1996 AMC 12 — Problems & Online Timed Exam

Timed

1:15:00

1.

The addition below is incorrect. What is the largest digit that can be changed to make the addition correct? $\begin{array}{r} 641\\ 852\\ {}+973\\ \hline 2456 \end{array}$

$4$

$5$

$6$

$7$

$8$

Answer: D

Concepts:addition place value

Difficulty rating: 800

Small Hint:

Compare the displayed sum with the actual sum of the three addends

Big Hint:

The correction must lower the total by one unit in a particular place

Solution:

The three addends total $641+852+973=2466,$ which is $10$ too large. Changing the tens digit $7$ in $973$ to $6$ lowers the sum by $10$ and gives $2456.$ No larger listed digit can make that change, so the correct answer is D.

2.

Each day Walter gets $\$3$ for doing his chores or $\$5$ for doing them exceptionally well. After $10$ days of doing his chores daily, Walter has received a total of $\$36.$ On how many days did Walter do them exceptionally well?

$3$

$4$

$5$

$6$

$7$

Answer: A

Concepts:linear equations rates

Difficulty rating: 800

Small Hint:

Begin with the amount Walter would earn at the ordinary rate on all ten days

Big Hint:

Each exceptionally good day adds the same amount to that baseline

Solution:

Ten ordinary days would pay $10\cdot3=30$ dollars. Each exceptional day adds $5-3=2$ dollars, and the actual total is $6$ dollars higher. Thus there were $\frac{6}{2}=3$ exceptional days, so the correct answer is A.

3.

$\displaystyle\frac{(3!)!}{3!}=$

$1$

$2$

$6$

$40$

$120$

Answer: E

Concepts:factorials arithmetic

Difficulty rating: 920

Small Hint:

Evaluate the inner factorial before the outer factorial

Big Hint:

Cancel the denominator from the product in the numerator

Solution:

Since $3!=6,$ the expression is $\frac{6!}{6}=5!=120.$ Thus the correct answer is E.

4.

Six numbers from a list of nine integers are $7,$ $8,$ $3,$ $5,$ $9,$ and $5.$ The largest possible value of the median of all nine numbers in this list is

$5$

$6$

$7$

$8$

$9$

Answer: D

Concepts:median extremal reasoning

Difficulty rating: 1020

Small Hint:

The median is the fifth number after all nine are sorted

Big Hint:

To maximize the median, take each of the three unspecified numbers as large as needed

Solution:

The six specified values in order are $3,5,5,7,8,9.$ Even if all three unspecified integers exceed $9,$ the fifth entry of the full sorted list is $8.$ This is attainable, so the correct answer is D.

5.

Given that $0\lt a\lt b\lt c\lt d,$ which of the following is the largest?

$\frac{a+b}{c+d}$

$\frac{a+d}{b+c}$

$\frac{b+c}{a+d}$

$\frac{b+d}{a+c}$

$\frac{c+d}{a+b}$

Answer: E

Concepts:inequalities fraction comparison

Difficulty rating: 1200

Small Hint:

A positive fraction grows when its numerator increases or its denominator decreases

Big Hint:

Compare every numerator and denominator with $c+d$ and $a+b$ , respectively

Solution:

Among the displayed numerators, $c+d$ is largest; among the denominators, $a+b$ is smallest. Therefore $\frac{c+d}{a+b}$ exceeds every other positive fraction listed, so the correct answer is E.

6.

If $f(x)=x^{x+1}(x+2)^{x+3},$ then $\begin{gathered} f(0)+f(-1)\\ {}+f(-2)+f(-3)= \end{gathered}$

$-\frac89$

$0$

$\frac89$

$1$

$\frac{10}9$

Answer: E

Concepts:exponents function evaluation

Difficulty rating: 1360

Small Hint:

Evaluate the four function values separately, paying attention to zero exponents

Big Hint:

The terms at $x=0$ and $x=-2$ vanish before any problematic power is needed

Solution:

Direct substitution gives $\begin{aligned} f(0)&=0,\\ f(-1)&=1,\\ f(-2)&=0,\\ f(-3)&=\frac19. \end{aligned}$ Their sum is $\frac{10}{9},$ so the correct answer is E.

7.

A father takes his twins and a younger child out to dinner on the twins’ birthday. The restaurant charges $\$4.95$ for the father and $\$0.45$ for each year of a child’s age, where age is defined as the age at the most recent birthday. If the bill is $\$9.45,$ which of the following could be the age of the youngest child?

$1$

$2$

$3$

$4$

$5$

Answer: B

Concepts:linear Diophantine equations age problems

Difficulty rating: 1100

Small Hint:

Subtract the father’s charge and divide the remainder by the price per year

Big Hint:

If the twins are $t$ years old and the younger child is $y$ , impose both $2t+y=10$ and $y\lt t$

Solution:

The children account for $9.45-4.95=4.50$ dollars, so their ages total $\frac{4.50}{0.45}=10.$ If each twin is $t$ and the younger child is $y,$ then $2t+y=10$ with $y\lt t.$ Of the choices, $y=2$ gives $t=4,$ which works. Thus the correct answer is B.

8.

If $3=k\cdot2^r$ and $15=k\cdot4^r,$ then $r=$

$-\log_2 5$

$\log_5 2$

$\log_{10}5$

$\log_2 5$

$\frac52$

Answer: D

Concepts:logarithms exponential equations

Difficulty rating: 1280

Small Hint:

Divide the second equation by the first to eliminate $k$

Big Hint:

Rewrite $\frac{4^r}{2^r}$ as a single power of $2$

Solution:

Dividing the equations gives $5=(\frac{4}{2})^r=2^r.$ Hence $r=\log_2 5,$ so the correct answer is D.

9.

Triangle $PAB$ and square $ABCD$ are in perpendicular planes. Given that $PA=3,$ $PB=4,$ and $AB=5,$ what is $PD?$

$5$

$\sqrt{34}$

$\sqrt{41}$

$2\sqrt{13}$

$8$

Answer: B

Concepts:three-dimensional geometry Pythagorean theorem

Difficulty rating: 1570

Small Hint:

First identify the right angle in the $3$ - $4$ - $5$ triangle $PAB$

Big Hint:

A line in one plane perpendicular to the planes’ intersection is perpendicular to the other plane

Solution:

Since $PA^2+PB^2$ $=3^2+4^2=5^2=AB^2,$ triangle $PAB$ is right at $P.$ In square $ABCD,$ $AD=5$ and $AD\perp AB.$ Because the two planes are perpendicular along $AB,$ $AD$ is perpendicular to the plane of $PAB,$ and hence to $AP.$ Thus $PD^2=PA^2+AD^2$ $=3^2+5^2=34.$ The correct answer is B.

10.

How many line segments have both their endpoints located at the vertices of a given cube?

$12$

$15$

$24$

$28$

$56$

Answer: D

Concepts:combinations solid geometry

Difficulty rating: 920

Small Hint:

A segment is determined by choosing two distinct cube vertices

Big Hint:

Count unordered pairs among the cube’s eight vertices

Solution:

Every unordered pair of the cube’s $8$ vertices determines one segment, including edges and diagonals. There are $\binom82=28,$ so the correct answer is D.

11.

Given a circle of radius $2,$ there are many line segments of length $2$ that are tangent to the circle at their midpoints. Find the area of the region consisting of all such line segments.

$\frac{\pi}{4}$

$4-\pi$

$\frac{\pi}{2}$

$\pi$

$2\pi$

Answer: D

Concepts:geometric loci annulus area

Difficulty rating: 1630

Small Hint:

Each tangent segment extends one unit in each direction from its midpoint

Big Hint:

Use a right triangle from the circle’s center to locate the inner and outer radii of the swept region

Solution:

Each segment is centered at a tangency point $2$ units from the circle’s center and extends $1$ unit along the tangent in both directions. As the tangency point rotates, the segments fill the annulus with inner radius $2$ and outer radius $\sqrt{2^2+1^2}=\sqrt5.$ Its area is $\pi(5-4)=\pi,$ so the correct answer is D.

12.

A function $f$ from the integers to the integers is defined as follows: $f(n)= \begin{cases} n+3 & \text{if $n$ is odd},\\ \frac{n}{2} & \text{if $n$ is even}. \end{cases}$ Suppose $k$ is odd and $f(f(f(k)))=27.$ What is the sum of the digits of $k?$

$3$

$6$

$9$

$12$

$15$

Answer: B

Concepts:piecewise functions parity

Difficulty rating: 1570

Small Hint:

Because $k$ is odd, the first iterate is $k+3$ , which is even

Big Hint:

Work backward from $27$ , checking the parity required by each inverse branch

Solution:

Since $k$ is odd, $f(k)=k+3$ is even, so $f(f(k))=\frac{k+3}{2}.$ If this value were odd, adding $3$ could not produce $27$ because that would require the value $24.$ Therefore it is even and is halved to $27,$ so $\frac{k+3}{2}=54.$ Hence $k=105,$ whose digits sum to $6.$ The correct answer is B.

13.

Sunny runs at a steady rate, and Moonbeam runs $m$ times as fast, where $m$ is a number greater than $1.$ If Moonbeam gives Sunny a head start of $h$ meters, how many meters must Moonbeam run to overtake Sunny?

$hm$

$\frac{h}{h+m}$

$\frac{h}{m-1}$

$\frac{hm}{m-1}$

$\frac{h+m}{m-1}$

Answer: D

Concepts:relative speed algebra

Difficulty rating: 1200

Small Hint:

Let Sunny’s speed be $v$ and Moonbeam’s speed be $mv$

Big Hint:

If Moonbeam runs $x$ meters, express Sunny’s distance during the same time in terms of $x$

Solution:

Suppose Moonbeam runs $x$ meters. The elapsed time is $\frac{x}{mv},$ so Sunny runs $\frac{v\cdot x}{mv}=\frac{x}{m}$ meters after the start. At the catch, $x=h+\frac{x}{m}.$ Solving gives $x=\frac{hm}{m-1},$ so the correct answer is D.

14.

Let $E(n)$ denote the sum of the even digits of $n.$ For example, $E(5681)=6+8=14.$ Find $\begin{gathered} E(1)+E(2)+E(3)+\cdots\\ {}+E(100). \end{gathered}$

$200$

$360$

$400$

$900$

$2250$

Answer: C

Concepts:digit sums place-value counting

Difficulty rating: 1360

Small Hint:

Include leading zeros and consider the integers from $00$ through $99$

Big Hint:

In each digit position, every digit occurs equally often

Solution:

From $00$ through $99,$ each digit occurs $10$ times in each of the two positions. The positive even digits sum to $2+4+6+8=20,$ so the total contribution is $2\cdot10\cdot20=400.$ The number $100$ contributes no even positive digit, so the correct answer is C.

15.

Two opposite sides of a rectangle are each divided into $n$ congruent segments, and the endpoints of one segment are joined to the center to form triangle $A.$ The other sides are each divided into $m$ congruent segments, and the endpoints of one of these segments are joined to the center to form triangle $B.$ [See figure for $n=5,$ $m=7.$ ] What is the ratio of the area of triangle $A$ to the area of triangle $B?$

$1$

$\frac mn$

$\frac nm$

$\frac{2m}{n}$

$\frac{2n}{m}$

Answer: B

Concepts:triangle area ratios

Difficulty rating: 1330

Small Hint:

Write the rectangle’s side lengths as $w$ and $h$

Big Hint:

Each triangle has a base that is one divided segment and an altitude equal to half the opposite side length

Solution:

Triangle $A$ has base $\frac{w}{n}$ and altitude $\frac{h}{2},$ so its area is $\frac{wh}{4n}.$ Triangle $B$ has base $\frac{h}{m}$ and altitude $\frac{w}{2},$ so its area is $\frac{wh}{4m}.$ Their ratio is $\frac{m}{n},$ and the correct answer is B.

16.

A fair standard six-sided dice is tossed three times. Given that the sum of the first two tosses equals the third, what is the probability that at least one $2$ is tossed?

$\frac16$

$\frac{91}{216}$

$\frac12$

$\frac8{15}$

$\frac7{12}$

Answer: D

Concepts:conditional probability case counting

Difficulty rating: 1730

Small Hint:

Under the stated condition, count ordered pairs for the first two tosses whose sum is at most $6$

Big Hint:

For the favorable count, separate a third-toss $2$ from pairs having a $2$ among the first two tosses

Solution:

For third tosses $2,3,4,5,6,$ the numbers of ordered first-two-toss pairs are $1,2,3,4,5,$ for $15$ equally likely conditional outcomes. A third toss of $2$ contributes $(1,1,2).$ A $2$ in the first position gives $4$ outcomes and a $2$ in the second gives $4,$ with $(2,2,4)$ counted twice. Thus there are $1+4+4-1=8$ favorable outcomes, and the correct answer is D.

17.

In rectangle $ABCD,$ angle $C$ is trisected by $\overline{CF}$ and $\overline{CE},$ where $E$ is on $\overline{AB},$ $F$ is on $\overline{AD},$ $BE=6,$ and $AF=2.$ Which of the following is closest to the area of the rectangle $ABCD?$

$110$

$120$

$130$

$140$

$150$

Answer: E

Concepts:trigonometry rectangle area

Difficulty rating: 1830

Small Hint:

Each of the three angles at $C$ is $30^\circ$

Big Hint:

Use the two right triangles adjacent to sides $CB$ and $CD$ to determine the rectangle’s height and width

Solution:

Let the rectangle have width $w$ and height $h.$ In right triangle $CBE,$ $\tan30^\circ=\frac{BE}{CB}=\frac{6}{h},$ so $h=6\sqrt3.$ In the triangle using $C,F,D,$ the horizontal run is $w$ and the vertical drop is $h-2,$ so $\tan60^\circ=\frac{w}{h-2}.$ Hence $w=\sqrt3(6\sqrt3-2)=18-2\sqrt3.$ The area is $6\sqrt3(18-2\sqrt3)$ $=108\sqrt3-36\approx151.1,$ closest to $150.$ Thus the correct answer is E.

18.

A circle of radius $2$ has center at $(2,0).$ A circle of radius $1$ has center at $(5,0).$ A line is tangent to the two circles at points in the first quadrant. Which of the following is closest to the $y$ -intercept of the line?

$\frac{\sqrt2}{4}$

$\frac{\sqrt8}{3}$

$1+\sqrt3$

$2\sqrt2$

$3$

Answer: D

Concepts:coordinate geometry common tangents

Difficulty rating: 1960

Small Hint:

Write the tangent as $y=mx+b$ and use point-to-line distance for each center

Big Hint:

Subtract the two distance equations to determine the slope before solving for the intercept

Solution:

Write the upper common tangent as $mx-y+b=0$ and let $s=\sqrt{m^2+1}.$ The two center-to-line distances give $2m+b=2s,\qquad 5m+b=s.$ Thus $3m=-s,$ so $9m^2=m^2+1$ and $m=-\frac{1}{\sqrt8}.$ From the second equation, $\begin{aligned} b&=s-5m\\ &=\frac3{\sqrt8}+\frac5{\sqrt8}=2\sqrt2. \end{aligned}$ Therefore the correct answer is D.

19.

The midpoints of the sides of a regular hexagon $ABCDEF$ are joined to form a smaller hexagon. What fraction of the area of $ABCDEF$ is enclosed by the smaller hexagon?

$\frac12$

$\frac{\sqrt3}{3}$

$\frac23$

$\frac34$

$\frac{\sqrt3}{2}$

Answer: D

Concepts:similar polygons area ratios

Difficulty rating: 1570

Small Hint:

Compare the distance between adjacent side midpoints with the original side length

Big Hint:

The smaller and larger hexagons are similar, so square their side-length ratio

Solution:

Two adjacent midpoints and their shared vertex form a triangle with two sides $\frac{s}{2}$ and included angle $120^\circ.$ Its opposite side has squared length $\begin{aligned} \frac{s^2}{4}+\frac{s^2}{4} &-2\left(\frac{s}{2}\right)^2\cos120^\circ\\ &=\frac{3s^2}{4}. \end{aligned}$ Thus the smaller hexagon has scale factor $\frac{\sqrt3}{2},$ and its area ratio is $\frac{3}{4}.$ The correct answer is D.

20.

In the $xy$ -plane, what is the length of the shortest path from $(0,0)$ to $(12,16)$ that does not go inside the circle $(x-6)^2+(y-8)^2=25?$

$10\sqrt3$

$10\sqrt5$

$10\sqrt3+\frac{5\pi}{3}$

$\frac{40\sqrt3}{3}$

$10+5\pi$

Answer: C

Concepts:shortest paths tangent lengths arc length

Difficulty rating: 1990

Small Hint:

The circle’s center is the midpoint of the two endpoints

Big Hint:

The shortest permitted path consists of two tangent segments and the shorter arc between their tangency points

Solution:

Each endpoint is $10$ units from the circle’s center, so each tangent segment has length $\sqrt{10^2-5^2}=5\sqrt3.$ In the right triangle at a tangency point, the angle at the center is $60^\circ.$ Since the endpoint rays are opposite, the intervening minor arc subtends $180^\circ-2(60^\circ)=60^\circ,$ so its length is $\frac{5\pi}{3}.$ The total is $10\sqrt3+\frac{5\pi}{3},$ and the correct answer is C.

21.

Triangles $ABC$ and $ABD$ are isosceles with $AB=AC=BD,$ and $\overline{BD}$ intersects $\overline{AC}$ at $E.$ If $\overline{BD}\perp\overline{AC},$ then $\angle C+\angle D$ is

$115^\circ$

$120^\circ$

$130^\circ$

$135^\circ$

not uniquely determined

Answer: D

Concepts:isosceles triangles angle chasing

Difficulty rating: 2030

Small Hint:

Let $\angle BAC=\alpha$ and express $\angle C$ using isosceles triangle $ABC$

Big Hint:

Use $BD\perp AC$ to find the vertex angle at $B$ of isosceles triangle $ABD$

Solution:

Let $\angle BAC=\alpha.$ Since $AB=AC,$ $\angle C=\frac{180^\circ-\alpha}{2}=90^\circ-\frac{\alpha}{2}.$ Because $BD\perp AC,$ the angle between $BA$ and $BD$ is $90^\circ-\alpha.$ In isosceles triangle $ABD,$ $AB=BD,$ so its two base angles are $\begin{aligned} \angle D &=\frac{180^\circ-(90^\circ-\alpha)}2\\ &=45^\circ+\frac{\alpha}{2}. \end{aligned}$ Their sum is $135^\circ,$ so the correct answer is D.

22.

Four distinct points, $A,$ $B,$ $C,$ and $D,$ are to be selected from $1996$ points evenly spaced around a circle. All quadruples are equally likely to be chosen. What is the probability that the chord $\overline{AB}$ intersects the chord $\overline{CD}?$

$\frac14$

$\frac13$

$\frac12$

$\frac23$

$\frac34$

Answer: B

Concepts:geometric probability symmetry

Difficulty rating: 1520

Small Hint:

Fix any four selected points and examine the three ways to pair them into two chords

Big Hint:

Exactly one pairing joins alternating points around the circle

Solution:

For any fixed four points, there are three ways to partition their labels into two unordered chord pairs. Exactly one pairing joins alternating points around the circle and therefore crosses. The named pair $\overline{AB},\overline{CD}$ is equally likely to be any of these three pairings, so the probability is $\frac{1}{3}.$ The correct answer is B.

23.

The sum of the lengths of the twelve edges of a rectangular box is $140,$ and the distance from one corner of the box to the farthest corner is $21.$ The total surface area of the box is

$776$

$784$

$798$

$800$

$812$

Answer: B

Concepts:rectangular solids symmetric identities

Difficulty rating: 1330

Small Hint:

If the side lengths are $a,b,c$ , translate the edge sum and space diagonal into equations

Big Hint:

Expand $(a+b+c)^2$ to obtain $2(ab+bc+ca)$ directly

Solution:

The edge sum gives $4(a+b+c)=140,$ so $a+b+c=35.$ The space diagonal gives $a^2+b^2+c^2=21^2=441.$ Therefore the surface area is $\begin{gathered} 2(ab+bc+ca)\\ {}=(a+b+c)^2\\ {}-(a^2+b^2+c^2)\\ {}=35^2-441=784. \end{gathered}$ Thus the correct answer is B.

24.

The sequence $\begin{gathered} 1,2,1,2,2,1,2,2,2,1,2,\\ 2,2,2,1,2,2,2,2,2,1,2,\ldots \end{gathered}$ consists of $1$ ’s separated by blocks of $2$ ’s with $n$ $2$ ’s in the $n$ th block. The sum of the first $1234$ terms of this sequence is

$1996$

$2419$

$2429$

$2439$

$2449$

Answer: B

Concepts:sequences triangular numbers

Difficulty rating: 1860

Small Hint:

Count the total number of terms through the end of the $k$ th block

Big Hint:

Find the last complete block before term $1234$ , then account for the partial next block

Solution:

Through block $k$ there are $k$ ones and $\frac{k(k+1)}{2}$ twos, hence $\frac{k(k+3)}{2}$ terms. For $k=48$ this is $1224.$ Their sum is $48+2\left(\frac{48\cdot49}{2}\right)=2400.$ The next $10$ terms are one $1$ and nine $2$ ’s, with sum $19.$ The requested sum is $2419,$ so the correct answer is B.

25.

Given that $x^2+y^2=14x+6y+6,$ what is the largest possible value that $3x+4y$ can have?

$72$

$73$

$74$

$75$

$76$

Answer: B

Concepts:circles linear optimization

Difficulty rating: 1590

Small Hint:

Complete the square to identify the circle’s center and radius

Big Hint:

The maximum of $3x+4y$ is its value at the center plus the radius times $\sqrt{3^2+4^2}$

Solution:

Completing the square gives $(x-7)^2+(y-3)^2=64.$ At the center, $3x+4y=3(7)+4(3)=33.$ Moving $8$ units in the direction of the vector $(3,4)$ increases the expression by $8\sqrt{3^2+4^2}=40.$ The maximum is $73,$ so the correct answer is B.

26.

An urn contains marbles of four colors: red, white, blue, and green. When four marbles are drawn without replacement, the following events are equally likely:

(a) the selection of four red marbles;

(b) the selection of one white and three red marbles;

(c) the selection of one white, one blue, and two red marbles; and

(d) the selection of one marble of each color.

What is the smallest number of marbles satisfying the given condition?

$19$

$21$

$46$

$69$

more than $69$

Answer: B

Concepts:combinations Diophantine equations

Difficulty rating: 2190

Small Hint:

Let the four color counts be $r,w,b,g$ and equate the combination counts for the four events

Big Hint:

Successive ratios determine $w,b,g$ in terms of $r$ ; then find the smallest $r$ making all three integers

Solution:

Equal probabilities have the same common denominator, so their favorable selection counts satisfy $\binom r4=w\binom r3$ $=wb\binom r2=wbgr.$ Successive ratios give $w=\frac{r-3}{4},$ $b=\frac{r-2}{3},$ and $g=\frac{r-1}{2}.$ The least $r\ge4$ making all three positive integers is $r=11.$ Then $(w,b,g)=(2,3,5),$ for $11+2+3+5=21$ marbles. Thus the correct answer is B.

27.

Consider two solid spherical balls, one centered at $(0,0,\frac{21}{2})$ with radius $6,$ and the other centered at $(0,0,1)$ with radius $\frac92.$ How many points $(x,y,z)$ with only integer coordinates (lattice points) are there in the intersection of the balls?

$7$

$9$

$11$

$13$

$15$

Answer: D

Concepts:lattice points spheres inequalities

Difficulty rating: 2100

Small Hint:

First intersect the possible integer ranges for the $z$ -coordinate in the two balls

Big Hint:

At the only possible height, reduce both sphere inequalities to a bound on $x^2+y^2$

Solution:

The first ball permits integer heights $5$ through $16,$ while the second permits $-3$ through $5,$ so an intersection lattice point must have $z=5.$ At that height the two bounds are $\begin{aligned} x^2+y^2 &\le36-\left(5-\frac{21}{2}\right)^2\\ &=\frac{23}{4},\\ x^2+y^2 &\le\frac{81}{4}-(5-1)^2\\ &=\frac{17}{4}. \end{aligned}$ Thus $x^2+y^2$ can be $0,1,2,$ or $4.$ These give $1,4,4,$ and $4$ ordered integer pairs, respectively, for $13$ points. The correct answer is D.

28.

On a $4\times4\times3$ rectangular parallelepiped, vertices $A,$ $B,$ and $C$ are adjacent to vertex $D.$ The perpendicular distance from $D$ to the plane containing $A,$ $B,$ and $C$ is closest to

$1.6$

$1.9$

$2.1$

$2.7$

$2.9$

Answer: C

Concepts:planes point-to-plane distance

Difficulty rating: 1990

Small Hint:

Place $D$ at the origin with the three adjacent edges along coordinate axes

Big Hint:

Write the intercept form of the plane through $(4,0,0),(0,4,0),(0,0,3)$

Solution:

Put $D=(0,0,0)$ and take the adjacent vertices as $(4,0,0),(0,4,0),(0,0,3).$ Their plane is $\frac{x}{4}+\frac{y}{4}+\frac{z}{3}=1.$ Its distance from the origin is $\begin{aligned} \frac1{\sqrt{\begin{gathered} (\frac{1}{4})^2+(\frac{1}{4})^2\\ {}+(\frac{1}{3})^2 \end{gathered}}} &=\frac{12}{\sqrt{34}}\\ &\approx2.06, \end{aligned}$ closest to $2.1.$ Thus the correct answer is C.

29.

If $n$ is a positive integer such that $2n$ has $28$ positive divisors and $3n$ has $30$ positive divisors, then how many positive divisors does $6n$ have?

$32$

$34$

$35$

$36$

$38$

Answer: C

Concepts:divisor function prime factorization

Difficulty rating: 2150

Small Hint:

Write $n=2^a3^bq$ , where $q$ is relatively prime to $6$ , and let $t$ be the number of divisors of $q$

Big Hint:

Use the two divisor-count equations to restrict $t$ to a divisor of both $28$ and $30$

Solution:

Write $n=2^a3^bq$ with $\gcd(q,6)=1,$ and let $t=\tau(q).$ Then $\begin{aligned} (a+2)(b+1)t&=28,\\ (a+1)(b+2)t&=30. \end{aligned}$ Hence $t$ is $1$ or $2.$ Checking the factor pairs of $14,15$ gives no solution when $t=2.$ Checking those of $28,30$ when $t=1$ gives the unique solution $a=5,b=3.$ Therefore $\begin{aligned} \tau(6n)&=(a+2)(b+2)t\\ &=7\cdot5=35, \end{aligned}$ so the correct answer is C.

30.

A hexagon inscribed in a circle has three consecutive sides each of length $3$ and three consecutive sides each of length $5.$ The chord of the circle that divides the hexagon into two trapezoids, one with three sides each of length $3$ and the other with three sides each of length $5,$ has length equal to $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

$309$

$349$

$369$

$389$

$409$

Answer: E

Concepts:cyclic polygons trigonometry triple-angle identities

Difficulty rating: 2330

Small Hint:

Let $\alpha$ and $\beta$ be the half-central angles subtended by sides $3$ and $5$

Big Hint:

Use $\alpha+\beta=60^\circ$ and the chord ratio to find $\sin^2\alpha$ , then apply $\frac{\sin3\alpha}{\sin\alpha}$

Solution:

Let the circle have radius $R,$ and let $\alpha,\beta$ be the half-central angles for the sides $3,5.$ Then $3=2R\sin\alpha,$ $5=2R\sin\beta,$ and $\alpha+\beta=60^\circ.$ Thus $\frac53=\frac{\sin(60^\circ-\alpha)}{\sin\alpha}$ $=\frac{\sqrt3}{2}\cot\alpha-\frac12,$ so $\tan\alpha=\frac{3\sqrt3}{13}$ and $\sin^2\alpha=\frac{27}{196}.$ The dividing chord spans the three consecutive sides of length $3,$ so $\frac L3=\frac{\sin3\alpha}{\sin\alpha}$ $=3-4\sin^2\alpha=3-\frac{27}{49}=\frac{120}{49}.$ Therefore $L=\frac{360}{49},$ so $m+n=409.$ The correct answer is E.