1977 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

If $y=2x$ and $z=2y,$ then $x+y+z$ equals

$x$

$3x$

$5x$

$7x$

$9x$

Concepts:substitution algebraic manipulation

Difficulty rating: 800

Small Hint:

Express every term in terms of $x$

Big Hint:

Use $y=2x$ first to obtain $z=4x$

Solution:

Since $y=2x,$ we have $z=2y=4x.$ Therefore $x+y+z=x+2x+4x=7x.$

Therefore, the correct answer is D.

2.

Which one of the following statements is false? All equilateral triangles are

equiangular

isosceles

regular polygons

congruent to each other

similar to each other

Concepts:equilateral triangle congruence (geometry)similarity

Difficulty rating: 890

Small Hint:

Separate properties determined by angles from properties determined by size

Big Hint:

Compare two equilateral triangles having different side lengths

Solution:

Every equilateral triangle is equiangular, isosceles, regular, and similar to every other equilateral triangle. Two equilateral triangles with different side lengths are not congruent, however.

Therefore, the correct answer is D.

3.

A man has $\$2.73$ in pennies, nickels, dimes, quarters and half dollars. If he has an equal number of coins of each kind, then the total number of coins he has is

$3$

$5$

$9$

$10$

$15$

Concepts:money linear equation

Difficulty rating: 1060

Small Hint:

Find the value of one coin of each of the five types

Big Hint:

The five coins in one complete set are worth $91$ cents

Solution:

One coin of each kind is worth $1+5+10+25+50=91$ cents. Since $\frac{273}{91}=3,$ there are three coins of each of the five kinds, or $5\cdot3=15$ coins.

Therefore, the correct answer is E.

4.

In triangle $ABC,$ $AB=AC$ and $\angle A=80^\circ.$ If points $D,$ $E$ and $F$ lie on sides $BC,$ $AC$ and $AB,$ respectively, and $CE=CD$ and $BF=BD,$ then $\angle EDF$ equals

$30^\circ$

$40^\circ$

$50^\circ$

$65^\circ$

none of these

Concepts:isosceles triangle angle chasing

Difficulty rating: 1690

Small Hint:

First determine the two base angles of $\triangle ABC$

Big Hint:

Use the two given equal-length pairs to find $\angle EDC$ and $\angle BDF$

Solution:

The base angles of $\triangle ABC$ are each $50^\circ.$ Since $CE=CD,$ triangle $CED$ has vertex angle $50^\circ,$ so $\angle EDC=65^\circ.$ Similarly, $BF=BD$ gives $\angle BDF=65^\circ.$ The three angles above the straight line $CB$ sum to $180^\circ,$ so $\begin{aligned} \angle EDF&=180^\circ-65^\circ-65^\circ\\ &=50^\circ. \end{aligned}$

Therefore, the correct answer is C.

5.

The set of all points $P$ such that the sum of the (undirected) distances from $P$ to two fixed points $A$ and $B$ equals the distance between $A$ and $B$ is

the line segment from $A$ to $B$

the line passing through $A$ and $B$

the perpendicular bisector of the line segment from $A$ to $B$

an ellipse having positive area

a parabola

Concepts:triangle inequality

Difficulty rating: 1140

Small Hint:

Apply the triangle inequality to $AP,$ $PB,$ and $AB$

Big Hint:

Recall when equality holds in the triangle inequality

Solution:

The triangle inequality gives $AP+PB\ge AB.$ Equality holds exactly when $A,$ $P,$ $B$ are collinear with $P$ between $A$ and $B,$ including the endpoints. Thus the locus is the segment from $A$ to $B.$

Therefore, the correct answer is A.

6.

If $x,$ $y$ and $2x+\frac y2$ are not zero, then $\left(2x+\frac y2\right)^{-1}\left[(2x)^{-1}+\left(\frac y2\right)^{-1}\right]$ equals

$1$

$xy^{-1}$

$x^{-1}y$

$(xy)^{-1}$

none of these

Concepts:algebraic manipulation fraction

Difficulty rating: 1410

Small Hint:

Rewrite the outside reciprocal using a single fraction

Big Hint:

Combine the two reciprocals inside the brackets over a common denominator

Solution:

We have $\begin{aligned} \left(2x+\frac y2\right)^{-1} &=\frac2{4x+y},\\ (2x)^{-1}+\left(\frac y2\right)^{-1} &=\frac1{2x}+\frac2y\\ &=\frac{y+4x}{2xy}. \end{aligned}$ Their product is $\frac{1}{xy}=(xy)^{-1}.$

Therefore, the correct answer is D.

7.

If $t=\dfrac1{1-\sqrt[4]{2}},$ then $t$ equals

$(1-\sqrt[4]{2})(2-\sqrt2)$

$(1-\sqrt[4]{2})(1+\sqrt2)$

$(1+\sqrt[4]{2})(1-\sqrt2)$

$(1+\sqrt[4]{2})(1+\sqrt2)$

$-(1+\sqrt[4]{2})(1+\sqrt2)$

Concepts:rationalizing denominator difference of squares radical

Difficulty rating: 1860

Small Hint:

Let $u=\sqrt[4]{2}$ and use $u^4=2$

Big Hint:

Factor $u^4-1$ into a linear factor, a second linear factor, and a quadratic factor

Solution:

Let $u=\sqrt[4]{2}.$ Since $\begin{aligned} &(1-u)\bigl(-(1+u)(1+u^2)\bigr)\\ &\qquad=-(1-u^2)(1+u^2)\\ &\qquad=u^4-1=1, \end{aligned}$ it follows that $t=-(1+\sqrt[4]{2})(1+\sqrt2).$

Therefore, the correct answer is E.

8.

For every triple $(a,b,c)$ of nonzero real numbers, form the number $\frac a{|a|}+\frac b{|b|}+\frac c{|c|}+\frac{abc}{|abc|}.$ The set of all numbers formed is

$\{0\}$

$\{-4,0,4\}$

$\{-4,-2,0,2,4\}$

$\{-4,-2,2,4\}$

none of these

Concepts:absolute value casework

Difficulty rating: 1440

Small Hint:

Each of the first three fractions is either $1$ or $-1$

Big Hint:

The sign of the final fraction is the product of the first three signs

Solution:

Let $r,$ $s,$ $t\in\{-1,1\}$ be the signs of $a,$ $b,$ $c.$ The expression is $r+s+t+rst.$ If all three signs are positive it is $4,$ and if all are negative it is $-4.$ If the signs are mixed, direct cancellation gives $0.$ Hence the set is $\{-4,0,4\}.$

Therefore, the correct answer is B.

9.

In the adjoining figure $\angle E=40^\circ$ and arc $AB,$ arc $BC$ and arc $CD$ all have equal length. Find the measure of $\angle ACD.$

$10^\circ$

$15^\circ$

$20^\circ$

$\left(\frac{45}2\right)^\circ$

$30^\circ$

Concepts:arc inscribed angle circle

Difficulty rating: 1690

Small Hint:

Assign one variable to each of the three equal arcs and another to arc $AD$

Big Hint:

Use both the full-circle arc sum and the external-secant angle theorem at $E$

Solution:

Let each of arcs $AB,$ $BC,$ $CD$ measure $x^\circ$ and let arc $AD$ measure $y^\circ.$ Then $\begin{aligned} 3x+y&=360,\\ \frac{x-y}{2}&=40. \end{aligned}$ Solving gives $y=30.$ The inscribed angle $\angle ACD$ subtends arc $AD,$ so it measures $\frac{y}{2}=15^\circ.$

Therefore, the correct answer is B.

10.

If $(3x-1)^7=a_7x^7+a_6x^6+\cdots+a_0,$ then $a_7+a_6+\cdots+a_0$ equals

$0$

$1$

$64$

$-64$

$128$

Concepts:polynomial substitution

Difficulty rating: 1440

Small Hint:

A polynomial’s coefficient sum is obtained by evaluating it at a particular input

Big Hint:

Substitute $x=1$ into the given identity

Solution:

Setting $x=1$ makes the right side the desired coefficient sum. Thus $\begin{aligned} a_7+a_6+\cdots+a_0 &=(3\cdot1-1)^7\\ &=2^7=128. \end{aligned}$

Therefore, the correct answer is E.

11.

For each real number $x,$ let $[x]$ be the largest integer not exceeding $x$ (i.e., the integer $n$ such that $n\le x\lt n+1$ ). Which of the following statements is (are) true?

$\mathrm{I}.$ $[x+1]=[x]+1$ for all $x$

$\mathrm{II}.$ $[x+y]=[x]+[y]$ for all $x$ and $y$

$\mathrm{III}.$ $[xy]=[x][y]$ for all $x$ and $y$

none

$\mathrm{I}$ only

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

$\mathrm{III}$ only

all

Concepts:floor and ceiling functions counterexample

Difficulty rating: 1530

Small Hint:

Prove the translation statement directly from $n\le x\lt n+1$

Big Hint:

Test the other two claims using noninteger values between $1$ and $2$

Solution:

If $n\le x\lt n+1,$ then $n+1\le x+1\lt n+2,$ so $\mathrm{I}$ is true. Taking $x=y=\frac{3}{2}$ gives $[x+y]=3\ne2=[x]+[y]$ and $[xy]=2\ne1=[x][y].$ Thus $\mathrm{II}$ and $\mathrm{III}$ are false.

Therefore, the correct answer is B.

12.

Al’s age is $16$ more than the sum of Bob’s age and Carl’s age, and the square of Al’s age is $1632$ more than the square of the sum of Bob’s age and Carl’s age. The sum of the ages of Al, Bob and Carl is

$64$

$94$

$96$

$102$

$140$

Concepts:difference of squares ages

Difficulty rating: 1360

Small Hint:

Let $s$ be the sum of Bob’s and Carl’s ages

Big Hint:

Factor $a^2-s^2$ and use the known value of $a-s$

Solution:

Let $a$ be Al’s age and $s$ the sum of the other two ages. Then $a-s=16$ and $\begin{aligned} 1632&=a^2-s^2\\ &=(a-s)(a+s)\\ &=16(a+s). \end{aligned}$ Hence the requested total $a+s$ is $\frac{1632}{16}=102.$

Therefore, the correct answer is D.

13.

If $a_1,$ $a_2,$ $a_3,$ $\ldots$ is a sequence of positive numbers such that $a_{n+2}=a_na_{n+1}$ for all positive integers $n,$ then the sequence $a_1,$ $a_2,$ $a_3,$ $\ldots$ is a geometric progression

for all positive values of $a_1$ and $a_2$

if and only if $a_1=a_2$

if and only if $a_1=1$

if and only if $a_2=1$

if and only if $a_1=a_2=1$

Concepts:recursion geometric sequence ratio and proportion

Difficulty rating: 1860

Small Hint:

Write out $a_3,$ $a_4,$ and $a_5$ in terms of $a_1,$ $a_2$

Big Hint:

Equate the first three successive ratios and use positivity

Solution:

The next terms are $a_3=a_1a_2,$ $a_4=a_1a_2^2,$ and $a_5=a_1^2a_2^3.$ If the sequence is geometric, then $\begin{aligned} \frac{a_2}{a_1}&=\frac{a_3}{a_2}=a_1,\\ \frac{a_4}{a_3}&=a_2=a_1. \end{aligned}$ Thus $a_2=a_1^2$ and $a_2=a_1.$ Positivity forces $a_1=a_2=1.$ Conversely, these initial values produce the constant geometric sequence $1,1,1,\ldots.$

Therefore, the correct answer is E.

14.

How many pairs $(m,n)$ of integers satisfy the equation $m+n=mn?$

$1$

$2$

$3$

$4$

more than $4$

Concepts:Diophantine Equation factoring

Difficulty rating: 1650

Small Hint:

Move all terms to one side and add $1$

Big Hint:

Factor the equation into a product of two integers equal to $1$

Solution:

Rearranging and completing the product gives $\begin{aligned} mn-m-n&=0,\\ (m-1)(n-1)&=1. \end{aligned}$ The integer factor pairs of $1$ are $(1,1)$ and $(-1,-1),$ producing $(m,n)=(2,2)$ and $(0,0).$ Thus there are two ordered pairs.

Therefore, the correct answer is B.

15.

Each of the three circles in the adjoining figure is externally tangent to the other two, and each side of the triangle is tangent to two of the circles. If each circle has radius three, then the perimeter of the triangle is

$36+9\sqrt2$

$36+6\sqrt3$

$36+9\sqrt3$

$18+18\sqrt3$

$45$

Concepts:tangent circles equilateral triangle incircle, incenter, and inradius

Difficulty rating: 2080

Small Hint:

The three circle centers form an equilateral triangle of side $6$

Big Hint:

View the outer sides as parallel offsets and compare the inradii of two equilateral triangles

Solution:

The circle centers form an equilateral triangle of side $6,$ whose inradius is $\frac{6\sqrt3}{6}=\sqrt3.$ Each side of the outer triangle is a parallel tangent line three units farther out, so the outer triangle has inradius $3+\sqrt3.$ An equilateral triangle of inradius $r$ has side $2\sqrt3r,$ hence each outer side is $2\sqrt3(3+\sqrt3)=6+6\sqrt3.$ The perimeter is $3(6+6\sqrt3)=18+18\sqrt3.$

Therefore, the correct answer is D.

16.

If $i^2=-1,$ then the sum $\begin{aligned} &\cos45^\circ+i\cos135^\circ+\cdots\\ &\quad+i^n\cos(45+90n)^\circ+\cdots\\ &\quad+i^{40}\cos3645^\circ \end{aligned}$ equals

$\frac{\sqrt2}{2}$

$-10i\sqrt2$

$\frac{21\sqrt2}{2}$

$\frac{\sqrt2}{2}(21-20i)$

$\frac{\sqrt2}{2}(21+20i)$

Concepts:complex number trigonometric identity summation

Difficulty rating: 2110

Small Hint:

Compare the term indexed by $n+2$ with the term indexed by $n$

Big Hint:

Count the even and odd indices from $0$ through $40$

Solution:

Let $u_n=i^n\cos(45+90n)^\circ.$ Then $\begin{aligned} u_{n+2} &=i^{n+2}\cos(225+90n)^\circ\\ &=(-i^n)(-\cos(45+90n)^\circ)\\ &=u_n. \end{aligned}$ Every even-indexed term equals $\frac{\sqrt2}{2},$ and every odd-indexed term equals $-\frac{i\sqrt2}{2}.$ There are $21$ even indices and $20$ odd indices, so the sum is $\frac{\sqrt2}{2}(21-20i).$

Therefore, the correct answer is D.

17.

Three fair dice are tossed at random (i.e., all faces have the same probability of coming up). What is the probability that the three numbers turned up can be arranged to form an arithmetic progression with common difference one?

$\frac16$

$\frac19$

$\frac1{27}$

$\frac1{54}$

$\frac7{36}$

Concepts:basic probability dice (probability)permutations

Difficulty rating: 1590

Small Hint:

List the possible three-element sets of consecutive die values

Big Hint:

Each qualifying set has six orderings among the $6^3$ ordered outcomes

Solution:

The possible sets are $\{1,2,3\},$ $\{2,3,4\},$ $\{3,4,5\},$ and $\{4,5,6\}.$ Each has $3!=6$ orderings, so $24$ of the $6^3=216$ outcomes work. The probability is $\frac{24}{216}=\frac{1}{9}.$

Therefore, the correct answer is B.

18.

If $\begin{aligned} y={}&(\log_2 3)(\log_3 4)\cdots\\ &(\log_n[n+1])\cdots(\log_{31}32), \end{aligned}$ then

$4\lt y\lt5$

$y=5$

$5\lt y\lt6$

$y=6$

$6\lt y\lt7$

Concepts:logarithm telescoping

Difficulty rating: 1860

Small Hint:

Rewrite every logarithm using the same base

Big Hint:

Adjacent numerators and denominators cancel in the product

Solution:

By change of base, $\begin{aligned} y&=\frac{\log 3}{\log2}\cdot \frac{\log4}{\log3}\cdots \frac{\log32}{\log31}\\ &=\frac{\log32}{\log2}\\ &=\log_2 32=5. \end{aligned}$

Therefore, the correct answer is B.

19.

Let $E$ be the point of intersection of the diagonals of convex quadrilateral $ABCD,$ and let $P,$ $Q,$ $R$ and $S$ be the centers of the circles circumscribing triangles $ABE,$ $BCE,$ $CDE$ and $ADE,$ respectively. Then

$PQRS$ is a parallelogram

$PQRS$ is a parallelogram if and only if $ABCD$ is a rhombus

$PQRS$ is a parallelogram if and only if $ABCD$ is a rectangle

$PQRS$ is a parallelogram if and only if $ABCD$ is a parallelogram

none of the above are true

Concepts:circumcircle, circumcenter, and circumradius perpendicular bisector parallelogram

Difficulty rating: 2220

Small Hint:

Two adjacent circumcenters lie on the perpendicular bisector of the side shared by their triangles

Big Hint:

Use the fact that $A,$ $E,$ $C$ are collinear and $B,$ $E,$ $D$ are collinear

Solution:

Both $P$ and $Q$ lie on the perpendicular bisector of $BE,$ while both $R$ and $S$ lie on the perpendicular bisector of $DE.$ Since $B,$ $E,$ $D$ are collinear, these bisectors are parallel, so $PQ\parallel RS.$ Likewise, $Q,$ $R$ lie on the perpendicular bisector of $CE,$ and $S,$ $P$ lie on that of $AE.$ Since $A,$ $E,$ $C$ are collinear, $QR\parallel SP.$ Therefore $PQRS$ is always a parallelogram.

Therefore, the correct answer is A.

20.

For how many paths consisting of a sequence of horizontal and/or vertical line segments, with each segment connecting a pair of adjacent letters in the diagram below, is the word CONTEST spelled out as the path is traversed from beginning to end?

$63$

$128$

$129$

$255$

none of these

Concepts:lattice paths symmetry basic counting

Difficulty rating: 2200

Small Hint:

Reverse each path, starting from the central $T$ in the bottom row

Big Hint:

Count the left-going and right-going families separately, then correct for their common path

Solution:

Reverse the paths and spell TSETNOC from the central bottom $T.$ In the family whose horizontal moves go left, each of the six steps has two choices: up or left. This gives $2^6=64$ paths. By symmetry, $64$ paths have horizontal moves going right. The all-vertical central path belongs to both families, so the total is $64+64-1=127.$ This number is not among the first four choices.

Therefore, the correct answer is E.

21.

For how many values of the coefficient $a$ do the equations $\begin{aligned} x^2+ax+1&=0,\\ x^2-x-a&=0 \end{aligned}$ have a common real solution?

$0$

$1$

$2$

$3$

infinitely many

Concepts:quadratic system of equations factoring

Difficulty rating: 2040

Small Hint:

Subtract the two equations to obtain a factored condition

Big Hint:

Check separately the cases $a=-1$ and $x=-1$

Solution:

Subtracting the second equation from the first gives $\begin{gathered} (a+1)x+(a+1)\\ =(a+1)(x+1)=0. \end{gathered}$ If $a=-1,$ the common equation is $x^2-x+1=0,$ which has no real roots. Otherwise $x=-1,$ and substitution into $x^2-x-a=0$ gives $a=2.$ This value works, so exactly one value of $a$ qualifies.

Therefore, the correct answer is B.

22.

If $f(x)$ is a real-valued function of the real variable $x,$ and $f(x)$ is not identically zero, and for all $a$ and $b$ $\begin{gathered} f(a+b)+f(a-b)\\ =2f(a)+2f(b), \end{gathered}$ then for all $x$ and $y$

$f(0)=1$

$f(-x)=-f(x)$

$f(-x)=f(x)$

$f(x+y)=f(x)+f(y)$

there is a positive number $T$ such that $f(x+T)=f(x)$

Concepts:functional equation function substitution

Difficulty rating: 1710

Small Hint:

First substitute $a=b=0$

Big Hint:

Next set $a=0$ and $b=x$

Solution:

Setting $a=b=0$ gives $2f(0)=4f(0),$ so $f(0)=0.$ Now set $a=0$ and $b=x.$ The equation becomes $\begin{aligned} f(x)+f(-x) &=2f(0)+2f(x)\\ &=2f(x), \end{aligned}$ and hence $f(-x)=f(x)$ for every real $x.$

Therefore, the correct answer is C.

23.

If the solutions of the equation $x^2+px+q=0$ are the cubes of the solutions of the equation $x^2+mx+n=0,$ then

$p=m^3+3mn$

$p=m^3-3mn$

$p+q=m^3$

$\left(\frac mn\right)^3=\frac pq$

none of these

Concepts:Vieta’s Formulas sum and difference of cubes quadratic

Difficulty rating: 2040

Small Hint:

Call the roots of the second equation $u$ and $v$ and apply Vieta’s formulas

Big Hint:

Use $u^3+v^3=(u+v)^3-3uv(u+v)$

Solution:

Let the roots of $x^2+mx+n$ be $u,$ $v.$ Then $u+v=-m$ and $uv=n,$ while the roots of the first equation are $u^3,$ $v^3.$ Thus $\begin{aligned} -p&=u^3+v^3\\ &=(u+v)^3-3uv(u+v)\\ &=-m^3+3mn. \end{aligned}$ Therefore $p=m^3-3mn.$

Therefore, the correct answer is B.

24.

Find the sum $\begin{aligned} &\frac1{1\cdot3}+\frac1{3\cdot5}+\cdots\\ &\quad+\frac1{(2n-1)(2n+1)}+\cdots\\ &\quad+\frac1{255\cdot257}. \end{aligned}$

$\frac{127}{255}$

$\frac{128}{255}$

$\frac12$

$\frac{128}{257}$

$\frac{129}{257}$

Concepts:telescoping partial fractions summation

Difficulty rating: 1690

Small Hint:

Decompose $\frac1{(2n-1)(2n+1)}$ into two unit fractions

Big Hint:

The last denominator corresponds to $n=128$

Solution:

For $1\le n\le128,$ $\begin{gathered} \frac1{(2n-1)(2n+1)}\\ =\frac1{2(2n-1)}-\frac1{2(2n+1)}. \end{gathered}$ Hence the sum telescopes to $\frac12\left(1-\frac1{257}\right)=\frac{128}{257}.$

Therefore, the correct answer is D.

25.

Determine the largest positive integer $n$ such that $1005!$ is divisible by $10^n.$

$102$

$112$

$249$

$502$

none of these

Concepts:factorial trailing zeros Legendre’s Formula

Difficulty rating: 1560

Small Hint:

Count the factors of $5$ in $1005!$

Big Hint:

Include contributions from multiples of $25,$ $125,$ and $625$

Solution:

There are more factors of $2$ than of $5,$ so the exponent of $10$ is $\begin{aligned} n={}&\left\lfloor\frac{1005}{5}\right\rfloor+ \left\lfloor\frac{1005}{25}\right\rfloor\\ &+\left\lfloor\frac{1005}{125}\right\rfloor+ \left\lfloor\frac{1005}{625}\right\rfloor\\ ={}&201+40+8+1=250. \end{aligned}$ Since $250$ is not listed, the correct choice is “none of these.”

Therefore, the correct answer is E.

26.

Let $a,$ $b,$ $c$ and $d$ be the lengths of sides $MN,$ $NP,$ $PQ$ and $QM,$ respectively, of quadrilateral $MNPQ.$ If $A$ is the area of $MNPQ,$ then

$A=\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)$ if and only if $MNPQ$ is convex

$A=\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)$ if and only if $MNPQ$ is a rectangle

$A\le\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)$ if and only if $MNPQ$ is a rectangle

$A\le\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)$ if and only if $MNPQ$ is a parallelogram

$A\ge\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)$ if and only if $MNPQ$ is a parallelogram

Concepts:area inequality rectangle

Difficulty rating: 2320

Small Hint:

Split the quadrilateral along each diagonal and bound every sine by $1$

Big Hint:

Combine the two resulting area bounds and analyze when every bound is an equality

Solution:

Splitting along diagonal $MP$ and using $\sin\theta\le1$ gives $A\le\frac{ab+cd}{2}.$ Splitting along $NQ$ similarly gives $A\le\frac{ad+bc}{2}.$ These bounds remain valid for a nonconvex quadrilateral by taking the appropriate difference of triangle areas. Adding them yields $\begin{aligned} 2A&\le\frac{ab+ad+bc+cd}{2}\\ &=\frac{(a+c)(b+d)}2, \end{aligned}$ or $A\le\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right).$ Equality requires equality in all four sine bounds, so all four angles are right angles; conversely a rectangle gives equality. Thus the equality holds exactly for rectangles.

Therefore, the correct answer is B.

27.

There are two spherical balls of different sizes lying in two corners of a rectangular room, each touching two walls and the floor. If there is a point on each ball which is $5$ inches from each wall which that ball touches and $10$ inches from the floor, then the sum of the diameters of the balls is

$20$ inches

$30$ inches

$40$ inches

$60$ inches

not determined by the given information

Concepts:sphere coordinate geometry quadratic

Difficulty rating: 2040

Small Hint:

Place the corner at the origin with the walls and floor as coordinate planes

Big Hint:

A sphere of radius $r$ tangent to all three planes has center $(r,r,r)$

Solution:

For radius $r,$ the center is $(r,r,r)$ and the given point is $(5,5,10).$ Thus $2(5-r)^2+(10-r)^2=r^2,$ which simplifies to $r^2-20r+75=0,$ or $(r-5)(r-15)=0.$ The two radii are $5$ and $15,$ so the sum of the diameters is $2(5+15)=40$ inches.

Therefore, the correct answer is C.

28.

Let $g(x)=x^5+x^4+x^3$ ${}+x^2+x+1.$ What is the remainder when the polynomial $g(x^{12})$ is divided by the polynomial $g(x)?$

$6$

$5-x$

$4-x+x^2$

$3-x+x^2-x^3$

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

Concepts:polynomial roots of unity

Difficulty rating: 2300

Small Hint:

Use $(x-1)g(x)=x^6-1$

Big Hint:

Evaluate the remainder at the five roots of $g$ and use its degree bound

Solution:

Let $R(x)$ be the remainder, so $\deg R\le4.$ The five roots $\alpha$ of $g$ satisfy $\alpha^6=1$ and $\alpha\ne1.$ Hence $\alpha^{12}=1$ and $R(\alpha)=g(\alpha^{12})=g(1)=6.$ Therefore $R(x)-6,$ a polynomial of degree at most $4,$ has five distinct roots. It must be identically zero, so $R(x)=6.$

Therefore, the correct answer is A.

29.

Find the smallest integer $n$ such that $\begin{aligned} (x^2+y^2+z^2)^2 &\le{}\\[-4pt] &n(x^4+y^4+z^4) \end{aligned}$ for all real numbers $x,$ $y$ and $z.$

$2$

$3$

$4$

$6$

There is no such integer $n.$

Concepts:Cauchy-Schwarz Inequality inequality

Difficulty rating: 1870

Small Hint:

Apply Cauchy-Schwarz to the three numbers $x^2,$ $y^2,$ $z^2$

Big Hint:

Use equal nonzero values of $x^2,$ $y^2,$ $z^2$ to test sharpness

Solution:

By Cauchy-Schwarz, $\begin{aligned} (x^2+y^2+z^2)^2&\le3Q,\\ Q&=x^4+y^4+z^4. \end{aligned}$ Equality occurs when $x^2=y^2=z^2\ne0,$ so no smaller value can work. Thus the smallest integer is $3.$

Therefore, the correct answer is B.

30.

If $a,$ $b$ and $d$ are the lengths of a side, a shortest diagonal and a longest diagonal, respectively, of a regular nonagon (see adjoining figure), then

$d=a+b$

$d^2=a^2+b^2$

$d^2=a^2+ab+b^2$

$b=\frac{a+d}{2}$

$b^2=ad$

Concepts:regular polygon chord trigonometric identity

Difficulty rating: 2300

Small Hint:

Write the three chord lengths using the nonagon’s circumradius

Big Hint:

Compare $\sin20^\circ+\sin40^\circ$ with $\sin80^\circ$ using the sum-to-product identity

Solution:

If the circumradius is $r,$ the three chords subtend central angles $40^\circ,$ $80^\circ,$ $160^\circ,$ respectively. Thus $\begin{aligned} a&=2r\sin20^\circ,\\ b&=2r\sin40^\circ,\\ d&=2r\sin80^\circ. \end{aligned}$ The sum-to-product identity gives $\begin{gathered} \sin20^\circ+\sin40^\circ\\ =2\sin30^\circ\cos10^\circ\\ =\cos10^\circ=\sin80^\circ. \end{gathered}$ Multiplying by $2r$ yields $a+b=d.$

Therefore, the correct answer is A.