Small Hint:

Write the two numbers in base $b:$ they are $b + 7$ and $9b + 7$

Big Hint:

Since $9(b + 7) - (9b + 7) = 56,$ the condition is that $b + 7$ divides $56$

Solution:

In base $b$ the two numbers are $17_b = b + 7$ and $97_b = 9b + 7.$ We need $9b + 7$ to be divisible by $b + 7,$ and since $b + 7$ certainly divides $9(b + 7) = 9b + 63,$ this is equivalent to $$\begin{gathered} (9b + 63) - (9b + 7) \\ = 56 \end{gathered}$$ being divisible by $b + 7.$

For $b \gt 9$ we have $b + 7 \gt 16,$ so $b + 7$ must be $28$ or $56,$ giving $b = 21$ or $b = 49.$ The sum is $21 + 49 = 70.$

Small Hint:

Since $AB = 28$ and $AC = 91,$ the points $D$ and $F$ sit $\frac{1}{7}$ of the way from $A,$ while $E$ and $G$ sit $\frac{5}{7}$ of the way, so $[DEGF]$ is a fixed fraction of $[ABC]$

Big Hint:

Write $M = 2F - D$ and $N = 2E - G$ as vectors from $A,$ then apply the shoelace formula to the heptagon; almost everything cancels

Solution:

Here $AB = 4 + 16 + 8 = 28$ and $AC = 13 + 52 + 26 = 91,$ so $D$ and $F$ lie $\frac{1}{7}$ of the way from $A$ along their sides while $E$ and $G$ lie $\frac{5}{7}$ of the way. Triangles sharing angle $A$ have areas proportional to the products of the adjacent sides, so $[ADF] = \frac{1}{49}[ABC]$ and $[AEG] = \frac{25}{49}[ABC].$ Therefore $$\begin{aligned} [DEGF] &= [AEG] - [ADF] \\ &= \frac{24}{49}[ABC] = 288, \end{aligned}$$ which gives $[ABC] = 588.$

Now set $\mathbf{b} = \overrightarrow{AB}$ and $\mathbf{c} = \overrightarrow{AC},$ so that $D = \frac{1}{7}\mathbf{b},$ $E = \frac{5}{7}\mathbf{b},$ $F = \frac{1}{7}\mathbf{c},$ $G = \frac{5}{7}\mathbf{c},$ and the reflections are $M = 2F - D = \frac{1}{7}(2\mathbf{c} - \mathbf{b})$ and $N = 2E - G = \frac{1}{7}(10\mathbf{b} - 5\mathbf{c}).$ The shoelace formula for $AFNBCEM$ sums cross products of consecutive vertices: the two terms at $A$ vanish, and $$\begin{aligned} F \times N &= -\tfrac{10}{49}\,\mathbf{b} \times \mathbf{c}, \\ N \times B &= \tfrac{5}{7}\,\mathbf{b} \times \mathbf{c}, \\ B \times C &= \mathbf{b} \times \mathbf{c}, \\ C \times E &= -\tfrac{5}{7}\,\mathbf{b} \times \mathbf{c}, \\ E \times M &= \tfrac{10}{49}\,\mathbf{b} \times \mathbf{c}. \end{aligned}$$

Everything cancels except the single term $\mathbf{b} \times \mathbf{c},$ so the heptagon’s area is $\frac{1}{2}\left|\mathbf{b} \times \mathbf{c}\right| = [ABC] = 588.$

Small Hint:

First find every triple of flavor counts: strictly decreasing positive integers summing to $9$

Big Hint:

For counts $(c, v, s),$ the number of ways to assign flavors to the nine distinct players is $\frac{9!}{c!\,v!\,s!}$

Solution:

Let $c \gt v \gt s \ge 1$ be the numbers of players choosing chocolate, vanilla, and strawberry, with $c + v + s = 9.$ Checking small values of $s$ shows the only possibilities are $(6, 2, 1),$ $(5, 3, 1),$ and $(4, 3, 2).$

Since the players are distinct, each triple of counts contributes a multinomial coefficient: $$\begin{aligned} \frac{9!}{6!\,2!\,1!} &= 252, \\ \frac{9!}{5!\,3!\,1!} &= 504, \\ \frac{9!}{4!\,3!\,2!} &= 1260. \end{aligned}$$ Thus $N = 252 + 504 + 1260 = 2016,$ and the remainder modulo $1000$ is $16.$

Small Hint:

The left side factors as $(3x + 2y)(4x - 3y)$

Big Hint:

One line gives $(x, y) = (3t, 4t)$ with $|t| \le 25;$ the other gives $(2t, -3t)$ with $|t| \le 33.$ Don’t count the origin twice.

Solution:

The equation factors as $$\begin{gathered} 12x^2 - xy - 6y^2 \\ = (3x + 2y)(4x - 3y) \\ = 0, \end{gathered}$$ so every solution has $4x = 3y$ or $3x = -2y.$

Integer solutions of $4x = 3y$ are $(x, y) = (3t, 4t);$ the constraint $|4t| \le 100$ gives $-25 \le t \le 25,$ or $51$ pairs. Integer solutions of $3x = -2y$ are $(x, y) = (2t, -3t);$ the constraint $|3t| \le 100$ gives $-33 \le t \le 33,$ or $67$ pairs. The families overlap only at $(0, 0),$ so the count is $51 + 67 - 1 = 117.$

Small Hint:

For divisibility by $11,$ the four digits in odd positions and the four in even positions must have equal sums

Big Hint:

Each side sums to $18,$ and there are $8$ such splits. Then count arrangements with an even units digit, sorted by how many even digits share the units digit’s side.

Solution:

The digits sum to $36.$ Divisibility by $11$ requires the alternating sum of digits to be a multiple of $11,$ so if the four digits in odd positions sum to $a,$ then $a - (36 - a) = 2a - 36$ must be a multiple of $11.$ Since $10 \le a \le 26,$ the only possibility is $a = 18:$ each block of four positions carries digit sum $18.$ The four-element subsets of $\{1, \ldots, 8\}$ with sum $18$ are $$\begin{gathered} \{1,2,7,8\},\ \{1,3,6,8\},\ \\ \{1,4,5,8\},\ \{1,4,6,7\},\ \\ \{2,3,5,8\},\ \{2,3,6,7\},\ \\ \{2,4,5,7\},\ \{3,4,5,6\}, \end{gathered}$$ eight in all, and they come in complementary pairs.

Choose which of the $8$ subsets occupies the even positions (which include the units place); the complement fills the odd positions. If that subset contains $k$ of the even digits, then the units digit can be chosen in $k$ ways, the rest of the even positions in $3!$ ways, and the odd positions in $4!$ ways, for $144k$ numbers. Complementary subsets have $k$-values summing to $4,$ so over all $8$ choices $\sum k = 16.$ Hence $N = 144 \cdot 16 = 2304,$ and $N - 2025 = 279.$

Small Hint:

The height is twice the inradius, so the area formula gives $r + s = 24$

Big Hint:

In a tangential quadrilateral the two pairs of opposite sides have equal sums, so each leg is $12;$ use the Pythagorean theorem to get $(r - s)^2$

Solution:

The circle is tangent to both parallel sides, so the height of the trapezoid is $2 \cdot 3 = 6.$ From the area, $\frac{r + s}{2} \cdot 6 = 72,$ so $r + s = 24.$ By the Pitot theorem the legs together also sum to $24,$ and since the trapezoid is isosceles each leg is $12.$

Dropping a perpendicular from an endpoint of the shorter base, the leg is the hypotenuse of a right triangle with legs $6$ and $\frac{|r - s|}{2}:$ $$144 = 36 + \left(\frac{r - s}{2}\right)^2,$$ so $(r - s)^2 = 432.$ Therefore $r^2 + s^2 = \frac{(r+s)^2 + (r-s)^2}{2}$ $= \frac{576 + 432}{2} = 504.$

Small Hint:

The last word’s first letter is the largest among the six smaller letters of the pairs

Big Hint:

Either $G$ pairs with a later letter and no two of $H$ through $L$ pair together, or the last word is $FG$ with $H$ through $L$ all paired into $A$ through $E$

Solution:

There are $11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1 = 10395$ ways to pair the letters. Each word begins with the smaller letter of its pair, so the last word alphabetically is the pair whose smaller letter is largest.

Case $1:$ $G$ is the smaller letter of the last word. Then $G$ pairs with one of $H,$ $I,$ $J,$ $K,$ or $L$ ($5$ ways), and no two of the remaining four late letters may pair together (such a pair would start with a letter after $G$). Those four letters must take distinct partners from $\{A, \ldots, F\},$ in $6 \cdot 5 \cdot 4 \cdot 3 = 360$ ways, and the two leftover early letters pair with each other. That gives $5 \cdot 360 = 1800$ pairings. Case $2:$ $G$ is the larger letter, paired with some $x$ before $G.$ Then none of $H, \ldots, L$ may pair together, so all five take partners among the other five early letters; the six smaller letters are then exactly $A$ through $F,$ and the largest is $F.$ For the last word to contain $G,$ its smaller letter must therefore be $F,$ so the last word is $FG,$ and $H, \ldots, L$ match with $A, \ldots, E$ in $5! = 120$ ways.

The probability is $\frac{1800 + 120}{10395} = \frac{1920}{10395} = \frac{128}{693},$ so $m + n = 128 + 693 = 821.$

Small Hint:

The first equation is a circle centered at $25 + 20i$ with radius $5;$ the second is the perpendicular bisector of the points $4 + k$ and $3i + k$

Big Hint:

Exactly one solution means the bisector is tangent to the circle: set the distance from $(25, 20)$ to the line equal to $5$

Solution:

The first equation says $z$ lies on the circle of radius $5$ centered at $(25, 20).$ The second says $z$ is equidistant from $P_1 = (k + 4, 0)$ and $P_2 = (k, 3),$ i.e. it lies on the perpendicular bisector of $\overline{P_1 P_2}.$ The system has exactly one solution precisely when this line is tangent to the circle.

The midpoint is $\left(k + 2, \frac{3}{2}\right)$ and $\overline{P_1 P_2}$ has slope $-\frac{3}{4},$ so the bisector has slope $\frac{4}{3}:$ in standard form $8x - 6y - (8k + 7) = 0.$ Tangency requires $$\begin{gathered} \frac{|8 \cdot 25 - 6 \cdot 20 - 8k - 7|}{\sqrt{8^2 + 6^2}} \\ = \frac{|73 - 8k|}{10} \\ = 5, \end{gathered}$$ so $8k = 73 \pm 50,$ giving $k = \frac{123}{8}$ or $k = \frac{23}{8}.$

The sum is $\frac{146}{8} = \frac{73}{4},$ so $m + n = 73 + 4 = 77.$

Small Hint:

$P$ lies on both curves exactly when $P$ and its rotation by $-60^\circ$ both lie on the original parabola, at equal distance from the origin

Big Hint:

Try making the rotated image the mirror point $(-x, y):$ that forces $y = -\sqrt{3}\,x,$ which you can substitute into the parabola

Solution:

A point $P$ lies on the image parabola exactly when its rotation by $-60^\circ,$ namely $$Q = \left(\frac{x + \sqrt{3}y}{2},\ \frac{-\sqrt{3}x + y}{2}\right),$$ lies on the original parabola. So we need $P$ and $Q$ both on $y = x^2 - 4.$ The parabola is symmetric in $x \mapsto -x,$ so we look for $P = (x, y)$ whose rotated image is the mirror point $Q = (-x, y).$

Matching $y$-coordinates gives $-\frac{\sqrt{3}}{2}x + \frac{y}{2} = y,$ i.e. $y = -\sqrt{3}\,x,$ and then the $x$-coordinate works automatically: $\frac{x}{2} + \frac{\sqrt{3}}{2}(-\sqrt{3}x) = -x.$ Substituting $y = -\sqrt{3}\,x$ into $y = x^2 - 4$ gives $x^2 + \sqrt{3}\,x - 4 = 0,$ whose positive root is $x = \frac{-\sqrt{3} + \sqrt{19}}{2}.$ Then $$y = -\sqrt{3}\,x = \frac{3 - \sqrt{57}}{2} \lt 0,$$ so this point is in the fourth quadrant, on both curves.

The problem guarantees the fourth-quadrant intersection is unique, so its $y$-coordinate is $\frac{3 - \sqrt{57}}{2},$ giving $a + b + c = 3 + 57 + 2 = 62.$

Small Hint:

Fill the left block in $9!$ ways, then decide which digits from each of its rows go into each row of the middle block

Big Hint:

The middle block’s row contents can be chosen in $56$ ways, the right block is then determined, and every row of both blocks can be ordered in $3!$ ways

Solution:

Fill the left block arbitrarily: $9!$ ways. Let $R_1,$ $R_2,$ and $R_3$ be the sets of three digits in its rows. In the middle block, row $i$ must avoid $R_i$ (those digits already appear in row $i$), and the block’s three rows must partition $\{1, \ldots, 9\}.$ Say its top row takes $j$ digits from $R_2$ and $3 - j$ from $R_3.$ Balancing the three rows then forces the middle row to take $3 - j$ digits from $R_1$ together with all $j$ remaining digits of $R_3,$ and the bottom row is determined. The number of content choices is $$\begin{gathered} \sum_{j=0}^{3} \binom{3}{j}\binom{3}{3-j}^2 \\ = 1 + 27 + 27 + 1 \\ = 56. \end{gathered}$$

The right block’s row contents are then forced (row $i$ takes whatever is missing from row $i$), and each of the six rows of the middle and right blocks can be ordered internally in $3!$ ways. The total is $$\begin{gathered} 9! \cdot 56 \cdot 6^6 \\ = (2^7 \cdot 3^4 \cdot 5 \cdot 7)(2^3 \cdot 7)(2^6 \cdot 3^6) \\ = 2^{16} \cdot 3^{10} \cdot 5^1 \cdot 7^2. \end{gathered}$$

Therefore $p \cdot a + q \cdot b + r \cdot c + s \cdot d$ $= 2 \cdot 16 + 3 \cdot 10 + 5 \cdot 1$ $+ 7 \cdot 2 = 81.$

Small Hint:

On the parabola $x = 34y^2,$ intersections are the solutions of $y = f(34y^2)$ with $-1 \le y \le 1,$ so $0 \le x \le 34$

Big Hint:

Each linear piece turns $y = f(34y^2)$ into a quadratic; add the roots with Vieta’s formulas, but check which roots actually land in the piece’s interval

Solution:

Since $f$ only takes values in $[-1, 1],$ any intersection has $-1 \le y \le 1$ and hence $x = 34y^2 \in [0, 34].$ On the rising pieces, $x \in [4k - 1, 4k + 1)$ with $f(x) = x - 4k,$ so $y = f(34y^2)$ becomes $34y^2 - y - 4k = 0;$ on the falling pieces, $x \in [4k + 1, 4k + 3)$ with $f(x) = 4k + 2 - x,$ giving $34y^2 + y - (4k + 2) = 0.$ In each case a root is valid exactly when it lies in $[-1, 1)$ (rising) or $(-1, 1]$ (falling), since then $x = 34y^2$ automatically falls in the correct interval.

For the rising pieces the roots are $\frac{1 \pm \sqrt{1 + 544k}}{68},$ and both are valid exactly when $\sqrt{1 + 544k} \le 67,$ i.e. for $k = 0, 1, \ldots, 8:$ nine quadratics, each contributing root sum $\frac{1}{34}$ by Vieta. For the falling pieces the roots are $\frac{-1 \pm \sqrt{544k + 273}}{68}.$ The root with the minus sign requires $\sqrt{544k + 273} \lt 67,$ which holds for $k = 0, \ldots, 7;$ those eight quadratics each contribute $-\frac{1}{34}.$ For $k = 8$ only the positive root $\frac{-1 + \sqrt{4625}}{68} = \frac{-1 + 5\sqrt{185}}{68}$ is valid.

The total is $$\begin{aligned} &\frac{9}{34} - \frac{8}{34} + \frac{-1 + 5\sqrt{185}}{68} \\ &\quad = \frac{1 + 5\sqrt{185}}{68}, \end{aligned}$$ and $185 = 5 \cdot 37$ is squarefree, so $a + b + c + d$ $= 1 + 5 + 185 + 68$ $= 259.$

Small Hint:

The first inequality rearranges to $(x - y)(1 + z) \lt 0,$ and the second to $(y - z)(1 + x) \lt 0$

Big Hint:

The bounded piece is $-1 \lt x \lt y \lt z$ on the plane; its corners come from the boundaries $x = -1,$ $x = y,$ and $y = z$

Solution:

Since $x - yz - (y - zx)$ $= (x - y) + z(x - y)$ $= (x - y)(1 + z),$ and similarly $y - zx - (z - xy)$ $= (y - z)(1 + x),$ the conditions are $$(x - y)(1 + z) \lt 0$$ and $$(y - z)(1 + x) \lt 0.$$ Each condition offers two sign patterns, giving four combinations. The combination $x \gt y,$ $z \lt -1,$ $y \gt z,$ $x \lt -1$ is impossible on the plane: $x, z \lt -1$ forces $y \gt 77,$ contradicting $x \gt y.$ Two of the remaining combinations allow a coordinate to run off to infinity, producing the two unbounded regions.

The bounded region is $x \lt y,$ $y \lt z,$ $x \gt -1$ (the fourth constraint $z \gt -1$ is then automatic): the set $-1 \lt x \lt y \lt z$ on the plane. Its closure is the triangle whose vertices come from intersecting the boundary lines pairwise: $x = -1,$ $x = y$ gives $(-1, -1, 77);$ $x = -1,$ $y = z$ gives $(-1, 38, 38);$ and $x = y = z$ gives $(25, 25, 25).$

With $A = (-1, -1, 77),$ the edge vectors are $B - A = (0, 39, -39)$ and $C - A = (26, 26, -52),$ whose cross product is $-1014\,(1, 1, 1),$ of length $1014\sqrt{3}.$ The area is $\frac{1014\sqrt{3}}{2} = 507\sqrt{3},$ so $a + b = 507 + 3 = 510.$