1981 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

If $\sqrt{x+2}=2,$ then $(x+2)^2$ equals

$\sqrt2$

$2$

$4$

$8$

$16$

Concepts:radical substitution

Difficulty rating: 900

Small Hint:

The equation already gives the value of $x+2$

Big Hint:

Square the known value once more

Solution:

From $\sqrt{x+2}=2,$ we get $x+2=4.$ Therefore $(x+2)^2=4^2=16.$

Therefore, the correct answer is E.

2.

Point $E$ is on side $AB$ of square $ABCD.$ If $EB$ has length one and $EC$ has length two, then the area of the square is

$\sqrt3$

$\sqrt5$

$3$

$2\sqrt3$

$5$

Concepts:square (geometry)Pythagorean Theorem area

Difficulty rating: 1160

Small Hint:

Triangle $EBC$ is a right triangle

Big Hint:

The unknown leg $BC$ is also the side length of the square

Solution:

Let the square’s side length be $s.$ In right triangle $EBC,$ $s^2+1^2=2^2.$ Hence $s^2=3,$ which is exactly the area of the square.

Therefore, the correct answer is C.

3.

For $x\ne0,$ $\frac1x+\frac1{2x}+\frac1{3x}$ equals

$\frac1{2x}$

$\frac1{6x}$

$\frac5{6x}$

$\frac{11}{6x}$

$\frac1{6x^3}$

Concepts:fraction algebraic manipulation

Difficulty rating: 920

Small Hint:

Use $6x$ as a common denominator

Big Hint:

Add the resulting numerators $6,$ $3,$ and $2$

Solution:

Using the common denominator $6x$ gives $\begin{aligned} \frac1x+\frac1{2x}+\frac1{3x} &=\frac6{6x}+\frac3{6x}\\ &\quad+\frac2{6x}\\ &=\frac{11}{6x}. \end{aligned}$

Therefore, the correct answer is D.

4.

If three times the larger of two numbers is four times the smaller and the difference between the numbers is $8,$ then the larger of the two numbers is

$16$

$24$

$32$

$44$

$52$

Concepts:linear equation ratio and proportion

Difficulty rating: 980

Small Hint:

Let $L$ and $S$ denote the larger and smaller numbers

Big Hint:

Use $3L=4S$ together with $L-S=8$

Solution:

The relation $3L=4S$ gives $S=\frac{3L}{4}.$ Thus $L-S=L-\frac{3L}{4}=\frac L4=8,$ so $L=32.$

Therefore, the correct answer is C.

5.

In trapezoid $ABCD,$ sides $AB$ and $CD$ are parallel, and diagonal $BD$ and side $AD$ have equal length. If $\angle DCB=110^\circ$ and $\angle CBD=30^\circ,$ then $\angle ADB=$

$80^\circ$

$90^\circ$

$100^\circ$

$110^\circ$

$120^\circ$

Concepts:trapezoid parallel lines isosceles triangle angle chasing

Difficulty rating: 1360

Small Hint:

Use the parallel bases to find the full interior angle at $B$

Big Hint:

Then use $AD=BD$ in triangle $ABD$

Solution:

Since $AB\parallel CD,$ consecutive interior angles give $\angle ABC=180^\circ-110^\circ=70^\circ.$ Hence $\angle ABD=70^\circ-30^\circ=40^\circ.$ Because $AD=BD,$ triangle $ABD$ has $\angle DAB=\angle ABD=40^\circ.$ Therefore $\begin{aligned} \angle ADB &=180^\circ-40^\circ-40^\circ\\ &=100^\circ. \end{aligned}$

Therefore, the correct answer is C.

6.

If $\frac{x}{x-1}=\frac{y^2+2y-1}{y^2+2y-2},$ then $x$ equals

$y^2+2y-1$

$y^2+2y-2$

$y^2+2y+2$

$y^2+2y+1$

$-y^2-2y+1$

Concepts:rational equation algebraic manipulation

Difficulty rating: 1210

Small Hint:

Notice that each denominator is one less than its numerator

Big Hint:

Set $A=y^2+2y-1$ and compare $\frac{x}{x-1}$ with $\frac A{A-1}$

Solution:

Let $A=y^2+2y-1.$ The equation becomes $\frac{x}{x-1}=\frac{A}{A-1}.$ Cross-multiplication gives $xA-x=Ax-A,$ so $x=A=y^2+2y-1.$

Therefore, the correct answer is A.

7.

How many of the first one hundred positive integers are divisible by all of the numbers $2,$ $3,$ $4,$ $5?$

$0$

$1$

$2$

$3$

$4$

Concepts:least common multiple divisibility

Difficulty rating: 980

Small Hint:

Find the least common multiple of $2,$ $3,$ $4,$ and $5$

Big Hint:

Count the multiples of that least common multiple through $100$

Solution:

The least common multiple is $\operatorname{lcm}(2,3,4,5)=60.$ Among the positive integers through $100,$ only $60$ is a multiple of $60.$ Thus the count is $1.$

Therefore, the correct answer is B.

8.

For all positive numbers $x,$ $y,$ $z,$ the product $\begin{aligned} &(x+y+z)^{-1}\\ &\quad\cdot(x^{-1}+y^{-1}+z^{-1})\\ &\quad\cdot(xy+yz+zx)^{-1}\\ &\quad\cdot\big((xy)^{-1}+(yz)^{-1}+(zx)^{-1}\big) \end{aligned}$ equals

$x^{-2}y^{-2}z^{-2}$

$x^{-2}+y^{-2}+z^{-2}$

$(x+y+z)^{-1}$

$\frac1{xyz}$

$\frac1{xy+yz+zx}$

Concepts:algebraic manipulation fraction

Difficulty rating: 1630

Small Hint:

Rewrite each sum of reciprocals over a common denominator

Big Hint:

The two reciprocal sums introduce factors $xy+yz+zx$ and $x+y+z$

Solution:

We have $x^{-1}+y^{-1}+z^{-1}=\frac{xy+yz+zx}{xyz}$ and $\begin{aligned} &(xy)^{-1}+(yz)^{-1}\\ &\quad+(zx)^{-1} =\frac{x+y+z}{xyz}. \end{aligned}$ All symmetric sum factors cancel, leaving $\frac{1}{x^2y^2z^2}=x^{-2}y^{-2}z^{-2}.$

Therefore, the correct answer is A.

9.

In the adjoining figure, $PQ$ is a diagonal of the cube. If $PQ$ has length $a,$ then the surface area of the cube is

$2a^2$

$2\sqrt2a^2$

$2\sqrt3a^2$

$3\sqrt3a^2$

$6a^2$

Concepts:cube geometry diagonal surface area

Difficulty rating: 1280

Small Hint:

Relate the cube’s space diagonal to its side length

Big Hint:

Substitute the side length into the formula for six square faces

Solution:

If the side length is $s,$ then the space diagonal is $s\sqrt3=a,$ so $s^2=\frac{a^2}{3}.$ The surface area is $6s^2=6(\frac{a^2}{3})=2a^2.$

Therefore, the correct answer is A.

10.

The lines $L$ and $K$ are symmetric to each other with respect to the line $y=x.$ If the equation of line $L$ is $y=ax+b$ with $a\ne0$ and $b\ne0,$ then the equation of $K$ is $y=$

$\frac1a x+b$

$-\frac1a x+b$

$-\frac1a x-\frac ba$

$\frac1a x+\frac ba$

$\frac1a x-\frac ba$

Concepts:coordinate geometry reflection (geometry)linear equation

Difficulty rating: 1300

Small Hint:

Reflection across $y=x$ interchanges the two coordinates

Big Hint:

Swap $x$ and $y$ in the original equation, then solve for $y$

Solution:

Interchanging $x$ and $y$ in $y=ax+b$ gives $x=ay+b.$ Solving, $y=\frac{x-b}{a}=\frac1a x-\frac ba.$

Therefore, the correct answer is E.

11.

The three sides of a right triangle have integral lengths which form an arithmetic progression. One of the sides could have length

$22$

$58$

$81$

$91$

$361$

Concepts:Pythagorean Triple arithmetic sequence divisibility

Difficulty rating: 1360

Small Hint:

Represent the three equally spaced sides as $s-d,$ $s,$ $s+d$

Big Hint:

Every such integral right triangle is a multiple of a $3$ - $4$ - $5$ triangle

Solution:

If the sides are $s-d,$ $s,$ $s+d,$ the Pythagorean equation gives $(s-d)^2+s^2=(s+d)^2,$ hence $s=4k$ and $d=k$ for some positive integer $k.$ Thus the sides are $3k,$ $4k,$ $5k.$ Of the choices, $81=3\cdot27$ can occur.

Therefore, the correct answer is C.

12.

If $p,$ $q,$ and $M$ are positive numbers and $q\lt100,$ then the number obtained by increasing $M$ by $p\%$ and decreasing the result by $q\%$ exceeds $M$ if and only if

$p\gt q$

$p\gt\frac q{100-q}$

$p\gt\frac q{1-q}$

$p\gt\frac{100q}{100+q}$

$p\gt\frac{100q}{100-q}$

Concepts:percentage inequality algebraic manipulation

Difficulty rating: 1450

Small Hint:

Write the two successive percentage changes as multiplication factors

Big Hint:

Use $q\lt100$ when dividing by $100-q$

Solution:

The final amount is $M\left(1+\frac p{100}\right) \left(1-\frac q{100}\right).$ It exceeds $M$ exactly when $(100+p)(100-q)\gt10000.$ Expanding gives $p(100-q)\gt100q.$ Since $100-q\gt0,$ this is $p\gt\frac{100q}{100-q}.$

Therefore, the correct answer is E.

13.

Suppose that at the end of any year, a unit of money has lost $10\%$ of the value it had at the beginning of that year. Find the smallest integer $n$ such that after $n$ years the unit of money will have lost at least $90\%$ of its value. (To the nearest thousandth $\log_{10}3$ is $0.477.$ )

$14$

$16$

$18$

$20$

$22$

Concepts:exponent logarithm inequality

Difficulty rating: 1660

Small Hint:

After $n$ years, the fraction of value remaining is $0.9^n$

Big Hint:

Use $\log_{10}0.9=2\log_{10}3-1$

Solution:

At least $90\%$ lost means $0.9^n\le0.1.$ Now $\begin{aligned} \log_{10}0.9 &=\log_{10}9-1\\ &=2(0.477)-1\\ &=-0.046. \end{aligned}$ Thus $n(-0.046)\le-1,$ so $n\ge21.739\ldots.$ The least integer is $22.$

Therefore, the correct answer is E.

14.

In a geometric sequence of real numbers, the sum of the first two terms is $7,$ and the sum of the first six terms is $91.$ The sum of the first four terms is

$28$

$32$

$35$

$49$

$84$

Concepts:geometric sequence quadratic

Difficulty rating: 1730

Small Hint:

Factor the first-six-term sum into the first-two-term sum times $1+r^2+r^4$

Big Hint:

Set $t=r^2$ and solve $1+t+t^2=13$

Solution:

If the first term is $a$ and ratio is $r,$ then $91=a(1+r)(1+r^2+r^4).$ Since $a(1+r)=7,$ we have $1+r^2+r^4=13.$ With $t=r^2\ge0,$ $t^2+t-12=0,$ so $t=3.$ The first four terms sum to $a(1+r)(1+r^2)=7(4)=28.$

Therefore, the correct answer is A.

15.

If $b\gt1,$ $x\gt0,$ and $(2x)^{\log_b2}-(3x)^{\log_b3}=0,$ then $x$ is

$\frac1{216}$

$\frac16$

$1$

$6$

not uniquely determined

Concepts:logarithm exponent substitution

Difficulty rating: 1860

Small Hint:

Take logarithms to base $b$ of the two equal positive powers

Big Hint:

Let $A=\log_b2,$ $B=\log_b3,$ and $u=\log_bx$

Solution:

Let $A=\log_b2,$ $B=\log_b3,$ and $u=\log_bx.$ Taking base- $b$ logarithms gives $A(A+u)=B(B+u).$ Since $A\ne B,$ $A+B+u=0.$ Therefore $u=-\log_b6=\log_b(\frac{1}{6}),$ so $x=\frac{1}{6}.$

Therefore, the correct answer is B.

16.

The base three representation of $x$ is $12112211122211112222.$ The first digit (on the left) of the base nine representation of $x$ is

$1$

$2$

$3$

$4$

$5$

Concepts:number base digits

Difficulty rating: 1450

Small Hint:

Since $9=3^2,$ group the base-three digits in pairs from the right

Big Hint:

Convert the leftmost pair $12_3$ into a base-nine digit

Solution:

The representation has an even number of base-three digits, so its leftmost base-nine digit comes from $12_3.$ That pair has value $1\cdot3+2=5,$ so the first base-nine digit is $5.$

Therefore, the correct answer is E.

17.

The function $f$ is not defined for $x=0,$ but, for all nonzero real numbers $x,$ $f(x)+2f\left(\frac1x\right)=3x.$ The equation $f(x)=f(-x)$ is satisfied by

exactly one real number

exactly two real numbers

no real numbers

infinitely many, but not all, nonzero real numbers

all nonzero real numbers

Concepts:functional equation system of equations symmetry (algebra)

Difficulty rating: 1940

Small Hint:

Apply the given relation again after replacing $x$ by $\frac{1}{x}$

Big Hint:

Solve the two equations for $f(x)$ , then compare its values at $x$ and $-x$

Solution:

Replacing $x$ by $\frac{1}{x}$ gives $f(\frac{1}{x})+2f(x)=\frac{3}{x}.$ Solving the two linear equations yields $f(x)=\frac2x-x.$ Hence $f(-x)=-f(x),$ so equality requires $f(x)=0.$ Thus $\frac{2}{x}-x=0,$ or $x^2=2.$ Both $x=\sqrt2$ and $x=-\sqrt2$ work, giving exactly two real numbers.

Therefore, the correct answer is B.

18.

The number of real solutions to the equation $\frac{x}{100}=\sin x$ is

$61$

$62$

$63$

$64$

$65$

Concepts:trigonometry counting intersections symmetry

Difficulty rating: 2130

Small Hint:

Use odd symmetry and first count the positive solutions

Big Hint:

Only positive sine humps below $x=100$ can meet the line, with special care for the first hump

Solution:

The equation is odd-symmetric and has the solution $x=0.$ On $(0,\pi)$ there is one further positive solution. For each $k=1,2,\ldots,15,$ the positive sine hump on $(2k\pi,(2k+1)\pi)$ rises above $\frac{x}{100}$ and then falls below it, giving two solutions. There are no positive solutions beyond $31\pi$ , because the next positive hump begins above $100$ while $|\sin x|\le1.$ Thus there are $1+2(15)=31$ positive solutions, the same number negative, and zero itself: $31+31+1=63.$

Therefore, the correct answer is C.

19.

In $\triangle ABC,$ $M$ is the midpoint of side $BC,$ $AN$ bisects $\angle BAC,$ $BN\perp AN,$ and $\theta$ is the measure of $\angle BAC.$ If sides $AB$ and $AC$ have lengths $14$ and $19,$ respectively, then length $MN$ equals

$2$

$\frac52$

$\frac52-\sin\theta$

$\frac52-\frac12\sin\theta$

$\frac52-\frac12\sin\left(\frac\theta2\right)$

Concepts:angle bisector reflection (geometry)midpoint

Difficulty rating: 1860

Small Hint:

Extend $BN$ through $N$ to meet $AC$

Big Hint:

Use congruence across the angle bisector, then apply the midpoint theorem

Solution:

Extend $BN$ to meet $AC$ at $E.$ The right triangles $ABN$ and $AEN$ are congruent because $AN$ is common and bisects the angle at $A.$ Hence $BE$ is bisected by $N$ and $AE=AB=14.$ Since $AC=19,$ $EC=5.$ In triangle $BEC,$ points $N$ and $M$ are the midpoints of $BE$ and $BC,$ so $MN=\frac{EC}{2}=\frac{5}{2}.$

Therefore, the correct answer is B.

20.

A ray of light originates from point $A$ and travels in a plane, being reflected $n$ times between lines $AD$ and $CD,$ before striking a point $B$ (which may be on $AD$ or $CD$ ) perpendicularly and retracing its path to $A.$ (At each point of reflection the light makes two equal angles as indicated in the adjoining figure. The figure shows the light path for $n=3.$ ) If $\angle CDA=8^\circ,$ what is the largest value $n$ can have?

$6$

$10$

$38$

$98$

There is no largest value.

Concepts:reflection (geometry)angle chasing inequality

Difficulty rating: 1950

Small Hint:

Track the acute angle between the ray and each successive reflecting line

Big Hint:

Each reflection advances that angle by the wedge angle $8^\circ$

Solution:

Let $\theta\gt0$ be the initial acute angle between the ray and $AD.$ Successive exterior-angle relations increase the corresponding acute angle by $8^\circ$ at each reflection. At the final perpendicular strike, $90^\circ=\theta+(8n+8)^\circ.$ Thus $\theta=82^\circ-8n^\circ\gt0,$ so $n\lt\frac{82}{8}=10.25.$ The greatest integer possible is $10,$ attained when $\theta=2^\circ.$

Therefore, the correct answer is B.

21.

In a triangle with sides of lengths $a,$ $b,$ and $c,$ $(a+b+c)(a+b-c)=3ab.$ The measure of the angle opposite the side of length $c$ is

$15^\circ$

$30^\circ$

$45^\circ$

$60^\circ$

$150^\circ$

Concepts:law of cosines difference of squares

Difficulty rating: 1450

Small Hint:

Expand the left side as a difference of squares

Big Hint:

Compare the resulting expression for $c^2$ with the law of cosines

Solution:

The condition gives $(a+b)^2-c^2=3ab,$ so $c^2=a^2+b^2-ab.$ If $\theta$ is opposite $c,$ the law of cosines says $c^2=a^2+b^2-2ab\cos\theta.$ Therefore $2ab\cos\theta=ab,$ so $\cos\theta=\frac{1}{2}$ and $\theta=60^\circ.$

Therefore, the correct answer is D.

22.

How many lines in a three-dimensional rectangular coordinate system pass through four distinct points of the form $(i,j,k),$ where $i,$ $j,$ and $k$ are positive integers not exceeding four?

$60$

$64$

$72$

$76$

$100$

Concepts:3D geometry lattice point basic counting

Difficulty rating: 2040

Small Hint:

Four collinear lattice points in this $4$ -by- $4$ -by- $4$ grid must advance by coordinate steps $0$ or $\pm1$

Big Hint:

Count axis-parallel lines, face-diagonal lines, and space diagonals separately

Solution:

There are $3\cdot4^2=48$ axis-parallel lines. For face-diagonal directions, choose the pair of varying coordinates in $3$ ways, choose one of $2$ diagonal slopes, and fix the remaining coordinate in $4$ ways, giving $3\cdot2\cdot4=24.$ Finally, the cube has $4$ space diagonals. Thus the total is $48+24+4=76.$

Therefore, the correct answer is D.

23.

Equilateral $\triangle ABC$ is inscribed in a circle. A second circle is tangent internally to the circumcircle at $T$ and tangent to sides $AB$ and $AC$ at points $P$ and $Q.$ If side $BC$ has length $12,$ then segment $PQ$ has length

$6$

$6\sqrt3$

$8$

$8\sqrt3$

$9$

Concepts:equilateral triangle tangent circles similarity

Difficulty rating: 1910

Small Hint:

Use symmetry to place both circle centers and $T$ on the altitude from $A$

Big Hint:

Relate the smaller circle’s radius to the circumradius using the $30^\circ$ half-angle at $A$

Solution:

The circumradius of the equilateral triangle is $R=\frac{12}{\sqrt3}=4\sqrt3.$ If the smaller radius is $r,$ its center lies on the altitude and is $2r$ from $A,$ because its distance to either side is $r$ and the half-angle is $30^\circ.$ Internal tangency at the bottom gives $2r=2R-r,$ so $r=\frac{2R}{3}=\frac{8\sqrt3}{3}.$ The tangency points on the two sides are separated by $PQ=r\sqrt3=8.$

Therefore, the correct answer is C.

24.

If $\theta$ is a constant such that $0\lt\theta\lt\pi$ and $x+\frac1x=2\cos\theta,$ then for each positive integer $n,$ $x^n+\frac1{x^n}$ equals

$2\cos\theta$

$2^n\cos\theta$

$2\cos^n\theta$

$2\cos n\theta$

$2^n\cos^n\theta$

Concepts:complex number De Moivre’s Theorem trigonometric identity

Difficulty rating: 1940

Small Hint:

Rewrite the condition as a quadratic equation in $x$

Big Hint:

Its two roots are $\cos\theta\pm i\sin\theta$

Solution:

The equation is $x^2-2x\cos\theta+1=0,$ whose roots are $e^{i\theta}$ and $e^{-i\theta}.$ Thus $x=e^{\pm i\theta}$ and $x^{-1}=e^{\mp i\theta}.$ By De Moivre’s theorem, $\begin{aligned} x^n+x^{-n} &=e^{in\theta}+e^{-in\theta}\\ &=2\cos n\theta. \end{aligned}$

Therefore, the correct answer is D.

25.

In triangle $ABC$ in the adjoining figure, $AD$ and $AE$ trisect $\angle BAC.$ The lengths of $BD,$ $DE,$ and $EC$ are $2,$ $3,$ and $6,$ respectively. The length of the shortest side of $\triangle ABC$ is

$2\sqrt{10}$

$11$

$6\sqrt6$

$6$

not uniquely determined by the given information

Concepts:angle bisector theorem law of cosines angle chasing

Difficulty rating: 2210

Small Hint:

Let the three equal angles be $\alpha$ , and apply the angle-bisector theorem to a suitable subtriangle

Big Hint:

Equate law-of-cosines expressions for $\cos\alpha$ in the three adjacent triangles

Solution:

Let $AB=c,$ $AD=y,$ $AC=b,$ and $AE=z.$ Since $AD$ bisects $\angle BAE,$ and $AE$ bisects $\angle DAC,$ the angle-bisector theorem gives $\frac{c}{z}=\frac{2}{3},\qquad \frac{y}{b}=\frac{3}{6},$ so $z=\frac{3c}{2}$ and $b=2y.$ Applying the law of cosines to triangles $ADB,$ $ADE,$ and $AEC$ and equating their common $\cos\alpha$ values gives $\begin{gathered} 3c^2-2y^2=12,\\ 9c^2-8y^2=-72. \end{gathered}$ Hence $y^2=54$ and $c^2=40.$ The sides are $AB=2\sqrt{10},$ $BC=11,$ and $AC=2\sqrt{54}=6\sqrt6,$ so the shortest is $2\sqrt{10}.$

Therefore, the correct answer is A.

26.

Alice, Bob, and Carol repeatedly take turns tossing a die. Alice begins; Bob always follows Alice; Carol always follows Bob; and Alice always follows Carol. Find the probability that Carol will be the first one to toss a six. (The probability of obtaining a six on any toss is $\frac16,$ independent of the outcome of any other toss.)

$\frac13$

$\frac29$

$\frac5{18}$

$\frac{25}{91}$

$\frac{36}{91}$

Concepts:dice (probability)geometric distribution independent events

Difficulty rating: 1680

Small Hint:

For Carol to win in a given round, Alice and Bob must fail before Carol succeeds

Big Hint:

If all three fail, the process restarts with the same probabilities

Solution:

Carol wins in the first round with probability $(\frac{5}{6})^2(\frac{1}{6})=\frac{25}{216}.$ A complete round with no six has probability $(\frac{5}{6})^3=\frac{125}{216}.$ Therefore the desired geometric series is $\frac{\frac{25}{216}}{1-\frac{125}{216}} =\frac{25}{91}.$

Therefore, the correct answer is D.

27.

In the adjoining figure triangle $ABC$ is inscribed in a circle. Point $D$ lies on $\overset{\frown}{AC}$ with $\overset{\frown}{DC}=30^\circ,$ and point $G$ lies on $\overset{\frown}{BA}$ with $\overset{\frown}{BG}\gt\overset{\frown}{GA}.$ Side $AB$ and side $AC$ each have length equal to the length of chord $DG,$ and $\angle CAB=30^\circ.$ Chord $DG$ intersects sides $AC$ and $AB$ at $E$ and $F,$ respectively. The ratio of the area of $\triangle AFE$ to the area of $\triangle ABC$ is

$\frac{2-\sqrt3}{3}$

$\frac{2\sqrt3-3}{3}$

$7\sqrt3-12$

$3\sqrt3-5$

$\frac{9-5\sqrt3}{3}$

Concepts:arc chord area ratio

Difficulty rating: 2260

Small Hint:

Use equal chords to identify the relevant equal arcs and isosceles triangles

Big Hint:

Normalize $AB=AC=DG=1$ and find $AE$ from a $30^\circ$ - $60^\circ$ - $90^\circ$ triangle

Solution:

Scale so $AB=AC=DG=1.$ The equal-chord arc relations show that triangle $DEC$ is $30^\circ$ - $60^\circ$ - $90^\circ$ and $AE=DE.$ Put $AE=DE=x.$ Then $CE=1-x=\frac{2x}{\sqrt3},$ so $x=2\sqrt3-3.$ Equal chords also give $AF=FG=EF=\frac{1-x}{2}.$ Hence $\begin{aligned} \frac{[AFE]}{[ABC]} &=\frac{x(1-x)}2\\ &=7\sqrt3-12. \end{aligned}$

Therefore, the correct answer is C.

28.

Consider the set of all equations $x^3+a_2x^2+a_1x+a_0=0,$ where $a_2,$ $a_1,$ $a_0$ are real constants and $|a_i|\le2$ for $i=0,$ $1,$ $2.$ Let $r$ be the largest positive real number which satisfies at least one of these equations. Then

$1\le r\lt\frac32$

$\frac32\le r\lt2$

$2\le r\lt\frac52$

$\frac52\le r\lt3$

$3\le r\lt\frac72$

Concepts:polynomial inequality bounding to limit cases

Difficulty rating: 2210

Small Hint:

For positive $x,$ the coefficients that push a root farthest right take their lower bounds

Big Hint:

Bracket the largest root of $x^3-2x^2-2x-2$ at two consecutive choice endpoints

Solution:

Write an allowed polynomial as $g.$ For $x\ge0,$ $\begin{aligned} g(x)&=x^3+a_2x^2\\ &\quad+a_1x+a_0\\ &\ge f(x), \end{aligned}$ where $f(x)=x^3-2x^2-2x-2.$ Thus no positive root can exceed the largest root $\rho$ of $f.$ Taking $a_2=a_1=a_0=-2$ gives $g=f,$ so $\rho$ itself is attained. Since $\begin{gathered} f\left(\frac52\right)=-\frac{31}{8}\lt0,\\ f(3)=1\gt0, \end{gathered}$ we have $\frac{5}{2}\lt\rho\lt3.$

Therefore, the correct answer is D.

29.

If $a\gt1,$ then the sum of the real solutions of $\sqrt{a-\sqrt{a+x}}=x$ is equal to

$\sqrt a-1$

$\frac{\sqrt a-1}{2}$

$\sqrt{a-1}$

$\frac{\sqrt{a-1}}2$

$\frac{\sqrt{4a-3}-1}{2}$

Concepts:radical quadratic casework

Difficulty rating: 2040

Small Hint:

The principal square root forces $x\ge0$

Big Hint:

After one squaring, factor a difference involving $\sqrt{a+x}-x$

Solution:

Since $x\ge0,$ squaring once gives $a-\sqrt{a+x}=x^2.$ Let $y=\sqrt{a+x}.$ Then $y=a-x^2$ and $a=y^2-x,$ so $0=(y+x)(y-x-1).$ Since $y+x\gt0,$ we have $y=x+1,$ or $\sqrt{a+x}=x+1.$ Therefore $a+x=x^2+2x+1$ and $x^2+x+1-a=0.$ The only nonnegative root is $x=\frac{\sqrt{4a-3}-1}{2},$ and it satisfies the original equation. It is therefore also the sum of all real solutions.

Therefore, the correct answer is E.

30.

If $a,$ $b,$ $c,$ $d$ are the solutions of the equation $x^4-bx-3=0,$ then an equation whose solutions are $\begin{gathered} \frac{a+b+c}{d^2},\quad \frac{a+b+d}{c^2},\\ \frac{a+c+d}{b^2},\quad \frac{b+c+d}{a^2} \end{gathered}$ is

$3x^4+bx+1=0$

$3x^4-bx+1=0$

$3x^4+bx^3-1=0$

$3x^4-bx^3-1=0$

none of these

Concepts:polynomial Vieta’s Formulas substitution

Difficulty rating: 2170

Small Hint:

Use the missing cubic term to find $a+b+c+d$

Big Hint:

Each listed expression becomes the negative reciprocal of one original root

Solution:

Vieta’s formulas give $a+b+c+d=0.$ Thus, for example, $\frac{a+b+c}{d^2}=-\frac1d,$ and similarly the new roots are $-\frac{1}{a},$ $-\frac{1}{b},$ $-\frac{1}{c},$ $-\frac{1}{d}.$ Put $x=-\frac{1}{r}$ in $r^4-br-3=0.$ Multiplying by $x^4$ gives $1+bx^3-3x^4=0,$ or $3x^4-bx^3-1=0.$

Therefore, the correct answer is D.