determines the product $\pi r^2$

Solution:

The condition is $4\left(\frac1{2\pi r}\right)=2r.$ Multiplying by $\pi r$ gives $2=2\pi r^2,$ so the area $\pi r^2$ equals $1.$

Therefore, the correct answer is C.

3.

For all nonzero numbers $x$ and $y$ such that $x=\frac{1}{y},$ $\left(x-\frac1x\right)\left(y+\frac1y\right)$ equals

$2x^2$

$2y^2$

$x^2+y^2$

$x^2-y^2$

$y^2-x^2$

Answer: D

Concepts:substitution difference of squares

Difficulty rating: 1210

Small Hint:

Use the reciprocal relation to replace $\frac{1}{x}$ and $\frac{1}{y}$

Big Hint:

The two factors become $x-y$ and $x+y$

Solution:

Since $x=\frac{1}{y},$ we also have $y=\frac{1}{x}.$ Thus the product is $(x-y)(y+x)=x^2-y^2.$

Therefore, the correct answer is D.

4.

If $a=1,$ $b=10,$ $c=100$ and $d=1000,$ then $\begin{aligned} &(a+b+c-d)+(a+b-c+d)\\ &\quad+(a-b+c+d)\\ &\quad+(-a+b+c+d) \end{aligned}$ is equal to

$1111$

$2222$

$3333$

$1212$

$4242$

Answer: B

Concepts:distributive property algebraic manipulation

Difficulty rating: 960

Small Hint:

Count how many times each variable appears with each sign

Big Hint:

The entire sum simplifies to twice $a+b+c+d$

Solution:

Each variable appears positively three times and negatively once, so the expression is $\begin{gathered} 2(a+b+c+d)\\ =2a+2b+2c+2d\\ =2+20+200+2000\\ =2222. \end{gathered}$

Therefore, the correct answer is B.

5.

Four boys bought a boat for $\$60.$ The first boy paid one half of the sum of the amounts paid by the other boys; the second boy paid one third of the sum of the amounts paid by the other boys; and the third boy paid one fourth of the sum of the amounts paid by the other boys. How much did the fourth boy pay?

$\$10$

$\$12$

$\$13$

$\$14$

$\$15$

Answer: C

Concepts:linear equation ratio and proportion money

Difficulty rating: 1280

Small Hint:

Replace each “sum paid by the other boys” by $60$ minus that boy’s payment

Big Hint:

Solve separately for the first three payments, then subtract their sum from $60$

Solution:

Let the first three payments be $w,$ $x,$ and $y.$ Then $\begin{aligned} w&=\frac12(60-w),\\ x&=\frac13(60-x),\\ y&=\frac14(60-y). \end{aligned}$ Hence $w=20,$ $x=15,$ $y=12.$ The fourth payment is $60-20-15-12=13$ dollars.

Therefore, the correct answer is C.

6.

The number of distinct pairs $(x,y)$ of real numbers satisfying both of the following equations: $\begin{aligned} x&=x^2+y^2,\\ y&=2xy. \end{aligned}$ is

$0$

$1$

$2$

$3$

$4$

Answer: E

Concepts:system of equations casework zero product property

Difficulty rating: 1590

Small Hint:

Factor the second equation as $y(1-2x)=0$

Big Hint:

Handle $y=0$ and $x=\frac12$ as separate cases

Solution:

If $y=0,$ the first equation gives $x=x^2,$ so $(x,y)=(0,0)$ or $(1,0).$ If $y\ne0,$ the second equation gives $x=\frac{1}{2}.$ The first then gives $y^2=\frac{1}{4},$ producing $(\frac{1}{2},\frac{1}{2})$ and $(\frac{1}{2},-\frac{1}{2}).$ There are four pairs.

Therefore, the correct answer is E.

7.

Opposite sides of a regular hexagon are $12$ inches apart. The length of each side, in inches, is

$7.5$

$6\sqrt2$

$5\sqrt2$

$\frac92\sqrt3$

$4\sqrt3$

Answer: E

Concepts:regular polygon special right triangle

Difficulty rating: 1440

Small Hint:

The distance between opposite sides is twice the apothem

Big Hint:

A regular hexagon of side $s$ has apothem $\frac{\sqrt3}{2}s$

Solution:

The distance between opposite sides is twice the apothem, hence $s\sqrt3.$ Therefore $s\sqrt3=12,$ so $s=\frac{12}{\sqrt3}=4\sqrt3.$

Therefore, the correct answer is E.

8.

If $x\ne y$ and the sequences $x,$ $a_1,$ $a_2,$ $y$ and $x,$ $b_1,$ $b_2,$ $b_3,$ $y$ each are in arithmetic progression, then $\frac{a_2-a_1}{b_2-b_1}$ equals

$\frac23$

$\frac34$

$1$

$\frac43$

$\frac32$

Answer: D

Concepts:arithmetic sequence ratio and proportion

Difficulty rating: 1330

Small Hint:

Count the equal steps from $x$ to $y$ in each sequence

Big Hint:

The two common differences are $\frac{y-x}{3}$ and $\frac{y-x}{4}$

Solution:

The first common difference is $\frac{y-x}{3},$ and the second is $\frac{y-x}{4}.$ Since $x\ne y,$ their ratio is $\frac{\frac{y-x}{3}}{\frac{y-x}{4}}=\frac43.$

Therefore, the correct answer is D.

9.

If $x\lt0,$ then $\left|x-\sqrt{(x-1)^2}\right|$ equals

$1$

$1-2x$

$-2x-1$

$1+2x$

$2x-1$

Answer: B

Concepts:absolute value radical casework

Difficulty rating: 1360

Small Hint:

Use $\sqrt{u^2}=|u|$

Big Hint:

When $x\lt0,$ determine the signs of $x-1$ and $2x-1$

Solution:

Because $x-1\lt0,$ $\sqrt{(x-1)^2}=|x-1|=1-x.$ Therefore $\begin{aligned} \left|x-(1-x)\right|&=|2x-1|\\ &=1-2x, \end{aligned}$ since $2x-1\lt0.$

Therefore, the correct answer is B.

10.

If $B$ is a point on circle $C$ with center $P,$ then the set of all points $A$ in the plane of circle $C$ such that the distance between $A$ and $B$ is less than or equal to the distance between $A$ and any other point on circle $C$ is

the line segment from $P$ to $B$

the ray beginning at $P$ and passing through $B$

a ray beginning at $B$

a circle whose center is $P$

a circle whose center is $B$

Answer: B

Concepts:circle extremal argument

Difficulty rating: 1510

Small Hint:

For $A\ne P,$ the nearest point of the circle lies on the ray from $P$ through $A$

Big Hint:

Require that this radial nearest point be the fixed point $B$

Solution:

For any $A\ne P,$ the closest point of the circle to $A$ is where the ray from $P$ through $A$ meets the circle. This point is $B$ exactly when $A$ lies on the ray from $P$ through $B.$ The center $P$ also qualifies because every point on the circle is equally distant from it. Thus the locus is that ray.

Therefore, the correct answer is B.

11.

If $r$ is positive and the line whose equation is $x+y=r$ is tangent to the circle whose equation is $x^2+y^2=r,$ then $r$ equals

$\frac12$

$1$

$2$

$\sqrt2$

$2\sqrt2$

Answer: C

Concepts:coordinate geometry circle tangent line

Difficulty rating: 1590

Small Hint:

The circle has center at the origin and radius $\sqrt r$

Big Hint:

Set the distance from the origin to the line equal to the circle’s radius

Solution:

The distance from the origin to $x+y-r=0$ is $\frac{r}{\sqrt2}.$ Tangency requires this to equal the radius $\sqrt r,$ so $\frac r{\sqrt2}=\sqrt r.$ Since $r\gt0,$ squaring and dividing by $r$ gives $r=2.$

Therefore, the correct answer is C.

12.

In $\triangle ADE,$ $\angle ADE=140^\circ,$ points $B$ and $C$ lie on sides $AD$ and $AE,$ respectively, and points $A,$ $B,$ $C,$ $D,$ $E$ are distinct. If lengths $AB,$ $BC,$ $CD,$ and $DE$ are all equal, then the measure of $\angle EAD$ is

$5^\circ$

$6^\circ$

$7.5^\circ$

$8^\circ$

$10^\circ$

Answer: E

Concepts:isosceles triangle angle chasing angle sum

Difficulty rating: 2040

Small Hint:

Name $\angle BAC=\angle BCA=x$ and use the successive isosceles triangles

Big Hint:

Express the other base angles in terms of $x,$ then use the angle sum in $\triangle ADE$

Solution:

Let $x=\angle BAC=\angle BCA,$ $y=\angle CBD=\angle CDB,$ and $z=\angle DCE=\angle DEC.$ The exterior-angle theorem applied successively gives $y=2x,\qquad z=x+y=3x.$ The angles of $\triangle ADE$ are $x,$ $140^\circ,$ and $z,$ so $x+140^\circ+3x=180^\circ.$ Hence $x=10^\circ.$

Therefore, the correct answer is E.

13.

If $a,$ $b,$ $c,$ and $d$ are nonzero numbers such that $c$ and $d$ are the solutions of $x^2+ax+b=0$ and $a$ and $b$ are the solutions of $x^2+cx+d=0,$ then $a+b+c+d$ equals

$0$

$-2$

$2$

$4$

$\frac{-1+\sqrt5}{2}$

Answer: B

Concepts:quadratic Vieta’s Formulas system of equations

Difficulty rating: 2040

Small Hint:

Apply Vieta’s formulas to both quadratics

Big Hint:

The two sum equations imply $b=d$ ; then use the product equations and nonzero condition

Solution:

Vieta’s formulas give $\begin{aligned} c+d&=-a,& cd&=b,\\ a+b&=-c,& ab&=d. \end{aligned}$ The two sum equations imply $b=d.$ Since $b=d\ne0,$ the product equations give $a=c=1.$ Then $a+b=-c$ gives $b=d=-2,$ and therefore $a+b+c+d=-2.$

Therefore, the correct answer is B.

14.

If an integer $n,$ greater than $8,$ is a solution of the equation $x^2-ax+b=0$ and the representation of $a$ in the base $n$ numeration system is $18,$ then the base $n$ representation of $b$ is

$18$

$20$

$80$

$81$

$280$

Answer: C

Concepts:number base quadratic Vieta’s Formulas

Difficulty rating: 1780

Small Hint:

Translate the base- $n$ numeral $18$ into $n+8$

Big Hint:

Use the sum and product of the two roots

Solution:

In ordinary notation, $a=(18)_n=n+8.$ Since one root is $n,$ the other root must be $8.$ Their product is $b=8n,$ whose base- $n$ representation is $80.$

Therefore, the correct answer is C.

15.

If $\sin x+\cos x=\frac15$ and $0\le x\lt\pi,$ then $\tan x$ is

$-\frac43$

$-\frac34$

$\frac34$

$\frac43$

not completely determined by the given information

Answer: A

Concepts:trigonometric identity quadratic casework

Difficulty rating: 1650

Small Hint:

Square the given equation to determine $\sin x\cos x$

Big Hint:

Treat $\sin x$ and $\cos x$ as roots of a quadratic, then use $0\le x\lt\pi$

Solution:

Squaring gives $1+2\sin x\cos x=\frac{1}{25},$ so $\sin x\cos x=-\frac{12}{25}.$ Thus $\sin x$ and $\cos x$ are the roots of $t^2-\frac15t-\frac{12}{25}=0,$ namely $\frac{4}{5}$ and $-\frac{3}{5}.$ Because $\sin x\ge0$ on the given interval, $\sin x=\frac{4}{5}$ and $\cos x=-\frac{3}{5}.$ Hence $\tan x=-\frac{4}{3}.$

Therefore, the correct answer is A.

16.

In a room containing $N$ people, $N\gt3,$ at least one person has not shaken hands with everyone else in the room. What is the maximum number of people in the room that could have shaken hands with everyone else?

$0$

$1$

$N-1$

$N$

none of these

Answer: E

Concepts:graph theory extremal argument

Difficulty rating: 1780

Small Hint:

A missed handshake always involves two people

Big Hint:

Find an upper bound, then realize it by omitting just one handshake

Solution:

If one person has missed a handshake, the other person in that missed pair also has not shaken hands with everyone. Thus at most $N-2$ people can have shaken hands with everyone. This is attainable when exactly two people fail to shake hands with each other and every other handshake occurs. Since $N-2$ is not listed, the answer is “none of these.”

Therefore, the correct answer is E.

17.

If $k$ is a positive number and $f$ is a function such that, for every positive number $x,$ $\left[f(x^2+1)\right]^{\sqrt x}=k,$ then, for every positive number $y,$ $\left[f\left(\frac{9+y^2}{y^2}\right)\right]^{\sqrt{\frac{12}{y}}}$ is equal to

$\sqrt k$

$2k$

$k\sqrt k$

$k^2$

$y\sqrt k$

Answer: D

Concepts:functional equation substitution exponent

Difficulty rating: 2040

Small Hint:

Choose $x$ so that $x^2+1=\frac{9+y^2}{y^2}$

Big Hint:

Compare $\sqrt{\frac{12}{y}}$ with $\sqrt x$ after making the substitution

Solution:

Set $x=\frac{3}{y},$ which is positive. Then $\begin{aligned} x^2+1&=\frac{9+y^2}{y^2},\\ \sqrt{\frac{12}{y}}&=2\sqrt x. \end{aligned}$ Therefore the requested expression is $\left(\left[f(x^2+1)\right]^{\sqrt x}\right)^2=k^2.$

Therefore, the correct answer is D.

18.

What is the smallest positive integer $n$ such that $\sqrt n-\sqrt{n-1}\lt0.01?$

$2499$

$2500$

$2501$

$10{,}000$

There is no such integer.

Answer: C

Concepts:radical rationalizing denominator inequality

Difficulty rating: 1860

Small Hint:

Rationalize $\sqrt n-\sqrt{n-1}$

Big Hint:

Compare the resulting denominator with $100$ near $n=2500$

Solution:

Rationalizing gives $\sqrt n-\sqrt{n-1} =\frac1{\sqrt n+\sqrt{n-1}}.$ For $n=2500,$ the denominator is $50+\sqrt{2499}\lt100,$ so the difference exceeds $0.01.$ For $n=2501,$ the denominator is $\sqrt{2501}+50\gt100,$ so the difference is less than $0.01.$ The denominator increases with $n,$ making $2501$ the least such integer.

Therefore, the correct answer is C.

19.

A positive integer $n$ not exceeding $100$ is chosen in such a way that if $n\le50,$ then the probability of choosing $n$ is $p,$ and if $n\gt50,$ then the probability of choosing $n$ is $3p.$ The probability that a perfect square is chosen is

$0.05$

$0.065$

$0.08$

$0.09$

$0.1$

Answer: C

Concepts:basic probability perfect square counting integers in a range

Difficulty rating: 1650

Small Hint:

First use the total probability to determine $p$

Big Hint:

Count the perfect squares at most $50$ and those from $51$ through $100$ separately

Solution:

The total probability is $50p+50(3p)=200p=1,$ so $p=0.005.$ There are seven perfect squares at most $50$ and three more, $64,$ $81,$ $100,$ above $50.$ Hence the desired probability is $7p+3(3p)=16p=0.08.$

Therefore, the correct answer is C.

20.

If $a,$ $b,$ $c$ are nonzero real numbers such that $\begin{aligned} \frac{a+b-c}{c} &=\frac{a-b+c}{b}\\ &=\frac{-a+b+c}{a}, \end{aligned}$ and $x=\frac{(a+b)(b+c)(c+a)}{abc},$ and $x\lt0,$ then $x$ equals

$-1$

$-2$

$-4$

$-6$

$-8$

Answer: A

Concepts:system of equations casework symmetry (algebra)

Difficulty rating: 2100

Small Hint:

Set the three equal fractions to $t$ and compare pairs of the resulting equations

Big Hint:

The comparison forces either $a=b=c$ or $a+b+c=0$ ; use the sign of $x$

Solution:

Let the common value be $t.$ The first two resulting equations imply $(b-c)(t+2)=0,$ and cyclic comparisons give the analogous relations. Thus either $a=b=c,$ which gives $x=8,$ or $t=-2.$ In the latter case $a+b=-c,$ $b+c=-a,$ and $c+a=-b,$ so $x=\frac{(-c)(-a)(-b)}{abc}=-1.$ The condition $x\lt0$ selects the latter value.

Therefore, the correct answer is A.

21.

For all positive numbers $x$ distinct from $1,$ $\frac1{\log_3x}+\frac1{\log_4x}+\frac1{\log_5x}$ equals

$\frac1{\log_{60}x}$

$\frac1{\log_x60}$

$\frac1{(\log_3x)(\log_4x)(\log_5x)}$

$\frac{12}{(\log_3x)+(\log_4x)+(\log_5x)}$

$\frac{\log_2x}{(\log_3x)(\log_5x)}$
${}+\frac{\log_3x}{(\log_2x)(\log_5x)}$
${}+\frac{\log_5x}{(\log_2x)(\log_3x)}$

Answer: A

Concepts:logarithm algebraic manipulation

Difficulty rating: 1780

Small Hint:

Use the reciprocal identity $\frac{1}{\log_bx}=\log_xb$

Big Hint:

Combine the resulting sum with the product rule for logarithms

Solution:

Let $S$ denote the given sum. Changing bases and combining logarithms gives $\begin{aligned} S&=\frac1{\log_3x}+\frac1{\log_4x}\\ &\quad+\frac1{\log_5x}\\ &=\log_x(3\cdot4\cdot5)\\ &=\log_x60\\ &=\frac1{\log_{60}x}. \end{aligned}$

Therefore, the correct answer is A.

22.

The following four statements, and only these, are found on a card:

On this card exactly one statement is false.
On this card exactly two statements are false.
On this card exactly three statements are false.
On this card exactly four statements are false.

(Assume each statement on the card is either true or false.) Among them the number of false statements is exactly

$0$

$1$

$2$

$3$

$4$

Answer: D

Concepts:logical deduction truth-tellers and liars

Difficulty rating: 1860

Small Hint:

Assume the actual number of false statements is $m$

Big Hint:

For each possible $m,$ count how many of the four displayed statements would then be true

Solution:

If the actual number $m$ of false statements is one of $1,$ $2,$ $3,$ $4,$ exactly one displayed statement—the one naming $m$ —is true. Therefore exactly three statements are false, forcing $m=3.$ This is consistent: the third statement is true and the other three are false.

Therefore, the correct answer is D.

23.

Vertex $E$ of equilateral triangle $ABE$ is in the interior of square $ABCD,$ and $F$ is the point of intersection of diagonal $BD$ and line segment $AE.$ If length $AB$ is $\sqrt{1+\sqrt3},$ then the area of $\triangle ABF$ is

$1$

$\frac{\sqrt2}{2}$

$\frac{\sqrt3}{2}$

$4-2\sqrt3$

$\frac12+\frac{\sqrt3}{4}$

Answer: C

Concepts:coordinate geometry equilateral triangle area

Difficulty rating: 2040

Small Hint:

Place $A=(0,0)$ and $B=(s,0),$ where $s=AB$

Big Hint:

Find the intersection of $AE:\ y=\sqrt3x$ and $BD:\ y=s-x$

Solution:

Let $A=(0,0),$ $B=(s,0),$ and $D=(0,s),$ where $s=\sqrt{1+\sqrt3}.$ The two lines containing $F$ have equations $\begin{aligned} AE:\quad y&=\sqrt3x,\\ BD:\quad y&=s-x. \end{aligned}$ Hence the altitude of $F$ above $AB$ is $\frac{s\sqrt3}{1+\sqrt3}.$ Therefore $[\triangle ABF] =\frac{s^2\sqrt3}{2(1+\sqrt3)} =\frac{\sqrt3}{2}.$

Therefore, the correct answer is C.

24.

If the distinct nonzero numbers $x(y-z),$ $y(z-x),$ $z(x-y)$ form a geometric progression with common ratio $r,$ then $r$ satisfies the equation

$r^2+r+1=0$

$r^2-r+1=0$

$r^4+r^2-1=0$

$(r+1)^4+r=0$

$(r-1)^4+r=0$

Answer: A

Concepts:geometric sequence algebraic manipulation

Difficulty rating: 2040

Small Hint:

Add the three given expressions

Big Hint:

Write the three nonzero terms as $u,$ $ur,$ $ur^2$

Solution:

The three expressions have sum $\begin{aligned} &x(y-z)+y(z-x)\\ &\qquad+z(x-y)=0. \end{aligned}$ Writing the nonzero geometric progression as $u,$ $ur,$ $ur^2$ gives $u(1+r+r^2)=0.$ Since $u\ne0,$ $r^2+r+1=0.$

Therefore, the correct answer is A.

25.

Let $a$ be a positive number. Consider the set $S$ of all points whose rectangular coordinates $(x,y)$ satisfy all of the following conditions: $\begin{aligned} &\text{(i) }\frac a2\le x\le2a\\ &\text{(ii) }\frac a2\le y\le2a\\ &\text{(iii) }x+y\ge a\\ &\text{(iv) }x+a\ge y\\ &\text{(v) }y+a\ge x. \end{aligned}$ The boundary of set $S$ is a polygon with

$3$ sides

$4$ sides

$5$ sides

$6$ sides

$7$ sides

Answer: D

Concepts:coordinate geometry inequality

Difficulty rating: 2100

Small Hint:

Begin with the square described by conditions $\text{(i)}$ and $\text{(ii)}$

Big Hint:

Determine which condition is redundant and which two cut off opposite corners

Solution:

The first two conditions form the square $[\frac{a}{2},2a]\times[\frac{a}{2},2a].$ Within this square, $x+y\ge a$ is automatic. The last two conditions are equivalent to $|x-y|\le a;$ their boundary lines cut off the corners $(2a,\frac{a}{2})$ and $(\frac{a}{2},2a).$ Cutting two opposite corners from a square produces a hexagon, so the boundary has six sides.

Therefore, the correct answer is D.

26.

In $\triangle ABC,$ $AB=10,$ $AC=8,$ and $BC=6.$ Circle $P$ is the circle with smallest radius which passes through $C$ and is tangent to $AB.$ Let $Q$ and $R$ be the points of intersection, distinct from $C,$ of circle $P$ with sides $AC$ and $BC,$ respectively. The length of segment $QR$ is

$4.75$

$4.8$

$5$

$4\sqrt2$

$3\sqrt3$

Answer: B

Concepts:right triangle tangent circles altitude inscribed angle

Difficulty rating: 2100

Small Hint:

The $6$ - $8$ - $10$ triangle is right at $C$ ; let $H$ be the foot from $C$ to $AB$

Big Hint:

The smallest circle has $CH$ as a diameter, and $\angle QCR=90^\circ$

Solution:

Let $H$ be the foot of the altitude from $C$ to $AB.$ Among circles through $C$ tangent to $AB,$ the least radius occurs when the tangency point is $H,$ so $CH$ is a diameter. The area of the right triangle gives $CH=\frac{AC\cdot BC}{AB}=\frac{8\cdot6}{10}=4.8.$ Also $\angle QCR=90^\circ,$ so $QR$ is a diameter of circle $P.$ Therefore $QR=CH=4.8.$

Therefore, the correct answer is B.

27.

There is more than one integer greater than $1$ which, when divided by any integer $k$ such that $2\le k\le11,$ has a remainder of $1.$ What is the difference between the two smallest such integers?

$2310$

$2311$

$27{,}720$

$27{,}721$

none of these

Answer: C

Concepts:least common multiple modular arithmetic

Difficulty rating: 1780

Small Hint:

Each desired integer is $1$ more than a common multiple of every integer from $2$ through $11$

Big Hint:

The difference of consecutive such integers is the least common multiple of those divisors

Solution:

A qualifying integer is congruent to $1$ modulo every integer from $2$ through $11,$ hence modulo $\begin{aligned} L&=\operatorname{lcm}(2,\ldots,11)\\ &=2^3\cdot3^2\cdot5\cdot7\cdot11\\ &=27720. \end{aligned}$ The two smallest qualifying integers greater than $1$ are $L+1$ and $2L+1,$ whose difference is $L=27720.$

Therefore, the correct answer is C.

28.

If $\triangle A_1A_2A_3$ is equilateral and $A_{n+3}$ is the midpoint of line segment $A_nA_{n+1}$ for all positive integers $n,$ then the measure of $\angle A_{44}A_{45}A_{43}$ equals

$30^\circ$

$45^\circ$

$60^\circ$

$90^\circ$

$120^\circ$

Answer: E

Concepts:midpoint recursion vector

Difficulty rating: 2200

Small Hint:

Let $\mathbf d_n=\overrightarrow{A_nA_{n+1}}$ and derive a recurrence for these vectors

Big Hint:

Show that $\mathbf d_{n+4}=-\frac14\mathbf d_n,$ reducing the requested angle to one among the first few points

Solution:

Let $\mathbf d_n=\overrightarrow{A_nA_{n+1}}.$ The midpoint rule gives $\mathbf d_{n+3}=\frac12(\mathbf d_n+\mathbf d_{n+1})$ and also $\mathbf d_n+\mathbf d_{n+1}+\mathbf d_{n+2}=\frac12\mathbf d_n.$ Consequently, $\begin{aligned} \mathbf d_{n+4} &=\frac12(\mathbf d_{n+1}+\mathbf d_{n+2})\\ &=-\frac14\mathbf d_n. \end{aligned}$ Thus $\mathbf d_{43}$ and $\mathbf d_{44}$ are the same positive scalar multiple of $\mathbf d_3$ and $\mathbf d_4,$ respectively, so $\angle A_{44}A_{45}A_{43}=\angle A_4A_5A_3.$ Since $A_4$ and $A_5$ are the midpoints of $A_1A_2$ and $A_2A_3,$ $A_4A_5\parallel A_1A_3.$ The equilateral-triangle angles then give $\angle A_4A_5A_3=120^\circ.$

Therefore, the correct answer is E.

29.

Sides $AB,$ $BC,$ $CD,$ and $DA,$ respectively, of convex quadrilateral $ABCD$ are extended past $B,$ $C,$ $D,$ and $A$ to points $B',$ $C',$ $D',$ and $A'.$ Also, $AB=BB'=6,$ $BC=CC'=7,$ $CD=DD'=8,$ and $DA=AA'=9;$ and the area of $ABCD$ is $10.$ The area of $A'B'C'D'$ is

$20$

$40$

$45$

$50$

$60$

Answer: D

Concepts:vector area decomposition

Difficulty rating: 2040

Small Hint:

In vector notation, $A'=2A-D,$ $B'=2B-A,$ and similarly for the other two vertices

Big Hint:

Substitute these expressions into the cross-product area formula for a polygon

Solution:

Using position vectors, the equal extensions give $\begin{aligned} A'&=2A-D,& B'&=2B-A,\\ C'&=2C-B,& D'&=2D-C. \end{aligned}$ Substitution into the oriented polygon-area sum shows that the diagonal cross terms cancel: $\begin{aligned} &A'\mathbin{\times}B' +B'\mathbin{\times}C'\\ &\quad+C'\mathbin{\times}D' +D'\mathbin{\times}A'\\ &=5A\mathbin{\times}B +5B\mathbin{\times}C\\ &\quad+5C\mathbin{\times}D +5D\mathbin{\times}A. \end{aligned}$ Hence the outer area is five times the original area, or $5\cdot10=50.$

Therefore, the correct answer is D.

30.

In a tennis tournament, $n$ women and $2n$ men play, and each player plays exactly one match with every other player. If there are no ties and the ratio of the number of matches won by women to the number of matches won by men is $\frac{7}{5},$ then $n$ equals

$2$

$4$

$6$

$7$

none of these

Answer: E

Concepts:basic counting divisibility bounding to limit cases

Difficulty rating: 2200

Small Hint:

Let $k$ be the number of mixed matches won by women and count all wins by women

Big Hint:

Use $0\le k\le2n^2$ together with the required $\frac{7}{12}$ share of all match wins

Solution:

There are $\frac{3n(3n-1)}{2}$ matches, so women must win $\frac{7n(3n-1)}{8}$ matches. If women win $k$ of the $2n^2$ mixed matches, then $\frac{n(n-1)}2+k=\frac{7n(3n-1)}8,$ giving $k=\frac{n(17n-3)}{8}.$ The bound $k\le2n^2$ forces $n\le3.$ For $n=1,$ $2,$ this formula is not an integer, while $n=3$ gives $k=18,$ which is possible. Thus $n=3,$ which is not among the listed numerical choices.

Therefore, the correct answer is E.