1999 AMC 12 — Problems & Online Timed Exam

Timed

1:15:00

1.

$\begin{aligned} &1 - 2 + 3 - 4 + \cdots - 98 \\ &\quad {}+ 99 = \, ? \end{aligned}$

$-50$

$-49$

$0$

$49$

$50$

Answer: E

Concepts:summation pairing and grouping

Difficulty rating: 800

Small Hint:

Group the terms in consecutive pairs

Big Hint:

Each pair $(2k-1) - 2k$ equals $-1$

Solution:

Pairing consecutive terms gives $\begin{aligned} &(1-2) + (3-4) + \cdots \\ &\quad {}+ (97-98) + 99. \end{aligned}$ There are $49$ pairs, each equal to $-1,$ so the sum is $-49 + 99 = 50.$

Thus, the correct answer is E.

2.

Which one of the following statements is false?

All equilateral triangles are congruent to each other.

All equilateral triangles are convex.

All equilateral triangles are equiangular.

All equilateral triangles are regular polygons.

All equilateral triangles are similar to each other.

Answer: A

Concepts:equilateral triangle congruence (geometry)similarity

Difficulty rating: 880

Small Hint:

Congruent means the same shape and the same size

Big Hint:

Equilateral triangles can have different side lengths

Solution:

Equilateral triangles with side lengths $1$ and $2$ have the same shape but different sizes, so they are similar but not congruent. Every equilateral triangle is convex, equiangular (all angles $60^\circ$ ), and a regular polygon, so the only false statement is that they are all congruent.

Thus, the correct answer is A.

3.

The number halfway between $\tfrac18$ and $\tfrac{1}{10}$ is

$\dfrac{1}{80}$

$\dfrac{1}{40}$

$\dfrac{1}{18}$

$\dfrac{1}{9}$

$\dfrac{9}{80}$

Answer: E

Concepts:fraction mean

Difficulty rating: 880

Small Hint:

The number halfway between two values is their average

Big Hint:

Compute $\dfrac12\left(\dfrac18 + \dfrac{1}{10}\right)$

Solution:

The halfway point is the average $\dfrac12\left(\dfrac18 + \dfrac{1}{10}\right) = \dfrac12 \cdot \dfrac{18}{80} = \dfrac{9}{80}.$

Thus, the correct answer is E.

4.

Find the sum of all prime numbers between $1$ and $100$ that are simultaneously $1$ greater than a multiple of $4$ and $1$ less than a multiple of $5.$

$118$

$137$

$158$

$187$

$245$

Answer: A

Concepts:modular arithmetic prime Chinese Remainder Theorem

Difficulty rating: 1240

Small Hint:

A number $1$ less than a multiple of $5$ has units digit $4$ or $9$

Big Hint:

Being also $\equiv 1 \pmod 4$ forces the number to be $\equiv 9 \pmod{20}$

Solution:

A number that is $1$ less than a multiple of $5$ ends in $4$ or $9,$ and one that is $1$ greater than a multiple of $4$ is odd. Together these give numbers $\equiv 9 \pmod{20},$ namely $9, 29, 49, 69, 89.$

Among these, only $29$ and $89$ are prime, and their sum is $29 + 89 = 118.$

Thus, the correct answer is A.

5.

The marked price of a book was $30\%$ less than the suggested retail price. Alice purchased the book for half the marked price at a Fiftieth Anniversary sale. What percent of the suggested retail price did Alice pay?

$25\%$

$30\%$

$35\%$

$60\%$

$65\%$

Answer: C

Concepts:percentage

Difficulty rating: 960

Small Hint:

Let the suggested retail price be $P$

Big Hint:

The marked price is $0.7P,$ and Alice pays half of that

Solution:

If the suggested retail price is $P,$ then the marked price is $0.7P.$ Alice pays half of this, $0.35P,$ which is $35\%$ of the suggested retail price.

Thus, the correct answer is C.

6.

What is the sum of the digits of the decimal form of the product $2^{1999} \cdot 5^{2001}?$

$2$

$4$

$5$

$7$

$10$

Answer: D

Concepts:exponent digits

Difficulty rating: 1170

Small Hint:

Pair each factor of $2$ with a factor of $5$ to make $10$

Big Hint:

$2^{1999} \cdot 5^{2001} = 5^2 \cdot 10^{1999}$

Solution:

Write $\begin{aligned} 2^{1999} \cdot 5^{2001} &= 2^{1999} \cdot 5^{1999} \cdot 5^2 \\ &= 25 \cdot 10^{1999}, \end{aligned}$ which is $25$ followed by $1999$ zeros. The sum of the digits is $2 + 5 = 7.$

Thus, the correct answer is D.

7.

What is the largest number of acute angles that a convex hexagon can have?

$2$

$3$

$4$

$5$

$6$

Answer: B

Concepts:angle sum bounding to limit cases

Difficulty rating: 1310

Small Hint:

The exterior angles of a convex polygon sum to $360^\circ$

Big Hint:

An acute interior angle gives an exterior angle greater than $90^\circ$

Solution:

Each acute interior angle corresponds to an exterior angle greater than $90^\circ.$ Since the exterior angles of a convex polygon sum to $360^\circ,$ at most three of them can exceed $90^\circ.$ This bound is attainable: take a hexagon with equal side lengths and exterior angles alternating $100^\circ$ and $20^\circ.$ Its edge directions split into two triples $120^\circ$ apart, so it closes, and its interior angles alternate between $80^\circ$ and $160^\circ.$ Hence the largest possible number is three.

Thus, the correct answer is B.

8.

At the end of $1994$ Walter was half as old as his grandmother. The sum of the years in which they were born is $3838.$ How old will Walter be at the end of $1999?$

$48$

$49$

$53$

$55$

$101$

Answer: D

Concepts:ages linear equation

Difficulty rating: 1240

Small Hint:

Let Walter be $w$ and his grandmother $2w$ at the end of $1994$

Big Hint:

Their birth years are $1994 - w$ and $1994 - 2w$

Solution:

Let Walter be $w$ years old at the end of $1994,$ so his grandmother is $2w.$ Their birth years are $1994 - w$ and $1994 - 2w,$ and $\begin{aligned} &(1994 - w) \\ &\quad {}+ (1994 - 2w) = 3838. \end{aligned}$ This gives $3988 - 3w = 3838,$ so $w = 50.$

At the end of $1999,$ Walter will be $50 + 5 = 55.$

Thus, the correct answer is D.

9.

Before Ashley started a three-hour drive, her car’s odometer reading was $29792,$ a palindrome. (A palindrome is a number that reads the same way from left to right as it does from right to left.) At her destination, the odometer reading was another palindrome. If Ashley never exceeded the speed limit of $75$ miles per hour, which of the following was her greatest possible average speed?

$33\tfrac13$

$53\tfrac13$

$66\tfrac23$

$70\tfrac13$

$74\tfrac13$

Answer: D

Concepts:palindrome distance rate and time bounding to limit cases

Difficulty rating: 1390

Small Hint:

List the palindromes just above $29792$

Big Hint:

The distance in $3$ hours cannot exceed $3 \cdot 75 = 225$ miles

Solution:

The palindromes after $29792$ are $29892, 29992, 30003,$ and $30103.$ In three hours Ashley can drive at most $3 \cdot 75 = 225$ miles.

Reaching $30103$ would require $30103 - 29792 = 311$ miles, which is too far. Reaching $30003$ requires $30003 - 29792 = 211$ miles, giving average speed $\dfrac{211}{3} = 70\tfrac13$ miles per hour.

Thus, the correct answer is D.

10.

A sealed envelope contains a card with a single digit on it. Three of the following statements are true, and the other is false.

I. The digit is $1.$

II. The digit is not $2.$

III. The digit is $3.$

IV. The digit is not $4.$

Which one of the following must necessarily be correct?

$\mathrm{I}$ is true.

$\mathrm{I}$ is false.

$\mathrm{II}$ is true.

$\mathrm{III}$ is true.

$\mathrm{IV}$ is false.

Answer: C

Concepts:logical deduction

Difficulty rating: 1370

Small Hint:

Statements I and III cannot both be true

Big Hint:

Exactly one statement is false, so the false one is I or III

Solution:

Statements I and III cannot both be true, so the single false statement is one of them. Therefore statements II and IV are both true, which makes “II is true” necessarily correct.

The digit is thus $1$ or $3.$ If it were $1,$ then (B) and (D) are false; if it were $3,$ then (A) is false; and (E) is always incorrect. Only (C) is guaranteed.

Thus, the correct answer is C.

11.

The student lockers at Olympic High are numbered consecutively beginning with locker number $1.$ The plastic digits used to number the lockers cost $2$ cents apiece. Thus, it costs $2$ cents to label locker number $9$ and $4$ cents to label locker number $10.$ If it costs $\$137.94$ to label all the lockers, how many lockers are there at the school?

$2001$

$2010$

$2100$

$2726$

$6897$

Answer: A

Concepts:digits counting integers in a range

Difficulty rating: 1450

Small Hint:

Divide the total cost by the cost per digit to get the number of digits

Big Hint:

Subtract the digits used by lockers $1$ - $9,$ $10$ - $99,$ and $100$ - $999$

Solution:

Labeling costs $\frac{\$137.94}{\$0.02} = 6897$ digits. Lockers $1$ - $9$ use $9$ digits, lockers $10$ - $99$ use $2 \cdot 90 = 180$ digits, and lockers $100$ - $999$ use $3 \cdot 900 = 2700$ digits.

The remaining digits number $6897 - 2700 - 180 - 9 = 4008,$ which label $\frac{4008}{4} = 1002$ four-digit lockers. In all there are $999 + 1002 = 2001$ lockers.

Thus, the correct answer is A.

12.

What is the maximum number of points of intersection of the graphs of two different fourth degree polynomial functions $y = p(x)$ and $y = q(x),$ each with leading coefficient $1?$

$1$

$2$

$3$

$4$

$8$

Answer: C

Concepts:polynomial counting intersections

Difficulty rating: 1510

Small Hint:

Intersections occur where $p(x) = q(x)$

Big Hint:

The leading $x^4$ terms cancel in $p(x) - q(x)$

Solution:

The $x$ -coordinates of the intersection points are the roots of $p(x) - q(x).$ Because both leading coefficients are $1,$ the $x^4$ terms cancel, so $p(x) - q(x)$ has degree at most $3$ and therefore at most $3$ roots. The bound is attainable, for example by taking $p(x) = x^4$ and $q(x) = x^4 - x(x-1)(x+1),$ whose graphs intersect at $x = -1, 0, 1.$

Thus, the correct answer is C.

13.

Define a sequence of real numbers $a_1,$ $a_2,$ $a_3,$ $\ldots$ by $a_1 = 1$ and $a_{n+1}^3 = 99 a_n^3$ for all $n \ge 1.$ Then $a_{100}$ equals

$33^{33}$

$33^{99}$

$99^{33}$

$99^{99}$

none of these

Answer: C

Concepts:geometric sequence exponent

Difficulty rating: 1420

Small Hint:

Take cube roots: $a_{n+1} = \sqrt[3]{99}\, a_n$

Big Hint:

This is geometric with ratio $\sqrt[3]{99},$ so $a_{100} = (\sqrt[3]{99})^{99}$

Solution:

Taking cube roots, $a_{n+1} = \sqrt[3]{99}\, a_n,$ so the sequence is geometric with first term $1$ and ratio $\sqrt[3]{99}.$ Then $a_{100} = \left(\sqrt[3]{99}\right)^{99} = 99^{33}.$

Thus, the correct answer is C.

14.

Four girls — Mary, Alina, Tina, and Hanna — sang songs in a concert as trios, with one girl sitting out each time. Hanna sang $7$ songs, which was more than any other girl, and Mary sang $4$ songs, which was fewer than any other girl. How many songs did these trios sing?

$7$

$8$

$9$

$10$

$11$

Answer: A

Concepts:divisibility double counting bounding to limit cases

Difficulty rating: 1610

Small Hint:

Each song has $3$ singers, so the total number of girl-appearances is $3N$

Big Hint:

Alina and Tina each sang $5$ or $6,$ and the total must be a multiple of $3$

Solution:

If $N$ songs are sung, the total number of girl-appearances is $3N.$ Alina and Tina each sang strictly between $4$ and $7,$ so each sang $5$ or $6.$

Then $3N = 7 + 4 + (\text{Alina}) + (\text{Tina}),$ which is $21, 22,$ or $23.$ Only $21$ is a multiple of $3,$ so $N = 7.$

Thus, the correct answer is A.

15.

Let $x$ be a real number such that $\sec x - \tan x = 2.$ What is $\sec x + \tan x?$

$0.1$

$0.2$

$0.3$

$0.4$

$0.5$

Answer: E

Concepts:trigonometric identity difference of squares

Difficulty rating: 1550

Small Hint:

Recall the identity $\sec^2 x - \tan^2 x = 1$

Big Hint:

Factor it as $(\sec x - \tan x)(\sec x + \tan x)$

Solution:

Since $\sec^2 x - \tan^2 x = 1,$ we have $\begin{aligned} &(\sec x - \tan x)(\sec x + \tan x) \\ &\quad = 1. \end{aligned}$ With $\sec x - \tan x = 2,$ it follows that $\sec x + \tan x = \tfrac12 = 0.5.$

Thus, the correct answer is E.

16.

What is the radius of a circle inscribed in a rhombus with diagonals of length $10$ and $24?$

$4$

$\dfrac{58}{13}$

$\dfrac{60}{13}$

$5$

$6$

Answer: C

Concepts:rhombus incircle, incenter, and inradius Pythagorean Theorem triangle area

Difficulty rating: 1610

Small Hint:

The diagonals cut the rhombus into four $5$ - $12$ - $13$ right triangles

Big Hint:

The inradius equals the altitude from the center to a side of the rhombus

Solution:

The half-diagonals are $5$ and $12,$ so each side of the rhombus is $\sqrt{5^2 + 12^2} = 13.$ One of the four right triangles formed by the diagonals has legs $5$ and $12$ and area $30.$

The altitude from the center to the side of length $13$ is $\dfrac{2 \cdot 30}{13} = \dfrac{60}{13},$ which is the inscribed circle’s radius.

Thus, the correct answer is C.

17.

Let $P(x)$ be a polynomial such that when $P(x)$ is divided by $x - 19,$ the remainder is $99,$ and when $P(x)$ is divided by $x - 99,$ the remainder is $19.$ What is the remainder when $P(x)$ is divided by $(x - 19)(x - 99)?$

$-x + 80$

$x + 80$

$-x + 118$

$x + 118$

$0$

Answer: C

Concepts:polynomial system of equations

Difficulty rating: 1680

Small Hint:

Dividing by a quadratic leaves a remainder of the form $ax + b$

Big Hint:

Use $P(19) = 99$ and $P(99) = 19$

Solution:

By the Remainder Theorem, $P(19) = 99$ and $P(99) = 19.$ Write $\begin{aligned} &P(x) = (x - 19)(x - 99)Q(x) \\ &\quad {}+ ax + b. \end{aligned}$ Then $19a + b = 99, \qquad 99a + b = 19.$

Subtracting gives $80a = -80,$ so $a = -1$ and $b = 118.$ The remainder is $-x + 118.$

Thus, the correct answer is C.

18.

How many zeros does $f(x) = \cos(\log x)$ have on the interval $0 \lt x \lt 1?$

$0$

$1$

$2$

$10$

infinitely many

Answer: E

Concepts:trigonometry logarithm function

Difficulty rating: 1770

Small Hint:

As $x$ ranges over $(0, 1),$ what values does $\log x$ take?

Big Hint:

$\log x$ covers all negative numbers, and cosine vanishes at infinitely many of them

Solution:

As $x$ ranges over $(0, 1),$ $\log x$ ranges over all negative real numbers. The cosine function is zero at $\tfrac{\pi}{2} - n\pi$ for every positive integer $n,$ all of which are negative, so $f$ has infinitely many zeros.

Thus, the correct answer is E.

19.

Consider all triangles $ABC$ satisfying the following conditions: $AB = AC,$ $D$ is a point on $\overline{AC}$ for which $\overline{BD} \perp \overline{AC},$ $AD$ and $CD$ are integers, and $BD^2 = 57.$ Among all such triangles, the smallest possible value of $AC$ is

$9$

$10$

$11$

$12$

$13$

Answer: C

Concepts:isosceles triangle Pythagorean Theorem Diophantine Equation

Difficulty rating: 1810

Small Hint:

Right triangle $ADB$ gives $AB^2 = AD^2 + BD^2$

Big Hint:

With $AB = AC = AD + CD,$ reduce to $CD(CD + 2\,AD) = 57$

Solution:

Let $AD = n$ and $CD = m.$ Since $\triangle ADB$ is right-angled at $D,$ $AB^2 = n^2 + 57.$ Also $AB = AC = m + n,$ so $(m + n)^2 = n^2 + 57,$ which simplifies to $m(m + 2n) = 57.$

The positive integer solutions are $m = 1, n = 28$ (giving $AC = 29$ ) and $m = 3, n = 8$ (giving $AC = 11$ ). The smallest possible value of $AC$ is $11.$

Thus, the correct answer is C.

20.

The sequence $a_1,$ $a_2,$ $a_3,$ $\ldots$ satisfies $a_1 = 19,$ $a_9 = 99,$ and, for all $n \ge 3,$ $a_n$ is the arithmetic mean of the first $n - 1$ terms. Find $a_2.$

$29$

$59$

$79$

$99$

$179$

Answer: E

Concepts:recursion mean invariant

Difficulty rating: 1740

Small Hint:

Show that $a_n$ is constant for all $n \ge 3$

Big Hint:

Then $a_3 = a_9 = 99$ and $a_3 = \dfrac{a_1 + a_2}{2}$

Solution:

For $n \ge 3,$ $(n - 1)a_n = a_1 + \cdots + a_{n-1}.$ Then $a_{n+1} = \dfrac{(n-1)a_n + a_n}{n} = a_n,$ so the sequence is constant from $a_3$ onward. Hence $a_3 = a_9 = 99.$

Since $a_3 = \dfrac{a_1 + a_2}{2} = \dfrac{19 + a_2}{2} = 99,$ we get $a_2 = 179.$

Thus, the correct answer is E.

21.

A circle is circumscribed about a triangle with sides $20,$ $21,$ and $29,$ thus dividing the interior of the circle into four regions. Let $A,$ $B,$ and $C$ be the areas of the non-triangular regions, with $C$ being the largest. Then

$A + B = C$

$A + B + 210 = C$

$A^2 + B^2 = C^2$

$20A + 21B = 29C$

$\dfrac{1}{A^2} + \dfrac{1}{B^2} = \dfrac{1}{C^2}$

Answer: B

Concepts:right triangle circle area area decomposition

Difficulty rating: 1810

Small Hint:

Check whether $20, 21, 29$ form a right triangle

Big Hint:

The hypotenuse is a diameter, so the largest region $C$ is a semicircle

Solution:

Since $20^2 + 21^2 = 841 = 29^2,$ the triangle is right-angled, and its hypotenuse of length $29$ is a diameter of the circle. Thus the largest region $C$ is the semicircle on one side of that diameter.

The other semicircle consists of the triangle together with regions $A$ and $B.$ Since the two semicircles are congruent and the triangle has area $\tfrac12 \cdot 20 \cdot 21 = 210,$ we get $A + B + 210 = C.$

Thus, the correct answer is B.

22.

The graphs of $y = -|x - a| + b$ and $y = |x - c| + d$ intersect at points $(2, 5)$ and $(8, 3).$ Find $a + c.$

$7$

$8$

$10$

$13$

$18$

Answer: C

Concepts:absolute value coordinate geometry midpoint

Difficulty rating: 1740

Small Hint:

The first graph peaks at $(a, b);$ the second bottoms out at $(c, d)$

Big Hint:

Rewrite the intersection equation as $|x-a|+|x-c|=b-d$ and use symmetry

Solution:

At an intersection, $-|x-a|+b=|x-c|+d,$ or $|x-a|+|x-c|=b-d.$ Because there are two isolated intersections, they lie on opposite sides of the interval with endpoints $a$ and $c.$ The two solutions are symmetric about $x=\tfrac{a+c}{2}.$ Their $x$ -coordinates are $2$ and $8,$ so $a+c=2+8=10.$

Thus, the correct answer is C.

23.

The equiangular convex hexagon $ABCDEF$ has $AB = 1,$ $BC = 4,$ $CD = 2,$ and $DE = 4.$ The area of the hexagon is

$\dfrac{15}{2}\sqrt{3}$

$9\sqrt{3}$

$16$

$\dfrac{39}{4}\sqrt{3}$

$\dfrac{43}{4}\sqrt{3}$

Answer: E

Concepts:equiangular polygon equilateral triangle area decomposition

Difficulty rating: 1980

Small Hint:

Extend alternate sides to enclose the hexagon in a large equilateral triangle

Big Hint:

Subtract three small equilateral corner triangles from the large one

Solution:

Each interior angle is $120^\circ,$ so extending sides $FA$ and $BC,$ $BC$ and $DE,$ and $DE$ and $FA$ cuts off three equilateral corner triangles and forms a large equilateral triangle.

Let $EF=e$ and $FA=f.$ Resolving the six sides in directions separated by $60^\circ$ gives $f-e=4$ and $e+f=6,$ so $e=1$ and $f=5.$ The corner triangles built on $AB, CD,$ and $EF$ are therefore equilateral. The large triangle has side $1+4+2=7,$ while the removed triangles have sides $1,2,$ and $1.$ The area is $\begin{aligned} &\frac{\sqrt3}{4}\left(7^2 - 1^2 - 2^2 - 1^2\right) \\ &\quad = \frac{43\sqrt3}{4}. \end{aligned}$

Thus, the correct answer is E.

24.

Six points on a circle are given. Four of the chords joining pairs of the six points are selected at random. What is the probability that the four chords form a convex quadrilateral?

$\dfrac{1}{15}$

$\dfrac{1}{91}$

$\dfrac{1}{273}$

$\dfrac{1}{455}$

$\dfrac{1}{1365}$

Answer: B

Concepts:basic probability combinations

Difficulty rating: 1880

Small Hint:

Any $4$ of the $6$ points determine exactly one convex quadrilateral

Big Hint:

Count all ways to choose $4$ chords from the $\binom{6}{2}$ chords

Solution:

There are $\binom{6}{2} = 15$ chords, so $\binom{15}{4} = 1365$ ways to select four of them. A convex quadrilateral arises exactly when the four chords are the sides of a quadrilateral on four of the six points, and each choice of $4$ points gives exactly one such quadrilateral.

Hence there are $\binom{6}{4} = 15$ favorable outcomes, and the probability is $\dfrac{15}{1365} = \dfrac{1}{91}.$

Thus, the correct answer is B.

25.

There are unique integers $a_2,$ $a_3,$ $a_4,$ $a_5,$ $a_6,$ $a_7$ such that

$\begin{aligned} &\frac{5}{7} = \frac{a_2}{2!} + \frac{a_3}{3!} + \frac{a_4}{4!} \\ &\quad {}+ \frac{a_5}{5!} + \frac{a_6}{6!} + \frac{a_7}{7!}, \end{aligned}$

where $0 \le a_i \lt i$ for $i = 2,$ $3,$ $\ldots,$ $7.$ Find $a_2 + a_3 + a_4 + a_5 + a_6 + a_7.$

$8$

$9$

$10$

$11$

$12$

Answer: B

Concepts:number base factorial modular arithmetic

Difficulty rating: 2030

Small Hint:

Multiply both sides by $7!$

Big Hint:

Reduce modulo $7,$ then $6,$ then $5, \ldots$ to peel off one digit at a time

Solution:

Multiplying by $7! = 5040$ gives $\begin{aligned} &3600 = 2520a_2 + 840a_3 \\ &\quad {}+ 210a_4 + 42a_5 \\ &\quad {}+ 7a_6 + a_7. \end{aligned}$ Reducing modulo $7,$ $a_7 = 2.$

Then $\dfrac{3600 - 2}{7} = 514$ $= 360a_2 + 120a_3$ $+ 30a_4 + 6a_5$ $+ a_6.$ Reducing modulo $6$ gives $a_6 = 4,$ and continuing this way yields $a_5 = 0, a_4 = 1, a_3 = 1, a_2 = 1.$

The sum is $1 + 1 + 1 + 0 + 4 + 2 = 9.$

Thus, the correct answer is B.

26.

Three non-overlapping regular plane polygons, at least two of which are congruent, all have sides of length $1.$ The polygons meet at a point $A$ in such a way that the sum of the three interior angles at $A$ is $360^\circ.$ Thus the three polygons form a new polygon with $A$ as an interior point. What is the largest possible perimeter that this polygon can have?

$12$

$14$

$18$

$21$

$24$

Answer: D

Concepts:regular polygon Diophantine Equation perimeter

Difficulty rating: 2090

Small Hint:

With two $a$ -gons and one $b$ -gon, $2 \cdot 180\left(1 - \tfrac2a\right)$ $+ 180\left(1 - \tfrac2b\right)$ $= 360$

Big Hint:

This simplifies to $(a - 4)(b - 2) = 8$

Solution:

Let two congruent $a$ -gons and one $b$ -gon meet at $A.$ Their interior angles satisfy $\begin{aligned} &2 \cdot 180\left(1 - \tfrac2a\right) \\ &\quad {}+ 180\left(1 - \tfrac2b\right) = 360, \end{aligned}$ which reduces to $(a - 4)(b - 2) = 8.$

The solutions $(a, b)$ are $(5, 10), (6, 6), (8, 4),$ and $(12, 3).$ The new polygon’s perimeter is $2a + b - 6,$ giving $14, 12, 14,$ and $21.$ The largest is $21.$

Thus, the correct answer is D.

27.

In triangle $ABC,$ $3\sin A + 4\cos B = 6$ and $4\sin B + 3\cos A = 1.$ Then $\angle C$ in degrees is

$30$

$60$

$90$

$120$

$150$

Answer: A

Concepts:trigonometric identity system of equations

Difficulty rating: 2120

Small Hint:

Square both equations and add them

Big Hint:

Use $\sin^2 + \cos^2 = 1$ to get $\sin(A + B),$ then rule out one case

Solution:

Squaring both equations and adding gives $\begin{aligned} &9 + 16 \\ &\quad {}+ 24\small(\sin A \cos B + \cos A \sin B) \\ &\quad = 37, \end{aligned}$ so $24\sin(A + B) = 12$ and $\sin(A + B) = \tfrac12.$

Then $\sin C = \sin(A + B) = \tfrac12,$ so $\angle C = 30^\circ$ or $150^\circ.$ If $\angle C = 150^\circ,$ then $A \lt 30^\circ,$ making $3\sin A + 4\cos B \lt 6,$ a contradiction. Hence $\angle C = 30^\circ.$

Thus, the correct answer is A.

28.

Let $x_1,$ $x_2,$ $\ldots,$ $x_n$ be a sequence of integers such that

(i) $-1 \le x_i \le 2,$ for $i = 1,$ $2,$ $3,$ $\ldots,$ $n$ ;

(ii) $x_1 + x_2 + \cdots + x_n = 19;$ and

(iii) $x_1^2 + x_2^2 + \cdots + x_n^2 = 99.$

Let $m$ and $M$ be the minimal and maximal possible values of $x_1^3 + x_2^3 + \cdots + x_n^3,$ respectively. What is $\dfrac{M}{m}?$

$3$

$4$

$5$

$6$

$7$

Answer: E

Concepts:system of equations optimization

Difficulty rating: 2240

Small Hint:

Let $a, b, c$ count the $-1$ s, $1$ s, and $2$ s (zeros do not matter)

Big Hint:

The sum of cubes is $-a + b + 8c = 19 + 6c;$ optimize over the allowed range of $c$

Solution:

Let $a, b, c$ be the numbers of $-1$ s, $1$ s, and $2$ s. Then $-a + b + 2c = 19$ and $a + b + 4c = 99,$ giving $a = 40 - c$ and $b = 59 - 3c$ with $0 \le c \le 19.$

The sum of cubes is $-a + b + 8c = 19 + 6c.$ The minimum is at $c = 0$ (value $19$ ) and the maximum at $c = 19$ (value $133$ ), so $\dfrac{M}{m} = \dfrac{133}{19} = 7.$

Thus, the correct answer is E.

29.

A tetrahedron with four equilateral triangular faces has a sphere inscribed within it and a sphere circumscribed about it. For each of the four faces, there is a sphere tangent externally to the face at its center and to the circumscribed sphere. A point $P$ is selected at random inside the circumscribed sphere. The probability that $P$ lies inside one of the five small spheres is closest to

$0$

$0.1$

$0.2$

$0.3$

$0.4$

Answer: C

Concepts:3D geometry sphere geometric probability volume

Difficulty rating: 2380

Small Hint:

For a regular tetrahedron the circumradius is $3$ times the inradius

Big Hint:

Each of the five small spheres has the same volume as the inscribed sphere

Solution:

Let $O$ be the common center of the inscribed and circumscribed spheres. Splitting the tetrahedron into four congruent pieces from $O$ shows the circumradius is $3$ times the inradius, so the circumscribed sphere has $27$ times the inscribed sphere’s volume $V.$

If one of the four other small spheres has radius $s,$ its center is $r+s$ from $O$ and also $3r-s$ from $O,$ so $s=r.$ These five spheres have disjoint interiors (the central sphere is tangent to each of the other four), so their union has volume $5V.$ The probability is $\dfrac{5V}{27V} = \dfrac{5}{27} \approx 0.185,$ closest to $0.2.$

Thus, the correct answer is C.

30.

The number of ordered pairs of integers $(m, n)$ for which $mn \ge 0$ and

$m^3 + n^3 + 99mn = 33^3$

is equal to

$2$

$3$

$33$

$35$

$99$

Answer: D

Concepts:factoring Diophantine Equation

Difficulty rating: 2460

Small Hint:

Since $99 = 3 \cdot 33,$ the equation is $m^3 + n^3 + (-33)^3$ $- 3mn(-33) = 0$

Big Hint:

Use $x^3 + y^3 + z^3$ $- 3xyz = (x + y + z)$ $\cdot (x^2 + y^2 + z^2 - xy - yz - zx)$

Solution:

Writing $z = -33,$ the equation becomes $m^3 + n^3 + z^3 - 3mnz = 0,$ which factors as $\begin{aligned} &(m + n - 33) \\ &\quad {}\cdot \scriptsize\left(m^2 + n^2 + 33^2 - mn + 33m + 33n\right) \\ &= 0. \end{aligned}$

The second factor equals $\scriptsize\tfrac12\big[(m - n)^2 + (m + 33)^2 + (n + 33)^2\big],$ which is $0$ only at $(m, n) = (-33, -33);$ this satisfies $mn \ge 0.$

Otherwise $m + n = 33.$ With $mn \ge 0$ both are nonnegative, giving $(0, 33), (1, 32), \ldots, (33, 0),$ which is $34$ pairs. Together there are $35$ solutions.

Thus, the correct answer is D.