Small Hint:

Write the counts as $a, a+d, \ldots, a+5d$ and turn the doubling condition into an equation between $a$ and $d$

Big Hint:

From $a + 5d = 2a$ we get $a = 5d,$ so the total $6a + 15d$ simplifies to $45d$

Solution:

Let the six counts be $a, a+d, \ldots, a+5d$ with common difference $d \ge 0.$ The greatest count is double the least, so $a + 5d = 2a,$ which gives $a = 5d.$ The total is $$\begin{aligned} 6a + 15d &= 30d + 15d \\ &= 45d = 990, \end{aligned}$$ so $d = 22.$

The greatest number of apples is $a + 5d = 10d = 220.$

Small Hint:

Any base-eight palindrome between $512$ and $999$ has four base-eight digits and looks like $\overline{1bb1}_8 = 513 + 72b$

Big Hint:

Three-digit base-eight numbers are at most $511,$ so just test $b = 6, 5, 4, \ldots$ for a base-ten palindrome

Solution:

A four-digit base-eight number lies between $512$ and $4095,$ so a base-eight palindrome less than $1000$ with four digits must have leading (and trailing) digit $1:$ it has the form $\overline{1bb1}_8 = 512 + 64b$ ${}+ 8b + 1$ $= 513 + 72b.$ Keeping this below $1000$ requires $b \le 6,$ giving the candidates $513, 585, 657, 729, 801, 873, 945.$

Checking from the top, the only one of these that is also a palindrome in base ten is $585 = 1111_8.$ Every base-eight palindrome with at most three digits is at most $777_8 = 511 \lt 585,$ so the answer is $585.$

Small Hint:

Let $\omega$ be the common angle. In triangle $APC$ the angles at $A$ and $C$ are $90^\circ - \omega$ and $\omega,$ so $\angle APC = 90^\circ$

Big Hint:

Compare the law of sines in triangle $ABP$ with $AC = \frac{AP}{\sin\omega};$ everything reduces to $\tan\omega = \frac{1}{2}$

Solution:

Let $\omega$ denote the common angle and $L = AB = AC.$ Since $\angle PAB = \omega,$ we have $\angle PAC = 90^\circ - \omega,$ and with $\angle PCA = \omega$ the angles of triangle $APC$ give $\angle APC = 90^\circ.$ Hence in right triangle $APC,$ $$L = AC = \frac{AP}{\sin\omega} = \frac{10}{\sin\omega}.$$

In triangle $ABP,$ the angle at $A$ is $\omega$ and the angle at $B$ is $45^\circ - \omega,$ so $\angle APB = 135^\circ.$ The law of sines gives $\frac{AP}{\sin(45^\circ - \omega)} = \frac{AB}{\sin 135^\circ},$ that is, $10 \sin 135^\circ = L \sin(45^\circ - \omega).$ Substituting $L = \frac{10}{\sin\omega}$ and expanding yields $$\begin{aligned} \sin\omega &= \sqrt{2}\,\sin(45^\circ - \omega) \\ &= \cos\omega - \sin\omega, \end{aligned}$$ so $\tan\omega = \frac{1}{2}$ and $\sin^2\omega = \frac{1}{5}.$

Therefore $L^2 = \frac{100}{\sin^2\omega} = 500,$ and the area is $\frac{1}{2}L^2 = 250.$

Small Hint:

Subtract the equations in pairs: the first two give $(y - 4)(x - z) = 0$

Big Hint:

When $x = z,$ eliminate $y$ to get $x^3 - 76x + 240 = 0,$ then look for small integer roots

Solution:

Subtracting the second equation from the first gives $xy - yz + 4z - 4x = 0,$ which factors as $(y - 4)(x - z) = 0.$ So $y = 4$ or $x = z.$

If $y = 4:$ the first equation becomes $4x + 4z = 60,$ so $x + z = 15,$ and the second becomes $4z + 4x = 60$ again while the third gives $zx = 44.$ Then $x$ and $z$ are roots of $t^2 - 15t + 44 = (t - 4)(t - 11),$ so $x \in \{4, 11\}.$

If $x = z:$ the first equation reads $x(y + 4) = 60,$ so $y = \frac{60}{x} - 4,$ and the third reads $x^2 + 4y = 60.$ Substituting, $$\begin{gathered} x^2 + \frac{240}{x} - 16 = 60 \\ \Longrightarrow x^3 - 76x + 240 = 0 \\ = (x - 4)(x - 6)(x + 10), \end{gathered}$$ so $x \in \{4, 6, -10\},$ each with real $y$ and $z.$ Hence $S = \{-10, 4, 6, 11\}$ and the sum of squares is $100 + 16 + 36 + 121 = 273.$

Small Hint:

Write $r = \frac{a}{b}$ in lowest terms; how $\frac{55a}{b}$ reduces is governed by $\gcd(b, 55),$ which is $1,$ $5,$ $11,$ or $55$

Big Hint:

The cases $\gcd = 5$ and $\gcd = 11$ give $2b = 25a$ and $5b = 22a;$ coprimality pins down exactly one fraction in each case

Solution:

Write $r = \frac{a}{b}$ in lowest terms and let $g = \gcd(b, 55).$ Then $55r$ in lowest terms is $\frac{\frac{55a}{g}}{\frac{b}{g}}$ (no further cancellation is possible since $\gcd(a, b) = 1$ and $55$ is squarefree). The condition is $$a + b = \frac{55a}{g} + \frac{b}{g}.$$ If $g = 1$ this forces $a = 55a,$ impossible; if $g = 55$ it forces $b = \frac{b}{55},$ impossible.

If $g = 5:$ $a + b = 11a + \frac{b}{5}$ gives $\frac{4b}{5} = 10a,$ so $2b = 25a.$ Since $\gcd(a, b) = 1,$ we need $a = 2$ and $b = 25,$ so $r = \frac{2}{25}$ (indeed $2 + 25 = 27 = 22 + 5$ from $55r = \frac{22}{5}$). If $g = 11:$ $a + b = 5a + \frac{b}{11}$ gives $\frac{10b}{11} = 4a,$ so $5b = 22a,$ forcing $a = 5,$ $b = 22$ and $r = \frac{5}{22}$ (with $5 + 22 = 27 = 25 + 2$ from $55r = \frac{25}{2}$).

Hence $S = \left\{\frac{2}{25}, \frac{5}{22}\right\}$ and the sum is $\frac{2}{25} + \frac{5}{22} = \frac{44 + 125}{550} = \frac{169}{550},$ already in lowest terms. The answer is $169 + 550 = 719.$

Small Hint:

The three squares plus the missing fourth square tile a $2 \times 2$ square, so the midpoint can fail only by landing in the missing square

Big Hint:

Failure needs one point in the top square and one in the right square, and then each coordinate condition, like $x_1 + x_2 \gt 2,$ holds with probability $\frac{1}{2}$

Solution:

Place the region as $[0,1]^2 \cup \bigl([0,1] \times [1,2]\bigr)$ $\cup \bigl([1,2] \times [0,1]\bigr),$ so it is the $2 \times 2$ square with the top-right unit square removed. Both coordinates of the midpoint are averages of numbers in $[0, 2],$ so the midpoint always lies in the $2 \times 2$ square; it fails to lie in the region exactly when it lands in the missing square, i.e. when $x_A + x_B \gt 2$ and $y_A + y_B \gt 2.$

If neither point is in the right square, then $x_A + x_B \le 2;$ if neither is in the top square, then $y_A + y_B \le 2.$ So failure requires one point in the top square and the other in the right square, which happens with probability $2 \cdot \frac{1}{3} \cdot \frac{1}{3} = \frac{2}{9}.$ In that case, one $x$-coordinate is uniform on $[0,1]$ and the other on $[1,2],$ so $x_A + x_B \gt 2$ with probability $\frac{1}{2},$ and independently $y_A + y_B \gt 2$ with probability $\frac{1}{2}.$

The failure probability is $\frac{2}{9} \cdot \frac{1}{4} = \frac{1}{18},$ so the desired probability is $\frac{17}{18}$ and $m + n = 17 + 18 = 35.$

Small Hint:

A rectangle inscribed in the circle must have both diagonals as diameters, so rectangles correspond to pairs of the $6$ antipodal pairs of vertices

Big Hint:

Classify each antipodal pair as both red, both blue, or mixed; a coloring works exactly when at most one pair is both red and at most one is both blue

Solution:

The twelve vertices lie on a circle, and a rectangle inscribed in a circle must have its diagonals pass through the center. So the rectangles with vertices among the twelve are exactly the pairs of distinct diameters, where the diameters join the $6$ antipodal pairs of vertices. A monochromatic rectangle appears exactly when two antipodal pairs are each colored solidly in the same color.

Each antipodal pair is independently both red ($1$ way), both blue ($1$ way), or mixed ($2$ ways). A coloring is valid exactly when at most one pair is both red and at most one pair is both blue. Counting by the numbers of solid red and solid blue pairs: $$\begin{gathered} 2^6 + 6 \cdot 2^5 \\ {}+ 6 \cdot 2^5 + 6 \cdot 5 \cdot 2^4 \\ = 64 + 192 + 192 + 480 \\ = 928. \end{gathered}$$

Small Hint:

The $k = 0$ factor equals $3;$ the rest is $\prod P(z)$ over the primitive seventh roots of unity, where $P(x) = x^3 + x + 1$

Big Hint:

$\prod_{z^7 = 1} P(z) = \prod (1 - \beta^7)$ over the roots $\beta$ of $P;$ reduce $\beta^7$ using $\beta^3 = -\beta - 1$

Solution:

Let $P(x) = x^3 + x + 1,$ so the product is $\prod_{k=0}^{6} P(\omega^k),$ where $\omega^0, \ldots, \omega^6$ are all seventh roots of unity. Since $x^7 - 1 = \prod_k (x - \omega^k),$ writing the factorization $P(x)$ $= (x - \beta_1)(x - \beta_2)(x - \beta_3)$ and swapping the order of the double product gives $$\begin{gathered} \prod_{k=0}^{6} P(\omega^k) \\ = \prod_{j=1}^{3} \prod_{k=0}^{6} (\omega^k - \beta_j) \\ = \prod_{j=1}^{3} \bigl(-(\beta_j^7 - 1)\bigr) \\ = \prod_{j=1}^{3} (1 - \beta_j^7). \end{gathered}$$

For a root $\beta$ of $P,$ repeatedly using $\beta^3 = -\beta - 1$ gives $\beta^4 = -\beta^2 - \beta,$ $\beta^5 = -\beta^2 + \beta + 1,$ $\beta^6 = \beta^2 + 2\beta + 1,$ and $\beta^7 = 2\beta^2 - 1.$ Hence $1 - \beta^7 = 2(1 - \beta)(1 + \beta),$ and $$\begin{gathered} \prod_{j} (1 - \beta_j^7) \\ = 2^3 \prod_j (1 - \beta_j) \prod_j (1 + \beta_j) \\ = 8 \cdot P(1) \cdot \bigl(-P(-1)\bigr) \\ = 8 \cdot 3 \cdot 1 = 24. \end{gathered}$$

So the requested product equals $24.$

Small Hint:

The tangent at $A$ is parallel to chord $XP,$ so the foot of the perpendicular from $A$ to line $XY$ is the midpoint of $XP;$ the same holds at $B$

Big Hint:

Line $QP$ meets $AB$ at its midpoint $M,$ and $MA^2 = MP(MP + 5)$ determines $MP,$ hence the height of the trapezoid

Solution:

Since the tangent to $\omega_1$ at $A$ is parallel to the chord $XP,$ the point $A$ is the midpoint of arc $XP,$ so the perpendicular from $A$ to line $XY$ lands at the midpoint of $\overline{XP};$ similarly the perpendicular from $B$ lands at the midpoint of $\overline{PY}.$ As $X$ and $Y$ are on opposite sides of $P,$ the parallel sides of the trapezoid are $XY = 10 + 14 = 24$ and $AB = \frac{10}{2} + \frac{14}{2} = 12.$

Line $PQ$ is the radical axis, so its intersection $M$ with the tangent line satisfies $MA^2 = MP \cdot MQ = MB^2:$ $M$ is the midpoint of $\overline{AB},$ and with $MA = 6$ and $MQ = MP + 5,$ $$\begin{gathered} 36 = MP(MP + 5), \\ MP = 4. \end{gathered}$$

Set up coordinates along $AB:$ the feet of $A$ and $B$ are the midpoints of $XP$ and $PY,$ so $P$ lies $5$ units from the first foot, while the distance from $M$ to $A$ is $6.$ Hence the horizontal offset between $M$ and $P$ is $6 - 5 = 1,$ and the height $h$ of the trapezoid satisfies $h^2 = MP^2 - 1 = 15.$ The area is $$\frac{24 + 12}{2}\sqrt{15} = 18\sqrt{15},$$ so $m + n = 18 + 15 = 33.$

Small Hint:

Cells sharing a side must have different residues mod $3;$ count the residue patterns first, then multiply by $(4!)^3$ to place the actual numbers

Big Hint:

Each column shows two distinct residues; from any column, each of the three unordered residue pairs can come next in exactly one orientation, and each pair must occur twice

Solution:

The condition says adjacent cells have different residues mod $3.$ Each residue class among $1, \ldots, 12$ has exactly $4$ members, so $N = K \cdot (4!)^3,$ where $K$ is the number of ways to fill the grid with residues $0, 1, 2,$ each used $4$ times, with adjacent cells different.

A column is an ordered pair $(a, b)$ of distinct residues. If the current column is $(a, b)$ and $e$ is the third residue, the next column must be one of $(b, a),$ $(b, e),$ $(e, a):$ each of the three unordered pairs $\{a,b\},$ $\{b,e\},$ $\{a,e\}$ occurs in exactly one allowed orientation. So a residue pattern is determined by the sequence of six unordered pairs together with the orientation of the first column. Since each residue must appear $4$ times, each of the three pairs must be used exactly twice, giving $\frac{6!}{2!\,2!\,2!} = 90$ sequences and $K = 2 \cdot 90 = 180.$

Therefore $N = 180 \cdot 24^3$ $= 2{,}488{,}320$ $= 2^{11} \cdot 3^5 \cdot 5,$ which has $12 \cdot 6 \cdot 2 = 144$ positive divisors.

Small Hint:

A set and its complement are disjoint, so the collection must contain exactly one set from each of the $16$ complementary pairs

Big Hint:

If a singleton is chosen, every set must contain it. Otherwise only the chosen $2$-element sets can conflict: they must pairwise share an element or form a triangle.

Solution:

The $32$ subsets split into $16$ complementary pairs $\{X, X^{\mathsf{c}}\},$ and no collection can contain both members of a pair (they are disjoint). A collection of $16$ pairwise-intersecting subsets must therefore contain exactly one member of every pair; in particular it contains $\{1,2,3,4,5\}$ and not $\emptyset.$

If some singleton $\{x\}$ is chosen, every member must meet $\{x\},$ i.e. contain $x.$ Exactly one set in each complementary pair contains $x,$ so the collection must be exactly the $16$ subsets containing $x:$ this gives $5$ collections. Otherwise no singleton is chosen, so all five $4$-element sets are in the collection. Any two $3$-element subsets of a $5$-element set intersect, a $4$-element set is disjoint only from its complement, and a chosen $2$-element set and a chosen $3$-element set are disjoint only if they are complements, which cannot both be chosen. So the only remaining condition is that the chosen $2$-element sets pairwise intersect.

Viewing $2$-element sets as edges of $K_5,$ a pairwise-intersecting collection of edges either has all edges through one common vertex or is a triangle. The number of such edge families is: the empty family ($1$), triangles ($\binom{5}{3} = 10$), and nonempty families within a star, $5(2^4 - 1) - 10 = 65$ (subtracting the $10$ single edges counted at both endpoints). That is $1 + 10 + 65 = 76$ collections, for a total of $5 + 76 = 81.$

Small Hint:

Use coordinates $B = (0, 0),$ $C = (14, 0),$ $A = (5, 12);$ power of the point $M$ gives $MA \cdot MP = MB \cdot MC = 49$

Big Hint:

Equate $\tan\angle PBQ$ and $\tan\angle PCQ$ using cross and dot products; the two cross products are proportional, leaving an equation linear in $Q$

Solution:

Place $B = (0, 0),$ $C = (14, 0),$ $A = (5, 12),$ so $M = (7, 0)$ and $AM = \sqrt{4 + 144} = 2\sqrt{37}.$ By power of the point $M$ in the circumcircle, $MA \cdot MP = MB \cdot MC = 49,$ so $MP = \frac{49}{2\sqrt{37}}$ and extending $A \to M$ by that length gives $P = \left(\frac{567}{74}, -\frac{147}{37}\right).$ The direction of $\overrightarrow{BP}$ is proportional to $(27, -14),$ and the direction of $\overrightarrow{CP}$ is proportional to $-(67, 42).$

Write $Q = (5 + 2t,\ 12 - 12t)$ for $t \in (0, 1),$ so that $AQ = t \cdot AM.$ Using $\tan\theta = \frac{|u \times v|}{u \cdot v}$ for the angle between rays, $$\begin{gathered} \tan\angle PBQ = \frac{394 - 296t}{222t - 33}, \\ \tan\angle PCQ = \frac{1182 - 888t}{370t + 99}, \end{gathered}$$ and the second numerator is exactly $3(394 - 296t).$ Setting the two tangents equal cancels this common factor and leaves $370t + 99 = 3(222t - 33),$ so $296t = 198$ and $t = \frac{99}{148}.$

Then $AQ = \frac{99}{148} \cdot 2\sqrt{37}$ $= \frac{99}{148}\sqrt{148}$ $= \frac{99}{\sqrt{148}},$ and since $\gcd(99, 148) = 1,$ the answer is $99 + 148 = 247.$

Small Hint:

With $s = \sec A$ and $t = \tan A,$ the condition says $st = 2,$ and always $s^2 - t^2 = 1,$ so $s^2 + t^2 = \sqrt{17}$

Big Hint:

$s^n + t^n$ is an integer only when $4$ divides $n,$ and $x_j = s^{4j} + t^{4j}$ satisfies $x_{j+1} = 9x_j - 16x_{j-1};$ track the units digits

Solution:

Let $s = \sec A$ and $t = \tan A.$ The hypothesis $\tan A = 2\cos A$ says $\tan A \sec A = 2,$ i.e. $st = 2,$ and always $s^2 - t^2 = 1.$ Then $(s^2 + t^2)^2$ $= (s^2 - t^2)^2 + 4s^2t^2$ $= 17,$ so $u = s^2$ and $v = t^2$ satisfy $u + v = \sqrt{17},$ $uv = 4.$ The sums $w_m = u^m + v^m$ obey $w_{m+1} = \sqrt{17}\,w_m - 4w_{m-1}$ with $w_0 = 2,$ $w_1 = \sqrt{17};$ by induction, when $m$ is even, $w_m$ is a positive integer, and when $m$ is odd, it is an integer times $\sqrt{17}.$