Solution:

From 10:34 am to 11:00 am is $26$ minutes, from 11:00 am to 1:00 pm is $2$ hours, and from 1:00 pm to 1:18 pm is $18$ minutes.

So the flight lasted $2$ hours and $26 + 18 = 44$ minutes. Thus $h = 2,$ $m = 44,$ and $h + m = 46.$

Thus, the correct answer is A.

Solution:

Starting inside, $1 + 1 = 2,$ so $1 + \dfrac{1}{2} = \dfrac{3}{2}.$ Then $\dfrac{1}{3/2} = \dfrac{2}{3},$ and $1 + \dfrac{2}{3} = \dfrac{5}{3}.$

Thus, the correct answer is C.

Solution:

The gap is $\dfrac{3}{4} - \dfrac{1}{4} = \dfrac{1}{2}.$ One third of the way adds $\dfrac{1}{3} \cdot \dfrac{1}{2} = \dfrac{1}{6}.$

So the number is $\dfrac{1}{4} + \dfrac{1}{6} = \dfrac{3}{12} + \dfrac{2}{12} = \dfrac{5}{12}.$

Thus, the correct answer is B.

Solution:

If the four coins include a penny, the total is not a multiple of $5,$ so it cannot equal any of the five listed multiples of $5.$ If there is no penny, every coin is worth at least $5$ cents, so the total is at least $20$ cents. Either way, $15$ is impossible.

The other amounts are attainable: $25 = 10 + 5 + 5 + 5,$ $35 = 10 + 10 + 10 + 5,$ $45 = 25 + 10 + 5 + 5,$ and $55 = 25 + 10 + 10 + 10.$

Thus, the correct answer is A.

Solution:

Let the cube have side length $x.$ The new solid has dimensions $x + 1,$ $x - 1,$ and $x,$ so its volume is $x(x+1)(x-1) = x^3 - x.$

Setting this equal to $x^3 - 5$ gives $x^3 - x = x^3 - 5,$ so $x = 5.$

The cube's volume is $5^3 = 125.$

Thus, the correct answer is D.

Solution:

Since $12 = 2^2 \cdot 3,$ $$12^{mn} = 2^{2mn} \cdot 3^{mn} = (2^m)^{2n}(3^n)^m = P^{2n}Q^m.$$

Thus, the correct answer is E.

Solution:

Equal consecutive differences give $(5x - 11) - (2x - 3) = (3x + 1) - (5x - 11),$ that is $3x - 8 = -2x + 12,$ so $x = 4.$

The first three terms are $5, 9, 13,$ with common difference $4.$

The $n$th term satisfies $2009 = 5 + (n - 1)\cdot 4,$ so $n - 1 = 501$ and $n = 502.$

Thus, the correct answer is B.

Solution:

Let the rectangles have shorter side $x$ and longer side $y.$ The outer square has side $x + y$ and the inner square has side $y - x.$

Since the outer area is $4$ times the inner area, the side ratio is $\sqrt{4} = 2,$ so $x + y = 2(y - x).$

This gives $y = 3x,$ so the ratio of longer to shorter side is $3.$

Thus, the correct answer is A.

Solution:

Note that $a + b + c = f(1).$

Using $f(x + 3) = 3x^2 + 7x + 4$ with $x = -2$ gives $$f(1) = f(-2 + 3) = 3(-2)^2 + 7(-2) + 4 = 12 - 14 + 4 = 2.$$

Thus, the correct answer is D.

Solution:

In $\triangle BCD,$ the triangle inequality gives $BD + CD \gt BC,$ so $BD + 5 \gt 17$ and $BD \gt 12.$

In $\triangle ABD,$ $AB + DA \gt BD,$ so $BD \lt 5 + 9 = 14.$

The only integer with $12 \lt BD \lt 14$ is $13.$

Thus, the correct answer is C.

Solution:

The outside square of $F_n$ has $4$ more diamonds than that of $F_{n-1},$ and the outside square of $F_2$ has $4,$ so the outside square of $F_n$ has $4(n - 1)$ diamonds.

Adding all the rings, $$1 + 4\big(1 + 2 + \cdots + (n - 1)\big) = 1 + 4\cdot\frac{(n - 1)n}{2} = 1 + 2(n - 1)n.$$

For $n = 20,$ this is $1 + 2\cdot 19\cdot 20 = 761.$

Thus, the correct answer is E.

Solution:

If $N = 6\cdot(\text{digit sum}),$ then since the digit sum of a number below $1000$ is at most $27,$ we have $N \le 162.$

For a two-digit number $10t + u = 6(t + u)$ gives $4t = 5u,$ forcing $t = 5$ and $u = 4,$ so $N = 54.$ A one-digit number would need $6u = u,$ impossible for $u \gt 0.$ A three-digit number $100h + 10t + u = 6(h + t + u)$ gives $94h + 4t = 5u,$ whose left side is at least $94$ while the right side is at most $45,$ so there is no solution.

Hence exactly one number, $54,$ works.

Thus, the correct answer is B.

Solution:

By the Law of Cosines, $$AC^2 = 10^2 + 20^2 - 2\cdot 10\cdot 20\cos(\angle ABC) = 500 - 400\cos(\angle ABC).$$

The ship turns through an angle between $45^\circ$ and $60^\circ,$ so the interior angle $\angle ABC$ lies between $120^\circ$ and $135^\circ.$

Since $\cos 120^\circ = -\dfrac{1}{2}$ and $\cos 135^\circ = -\dfrac{\sqrt{2}}{2},$ $$700 = 500 + 200 \le AC^2 \le 500 + 200\sqrt{2} \lt 800.$$

So $AC^2$ lies in $[700, 800].$

Thus, the correct answer is D.

Solution:

The line $y = mx$ passes through the vertex $(0, 0),$ so it bisects the triangle's area exactly when it passes through the midpoint of the opposite side, joining $(1, 1)$ and $(6m, 0).$ That midpoint is $\left(\dfrac{6m + 1}{2}, \dfrac{1}{2}\right).$

Requiring it to satisfy $y = mx$ gives $$\frac{1}{2} = m\cdot\frac{6m + 1}{2},$$ so $6m^2 + m - 1 = 0,$ that is $(3m - 1)(2m + 1) = 0.$

The possible values are $m = \dfrac{1}{3}$ and $m = -\dfrac{1}{2},$ whose sum is $-\dfrac{1}{6}.$

Thus, the correct answer is B.

Solution:

For $k$ a multiple of $4,$ $$(k + 1)i^{k+1} + (k + 2)i^{k+2} + (k + 3)i^{k+3} + (k + 4)i^{k+4} = (k + 1)i - (k + 2) - (k + 3)i + (k + 4) = 2 - 2i.$$

Summing the first $96$ terms (that is $24$ blocks) gives $24(2 - 2i) = 48 - 48i.$

Adding the next term $97i^{97} = 97i$ yields $48 - 48i + 97i = 48 + 49i.$ So $n = 97.$

Thus, the correct answer is D.

Solution:

A circle tangent to both positive axes with radius $r$ has center $(r, r).$ External tangency to the circle at $(3, 0)$ of radius $1$ means the distance between centers is $r + 1$: $$(r - 3)^2 + r^2 = (r + 1)^2.$$

Expanding gives $r^2 - 8r + 8 = 0.$ Both roots $r = 4 \pm 2\sqrt{2}$ are positive, and by Vieta's formulas their sum is $8.$

Thus, the correct answer is D.

Solution:

For a series with first term $a$ and ratio $r,$ the sum is $\dfrac{a}{1 - r} = r,$ so $r^2 - r + a = 0.$

Both $r_1$ and $r_2$ satisfy this same quadratic, and since the two series are different, $r_1 \ne r_2,$ so they are its two distinct roots. By Vieta's formulas, $r_1 + r_2 = 1.$

Thus, the correct answer is C.

Solution:

Note that $I_k = 10^{k+2} + 64 = 2^{k+2}5^{k+2} + 2^6.$

For $k \lt 4$ the first term has fewer than $6$ factors of $2,$ so $N(k) \lt 6.$ For $k \gt 4$ the first term is divisible by $2^7$ but the $2^6$ term is not, so $N(k) \lt 7.$

For $k = 4,$ $I_4 = 2^6(5^6 + 1).$ Since $5^6 + 1 = (5^2 + 1)\big((5^2)^2 - 5^2 + 1\big) = 26\cdot 601,$ and $26 = 2\cdot 13$ contributes exactly one more factor of $2,$ we get $N(4) = 7.$

So the maximum value is $7.$

Thus, the correct answer is B.

Solution:

For a regular polygon with side length $2,$ let $O$ be the center, $M$ the midpoint of a side, and $N$ an endpoint of that side. Then $\triangle OMN$ has a right angle at $M,$ with $MN = 1,$ $OM = r$ (inradius), and $ON = R$ (circumradius).

So $R^2 - r^2 = 1,$ and the area between the circles is $\pi(R^2 - r^2) = \pi$ for any number of sides. Hence $A = B.$

Thus, the correct answer is C.

Solution:

Adding $\triangle CED$ to each of $\triangle AED$ and $\triangle BEC$ shows $\triangle ACD$ and $\triangle BCD$ have equal areas. They share base $CD,$ so $A$ and $B$ are equidistant from line $CD,$ meaning $AB \parallel CD.$

Then $\triangle ABE \sim \triangle CDE$ with ratio $\dfrac{AB}{CD} = \dfrac{9}{12} = \dfrac{3}{4},$ so $\dfrac{AE}{EC} = \dfrac{3}{4}.$

Writing $AE = 3x$ and $EC = 4x,$ we get $7x = AC = 14,$ so $x = 2$ and $AE = 6.$

Thus, the correct answer is E.

Solution:

Since $x^{12} + ax^8 + bx^4 + c = p(x^4),$ a value is a zero exactly when $x^4$ equals one of the roots of $p,$ namely $2009 + 9002\pi i,$ $2009,$ or $9002.$

The equation $x^4 = 2009 + 9002\pi i$ has four distinct nonreal roots. Each of $x^4 = 2009$ and $x^4 = 9002$ has two real roots and two nonreal roots.

So the nonreal zeros number $4 + 2 + 2 = 8.$

Thus, the correct answer is C.

Solution:

Let the two parallel faces be triangles. The plane passes through the midpoints of the six edges not on those faces, forming an equilateral hexagon of side $\dfrac{1}{2},$ which by symmetry is also equiangular and hence regular.

A regular hexagon is six equilateral triangles, so its area is $$6\cdot\frac{\sqrt{3}}{4}\left(\frac{1}{2}\right)^2 = \frac{3\sqrt{3}}{8}.$$

Thus $a = 3,$ $b = 3,$ $c = 8,$ and $a + b + c = 14.$

Thus, the correct answer is E.

Solution:

Because $g(x) = -f(100 - x),$ the graphs of $f$ and $g$ are reflections of each other through the point $(50, 0),$ so the four intercepts pair up with $x_2 + x_3 = x_1 + x_4 = 100.$

With $x_3 - x_2 = 150,$ we get $x_2 = -25$ and $x_3 = 125.$

Take $x_1, x_3$ as the roots of $f,$ whose vertex has $x$-coordinate $h = \dfrac{x_1 + x_3}{2},$ so $x_1 = 2h - 125.$ The condition that the vertex of $f$ lies on the graph of $g$ gives $$1 = \frac{f(h)}{g(h)} = \frac{(125 - h)(h - 125)}{-(h + 25)(3h - 225)},$$ which solves to $h = -25 - 75\sqrt{2}.$

Then $x_4 = 100 - x_1,$ so $$x_4 - x_1 = 350 - 4h = 450 + 300\sqrt{2}.$$ Hence $m + n + p = 450 + 300 + 2 = 752.$

Thus, the correct answer is D.

Solution:

Since $\log_2 T(n + 1) = T(n),$ each application of $\log_2$ strips one $2$ off the top of a tower of twos.

Reducing $B = (T(2009))^A$ with $A = (T(2009))^{T(2009)},$ one finds $\log_2 B = A\cdot T(2008),$ $\log_2^2 B = T(2009)T(2008) + T(2007),$ and in general the dominant term after $k + 3$ logs is $T(2008 - k).$

So after $2012$ applications of $\log_2$ the result is still positive, meaning a $2013$th $\log_2$ is defined. A matching upper bound shows the result becomes negative after $2013$ applications, so a $2014$th $\log_2$ is undefined. Hence the largest $k$ is $2013.$

Thus, the correct answer is E.

Solution:

The recursion is exactly the tangent addition formula, and $a_1 = \tan\dfrac{\pi}{4},$ $a_2 = \tan\dfrac{\pi}{6}.$

Writing $a_n = \tan\dfrac{\pi c_n}{12}$ with $c_1 = 3,$ $c_2 = 2,$ and $c_{n+2} \equiv c_n + c_{n+1} \pmod{12},$ the sequence $c_n$ is $$3, 2, 5, 7, 0, 7, 7, 2, 9, 11, 8, 7, 3, 10, 1, 11, 0, 11, 11, 10, 9, 7, 4, 11, \ldots$$ which is periodic with period $24.$

Since $2009 = 24\cdot 83 + 17,$ $c_{2009} = c_{17} = 0,$ so $a_{2009} = \tan 0 = 0$ and $|a_{2009}| = 0.$

Thus, the correct answer is A.