Solution:

Let $x = AB,$ so $BC = 1800 - x,$ and let Eve's speed be $v,$ so Ina runs at $2v$ and Paul at $4v.$ Paul and Eve start at $B$ and $C$ running toward each other, so together they cover the $1800 - x$ meters between them, with Paul covering $\frac{4}{5}$ of it. Paul then retraces that distance back to $B,$ so when he reaches $B$ he has run $\frac{8}{5}(1800 - x)$ meters in total.

Ina reaches $B$ at the same moment, having run $x$ meters. Since Paul runs twice as fast as Ina, he has run $2x$ meters in that time. Therefore $$\frac{8}{5}(1800 - x) = 2x,$$ which gives $8 \cdot 1800 = 18x,$ so $x = 800.$

Solution:

Computing successive terms gives $$2,\ 5,\ 8,\ 5,\ 6,\ 10,\ 7,\ 4,\ 7,\ 6,\ 2,\ 5,\ 8,\ \ldots$$ Since $(a_{10}, a_{11}, a_{12}) = (2, 5, 8) = (a_0, a_1, a_2)$ and each term depends only on the previous three, the sequence is periodic with period $10.$

Therefore $a_{2018} = a_8 = 7,$ $a_{2020} = a_0 = 2,$ and $a_{2022} = a_2 = 8,$ so the product is $7 \cdot 2 \cdot 8 = 112.$

Solution:

The conditions say $3b + 6$ is a perfect square and $2b + 7$ is a perfect cube. Since $2b + 7$ is odd and $b \lt 1000$ forces $2b + 7 \lt 2007,$ the cube must be one of $1,$ $27,$ $125,$ $343,$ $729,$ $1331,$ giving $b = -3,$ $10,$ $59,$ $168,$ $361,$ $662.$

The corresponding values of $3b + 6$ for the positive candidates are $36,$ $183,$ $510,$ $1089,$ $1992,$ and only $36 = 6^2$ (for $b = 10$) and $1089 = 33^2$ (for $b = 361$) are perfect squares. The requested sum is $10 + 361 = 371.$

Solution:

Since the interior angles are all $135^\circ$ and the $\sqrt{2}$ sides are diagonals of unit squares, the octagon fits on a lattice: $C = (0, 0),$ $A = (1, 1),$ $R = (2, 1),$ $O = (3, 0),$ $L = (3, -1),$ $I = (2, -2),$ $N = (1, -2),$ $E = (0, -1).$ The path $CORNELIA$ is carried to itself by the $180^\circ$ rotation about $(\tfrac{3}{2}, -\tfrac{1}{2}).$ Let $Y$ and $Z$ be the points where $AI$ and $RN$ cross $CO,$ and let $W = AI \cap RN.$ Segment $AI$ has slope $-3,$ so $Y = (\tfrac{4}{3}, 0),$ and by symmetry $Z = (\tfrac{5}{3}, 0)$ and $W = (\tfrac{3}{2}, -\tfrac{1}{2}).$

The six enclosed regions are the four congruent corner triangles like $CAY$ and the two small congruent triangles like $YZW.$ Triangle $CAY$ has base $CY = \tfrac{4}{3}$ and height $1,$ so its area is $\tfrac{2}{3}.$ Triangle $YZW$ has base $YZ = \tfrac{1}{3}$ and height $\tfrac{1}{2},$ so its area is $\tfrac{1}{12}.$ Therefore $$K = 4 \cdot \frac{2}{3} + 2 \cdot \frac{1}{12} = \frac{8}{3} + \frac{1}{6} = \frac{17}{6},$$ and $a + b = 17 + 6 = 23.$

Solution:

Multiplying the three equations gives $$(xyz)^2 = (-80 - 320i)(60)(-96 + 24i) = 80 \cdot 60 \cdot 24 \,(-1 - 4i)(-4 + i) = 240^2 (16 + 30i).$$ Since $16 + 30i = (5 + 3i)^2,$ we get $xyz = \pm 240(5 + 3i).$

Dividing $xyz$ by each given product yields $$x = \frac{xyz}{yz} = \pm(20 + 12i), \qquad y = \frac{xyz}{zx} = \pm(-10 - 10i), \qquad z = \frac{xyz}{xy} = \pm(-3 + 3i),$$ with matching signs. Hence $x + y + z = \pm(7 + 5i),$ so $(a, b) = \pm(7, 5)$ and $a^2 + b^2 = 49 + 25 = 74.$

Solution:

Group the terms by whether they involve $a:$ $$(x^4 - 2x^2 + 3x - 2) + 2a(x^3 + x^2 - 2x) = (x - 1)(x + 2)(x^2 - x + 1) + 2ax(x - 1)(x + 2),$$ so the polynomial factors as $(x - 1)(x + 2)\left(x^2 + (2a - 1)x + 1\right).$

All four roots are real exactly when the quadratic factor has real roots, i.e. when $(2a - 1)^2 - 4 \ge 0,$ which means $a \le -\frac{1}{2}$ or $a \ge \frac{3}{2}.$ The excluded interval $\left(-\frac{1}{2}, \frac{3}{2}\right)$ has length $2$ inside $[-20, 18],$ which has length $38,$ so the probability is $\frac{36}{38} = \frac{18}{19}.$ The requested sum is $18 + 19 = 37.$

Solution:

Since the $2450$ regions have equal areas, triangle $AP_kQ_k$ (the union of the first $k$ regions) has area $\frac{k}{2450}$ of triangle $ABC.$ Each triangle $AP_kQ_k$ is similar to $ABC,$ and lengths scale as the square root of areas, so $$P_kQ_k = 5\sqrt{3}\,\sqrt{\frac{k}{2450}} = 5\sqrt{3} \cdot \frac{\sqrt{k}}{35\sqrt{2}} = \frac{\sqrt{6k}}{14}.$$

This is rational exactly when $6k$ is a perfect square, which happens exactly when $k = 6j^2$ for a positive integer $j.$ The condition $6j^2 \le 2450$ gives $j^2 \le 408,$ so $j = 1, 2, \ldots, 20.$ There are $20$ such segments.

Solution:

The horizontal jumps are steps of $1$ or $2$ summing to $4,$ so as a multiset they are $\{1,1,1,1\},$ $\{1,1,2\},$ or $\{2,2\},$ and the same holds for the vertical jumps. For any choice of the two multisets, every ordering of all the jumps is a valid sequence, and the number of orderings is the multinomial coefficient of the combined multiset.

The nine cases give $$\binom{8}{4} = 70, \quad \frac{7!}{4!\,2!} = 105 \text{ (twice)}, \quad \frac{6!}{4!\,2!} = 15 \text{ (twice)},$$ $$\frac{6!}{2!\,2!} = 180, \quad \frac{5!}{2!\,2!} = 30 \text{ (twice)}, \quad \binom{4}{2} = 6.$$

The total is $70 + 2 \cdot 105 + 2 \cdot 15 + 180 + 2 \cdot 30 + 6 = 556.$

Solution:

Each of the $7$ triangles has $J$ as a vertex, and the centroid of a triangle $JVW$ lies on the segment from $J$ to the midpoint of $\overline{VW},$ two-thirds of the way out. So the centroid heptagon is the image of the heptagon $S$ formed by the midpoints of $\overline{AB}, \overline{BC}, \ldots, \overline{GH}$ under a dilation centered at $J$ with ratio $\frac{2}{3},$ and its area is $\frac{4}{9}[S].$

Place the rectangle with $A = (0, 6),$ $B = (8, 0),$ $C = (19, 0),$ $D = (27, 6),$ $E = (27, 17),$ $F = (19, 23),$ $G = (8, 23),$ $H = (0, 17),$ so $J = (0, \tfrac{23}{2}).$ The midpoints are $(4, 3),$ $(\tfrac{27}{2}, 0),$ $(23, 3),$ $(27, \tfrac{23}{2}),$ $(23, 20),$ $(\tfrac{27}{2}, 23),$ $(4, 20).$ The vertical segments at $x = 4,$ $x = \tfrac{27}{2},$ and $x = 23$ have lengths $17,$ $23,$ and $17,$ cutting $S$ into two trapezoids of height $\tfrac{19}{2}$ and a triangle of height $4:$ $$[S] = 2 \cdot \frac{17 + 23}{2} \cdot \frac{19}{2} + \frac{17 \cdot 4}{2} = 380 + 34 = 414.$$

The requested area is $\frac{4}{9} \cdot 414 = 184.$

Solution:

Applying $f$ to $f(f(x)) = f(f(f(x)))$ repeatedly shows the condition means that $f(f(x))$ is a fixed point of $f$ for every $x.$ So the elements organize into levels: a nonempty set of $i$ fixed points, then $j$ elements whose image is a fixed point (but which are not fixed), and the remaining $5 - i - j$ elements, each of which must map to one of the $j$ middle elements.

For given $i$ and $j$ there are $\binom{5}{i}$ choices of fixed points, $\binom{5-i}{j}$ choices of the middle level, $i^j$ maps from the middle level to the fixed points, and $j^{\,5-i-j}$ maps for the rest. Summing $\binom{5}{i}\binom{5-i}{j}\, i^j \, j^{\,5-i-j}$ over the valid pairs (all $i \ge 1,$ $j \ge 1,$ plus the identity case $i = 5$) gives $$20 + 120 + 60 + 5 + 60 + 240 + 80 + 60 + 90 + 20 + 1 = 756.$$

Solution:

The condition fails exactly when the first $k$ terms are a permutation of $\{1, \ldots, k\}$ for some $k \le 5.$ For a permutation of $1, \ldots, n,$ let $k$ be the smallest length for which the prefix is $\{1, \ldots, k\}$ (the full length $n$ always works), and let $c_n$ be the number of permutations whose smallest such $k$ is $n.$ We want $c_6.$

Every permutation of $1, \ldots, n$ decomposes uniquely as a minimal prefix of length $k$ ($c_k$ choices) followed by any arrangement of the remaining $n - k$ values, so $$\sum_{k=1}^{n} c_k \,(n-k)! = n!.$$ Starting from $c_1 = 1,$ this gives $c_2 = 1,$ $c_3 = 3,$ $c_4 = 13,$ $c_5 = 71,$ and $$c_6 = 720 - (120 \cdot 1 + 24 \cdot 1 + 6 \cdot 3 + 2 \cdot 13 + 1 \cdot 71) = 720 - 259 = 461.$$

Solution:

Let $a = AP,$ $b = BP,$ $c = CP,$ $d = DP,$ and let $\theta = \angle CPD.$ Since $\sin(\pi - \theta) = \sin\theta,$ the equal-area condition $\frac{1}{2}(ab + cd)\sin\theta = \frac{1}{2}(ad + bc)\sin\theta$ simplifies to $(a - c)(d - b) = 0.$ By symmetry assume $a = c.$

The law of cosines in triangles $BPC$ and $APB$ (whose angles at $P$ are supplementary) gives $a^2 + b^2 + 2ab\cos\theta = 196$ and $a^2 + b^2 - 2ab\cos\theta = 100,$ so $a^2 + b^2 = 148$ and $ab\cos\theta = 24.$ Similarly triangles $APD$ and $CPD$ give $a^2 + d^2 = 180$ and $ad\cos\theta = 40.$ Dividing, $\frac{d}{b} = \frac{5}{3},$ while subtracting gives $d^2 - b^2 = 32;$ hence $b = 3\sqrt{2},$ $d = 5\sqrt{2},$ $a^2 = 130,$ and $\cos^2\theta = \frac{24^2}{130 \cdot 18} = \frac{16}{65},$ so $\sin\theta = \frac{7}{\sqrt{65}}.$

The total area is $$\frac{1}{2}(a + c)(b + d)\sin\theta = a(b + d)\sin\theta = \sqrt{130} \cdot 8\sqrt{2} \cdot \frac{7}{\sqrt{65}} = 112.$$

Solution:

Let $a$ be the probability that the total number of rolls is odd; let $b$ be that probability given that the first roll is a $1,$ and $c$ given that the first two rolls are $1$-$2$ (in each case counting all rolls). Condition on the next roll, noting that whenever the count restarts, the rolls already used flip the required parity. Starting fresh: a $1$ leads to state $b;$ anything else uses one roll, after which an even continuation is needed. After a $1:$ another $1$ means the first roll is wasted, needing an even continuation of the $b$-type; a $2$ leads to $c;$ anything else wastes both rolls. After $1$-$2:$ a $3$ finishes in $3$ rolls (odd); a $1$ restarts at the $b$-state with two wasted rolls; anything else wastes all three. Thus $$a = \frac{1}{6}b + \frac{5}{6}(1 - a), \qquad b = \frac{1}{6}(1 - b) + \frac{1}{6}c + \frac{4}{6}a, \qquad c = \frac{1}{6}b + \frac{1}{6} + \frac{4}{6}(1 - a).$$

The first equation gives $b = 11a - 5;$ substituting the third into the second yields $41b = 11 + 20a,$ so $41(11a - 5) = 11 + 20a,$ giving $431a = 216$ and $a = \frac{216}{431}.$ Since $431$ is prime, $m + n = 216 + 431 = 647.$

Solution:

Let $\omega$ touch $\overline{AB}$ at $Z$ and $\overline{AC}$ at $W,$ and set $\alpha = \angle BAX$ and $\beta = \angle AXC.$ The tangent-chord angle between $PQ$ and chord $XY$ equals the one between $BC$ and $XY,$ so $\angle QYX = \angle YXC = \beta,$ and vertical angles give $\angle AYP = \beta.$ In triangle $APY$ the law of sines gives $PY = AP\,\frac{\sin\alpha}{\sin\beta},$ and by equal tangents $PZ = PY,$ so $\frac{AZ}{AP} = 1 + \frac{PY}{AP} = 1 + \frac{\sin\alpha}{\sin\beta}.$ In triangle $ABX,$ since $\angle AXB = 180^\circ - \beta,$ similarly $BX = AB\,\frac{\sin\alpha}{\sin\beta},$ and $BZ = BX$ gives $\frac{AZ}{AB} = 1 - \frac{\sin\alpha}{\sin\beta}.$

Adding the two relations, $\frac{AZ}{AP} + \frac{AZ}{AB} = 2,$ so with $AP = 3$ and $AB = 7$ we get $AZ\left(\frac{1}{3} + \frac{1}{7}\right) = 2,$ hence $AZ = \frac{21}{5}.$ The identical argument on side $AC$ (using $\angle XAC$ in triangles $AQY$ and $ACX$) gives $\frac{AW}{AQ} + \frac{AW}{AC} = 2,$ and $AW = AZ = \frac{21}{5}$ by equal tangents from $A.$ Therefore $$\frac{1}{AQ} = \frac{10}{21} - \frac{1}{8} = \frac{59}{168},$$ so $AQ = \frac{168}{59}$ and $m + n = 168 + 59 = 227.$

Solution:

Let $d_i = f(i) - f(i-1),$ so each $|d_i| \in \{1, 2, 3\}$ and $d_1 + \cdots + d_6 = 12.$ If $k$ of the differences were negative, the sum would be at most $3(6 - k) - k = 18 - 4k,$ so $k \le 1.$ If no difference is negative, the solutions of $a + b + c = 6,$ $a + 2b + 3c = 12$ (with $a, b, c$ counting $1$s, $2$s, $3$s) are $(0,6,0),$ $(1,4,1),$ $(2,2,2),$ $(3,0,3),$ and all such orderings satisfy every pair condition, giving $$\binom{6}{0,6,0} + \binom{6}{1,4,1} + \binom{6}{2,2,2} + \binom{6}{3,0,3} = 1 + 30 + 90 + 20 = 141.$$

If $d_1 \lt 0,$ then $|f(2)| \ge 2$ with $f(2) \le d_1 + 3$ forces $d_1 = -1$ and $d_2 = 3;$ the remaining four differences are positive and sum to $10,$ giving $\binom{4}{1,0,3} + \binom{4}{0,2,2} = 4 + 6 = 10$ functions, and the case $d_6 \lt 0$ is symmetric: $20$ in all. Finally, if $d_{n+1} \lt 0$ for some $n = 1, 2, 3, 4,$ the pair conditions $|f(n+1) - f(n-1)| \ge 2$ and $|f(n+2) - f(n)| \ge 2$ force $d_{n+1} = -1$ and $d_n = d_{n+2} = 3.$ The other three differences are positive and sum to $7,$ achievable as two $3$s and a $1$ or a $3$ and two $2$s, each in $3$ orders: $6$ ways for each of the $4$ positions, or $24$ functions.

The total is $141 + 20 + 24 = 185.$