1979 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

If rectangle $ABCD$ has area $72$ square meters and $E$ and $G$ are the midpoints of sides $AD$ and $CD,$ respectively, then the area of rectangle $DEFG$ in square meters is

$8$

$9$

$12$

$18$

$24$

Concepts:rectangle midpoint area

Difficulty rating: 840

Small Hint:

The two midpoint conditions halve the relevant side lengths

Big Hint:

Compare the area scale factor of $DEFG$ with that of $ABCD$

Solution:

The side lengths of $DEFG$ are one half the corresponding side lengths of $ABCD.$ Its area is therefore $(\frac{1}{2})(\frac{1}{2})=\frac{1}{4}$ of $72,$ which is $18.$

Therefore, the correct answer is D.

2.

For all nonzero real numbers $x$ and $y$ such that $x-y=xy,$ $\frac1x-\frac1y$ equals

$\frac1{xy}$

$\frac1{x-y}$

$0$

$-1$

$y-x$

Concepts:fraction algebraic manipulation

Difficulty rating: 960

Small Hint:

Combine the two reciprocals over the denominator $xy$

Big Hint:

Use the given equation to replace $y-x$

Solution:

We have $\begin{aligned} \frac1x-\frac1y&=\frac{y-x}{xy}\\ &=-\frac{x-y}{xy}\\ &=-1, \end{aligned}$ because $x-y=xy.$

Therefore, the correct answer is D.

3.

In the adjoining figure, $ABCD$ is a square, $ABE$ is an equilateral triangle and point $E$ is outside square $ABCD.$ What is the measure of $\angle AED$ in degrees?

$10$

$12.5$

$15$

$20$

$25$

Concepts:square (geometry)equilateral triangle isosceles triangle angle chasing

Difficulty rating: 1410

Small Hint:

Compare $DA,$ $AB,$ and $AE$

Big Hint:

Find $\angle DAE$ , then use the base angles of isosceles $\triangle DAE$

Solution:

The square and equilateral triangle give $DA=AB=AE.$ Also $\begin{aligned} \angle DAE &=\angle DAB+\angle BAE\\ &=90^\circ+60^\circ\\ &=150^\circ. \end{aligned}$ Thus $\triangle DAE$ is isosceles, and each base angle is $\frac{180^\circ-150^\circ}{2}=15^\circ.$

Therefore, the correct answer is C.

4.

For all real numbers $x,$ $x[x\{x(2-x)-4\}+10]+1$ equals

$-x^4+2x^3+4x^2+10x+1$

$-x^4-2x^3+4x^2+10x+1$

$-x^4-2x^3-4x^2+10x+1$

$-x^4-2x^3-4x^2-10x+1$

$-x^4+2x^3-4x^2+10x+1$

Concepts:polynomial distributive property algebraic manipulation

Difficulty rating: 1410

Small Hint:

Expand from the innermost parentheses outward

Big Hint:

Track the outer two factors of $x$ separately

Solution:

Expanding from the inside gives $\begin{aligned} &x[x\{x(2-x)-4\}+10]+1\\ &\quad=x[x(2x-x^2-4)+10]+1\\ &\quad=-x^4+2x^3-4x^2\\ &\qquad+10x+1. \end{aligned}$

Therefore, the correct answer is E.

5.

Find the sum of the digits of the largest even three digit number (in base ten representation) which is not changed when its units and hundreds digits are interchanged.

$22$

$23$

$24$

$25$

$26$

Concepts:digits place value parity

Difficulty rating: 960

Small Hint:

The hundreds and units digits must be equal

Big Hint:

Maximize the equal outer digits subject to the number being even, then maximize the tens digit

Solution:

The equal hundreds and units digit must be even. The largest possible nonzero such digit is $8,$ and the tens digit can be $9.$ Thus the largest number is $898,$ whose digit sum is $8+9+8=25.$

Therefore, the correct answer is D.

6.

$\frac32+\frac54+\frac98+\frac{17}{16}$ $+\frac{33}{32}+\frac{65}{64}-7=$

$-\frac1{64}$

$-\frac1{16}$

$0$

$\frac1{16}$

$\frac1{64}$

Concepts:fraction geometric sequence summation

Difficulty rating: 1540

Small Hint:

Write each fraction as $1$ plus a reciprocal power of $2$

Big Hint:

Sum $\frac12+\frac14+\cdots+\frac1{64}$ as a finite geometric series

Solution:

Each numerator is one more than its denominator, so the expression is $\begin{aligned} 6+\biggl(&\frac12+\frac14+\frac18\\ &+\frac1{16}+\frac1{32}+\frac1{64}\biggr)-7. \end{aligned}$ The parenthesized geometric sum is $1-\frac{1}{64}=\frac{63}{64}.$ Therefore the value is $6+\frac{63}{64}-7=-\frac{1}{64}.$

Therefore, the correct answer is A.

7.

The square of an integer is called a perfect square. If $x$ is a perfect square, the next larger perfect square is

$x+1$

$x^2+1$

$x^2+2x+1$

$x^2+x$

$x+2\sqrt x+1$

Concepts:perfect square algebraic manipulation pattern recognition

Difficulty rating: 1100

Small Hint:

Write $x=n^2$ for a nonnegative integer $n$

Big Hint:

Expand the next square $(n+1)^2$ and replace $n$ by $\sqrt x$

Solution:

Let $x=n^2$ with $n\ge0.$ The next larger perfect square is $\begin{aligned} (n+1)^2&=n^2+2n+1\\ &=x+2\sqrt x+1. \end{aligned}$

Therefore, the correct answer is E.

8.

Find the area of the smallest region bounded by the graphs of $y=|x|$ and $x^2+y^2=4.$

$\frac\pi4$

$\frac{3\pi}4$

$\pi$

$\frac{3\pi}2$

$2\pi$

Concepts:circle absolute value sector circle area

Difficulty rating: 1540

Small Hint:

The two rays of $y=|x|$ meet at the center of the circle

Big Hint:

Find the smaller central angle between the rays $y=x$ and $y=-x$

Solution:

The two rays meet the circle at angles $45^\circ$ and $135^\circ,$ so the smallest bounded region is a $90^\circ$ sector of the radius- $2$ circle. Its area is $\frac{90^\circ}{360^\circ}\pi(2)^2=\pi.$

Therefore, the correct answer is C.

9.

The product of $\sqrt[3]{4}$ and $\sqrt[4]{8}$ equals

$\sqrt[7]{12}$

$2\sqrt[7]{12}$

$\sqrt[7]{32}$

$\sqrt[12]{32}$

$2\sqrt[12]{32}$

Concepts:radical exponent algebraic manipulation

Difficulty rating: 1590

Small Hint:

Rewrite both radicands as powers of $2$

Big Hint:

Add the exponents $\frac23$ and $\frac34$ , then separate the integer part

Solution:

Writing both radicals as powers of $2$ gives $\begin{aligned} 4^{\frac{1}{3}}8^{\frac{1}{4}} &=2^{\frac{2}{3}}2^{\frac{3}{4}}\\ &=2^{\frac{17}{12}}\\ &=2\sqrt[12]{2^5}\\ &=2\sqrt[12]{32}. \end{aligned}$

Therefore, the correct answer is E.

10.

If $P_1P_2P_3P_4P_5P_6$ is a regular hexagon whose apothem (distance from the center to the midpoint of a side) is $2,$ and $Q_i$ is the midpoint of side $P_iP_{i+1}$ for $i=1,$ $2,$ $3,$ $4,$ then the area of quadrilateral $Q_1Q_2Q_3Q_4$ is

$6$

$2\sqrt6$

$\frac{8\sqrt3}{3}$

$3\sqrt3$

$4\sqrt3$

Concepts:regular polygon equilateral triangle area decomposition

Difficulty rating: 1780

Small Hint:

The points $Q_i$ lie on a circle of radius equal to the apothem

Big Hint:

Join the center to the four consecutive side midpoints and decompose the quadrilateral

Solution:

Let $O$ be the center. Consecutive apothems $OQ_i$ and $OQ_{i+1}$ have length $2$ and form a $60^\circ$ angle, so each $\triangle OQ_iQ_{i+1}$ is equilateral of side $2.$ The quadrilateral is the union of three such triangles, and hence has area $3\left(\frac{\sqrt3}{4}\cdot2^2\right)=3\sqrt3.$

Therefore, the correct answer is D.

11.

Find a positive integral solution to the equation $\frac{\substack{1+3+5+\cdots\\{}+(2n-1)}} {\substack{2+4+6+\cdots\\{}+2n}}=\frac{115}{116}.$

$110$

$115$

$116$

$231$

The equation has no positive integral solutions.

Concepts:sum of first n odd numbers summation linear equation

Difficulty rating: 1540

Small Hint:

Use the sums of the first $n$ odd and first $n$ even positive integers

Big Hint:

The fraction simplifies to $\frac{n}{n+1}$

Solution:

The numerator is $n^2,$ and the denominator is $n(n+1).$ Thus $\frac{n}{n+1}=\frac{115}{116},$ which gives $116n=115n+115,$ so $n=115.$

Therefore, the correct answer is B.

12.

In the adjoining figure, $CD$ is the diameter of a semicircle with center $O.$ Point $A$ lies on the extension of $DC$ past $C;$ point $E$ lies on the semicircle, and $B$ is the point of intersection (distinct from $E$ ) of line segment $AE$ with the semicircle. If length $AB$ equals length $OD,$ and the measure of $\angle EOD$ is $45^\circ,$ then the measure of $\angle BAO$ is

$10^\circ$

$15^\circ$

$20^\circ$

$25^\circ$

$30^\circ$

Concepts:circle isosceles triangle angle chasing

Difficulty rating: 1980

Small Hint:

Let $y=\angle BAO$ and use $AB=OB=OE$

Big Hint:

Express $\angle BOE$ in two ways using the isosceles triangles and the $45^\circ$ central angle

Solution:

Let $y=\angle BAO.$ Since $AB=OB,$ $\angle AOB=y,$ so $\angle EBO=2y.$ Also $OB=OE,$ hence $\angle BEO=2y$ and $\angle BOE=180^\circ-4y.$ On the other hand, $\angle BOD=180^\circ-y,$ so $\angle BOE=135^\circ-y.$ Therefore $180^\circ-4y=135^\circ-y,$ giving $y=15^\circ.$

Therefore, the correct answer is B.

13.

The inequality $y-x\lt\sqrt{x^2}$ is satisfied if and only if

$y\lt0$ or $y\lt2x$ (or both inequalities hold)

$y\gt0$ or $y\lt2x$ (or both inequalities hold)

$y^2\lt2xy$

$y\lt0$

$x\gt0$ and $y\lt2x$

Concepts:absolute value inequality casework

Difficulty rating: 1650

Small Hint:

Replace $\sqrt{x^2}$ by $|x|$

Big Hint:

Separate the cases $x\ge0$ and $x\lt0$ , then combine the resulting regions

Solution:

If $x\ge0,$ the inequality becomes $y-x\lt x,$ or $y\lt2x.$ If $x\lt0,$ it becomes $y-x\lt-x,$ or $y\lt0.$ Combining the two cases gives exactly $y\lt0$ or $y\lt2x,$ with overlap allowed.

Therefore, the correct answer is A.

14.

In a certain sequence of numbers, the first number is $1,$ and, for all $n\ge2,$ the product of the first $n$ numbers in the sequence is $n^2.$ The sum of the third and the fifth numbers in the sequence is

$\frac{25}{9}$

$\frac{31}{15}$

$\frac{61}{16}$

$\frac{576}{225}$

$34$

Concepts:algebraic manipulation fraction pattern recognition

Difficulty rating: 1590

Small Hint:

Divide the product of the first $n$ terms by the product of the first $n-1$ terms

Big Hint:

This gives the $n$ th term directly as a ratio of two squares

Solution:

For $n\ge2,$ the $n$ th term is $a_n=\frac{n^2}{(n-1)^2}.$ Therefore $a_3=\frac{9}{4}$ and $a_5=\frac{25}{16},$ whose sum is $\frac{36}{16}+\frac{25}{16}=\frac{61}{16}.$

Therefore, the correct answer is C.

15.

Two identical jars are filled with alcohol solutions, the ratio of the volume of alcohol to the volume of water being $p:1$ in one jar and $q:1$ in the other jar. If the entire contents of the two jars are mixed together, the ratio of the volume of alcohol to the volume of water in the mixture is

$\frac{p+q}{2}$

$\frac{p^2+q^2}{p+q}$

$\frac{2pq}{p+q}$

$\frac{2(p^2+pq+q^2)}{3(p+q)}$

$\frac{p+q+2pq}{p+q+2}$

Concepts:mixture ratio and proportion fraction

Difficulty rating: 1860

Small Hint:

Assign the same total volume to each jar and convert each ratio to alcohol and water fractions

Big Hint:

Add the two alcohol amounts and the two water amounts before forming their ratio

Solution:

For equal total volumes, the alcohol fractions are $\frac{p}{p+1}$ and $\frac{q}{q+1},$ while the water fractions are $\frac{1}{p+1}$ and $\frac{1}{q+1}.$ Their ratio is $\frac{\frac p{p+1}+\frac q{q+1}} {\frac1{p+1}+\frac1{q+1}} =\frac{p+q+2pq}{p+q+2}.$

Therefore, the correct answer is E.

16.

A circle with area $A_1$ is contained in the interior of a larger circle with area $A_1+A_2.$ If the radius of the larger circle is $3,$ and if $A_1,$ $A_2,$ $A_1+A_2$ is an arithmetic progression, then the radius of the smaller circle is

$\frac{\sqrt3}{2}$

$1$

$\frac2{\sqrt3}$

$\frac32$

$\sqrt3$

Concepts:circle area arithmetic sequence algebraic manipulation

Difficulty rating: 1590

Small Hint:

Use the middle-term condition for the three areas in arithmetic progression

Big Hint:

Express the larger circle’s area as a multiple of $A_1$

Solution:

The progression condition gives $2A_2=A_1+(A_1+A_2),$ so $A_2=2A_1.$ Hence the larger area is $3A_1=9\pi,$ and $A_1=3\pi.$ If the smaller radius is $r,$ then $\pi r^2=3\pi,$ so $r=\sqrt3.$

Therefore, the correct answer is E.

17.

Points $A,$ $B,$ $C,$ and $D$ are distinct and lie, in the given order, on a straight line. Line segments $AB,$ $AC,$ and $AD$ have lengths $x,$ $y,$ and $z,$ respectively. If line segments $AB$ and $CD$ may be rotated about points $B$ and $C,$ respectively, so that points $A$ and $D$ coincide, to form a triangle with positive area, then which of the following three inequalities must be satisfied? $\begin{aligned} &\text{I. }x\lt\frac z2\\ &\text{II. }y\lt x+\frac z2\\ &\text{III. }y\lt\frac z2 \end{aligned}$

$\mathrm{I}$ only

$\mathrm{II}$ only

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

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

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

Concepts:triangle inequality casework logical deduction

Difficulty rating: 2040

Small Hint:

After the rotations, identify the three side lengths of the new triangle

Big Hint:

Apply all three strict triangle inequalities to $x,$ $y-x,$ and $z-y$

Solution:

The new triangle has side lengths $x,$ $y-x,$ and $z-y.$ Its triangle inequalities give $\begin{aligned} x&\lt(y-x)+(z-y),\\ &\Longrightarrow x\lt \frac{z}{2};\\ y-x&\lt x+(z-y),\\ &\Longrightarrow y\lt x+\frac{z}{2};\\ z-y&\lt x+(y-x),\\ &\Longrightarrow y\gt \frac{z}{2}. \end{aligned}$ Thus I and II must hold, while III must fail.

Therefore, the correct answer is C.

18.

To the nearest thousandth, $\log_{10}2$ is $0.301$ and $\log_{10}3$ is $0.477.$ Which of the following is the best approximation of $\log_5 10?$

$\frac87$

$\frac97$

$\frac{10}{7}$

$\frac{11}{7}$

$\frac{12}{7}$

Concepts:logarithm estimation fraction

Difficulty rating: 1540

Small Hint:

Use $\log_{10}5=1-\log_{10}2$

Big Hint:

Apply change of base, then compare the decimal result with the five fractions

Solution:

We have $\log_{10}5\approx1-0.301=0.699.$ Therefore $\begin{aligned} \log_5 10&=\frac1{\log_{10}5}\\ &\approx\frac1{0.699}\\ &\approx1.431. \end{aligned}$ Of the choices, $\frac{10}{7}\approx1.429$ is closest.

Therefore, the correct answer is C.

19.

Find the sum of the squares of all real numbers satisfying the equation $x^{256}-256^{32}=0.$

$8$

$128$

$512$

$65{,}536$

$2(256^{32})$

Concepts:exponent perfect power zero product property

Difficulty rating: 1650

Small Hint:

Rewrite $256^{32}$ as a single power with exponent $256$

Big Hint:

Determine the two real values whose $256$ th powers equal $2^{256}$

Solution:

Since $256=2^8,$ $256^{32}=2^{256}.$ Thus $(\frac{x}{2})^{256}=1.$ The real solutions are $x=2$ and $x=-2,$ and the sum of their squares is $4+4=8.$

Therefore, the correct answer is A.

20.

If $a=\frac12$ and $(a+1)(b+1)=2,$ then the radian measure of $\operatorname{Arctan}a+\operatorname{Arctan}b$ equals

$\frac\pi2$

$\frac\pi3$

$\frac\pi4$

$\frac\pi5$

$\frac\pi6$

Concepts:trigonometric identity algebraic manipulation

Difficulty rating: 1780

Small Hint:

First solve the given product equation for $b$

Big Hint:

Use the tangent addition formula on the two positive inverse-tangent angles

Solution:

Substituting $a=\frac{1}{2}$ gives $b=\frac{1}{3}.$ If $u=\operatorname{Arctan}a$ and $v=\operatorname{Arctan}b,$ then $\begin{aligned} \tan(u+v)&=\frac{a+b}{1-ab}\\ &=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}\\ &=1. \end{aligned}$ Both angles are positive and their sum is less than $\frac{\pi}{2},$ so $u+v=\frac{\pi}{4}.$

Therefore, the correct answer is C.

21.

The length of the hypotenuse of a right triangle is $h,$ and the radius of the inscribed circle is $r.$ The ratio of the area of the circle to the area of the triangle is

$\frac{\pi r}{h+2r}$

$\frac{\pi r}{h+r}$

$\frac{\pi r}{2h+r}$

$\frac{\pi r^2}{h^2+r^2}$

none of these

Concepts:right triangle incircle, incenter, and inradius triangle area

Difficulty rating: 1870

Small Hint:

Use the formula $K=rs$ for a triangle’s area

Big Hint:

For a right triangle, relate the sum of the legs to $h$ and $r$

Solution:

If the legs are $x$ and $y,$ then the right-triangle inradius identity is $x+y=h+2r.$ Hence the semiperimeter is $s=\frac{x+y+h}{2}=h+r.$ The triangle’s area is $rs=r(h+r),$ while the circle’s area is $\pi r^2.$ Their ratio is $\frac{\pi r}{h+r}.$

Therefore, the correct answer is B.

22.

Find the number of pairs $(m,n)$ of integers which satisfy the equation $\begin{aligned} m^3+6m^2+5m &=27n^3+9n^2\\ &\quad+9n+1. \end{aligned}$

$0$

$1$

$3$

$9$

infinitely many

Concepts:divisibility modular arithmetic factoring

Difficulty rating: 1860

Small Hint:

Factor the left side as $m(m+1)(m+5)$

Big Hint:

Compare the two sides modulo $3$

Solution:

The left side is $\begin{gathered} m(m+1)(m+5)\\ =m(m+1)(m+2)\\ \quad+3m(m+1), \end{gathered}$ which is divisible by $3$ for every integer $m.$ The right side is $3(9n^3+3n^2+3n)+1,$ which is congruent to $1\pmod3.$ Therefore no integer pair satisfies the equation.

Therefore, the correct answer is A.

23.

The edges of a regular tetrahedron with vertices $A,$ $B,$ $C,$ and $D$ each have length one. Find the least possible distance between a pair of points $P$ and $Q,$ where $P$ is on edge $AB$ and $Q$ is on edge $CD.$

$\frac12$

$\frac34$

$\frac{\sqrt2}{2}$

$\frac{\sqrt3}{2}$

$\frac{\sqrt3}{3}$

Concepts:3D geometry distance formula optimization

Difficulty rating: 2100

Small Hint:

Use symmetric coordinates for a regular tetrahedron with vertices $(\pm1,\pm1,\pm1)$

Big Hint:

Parameterize points on the two opposite edges and minimize the squared distance

Solution:

Before scaling, take $\begin{aligned} A&=(1,1,1),\\ B&=(1,-1,-1),\\ C&=(-1,1,-1),\\ D&=(-1,-1,1). \end{aligned}$ These edges have length $2\sqrt2,$ so scale by $\frac{1}{2\sqrt2}.$ Points on $AB$ and $CD$ have forms $P=(1,t,t)$ and $Q=(-1,u,-u).$ Before scaling, $\begin{aligned} PQ^2&=4+(t-u)^2+(t+u)^2\\ &=4+2t^2+2u^2, \end{aligned}$ minimized at $t=u=0.$ After scaling, the minimum squared distance is $\frac{4}{8}=\frac{1}{2},$ so the distance is $\frac{\sqrt2}{2}.$

Therefore, the correct answer is C.

24.

A polygon is called “simple” if it is not self-intersecting. Sides $AB,$ $BC,$ and $CD$ of simple quadrilateral $ABCD$ have lengths $4,$ $5,$ and $20,$ respectively. If vertex angles $B$ and $C$ are obtuse and $\sin C=-\cos B=\frac35,$ then side $AD$ has length

$24$

$24.5$

$24.6$

$24.8$

$25$

Concepts:coordinate geometry trigonometry distance formula

Difficulty rating: 2040

Small Hint:

Place $B=(0,0)$ and $C=(5,0)$ , then use the signs forced by the obtuse angles

Big Hint:

Determine coordinates for $A$ and $D$ from the $3$ - $4$ - $5$ ratios

Solution:

Put $B=(0,0)$ and $C=(5,0).$ Since $B$ is obtuse and $-\cos B=\frac{3}{5},$ we may take $\begin{aligned} A&=4(-\frac{3}{5},\frac{4}{5})\\ &=(-\frac{12}{5},\frac{16}{5}). \end{aligned}$ The simple configuration and the obtuse angle at $C$ give $\begin{aligned} D&=C+20(\frac{4}{5},\frac{3}{5})\\ &=(21,12). \end{aligned}$ Thus $\begin{aligned} AD&=\sqrt{\begin{gathered} (21+\frac{12}{5})^2\\ {}+(12-\frac{16}{5})^2 \end{gathered}}\\ &=\sqrt{(\frac{117}{5})^2+(\frac{44}{5})^2}\\ &=25. \end{aligned}$

Therefore, the correct answer is E.

25.

If $q_1(x)$ and $r_1$ are the quotient and remainder, respectively, when the polynomial $x^8$ is divided by $x+\frac12,$ and if $q_2(x)$ and $r_2$ are the quotient and remainder, respectively, when $q_1(x)$ is divided by $x+\frac12,$ then $r_2$ equals

$\frac1{256}$

$-\frac1{16}$

$1$

$-16$

$256$

Concepts:polynomial factoring substitution

Difficulty rating: 2040

Small Hint:

Let $a=-\frac12$ and express the first quotient using $x^8-a^8$

Big Hint:

Apply the remainder theorem to $q_1(x)$ at $x=a$

Solution:

Let $a=-\frac{1}{2}.$ Since the first remainder is $a^8,$ $\begin{aligned} q_1(x)&=\frac{x^8-a^8}{x-a}\\ &=x^7+ax^6+\cdots+a^7. \end{aligned}$ Therefore the second remainder is $\begin{aligned} q_1(a)&=8a^7\\ &=8(-\frac{1}{2})^7\\ &=-\frac{1}{16}. \end{aligned}$

Therefore, the correct answer is B.

26.

The function $f$ satisfies the functional equation $f(x)+f(y)=f(x+y)-xy-1$ for every pair $x,$ $y$ of real numbers. If $f(1)=1,$ then the number of integers $n\ne1$ for which $f(n)=n$ is

$0$

$1$

$2$

$3$

infinite

Concepts:functional equation recursion quadratic

Difficulty rating: 2200

Small Hint:

Set one variable equal to $1$ to get a recurrence for consecutive integer inputs

Big Hint:

Find $f(0)$ , solve the recurrence for all integers, and then impose $f(n)=n$

Solution:

Setting $x=1$ gives $f(y+1)-f(y)=y+2.$ Setting $x=y=0$ gives $f(0)=-1.$ Extending the recurrence in both directions yields, for every integer $n,$ $f(n)=\frac{n^2+3n-2}{2}.$ Thus $f(n)=n$ is equivalent to $\begin{aligned} n^2+n-2&=(n+2)(n-1)\\ &=0. \end{aligned}$ The integer solutions are $n=-2$ and $n=1,$ and after excluding $1$ exactly one remains.

Therefore, the correct answer is B.

27.

An ordered pair $(b,c)$ of integers, each of which has absolute value less than or equal to five, is chosen at random, with each such ordered pair having an equal likelihood of being chosen. What is the probability that the equation $x^2+bx+c=0$ will not have distinct positive real roots?

$\frac{106}{121}$

$\frac{108}{121}$

$\frac{110}{121}$

$\frac{112}{121}$

none of these

Concepts:quadratic complementary probability systematic listing

Difficulty rating: 2200

Small Hint:

Distinct positive roots require $b\lt0,$ $c\gt0,$ and $b^2\gt4c$

Big Hint:

Count the qualifying $c$ values for each $b=-1,$ $b=-2,$ $\ldots,$ $b=-5,$ then take the complement among $121$ pairs

Solution:

There are $11^2=121$ ordered pairs. Distinct positive roots require $b\lt0,$ $c\gt0,$ and $b^2\gt4c.$ For $b=-1$ and $b=-2$ there are no choices; for $b=-3,$ $b=-4,$ and $b=-5$ there are respectively $2,$ $3,$ and $5$ choices of $c.$ Thus only $10$ pairs produce distinct positive roots, so the requested probability is $1-\frac{10}{121}=\frac{111}{121},$ which is not listed.

Therefore, the correct answer is E.

28.

Circles with centers $A,$ $B,$ and $C$ each have radius $r,$ where $1\lt r\lt2.$ The distance between each pair of centers is $2.$ If $B'$ is the point of intersection of circle $A$ and circle $C$ which is outside circle $B,$ and if $C'$ is the point of intersection of circle $A$ and circle $B$ which is outside circle $C,$ then length $B'C'$ equals

$3r-2$

$r^2$

$r+\sqrt{3(r-1)}$

$1+\sqrt{3(r^2-1)}$

none of these

Concepts:circle equilateral triangle similarity symmetry

Difficulty rating: 2200

Small Hint:

Introduce the analogous point $A'$ outside circle $A$ ; symmetry makes $A'B'C'$ equilateral

Big Hint:

Use the common centroid and the altitude $\sqrt{r^2-1}$ of an isosceles triangle with sides $r,$ $r,$ and $2$

Solution:

Let $A'$ be the analogous outer intersection of the circles centered at $B$ and $C.$ Then $A'B'C'$ and $ABC$ are concentric equilateral triangles. If $K$ is their common centroid and $M$ is the midpoint of $BC,$ then $\begin{aligned} A'M&=\sqrt{r^2-1},\\ MK&=\frac{\sqrt3}{3},\\ AK&=\frac{2\sqrt3}{3}. \end{aligned}$ Similarity gives $\begin{aligned} \frac{B'C'}2&=\frac{A'K}{AK}\\ &=\frac{\sqrt{r^2-1}+\frac{\sqrt3}{3}} {\frac{2\sqrt3}{3}}. \end{aligned}$ Hence $B'C'=1+\sqrt{3(r^2-1)}.$

Therefore, the correct answer is D.

29.

For each positive number $x,$ let $f(x)= \frac{\begin{gathered} \left(x+\frac1x\right)^6\\ {}-\left(x^6+\frac1{x^6}\right)-2 \end{gathered}} {\begin{gathered} \left(x+\frac1x\right)^3\\ {}+\left(x^3+\frac1{x^3}\right) \end{gathered}}.$ The minimum value of $f(x)$ is

$1$

$2$

$3$

$4$

$6$

Concepts:algebraic manipulation difference of squares AM-GM Inequality

Difficulty rating: 2100

Small Hint:

Set $t=x+\frac{1}{x}$ and write $x^3+x^{-3}$ in terms of $t$

Big Hint:

Factor the numerator as a difference of squares, then use $x+\frac{1}{x}\ge2$

Solution:

Let $t=x+\frac{1}{x}$ and $u=x^3+x^{-3}=t^3-3t.$ Since $x^6+x^{-6}=u^2-2,$ the numerator is $\begin{aligned} t^6-u^2&=(t^3-u)(t^3+u)\\ &=3t(t^3+u). \end{aligned}$ The denominator is $t^3+u,$ so $f(x)=3t=3(x+\frac{1}{x}).$ For $x\gt0,$ $x+\frac{1}{x}\ge2,$ with equality at $x=1.$ Thus the minimum is $6.$

Therefore, the correct answer is E.

30.

In $\triangle ABC,$ $E$ is the midpoint of side $BC$ and $D$ is on side $AC.$ If the length of $AC$ is $1$ and $\angle BAC=60^\circ,$ $\angle ABC=100^\circ,$ $\angle ACB=20^\circ,$ and $\angle DEC=80^\circ,$ then the area of $\triangle ABC$ plus twice the area of $\triangle CDE$ equals

$\frac14\cos10^\circ$

$\frac{\sqrt3}{8}$

$\frac14\cos40^\circ$

$\frac14\cos50^\circ$

$\frac18$

Concepts:triangle area congruence (geometry)similarity equilateral triangle

Difficulty rating: 2200

Small Hint:

Extend $AB$ to $F$ so that $AF=AC$ , creating an equilateral triangle

Big Hint:

Choose $G$ on $BF$ with $\angle BCG=20^\circ$ ; compare $\triangle BCG$ with $\triangle DCE$

Solution:

Extend $AB$ through $B$ to $F$ with $AF=AC=1.$ Then $\triangle ACF$ is equilateral. Choose $G$ on $BF$ so that $\angle BCG=20^\circ.$ The angle conditions give $\triangle FGC\cong\triangle ABC,$ while $\triangle BCG\sim\triangle DCE.$ Since $E$ is the midpoint of $BC,$ the similarity scale is $2,$ so $[BCG]=4[CDE].$ Decomposing the equilateral triangle gives $\frac{\sqrt3}{4} =2[ABC]+4[CDE].$ Dividing by $2$ yields $[ABC]+2[CDE]=\frac{\sqrt3}{8}.$

Therefore, the correct answer is B.