2018 AIME I Exam Problems

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1.

Let $S$ be the number of ordered pairs of integers $(a, b),$ with $1 \le a \le 100$ and $b \ge 0,$ such that the polynomial $x^2 + ax + b$ can be factored into the product of two (not necessarily distinct) linear factors with integer coefficients. Find the remainder when $S$ is divided by $1000.$

Answer

Answer: 600

Concepts:quadratic perfect square parity

Difficulty rating: 2260

Solution:

The polynomial factors into integer linear factors exactly when its roots are integers, that is, when the discriminant $a^2 - 4b$ equals $c^2$ for some integer $c \ge 0.$ Given $a,$ such a $b \ge 0$ exists exactly when $4b = a^2 - c^2 = (a-c)(a+c)$ for some $c$ with $0 \le c \le a$ and $c \equiv a \pmod 2,$ and distinct such $c$ give distinct values $b = \frac{a^2 - c^2}{4}.$

For odd $a$ the valid $c$ are $1, 3, \ldots, a,$ which is $\frac{a+1}{2}$ choices; for even $a$ they are $0, 2, \ldots, a,$ which is $\frac{a}{2} + 1$ choices.

Summing over $a = 1, \ldots, 100:$ the odd $a$ contribute $1 + 2 + \cdots + 50 = 1275,$ and the even $a$ contribute $2 + 3 + \cdots + 51 = 1325.$ Thus $S = 2600,$ and the remainder is $600.$

2.

The number $n$ can be written in base $14$ as $\underline{a}\,\underline{b}\,\underline{c},$ can be written in base $15$ as $\underline{a}\,\underline{c}\,\underline{b},$ and can be written in base $6$ as $\underline{a}\,\underline{c}\,\underline{a}\,\underline{c},$ where $a \gt 0.$ Find the base- $10$ representation of $n.$

Answer

Answer: 925

Concepts:number base place value Diophantine Equation

Difficulty rating: 2180

Solution:

Writing out the place values, $n = 196a + 14b + c = 225a + 15c + b = 222a + 37c,$ where $a$ and $c$ are base- $6$ digits with $1 \le a \le 5$ and $0 \le c \le 5,$ and $0 \le b \le 13.$

Equating the last two expressions gives $b = 22c - 3a.$ Substituting into $196a + 14b + c = 225a + 15c + b$ (which says $13b = 29a + 14c$ ) yields $13(22c - 3a) = 29a + 14c,$ so $272c = 68a,$ that is $a = 4c.$ The digit bounds force $c = 1,$ $a = 4,$ and then $b = 22 - 12 = 10,$ which is a valid digit in bases $14$ and $15.$

Therefore $n = 222 \cdot 4 + 37 \cdot 1 = 925.$ Indeed $925 = 196 \cdot 4 + 14 \cdot 10 + 1,$ confirming the base- $14$ form. The answer is $925.$

3.

Kathy has $5$ red cards and $5$ green cards. She shuffles the $10$ cards and lays out $5$ of the cards in a row in a random order. She will be happy if and only if all the red cards laid out are adjacent and all the green cards laid out are adjacent. For example, card orders RRGGG, GGGGR, or RRRRR will make Kathy happy, but RRRGR will not. The probability that Kathy will be happy is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Answer

Answer: 157

Concepts:basic probability arrangements with restrictions casework

Difficulty rating: 2400

Solution:

There are $10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 = 30240$ equally likely ordered layouts of $5$ of the $10$ distinct cards. Kathy is happy exactly when the color pattern consists of one block of reds and one block of greens: the patterns are RRRRR, GGGGG, and the eight mixed patterns $\text{R}^r\text{G}^{5-r}$ and $\text{G}^{5-r}\text{R}^r$ for $r = 1, 2, 3, 4.$

A pattern using $r$ red and $5 - r$ green positions can be filled in $\frac{5!}{(5-r)!} \cdot \frac{5!}{r!}$ ways (ordered choices of which red cards and which green cards appear). For $r = 5, 4, 3, 2, 1, 0$ these counts are $120,$ $600,$ $1200,$ $1200,$ $600,$ $120.$ The happy layouts number $120 + 120 + 2\,(600 + 1200 + 1200 + 600) = 7440.$

The probability is $\frac{7440}{30240} = \frac{31}{126},$ so $m + n = 31 + 126 = 157.$

4.

In $\triangle ABC,$ $AB = AC = 10$ and $BC = 12.$ Point $D$ lies strictly between $A$ and $B$ on $\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\overline{AC}$ so that $AD = DE = EC.$ Then $AD$ can be expressed in the form $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p + q.$

Answer

Answer: 289

Concepts:law of cosines isosceles triangle

Difficulty rating: 2410

Solution:

By the law of cosines in $\triangle ABC,$ $\cos A = \frac{10^2 + 10^2 - 12^2}{2 \cdot 10 \cdot 10} = \frac{56}{200} = \frac{7}{25}.$

Let $x = AD = DE = EC,$ so $AE = 10 - x.$ The law of cosines in $\triangle ADE$ gives $x^2 = x^2 + (10 - x)^2 - 2x(10 - x)\cdot\frac{7}{25},$ so $(10 - x)^2 = \frac{14}{25}\,x(10 - x).$ Since $x \lt 10,$ we may divide by $10 - x$ to get $10 - x = \frac{14x}{25},$ hence $250 = 39x$ and $x = \frac{250}{39}.$

As $\gcd(250, 39) = 1,$ the answer is $250 + 39 = 289.$

5.

For each ordered pair of real numbers $(x, y)$ satisfying $\log_2(2x + y) = \log_4(x^2 + xy + 7y^2),$ there is a real number $K$ such that $\log_3(3x + y) = \log_9(3x^2 + 4xy + Ky^2).$ Find the product of all possible values of $K.$

Answer

Answer: 189

Concepts:logarithm factoring casework

Difficulty rating: 2510

Solution:

Because $\log_4 u = \log_2 \sqrt{u},$ the first equation is equivalent to $(2x + y)^2 = x^2 + xy + 7y^2$ together with $2x + y \gt 0.$ Expanding gives $3x^2 + 3xy - 6y^2 = 0,$ which factors as $3(x - y)(x + 2y) = 0.$ So $x = y$ or $x = -2y,$ with $(x, y) \ne (0, 0).$

Similarly the second equation says $(3x + y)^2 = 3x^2 + 4xy + Ky^2,$ that is $6x^2 + 2xy + y^2 = Ky^2.$ If $x = y$ (taking $x \gt 0$ so both logarithms are defined), then $K = 6 + 2 + 1 = 9.$ If $x = -2y$ (taking $y \lt 0,$ so $2x + y = -3y \gt 0$ and $3x + y = -5y \gt 0$ ), then $24y^2 - 4y^2 + y^2 = Ky^2,$ so $K = 21.$

Both cases occur, so the product of all possible values is $9 \cdot 21 = 189.$

6.

Let $N$ be the number of complex numbers $z$ with the properties that $|z| = 1$ and $z^{6!} - z^{5!}$ is a real number. Find the remainder when $N$ is divided by $1000.$

Answer

Answer: 440

Concepts:complex number trigonometry casework

Difficulty rating: 2720

Solution:

Write $z = e^{i\theta}$ with $\theta \in [0, 2\pi).$ Then $z^{720} - z^{120}$ is real exactly when $\sin 720\theta = \sin 120\theta,$ which happens when the angles are equal or supplementary modulo $2\pi:$ either $720\theta = 120\theta + 2\pi k,$ giving $\theta = \frac{\pi k}{300},$ or $720\theta = \pi - 120\theta + 2\pi k,$ giving $\theta = \frac{(2k+1)\pi}{840}.$

The first family has $600$ values in $[0, 2\pi)$ and the second has $840.$ They cannot coincide: $\frac{\pi k}{300} = \frac{(2j+1)\pi}{840}$ would give $14k = 5(2j + 1),$ equating an even number with an odd one.

Hence $N = 600 + 840 = 1440,$ and the remainder is $440.$

7.

A right hexagonal prism has height $2.$ The bases are regular hexagons with side length $1.$ Any $3$ of the $12$ vertices determine a triangle. Find the number of these triangles that are isosceles (including equilateral triangles).

Answer

Answer: 52

Concepts:counting shapes in figures 3D geometry casework

Difficulty rating: 2840

Solution:

The chords of a unit regular hexagon have lengths $1,$ $\sqrt{3},$ and $2.$ Among the $\binom{6}{3} = 20$ triangles in one hexagon, $6$ have sides $1, 1, \sqrt{3}$ and $2$ are equilateral with side $\sqrt{3};$ the other $12,$ with sides $1, \sqrt{3}, 2,$ are scalene. So each base contributes $8$ isosceles triangles, for $16$ in all.

Otherwise two vertices lie on one base ( $2$ choices of that base) and one on the other. A vertex of the top base at horizontal distance $d$ from a bottom vertex is at distance $\sqrt{d^2 + 4} \ge 2$ from it. If the bottom pair is adjacent (chord $1$ ): the perpendicular bisector of a hexagon edge passes through no vertices, and no slant side can equal $1,$ so there are no isosceles triangles. If the pair has one vertex between them (chord $\sqrt{3},$ $6$ pairs): the top vertices above that middle vertex and above the opposite vertex are equidistant from the pair, giving $6 \cdot 2 = 12.$ If the pair is diametrically opposite (chord $2,$ $3$ pairs): no vertex lies above the perpendicular bisector, but the top vertex directly above either endpoint gives a slant side $\sqrt{0 + 4} = 2$ equal to the chord, giving $3 \cdot 2 = 6.$

The total is $16 + 2\,(12 + 6) = 52.$

8.

Let $ABCDEF$ be an equiangular hexagon such that $AB = 6,$ $BC = 8,$ $CD = 10,$ and $DE = 12.$ Denote by $d$ the diameter of the largest circle that fits inside the hexagon. Find $d^2.$

Answer

Answer: 147

Concepts:equiangular polygon parallel lines optimization

Difficulty rating: 2920

Solution:

All interior angles are $120^\circ,$ so opposite sides are parallel. Attaching equilateral triangles to two opposite sides produces a parallelogram, which forces $AB + BC = DE + EF$ and $FA + AB = CD + DE.$ Hence $EF = 2$ and $FA = 16.$

Walking from one side to the opposite side along the two connecting sides shows that the distance between a pair of opposite sides is $\frac{\sqrt{3}}{2}$ times the sum of those two connecting sides: the strips have widths $\frac{\sqrt{3}}{2}(BC + CD) = 9\sqrt{3}$ between $AB$ and $DE,$ $\frac{\sqrt{3}}{2}(CD + DE) = 11\sqrt{3}$ between $BC$ and $EF,$ and $\frac{\sqrt{3}}{2}(DE + EF) = 7\sqrt{3}$ between $CD$ and $FA.$ Any circle inside the hexagon fits in the narrowest strip, so $d \le 7\sqrt{3}.$

A circle of diameter $7\sqrt{3}$ tangent to lines $CD$ and $FA$ can be centered so that it also touches $DE$ exactly and has distances $6\sqrt{3},$ $5\sqrt{3},$ and $\frac{11\sqrt{3}}{2}$ from lines $EF,$ $BC,$ and $AB,$ all more than its radius $\frac{7\sqrt{3}}{2},$ so it fits inside the hexagon. Therefore $d = 7\sqrt{3}$ and $d^2 = 147.$

9.

Find the number of four-element subsets of $\{1, 2, 3, 4, \ldots, 20\}$ with the property that two distinct elements of the subset have a sum of $16,$ and two distinct elements of the subset have a sum of $24.$ For example, $\{3, 5, 13, 19\}$ and $\{6, 10, 20, 18\}$ are two such subsets.

Answer

Answer: 210

Concepts:subsets counting pairs casework

Difficulty rating: 2990

Solution:

The pairs of distinct elements summing to $16$ are $\{1,15\}, \{2,14\}, \ldots, \{7,9\}$ (seven pairs), and those summing to $24$ are $\{4,20\}, \{5,19\}, \ldots, \{11,13\}$ (eight pairs). First count subsets containing a $16$ -pair and a $24$ -pair that are disjoint. Of the $7 \cdot 8 = 56$ combinations, the ones sharing an element $x$ require $16 - x,$ $x,$ and $24 - x$ all to be valid, which happens for the $10$ values $x \in \{4, \ldots, 15\}$ other than $8$ and $12.$ No four-element set arises from two different disjoint combinations (a second decomposition would force a $16$ -pair to coincide with a $24$ -pair), so this case gives $56 - 10 = 46$ subsets.

In the remaining subsets every $16$ -pair meets every $24$ -pair, so some center $a$ has both $b = 16 - a$ and $c = 24 - a$ in the subset. There are $10$ possible centers ( $a \in \{4, \ldots, 15\}$ with $a \ne 8, 12$ ), and the fourth element can be any of the $17$ remaining numbers, giving $170$ center–subset counts. Exactly $6$ subsets admit two centers and are counted twice: $\{1,7,9,15\},$ $\{2,6,10,14\},$ $\{3,5,11,13\},$ $\{5,11,13,19\},$ $\{6,10,14,18\},$ and $\{7,9,15,17\}.$ This case gives $170 - 6 = 164$ subsets, none of which contain disjoint pairs.

The total is $46 + 164 = 210.$

10.

The wheel shown below consists of two circles and five spokes, with a label at each point where a spoke meets a circle. A bug walks along the wheel, starting at point $A.$ At every step of the process, the bug walks from one labeled point to an adjacent labeled point. Along the inner circle the bug only walks in a counterclockwise direction, and along the outer circle the bug only walks in a clockwise direction. For example, the bug could travel along the path $AJABCHCHIJA,$ which has $10$ steps. Let $n$ be the number of paths with $15$ steps that begin and end at point $A.$ Find the remainder when $n$ is divided by $1000.$

Answer

Answer: 4

Concepts:bijection combinations invariant

Difficulty rating: 3060

Solution:

From any inner point the bug has exactly two moves, counterclockwise along the inner circle or outward along a spoke; from any outer point it has exactly two, clockwise along the outer circle or inward along a spoke. Call a move $X$ if it is counterclockwise or inward and $Y$ if it is clockwise or outward. Then every string in $\{X, Y\}^{15}$ describes exactly one $15$ -step path from $A.$

A step arrives on the inner circle exactly when it is an $X,$ so the path ends on the inner circle exactly when its last move is an $X;$ in that case the numbers of inward and outward moves are equal. Measuring angular position in fifths of a turn (counterclockwise $+1,$ clockwise $-1,$ spokes $0$ ), the path returns to $A$ exactly when it ends on the inner circle and the net rotation is a multiple of $5,$ that is, when the last move is $X$ and $\#X - \#Y \equiv 0 \pmod 5.$ With $15$ moves this means the number of $X$ s is $5,$ $10,$ or $15.$

Fixing the last move as $X,$ the first $14$ moves contain $4,$ $9,$ or $14$ $X$ s, so $n = \binom{14}{4} + \binom{14}{9} + \binom{14}{14} = 1001 + 2002 + 1 = 3004,$ and the remainder is $4.$

11.

Find the least positive integer $n$ such that when $3^n$ is written in base $143,$ its two right-most digits in base $143$ are $01.$

Answer

Answer: 195

Concepts:multiplicative order modular arithmetic binomial theorem least common multiple

Difficulty rating: 2990

Solution:

The last two base- $143$ digits are $01$ exactly when $3^n \equiv 1 \pmod{143^2},$ and since $143^2 = 11^2 \cdot 13^2,$ this holds exactly when $3^n \equiv 1$ modulo both $11^2$ and $13^2.$

Modulo $121:$ $3^5 = 243 = 2 \cdot 121 + 1 \equiv 1,$ and since $5$ is prime and $3 \not\equiv 1,$ the order of $3$ is exactly $5.$ Modulo $169:$ the order of $3$ modulo $13$ is $3,$ so the order modulo $169$ is a multiple of $3.$ Writing $3^3 = 27 = 1 + 26$ and noting $26^2 = 4 \cdot 169 \equiv 0 \pmod{169},$ the binomial theorem gives $3^{3k} = (1 + 26)^k \equiv 1 + 26k \pmod{169},$ which is $1$ exactly when $13 \mid k.$ So the order of $3$ modulo $169$ is $39.$

Therefore $n$ must be a common multiple of $5$ and $39,$ and the least is $\operatorname{lcm}(5, 39) = 195.$

12.

For each subset $T$ of $U = \{1, 2, 3, \ldots, 18\},$ let $s(T)$ be the sum of the elements of $T,$ with $s(\emptyset)$ defined to be $0.$ If $T$ is chosen at random among all subsets of $U,$ the probability that $s(T)$ is divisible by $3$ is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m.$

Answer

Answer: 683

Concepts:subsets modular arithmetic combinations Vandermonde’s Convolution

Difficulty rating: 3060

Solution:

The set $U$ contains six elements in each residue class modulo $3.$ If $T$ contains $a$ elements $\equiv 1$ and $b$ elements $\equiv 2 \pmod 3,$ then $s(T) \equiv a + 2b \equiv a - b \pmod 3,$ so $3 \mid s(T)$ exactly when $a \equiv b \pmod 3;$ the six multiples of $3$ may be included freely, contributing a factor $2^6$ to both the favorable and total counts.

By Vandermonde's identity, the number of ways to choose the $a$ s and $b$ s with $a - b = 0$ is $\sum_a \binom{6}{a}^2 = \binom{12}{6} = 924;$ with $a - b = \pm 3$ it is $2\sum_a \binom{6}{a}\binom{6}{a - 3} = 2\binom{12}{9} = 440;$ and with $a - b = \pm 6$ it is $2.$ The favorable choices number $924 + 440 + 2 = 1366$ out of $2^{12}.$

The probability is $\frac{1366}{4096} = \frac{683}{2048},$ which is in lowest terms since $683$ is odd. Thus $m = 683.$

13.

Let $\triangle ABC$ have side lengths $AB = 30,$ $BC = 32,$ and $AC = 34.$ Point $X$ lies in the interior of $\overline{BC},$ and points $I_1$ and $I_2$ are the incenters of $\triangle ABX$ and $\triangle ACX,$ respectively. Find the minimum possible area of $\triangle AI_1I_2$ as $X$ varies along $\overline{BC}.$

Answer

Answer: 126

Concepts:incircle, incenter, and inradius law of sines trigonometric identity optimization

Difficulty rating: 3270

Solution:

Since $AI_1$ and $AI_2$ bisect angles $BAX$ and $XAC,$ $\angle I_1AI_2 = \frac{1}{2}\angle BAX + \frac{1}{2}\angle XAC = \frac{A}{2},$ a constant. Let $\alpha = \angle AXB.$ The incenter angle formula gives $\angle AI_1B = 90^\circ + \frac{\alpha}{2},$ so the law of sines in $\triangle ABI_1$ yields $AI_1 = \frac{AB \sin\frac{B}{2}}{\cos\frac{\alpha}{2}},$ and similarly, since $\angle AXC = 180^\circ - \alpha,$ $AI_2 = \frac{AC \sin\frac{C}{2}}{\sin\frac{\alpha}{2}}.$

Therefore $[\triangle AI_1I_2] = \frac{1}{2}\,AI_1 \cdot AI_2 \sin\frac{A}{2} = \frac{AB \cdot AC \sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}{\sin\alpha},$ which is minimized when $\alpha = 90^\circ,$ that is, when $X$ is the foot of the altitude from $A.$

With $a = 32,$ $b = 34,$ $c = 30,$ and $s = 48,$ the half-angle formulas give $\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} = \frac{(s-a)(s-b)(s-c)}{abc},$ so the minimum area is $bc \cdot \frac{(s-a)(s-b)(s-c)}{abc} = \frac{16 \cdot 14 \cdot 18}{32} = 126.$

14.

Let $SP_1P_2P_3EP_4P_5$ be a heptagon. A frog starts jumping at vertex $S.$ From any vertex of the heptagon except $E,$ the frog may jump to either of the two adjacent vertices. When it reaches vertex $E,$ the frog stops and stays there. Find the number of distinct sequences of jumps of no more than $12$ jumps that end at $E.$

Answer

Answer: 351

Concepts:recursive counting symmetry

Difficulty rating: 3160

Solution:

Group the vertices into classes $\mathcal{A} = \{S, P_1\},$ $\mathcal{B} = \{P_2, P_5\},$ and $\mathcal{C} = \{P_3, P_4\}.$ Each vertex of $\mathcal{A}$ adjoins one vertex of $\mathcal{A}$ and one of $\mathcal{B};$ each vertex of $\mathcal{B}$ adjoins one of $\mathcal{A}$ and one of $\mathcal{C};$ and each vertex of $\mathcal{C}$ adjoins one of $\mathcal{B}$ and the absorbing vertex $E.$ Hence if $a_n,$ $b_n,$ $c_n$ count the $n$ -jump paths from $S$ that have not yet reached $E$ and end in each class, $a_{n+1} = a_n + b_n, \qquad b_{n+1} = a_n + c_n, \qquad c_{n+1} = b_n,$ and exactly $c_n$ paths reach $E$ for the first time on jump $n + 1.$

Starting from $(a_0, b_0, c_0) = (1, 0, 0),$ the values of $c_n$ for $n = 0, 1, \ldots, 11$ are $0, 0, 1, 1, 3, 4, 9, 14, 28, 47, 89, 155.$

The number of sequences of at most $12$ jumps ending at $E$ is $c_0 + c_1 + \cdots + c_{11} = 351.$

15.

David found four sticks of different lengths that can be used to form three non-congruent convex cyclic quadrilaterals, $A,$ $B,$ $C,$ which can each be inscribed in a circle with radius $1.$ Let $\varphi_A$ denote the measure of the acute angle made by the diagonals of quadrilateral $A,$ and define $\varphi_B$ and $\varphi_C$ similarly. Suppose that $\sin\varphi_A = \frac{2}{3},$ $\sin\varphi_B = \frac{3}{5},$ and $\sin\varphi_C = \frac{6}{7}.$ All three quadrilaterals have the same area $K,$ which can be written in the form $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Answer

Answer: 59

Concepts:cyclic quadrilateral arc trigonometry area

Difficulty rating: 3500

Solution:

The four sticks are chords of the unit circle subtending fixed arcs $\alpha,$ $\beta,$ $\gamma,$ $\delta$ with $\alpha + \beta + \gamma + \delta = 360^\circ.$ The three quadrilaterals are the three distinct cyclic orders of the sides: say $A$ has arcs in order $\alpha, \beta, \gamma, \delta;$ then $B$ (order $\alpha, \gamma, \beta, \delta$ ) and $C$ (order $\alpha, \beta, \delta, \gamma$ ) are the other two. The angle between the diagonals of a cyclic quadrilateral is half the sum of the arcs subtended by either pair of opposite sides, so $\sin\varphi_B = \sin\frac{\alpha + \beta}{2}$ and $\sin\varphi_C = \sin\frac{\alpha + \delta}{2} = \sin\frac{\beta + \gamma}{2}.$

In a circle of radius $1,$ a chord spanning an arc $\theta$ has length $2\sin\frac{\theta}{2}.$ The diagonals of $A$ span the arcs $\alpha + \beta$ and $\beta + \gamma,$ so their lengths are $2\sin\varphi_B$ and $2\sin\varphi_C.$ Hence $K = \frac{1}{2}\,d_1 d_2 \sin\varphi_A = 2\sin\varphi_A \sin\varphi_B \sin\varphi_C,$ a formula symmetric in the three quadrilaterals, which is why all three areas are equal.

Therefore $K = 2 \cdot \frac{2}{3} \cdot \frac{3}{5} \cdot \frac{6}{7} = \frac{24}{35},$ and $m + n = 24 + 35 = 59.$