1984 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

$\frac{1000^2}{252^2-248^2}$ equals

$62{,}500$

$1000$

$500$

$250$

$\frac12$

Concepts:difference of squares algebraic manipulation

Difficulty rating: 1210

Small Hint:

Factor the denominator as a difference of squares

Big Hint:

Use $252+248=500$ and $252-248=4$

Solution:

Factoring the denominator gives $\begin{aligned} 252^2-248^2 &=(252+248)\\ &\quad{}\cdot(252-248)\\ &=500\cdot4=2000. \end{aligned}$ Thus the given quotient is $\frac{1000^2}{2000}=500.$ Therefore, the correct answer is C.

2.

If $x,$ $y$ and $y-\frac1x$ are not $0,$ then $\frac{x-\frac1y}{y-\frac1x}$ equals

$1$

$\frac{x}{y}$

$\frac{y}{x}$

$\frac{x}{y}-\frac{y}{x}$

$xy-\frac1{xy}$

Concepts:fraction algebraic manipulation

Difficulty rating: 1330

Small Hint:

Write the numerator and denominator as single fractions

Big Hint:

Both parts contain the common factor $xy-1$

Solution:

We have $x-\frac1y=\frac{xy-1}{y}$ and $y-\frac1x=\frac{xy-1}{x}.$ The hypotheses make the needed quantities nonzero, so their quotient is $\frac{\frac{xy-1}{y}}{\frac{xy-1}{x}}=\frac{x}{y}.$ Therefore, the correct answer is B.

3.

Let $n$ be the smallest nonprime integer greater than $1$ with no prime factor less than $10.$ Then

$100\lt n\le110$

$110\lt n\le120$

$120\lt n\le130$

$130\lt n\le140$

$140\lt n\le150$

Concepts:prime factorization extremal argument

Difficulty rating: 1770

Small Hint:

The smallest permitted prime factor is $11$

Big Hint:

To make the smallest composite, use the two smallest permitted prime factors

Solution:

Every prime factor of $n$ is at least $11.$ The smallest composite with that property is $11\cdot11=121,$ which lies in $120\lt n\le130.$

Therefore, the correct answer is C.

4.

A rectangle intersects a circle as shown: $AB=4,$ $BC=5$ and $DE=3.$ Then $EF$ equals

$6$

$7$

$\frac{20}{3}$

$8$

$9$

Concepts:chord perpendicular bisector rectangle

Difficulty rating: 1720

Small Hint:

The perpendicular bisectors of the two parallel chords pass through the same center

Big Hint:

Therefore the midpoints of $BC$ and $EF$ have the same horizontal position

Solution:

The line through the circle’s center perpendicular to the parallel chords $BC$ and $EF$ bisects both. Measured from the rectangle’s left side, the midpoint of $BC$ is $4+\frac52,$ while the midpoint of $EF$ is $3+\frac{EF}{2}.$ Thus $4+\frac52=3+\frac{EF}{2},$ giving $EF=7.$

Therefore, the correct answer is B.

5.

The largest integer $n$ for which $n^{200}\lt5^{300}$ is

$8$

$9$

$10$

$11$

$12$

Concepts:exponent perfect square inequality

Difficulty rating: 1630

Small Hint:

Take the positive one-hundredth root of both sides

Big Hint:

Compare $n^2$ with $5^3$

Solution:

Taking positive one-hundredth roots gives $n^2\lt5^3=125.$ Since $11^2=121\lt125$ but $12^2=144\gt125,$ the largest possible integer is $11.$

Therefore, the correct answer is D.

6.

In a certain school, there are three times as many boys as girls and nine times as many girls as teachers. Using the letters $b,$ $g,$ $t$ to represent the number of boys, girls and teachers, respectively, then the total number of boys, girls and teachers can be represented by the expression

$31b$

$\frac{37}{27}b$

$13g$

$\frac{37}{27}g$

$\frac{37}{27}t$

Concepts:ratio and proportion algebraic manipulation

Difficulty rating: 1290

Small Hint:

Express both $g$ and $t$ in terms of $b$

Big Hint:

The given ratios imply $g=\frac{b}{3}$ and $t=\frac{b}{27}$

Solution:

Because $b=3g$ and $g=9t,$ we have $g=\frac{b}{3}$ and $t=\frac{b}{27}.$ Hence $b+g+t=b+\frac b3+\frac b{27}=\frac{37}{27}b.$ Therefore, the correct answer is B.

7.

When Dave walks to school, he averages $90$ steps per minute, each of his steps $75$ cm long. It takes him $16$ minutes to get to school. His brother, Jack, going to the same school by the same route, averages $100$ steps per minute, but his steps are only $60$ cm long. How long does it take Jack to get to school?

$14\frac29$ min.

$15$ min.

$18$ min.

$20$ min.

$22\frac29$ min.

Concepts:distance rate and time unit conversion

Difficulty rating: 1290

Small Hint:

Find the route’s length from Dave’s step rate, step length, and time

Big Hint:

Jack covers $100\cdot60$ centimeters each minute

Solution:

The route is $16\cdot90\cdot75$ centimeters long. Jack covers $100\cdot60$ centimeters per minute, so his time is $\frac{16\cdot90\cdot75}{100\cdot60}=18$ minutes. Therefore, the correct answer is C.

8.

Figure $ABCD$ is a trapezoid with $AB\parallel DC,$ $AB=5,$ $BC=3\sqrt2,$ $\angle BCD=45^\circ$ and $\angle CDA=60^\circ.$ The length of $DC$ is

$7+\frac23\sqrt3$

$8$

$9\frac12$

$8+\sqrt3$

$8+3\sqrt3$

Concepts:trapezoid special right triangle area decomposition

Difficulty rating: 1800

Small Hint:

Drop perpendiculars from $A$ and $B$ to $DC$

Big Hint:

The right-hand triangle is $45^\circ$ - $45^\circ$ - $90^\circ,$ and the left-hand triangle is $30^\circ$ - $60^\circ$ - $90^\circ$

Solution:

Drop perpendiculars from $A$ and $B$ to $DC.$ Since $BC=3\sqrt2$ and the angle at $C$ is $45^\circ,$ both the height and the right horizontal offset are $3.$ The left $30^\circ$ - $60^\circ$ - $90^\circ$ triangle has height $3,$ so its horizontal offset is $\sqrt3.$ Therefore $DC=\sqrt3+AB+3=8+\sqrt3.$ The correct answer is D.

9.

The number of digits in $4^{16}5^{25}$ (when written in the usual base $10$ form) is

$31$

$30$

$29$

$28$

$27$

Concepts:exponent digits power of 2

Difficulty rating: 1750

Small Hint:

Rewrite the power of $4$ as a power of $2$

Big Hint:

Pair $25$ factors of $2$ with the $25$ factors of $5$

Solution:

We have $\begin{aligned} 4^{16}5^{25} &=2^{32}5^{25}\\ &=2^7\cdot10^{25}\\ &=128\cdot10^{25}. \end{aligned}$ This is $128$ followed by $25$ zeros, so it has $3+25=28$ digits. Therefore, the correct answer is D.

10.

Four complex numbers lie at the vertices of a square in the complex plane. Three of the numbers are $1+2i,$ $-2+i$ and $-1-2i.$ The fourth number is

$2+i$

$2-i$

$1-2i$

$-1+2i$

$-2-i$

Concepts:complex number square (geometry)symmetry

Difficulty rating: 1680

Small Hint:

Two of the listed vertices are opposites

Big Hint:

Their midpoint is the square’s center, so reflect the remaining given vertex through that point

Solution:

The points $1+2i$ and $-1-2i$ are opposite vertices, so their midpoint $0$ is the center of the square. The fourth vertex is therefore the reflection of $-2+i$ through the origin, namely $2-i.$

Therefore, the correct answer is B.

11.

A calculator has a key which replaces the displayed entry with its square, and another key which replaces the displayed entry with its reciprocal. Let $y$ be the final result if one starts with an entry $x\ne0$ and alternately squares and reciprocates $n$ times each. Assuming the calculator is completely accurate (e.g., no roundoff or overflow), then $y$ equals

$x^{(-2)^n}$

$x^{2n}$

$x^{-2n}$

$x^{-2^n}$

$x^{(-1)^n2n}$

Concepts:exponent process simulation pattern recognition

Difficulty rating: 1750

Small Hint:

Track only the exponent of $x$

Big Hint:

One squaring-reciprocating pair multiplies the exponent by $-2$

Solution:

Write the displayed value as $x^e.$ Squaring changes $e$ to $2e,$ and then reciprocating changes it to $-2e.$ Starting from $e=1,$ after $n$ such pairs the exponent is $(-2)^n.$ Thus $y=x^{(-2)^n}.$

Therefore, the correct answer is A.

12.

If the sequence $\{a_n\}$ is defined by $\begin{aligned} a_1&=2,\\ a_{n+1}&=a_n+2n\quad(n\ge1), \end{aligned}$ then $a_{100}$ equals

$9900$

$9902$

$9904$

$10100$

$10102$

Concepts:recursion arithmetic sequence summation

Difficulty rating: 1800

Small Hint:

Add all increments from $a_1$ through $a_{100}$

Big Hint:

Use $a_{100}=2+2(1+2+\cdots+99)$

Solution:

Telescoping the recurrence gives $\begin{aligned} a_{100} &=a_1+2\sum_{n=1}^{99}n\\ &=2+2\cdot\frac{99\cdot100}{2}\\ &=9902. \end{aligned}$ Therefore, the correct answer is B.

13.

$\frac{2\sqrt6}{\sqrt2+\sqrt3+\sqrt5}$ equals

$\sqrt2+\sqrt3-\sqrt5$

$4-\sqrt2-\sqrt3$

$\sqrt2+\sqrt3+\sqrt6-5$

$\frac12(\sqrt2+\sqrt5-\sqrt3)$

$\frac13(\sqrt3+\sqrt5-\sqrt2)$

Concepts:radical rationalizing denominator algebraic manipulation

Difficulty rating: 2150

Small Hint:

Group the first two radicals in the denominator

Big Hint:

Multiply $\sqrt2+\sqrt3+\sqrt5$ by $\sqrt2+\sqrt3-\sqrt5$

Solution:

Observe that $\begin{aligned} &(\sqrt2+\sqrt3+\sqrt5)\\ &\quad{}\cdot(\sqrt2+\sqrt3-\sqrt5)\\ &=(\sqrt2+\sqrt3)^2-5\\ &=2\sqrt6. \end{aligned}$ Hence the quotient is $\sqrt2+\sqrt3-\sqrt5.$

Therefore, the correct answer is A.

14.

The product of all real roots of the equation $x^{\log_{10}x}=10$ is

$1$

$-1$

$10$

$10^{-1}$

none of these

Concepts:logarithm quadratic substitution

Difficulty rating: 1960

Small Hint:

Take the base- $10$ logarithm of both sides

Big Hint:

Let $t=\log_{10}x$ and solve the resulting equation in $t$

Solution:

The logarithm requires $x\gt0.$ Taking base- $10$ logarithms gives $(\log_{10}x)^2=1.$ Thus $\log_{10}x=\pm1,$ so the two roots are $10$ and $10^{-1},$ whose product is $1.$

Therefore, the correct answer is A.

15.

If $\sin2x\sin3x=\cos2x\cos3x,$ then one value for $x$ is

$18^\circ$

$30^\circ$

$36^\circ$

$45^\circ$

$60^\circ$

Concepts:trigonometric identity angle sum

Difficulty rating: 1800

Small Hint:

Move all terms to one side and recognize a cosine addition identity

Big Hint:

Use $\cos2x\cos3x-\sin2x\sin3x$ $=\cos5x$

Solution:

The equation is equivalent to $\begin{aligned} &\cos2x\cos3x\\ &\quad-\sin2x\sin3x\\ &=\cos5x=0. \end{aligned}$ Therefore $5x=90^\circ+180^\circ k.$ Taking $k=0$ gives $x=18^\circ,$ the listed value.

Therefore, the correct answer is A.

16.

The function $f(x)$ satisfies $f(2+x)=f(2-x)$ for all real numbers $x.$ If the equation $f(x)=0$ has exactly four distinct real roots, then the sum of these roots is

$0$

$2$

$4$

$6$

$8$

Concepts:functional equation symmetry (algebra)pairing and grouping

Difficulty rating: 1800

Small Hint:

The equation makes the graph symmetric about $x=2$

Big Hint:

Pair each root $2+r$ with its reflected root $2-r$

Solution:

The relation shows that every root $2+r$ is paired with $2-r.$ Four distinct roots therefore form two such pairs, and each pair has sum $4.$ The sum of all four roots is $2\cdot4=8.$

Therefore, the correct answer is E.

17.

A right triangle $ABC$ with hypotenuse $AB$ has side $AC=15.$ Altitude $CH$ divides $AB$ into segments $AH$ and $HB,$ with $HB=16.$ The area of $\triangle ABC$ is

$120$

$144$

$150$

$216$

$144\sqrt5$

Concepts:right triangle altitude similarity triangle area

Difficulty rating: 1920

Small Hint:

Use the right-triangle projection relation $AC^2=AH\cdot AB$

Big Hint:

After finding $AH,$ use $CH^2=AH\cdot HB$

Solution:

Let $AH=h.$ Similarity in the right triangle gives $\begin{aligned} AC^2&=AH\cdot AB,\\ 225&=h(h+16), \end{aligned}$ so $h=9.$ Hence $AB=25.$ Also $CH=\sqrt{AH\cdot HB},$ so $CH=\sqrt{9\cdot16}=12.$ The area is $\frac12\cdot25\cdot12=150.$

Therefore, the correct answer is C.

18.

A point $(x,y)$ is to be chosen in the coordinate plane so that it is equally distant from the $x$ -axis, the $y$ -axis, and the line $x+y=2.$ Then $x$ is

$\sqrt2-1$

$\frac12$

$2-\sqrt2$

$1$

not uniquely determined

Concepts:distance formula angle bisector counterexample

Difficulty rating: 2130

Small Hint:

Equal distance from the two axes forces $|x|=|y|$

Big Hint:

Check both lines $y=x$ and $y=-x$

Solution:

Equal distance from the axes gives $y=x$ or $y=-x.$ On $y=-x,$ the distance to the line $x+y=2$ is $\sqrt2,$ so both $(\sqrt2,-\sqrt2)$ and $(-\sqrt2,\sqrt2)$ satisfy all three distance conditions. Their $x$ -coordinates differ, so $x$ is not uniquely determined.

Therefore, the correct answer is E.

19.

A box contains $11$ balls, numbered $1,$ $2,$ $3,$ $\ldots,$ $11.$ If $6$ balls are drawn simultaneously at random, what is the probability that the sum of the numbers on the balls drawn is odd?

$\frac{100}{231}$

$\frac{115}{231}$

$\frac12$

$\frac{118}{231}$

$\frac6{11}$

Concepts:combinations parity sampling without replacement

Difficulty rating: 2070

Small Hint:

There are $6$ odd-numbered balls and $5$ even-numbered balls

Big Hint:

Count selections containing $1,$ $3,$ or $5$ odd balls

Solution:

An odd sum requires an odd number of the six selected balls to be odd. There are $6$ odd and $5$ even balls, so the favorable count is $\begin{aligned} &\binom61\binom55+\binom63\binom53\\ &\quad+\binom65\binom51\\ &=6+200+30=236. \end{aligned}$ Of the $\binom{11}{6}=462$ selections, the desired probability is $\frac{236}{462}=\frac{118}{231}.$

Therefore, the correct answer is D.

20.

The number of distinct solutions of the equation $\left|x-|2x+1|\right|=3$ is

$0$

$1$

$2$

$3$

$4$

Concepts:absolute value casework linear equation

Difficulty rating: 1720

Small Hint:

Split the outer absolute value into two equations

Big Hint:

For each equation, isolate $|2x+1|$ before checking its two cases

Solution:

If $x-|2x+1|=3,$ then $|2x+1|=x-3,$ which requires $x\ge3$ and yields no solution. If $x-|2x+1|=-3,$ then $|2x+1|=x+3.$ Its two linear cases give $x=2$ and $x=-\frac{4}{3},$ both valid. Thus there are $2$ distinct solutions.

Therefore, the correct answer is C.

21.

The number of triples $(a,b,c)$ of positive integers which satisfy the simultaneous equations $\begin{aligned} ab+bc&=44,\\ ac+bc&=23. \end{aligned}$ is

$0$

$1$

$2$

$3$

$4$

Concepts:Diophantine Equation factor substitution

Difficulty rating: 1960

Small Hint:

Factor the second equation as $c(a+b)=23$

Big Hint:

Use the primality of $23$ and positivity to determine $c$ and $a+b$

Solution:

Since $c(a+b)=23$ and all variables are positive integers, $c=1$ and $a+b=23.$ Put $b=23-a$ in $ab+b=44.$ Then $(23-a)(a+1)=44,$ or $a^2-22a+21=0,$ so $a=1$ or $21.$ Each gives a positive $b,$ producing exactly two triples.

Therefore, the correct answer is C.

22.

Let $a$ and $c$ be fixed positive numbers. For each real number $t$ let $(x_t,y_t)$ be the vertex of the parabola $y=ax^2+tx+c.$ If the set of vertices $(x_t,y_t)$ for all real values of $t$ is graphed in the plane, the graph is

a straight line

a parabola

part, but not all, of a parabola

one branch of a hyperbola

none of these

Concepts:parabola completing the square coordinate geometry

Difficulty rating: 1960

Small Hint:

Write the vertex coordinates in terms of $t$

Big Hint:

Use $x_t=-\frac{t}{2a}$ to eliminate $t$ from $y_t$

Solution:

The vertex coordinates are $x_t=-\frac{t}{2a},\qquad y_t=c-\frac{t^2}{4a}.$ Since $t=-2ax_t,$ eliminating $t$ gives $y_t=c-ax_t^2.$ As $t$ ranges over all reals, so does $x_t,$ so the entire parabola is traced.

Therefore, the correct answer is B.

23.

$\frac{\sin10^\circ+\sin20^\circ}{\cos10^\circ+\cos20^\circ}$ equals

$\tan10^\circ+\tan20^\circ$

$\tan30^\circ$

$\frac12(\tan10^\circ+\tan20^\circ)$

$\tan15^\circ$

$\frac14\tan60^\circ$

Concepts:trigonometric identity algebraic manipulation

Difficulty rating: 2130

Small Hint:

Apply the sum-to-product identities to both numerator and denominator

Big Hint:

Both transformed sums contain the same $\cos5^\circ$ factor

Solution:

Let $R$ denote the given ratio. By the sum-to-product identities, $\begin{aligned} R&=\frac{\sin10^\circ+\sin20^\circ} {\cos10^\circ+\cos20^\circ},\\ R&=\frac{2\sin15^\circ\cos5^\circ} {2\cos15^\circ\cos5^\circ}\\ &=\tan15^\circ. \end{aligned}$ Therefore, the correct answer is D.

24.

If $a$ and $b$ are positive real numbers and each of the equations $\begin{aligned} x^2+ax+2b&=0,\\ x^2+2bx+a&=0 \end{aligned}$ has real roots, then the smallest possible value of $a+b$ is

$2$

$3$

$4$

$5$

$6$

Concepts:quadratic inequality

Difficulty rating: 2240

Small Hint:

Require both quadratic discriminants to be nonnegative

Big Hint:

Combine $a^2\ge8b$ and $b^2\ge a$ to bound $a$ and $b$

Solution:

The two discriminants give $a^2\ge8b$ and $b^2\ge a.$ Therefore $a^4\ge64b^2\ge64a.$ Since $a\gt0,$ this yields $a\ge4,$ and then $b^2\ge a\ge4$ gives $b\ge2.$ Thus $a+b\ge6.$ Equality occurs at $a=4,b=2,$ for which both discriminants are zero.

Therefore, the correct answer is E.

25.

The total area of all the faces of a rectangular solid is $22\text{ cm}^2,$ and the total length of all its edges is $24\text{ cm}.$ Then the length in cm of any one of its internal diagonals is

$\sqrt{11}$

$\sqrt{12}$

$\sqrt{13}$

$\sqrt{14}$

not uniquely determined

Concepts:rectangular prism surface area diagonal

Difficulty rating: 1770

Small Hint:

Let the side lengths be $x,y,z$ and translate both totals into equations

Big Hint:

Expand $(x+y+z)^2$ to find $x^2+y^2+z^2$

Solution:

The conditions give $2(xy+xz+yz)=22$ and $4(x+y+z)=24,$ so $xy+xz+yz=11$ and $x+y+z=6.$ Hence the square of an internal diagonal is $x^2+y^2+z^2=6^2-2(11)=14.$ Its length is $\sqrt{14}.$

Therefore, the correct answer is D.

26.

In the obtuse triangle $ABC,$ $AM=MB,$ $MD\perp BC,$ $EC\perp BC.$ If the area of $\triangle ABC$ is $24,$ then the area of $\triangle BED$ is

$9$

$12$

$15$

$18$

not uniquely determined

Concepts:triangle area midpoint area decomposition

Difficulty rating: 1960

Small Hint:

Draw segment $MC$ and compare $\triangle DMC$ with $\triangle DME$

Big Hint:

Those triangles have the same base $MD$ and equal altitudes; also $M$ is the midpoint of $AB$

Solution:

Draw $MC.$ Since $EC\parallel MD,$ points $E$ and $C$ have equal perpendicular distances from $MD,$ so $[DME]=[DMC].$ Therefore $\begin{aligned} [BED]&=[BMD]+[DME]\\ &=[BMD]+[DMC]\\ &=[BMC]. \end{aligned}$ Because $M$ is the midpoint of $AB,$ triangle $BMC$ has half the area of $ABC.$ Thus $[BED]=12.$

Therefore, the correct answer is B.

27.

In $\triangle ABC,$ $D$ is on $AC$ and $F$ is on $BC.$ Also, $AB\perp AC,$ $AF\perp BC,$ and $BD=DC=FC=1.$ Find $AC.$

$\sqrt2$

$\sqrt3$

$\sqrt[3]{2}$

$\sqrt[3]{3}$

$\sqrt[4]{3}$

Concepts:right triangle similarity law of cosines

Difficulty rating: 2350

Small Hint:

Let $AC=s$ and use the altitude-to-hypotenuse similarity relation

Big Hint:

Then compare $\cos\angle ACB$ in right triangle $ABC$ and isosceles triangle $BDC$

Solution:

Let $AC=s.$ The right-triangle projection relation gives $AC^2=FC\cdot BC,$ so $BC=s^2.$ Since $BD=DC=1,$ the numerator and denominator in the Law of Cosines satisfy $\begin{aligned} BC^2+DC^2-BD^2&=s^4,\\ 2(BC)(DC)&=2s^2. \end{aligned}$ Hence $\cos\angle BCD=\frac{s^2}{2}.$ But $D$ lies on $AC,$ and in right triangle $ABC,$ $\cos\angle BCA=\frac{AC}{BC}=\frac{1}{s}.$ Thus $\frac{s^2}{2}=\frac{1}{s},$ so $s^3=2$ and $AC=\sqrt[3]{2}.$

Therefore, the correct answer is C.

28.

The number of distinct pairs of integers $(x,y)$ such that $\begin{aligned} 0&\lt x\lt y,\\ \sqrt{1984}&=\sqrt{x}+\sqrt{y} \end{aligned}$ is

$0$

$1$

$2$

$3$

$7$

Concepts:radical perfect square counting pairs

Difficulty rating: 2380

Small Hint:

Use $1984=31\cdot8^2$ and write both radicands with a common squarefree part

Big Hint:

Set $x=31u^2,\ y=31v^2$ and count positive integers $u\lt v$ with $u+v=8$

Solution:

Because $\sqrt{x}+\sqrt{y}=8\sqrt{31},$ the two radicals must have common squarefree part $31.$ Write $x=31u^2$ and $y=31v^2$ for positive integers $u,v.$ Then $u+v=8,\qquad u\lt v.$ The possibilities are $(u,v)=(1,7),$ $(2,6),$ and $(3,5),$ producing three distinct pairs $(x,y).$

Therefore, the correct answer is D.

29.

Find the largest value of $\frac{y}{x}$ for pairs of real numbers $(x,y)$ which satisfy $(x-3)^2+(y-3)^2=6.$

$3+2\sqrt2$

$2+\sqrt3$

$3\sqrt3$

$6$

$6+2\sqrt3$

Concepts:circle slope tangent line optimization

Difficulty rating: 2240

Small Hint:

Interpret $\frac{y}{x}$ as the slope of a line through the origin

Big Hint:

At an extreme slope, the line $y=mx$ is tangent to the circle

Solution:

The circle lies entirely where $x\gt0,$ so $\frac{y}{x}$ is the slope $m$ of the line $y=mx$ through a point on it. At an extreme, this line is tangent. The distance from the center $(3,3)$ to $mx-y=0$ must equal $\sqrt6,$ so $\frac{|3m-3|}{\sqrt{m^2+1}}=\sqrt6.$ Squaring and simplifying gives $m^2-6m+1=0,$ hence $m=3\pm2\sqrt2.$ The larger value is $3+2\sqrt2.$

Therefore, the correct answer is A.

30.

For any complex number $w=a+bi,$ $|w|$ is defined to be the real number $\sqrt{a^2+b^2}.$ If $w=\cos40^\circ+i\sin40^\circ,$ then $|w+2w^2+3w^3+\cdots+9w^9|^{-1}$ equals

$\frac19\sin40^\circ$

$\frac29\sin20^\circ$

$\frac19\cos40^\circ$

$\frac1{18}\cos20^\circ$

none of these

Concepts:complex number roots of unity arithmetico-geometric series

Difficulty rating: 2460

Small Hint:

Let $S=w+2w^2+\cdots+9w^9$ and subtract $wS$ from $S$

Big Hint:

Use $w^9=1$ and $|1-e^{i\theta}|=2\sin(\frac{\theta}{2})$

Solution:

Let $S=\sum_{k=1}^9kw^k.$ Since $w^9=1$ and $w^{10}=w,$ $\begin{aligned} (1-w)S &=(w+w^2+\cdots+w^9)\\ &\quad-9w^{10}\\ &=-9w. \end{aligned}$ Therefore $|S|=\frac{9}{|1-w|},$ because $|w|=1.$ Thus $|S|^{-1}=\frac{|1-w|}{9} =\frac{2}{9}\sin20^\circ.$ Therefore, the correct answer is B.