Solution:

Triangle $ABC$ is right-angled at $B,$ so $AC^2 = 18^2 + 21^2 = 765.$ Triangle $ACD$ is right-angled at $C,$ so $DA^2 = AC^2 + CD^2 = 765 + 196 = 961,$ giving $DA = 31.$

The perimeter is $18 + 21 + 14 + 31 = 84.$

Solution:

The smallest possible sum is $1 + 2 + \cdots + 90 = 4095,$ and the largest is $11 + 12 + \cdots + 100 = 4995.$

Every integer in between also occurs. Suppose $\mathcal{A}$ has sum $S \le 4994,$ and let $k$ be the smallest element of $\mathcal{A}$ with $k + 1 \notin \mathcal{A}.$ If $k$ were $100,$ then $\mathcal{A}$ would be a block of consecutive integers ending at $100,$ namely $\{11, \ldots, 100\},$ whose sum exceeds $4994.$ So $k \ne 100,$ and replacing $k$ by $k + 1$ produces a $90$-element subset with sum $S + 1.$

Hence $S$ takes every value from $4095$ to $4995,$ for $4995 - 4095 + 1 = 901$ possible values.

Solution:

Let $d$ be the leftmost digit and $n$ the integer that remains after deleting it, so the original integer is $d \cdot 10^p + n$ for some positive integer $p.$ The condition says $d \cdot 10^p + n = 29n,$ so $d \cdot 10^p = 28n.$

Since $7 \mid 28n$ but $7 \nmid 10^p,$ the digit $d$ must be a multiple of $7,$ so $d = 7.$ Then $10^p = 4n,$ giving $n = 25 \cdot 10^{p-2},$ which requires $p \ge 2.$ The smallest case is $p = 2,$ $n = 25.$

The least such integer is $725,$ and indeed $725 = 29 \cdot 25.$

Solution:

Factors of $2$ are plentiful, so $N$ is the exponent of $5$ in the product. Each integer $j$ with $1 \le j \le 100$ appears as a factor in exactly $101 - j$ of the factorials, namely $j!, (j+1)!, \ldots, 100!.$

Every multiple of $5$ contributes one factor of $5$ per appearance, and every multiple of $25$ contributes one more. Over $j = 5, 10, \ldots, 100$ the appearances total $96 + 91 + \cdots + 1 = \frac{20 \cdot 97}{2} = 970,$ and over $j = 25, 50, 75, 100$ they total $76 + 51 + 26 + 1 = 154.$

Hence $N = 970 + 154 = 1124,$ and the remainder upon division by $1000$ is $124.$

Solution:

Squaring $a\sqrt{2} + b\sqrt{3} + c\sqrt{5}$ gives $$2a^2 + 3b^2 + 5c^2 + 2ab\sqrt{6} + 2ac\sqrt{10} + 2bc\sqrt{15}.$$ Matching coefficients yields $2ab = 104,$ $2ac = 468,$ $2bc = 144,$ that is $ab = 52,$ $ac = 234,$ $bc = 72,$ along with $2a^2 + 3b^2 + 5c^2 = 2006.$

Then $(abc)^2 = (ab)(ac)(bc) = 52 \cdot 234 \cdot 72 = 876096 = 936^2,$ so $abc = 936.$ As a check, $a = \frac{abc}{bc} = 13,$ $b = \frac{abc}{ac} = 4,$ $c = \frac{abc}{ab} = 18,$ and $2 \cdot 169 + 3 \cdot 16 + 5 \cdot 324 = 338 + 48 + 1620 = 2006,$ as required.

Therefore $a \cdot b \cdot c = 936.$

Solution:

Each element equals $0.\overline{abc} = \frac{100a + 10b + c}{999},$ and there are $10 \cdot 9 \cdot 8 = 720$ ordered triples of distinct digits. By symmetry, each digit $0$ through $9$ appears in each of the three positions exactly $\frac{720}{10} = 72$ times.

The numerators therefore total $72 (0 + 1 + \cdots + 9)(100 + 10 + 1) = 72 \cdot 45 \cdot 111 = 359640,$ so the sum of the elements is $\frac{359640}{999} = 360.$

Solution:

Take the spacing between consecutive lines as the unit, and let $x$ be the distance from the vertex of the angle to the first line. The triangle cut off by the $j$th line is similar to triangle $\mathcal{A}$ with ratio $\frac{x + j - 1}{x},$ so its area is proportional to $(x + j - 1)^2,$ and the strip between lines $j$ and $j + 1$ has area proportional to $(x + j)^2 - (x + j - 1)^2 = 2x + 2j - 1.$

Regions $\mathcal{B},$ $\mathcal{C},$ and $\mathcal{D}$ are the strips beginning at lines $2,$ $4,$ and $6,$ with areas proportional to $2x + 3,$ $2x + 7,$ and $2x + 11.$ The given ratio yields $\frac{2x+7}{2x+3} = \frac{11}{5},$ so $10x + 35 = 22x + 33$ and $x = \frac{1}{6}.$

Since $\mathcal{A}$ has area proportional to $x^2,$ the requested ratio is $\frac{2x + 11}{x^2} = \frac{34/3}{1/36} = 408.$

Solution:

Since $\mathcal{T}$ shares a side with each of the other rhombuses, all five have the same side length $z.$ Let $Y$ be the vertex of $\mathcal{T}$ on $\overline{AB},$ and let $\alpha$ be the angle of $\mathcal{P}$ at $Y.$ Then each congruent rhombus has area $z^2 \sin\alpha = \sqrt{2006}.$ The angles of $\mathcal{P},$ $\mathcal{T},$ and $\mathcal{Q}$ at $Y$ lie along the line $AB,$ and by symmetry $\mathcal{Q}$'s angle there also equals $\alpha,$ so $\mathcal{T}$'s angle is $180^\circ - 2\alpha.$ Hence $$K = z^2 \sin(180^\circ - 2\alpha) = z^2 \sin 2\alpha = 2 z^2 \sin\alpha \cos\alpha = 2\sqrt{2006}\,\cos\alpha.$$

As $\alpha$ ranges over $(0^\circ, 90^\circ),$ the value $\cos\alpha$ takes every value in $(0, 1),$ so $K$ takes every value in $\left(0, \sqrt{8024}\right).$ Since $89^2 = 7921 \lt 8024 \lt 8100 = 90^2,$ the possible positive integer values are $1, 2, \ldots, 89:$ there are $89$ of them.

Solution:

The sum of the logarithms is $\log_8 (a_1 a_2 \cdots a_{12}) = \log_8\!\left(a^{12} r^{66}\right),$ so $a^{12} r^{66} = 8^{2006} = 2^{6018},$ which gives $a^2 r^{11} = 2^{1003}.$

Thus $a$ and $r$ are powers of $2:$ write $a = 2^x$ and $r = 2^y$ with integers $x, y \ge 0$ and $2x + 11y = 1003.$ Since $2x$ is even and $1003$ is odd, $y$ must be odd, say $y = 2k - 1$ for $k \ge 1.$ Then $x = 507 - 11k \ge 0$ exactly when $k \le \lfloor 507/11 \rfloor = 46.$

Each $k = 1, 2, \ldots, 46$ gives one pair, so there are $46$ ordered pairs $(a, r).$

Solution:

The circles have radius $\frac{1}{2}$ and centers at $\left(\frac{1}{2}, \frac{1}{2}\right),$ $\left(\frac{3}{2}, \frac{1}{2}\right),$ $\left(\frac{5}{2}, \frac{1}{2}\right),$ $\left(\frac{1}{2}, \frac{3}{2}\right),$ $\left(\frac{3}{2}, \frac{3}{2}\right),$ $\left(\frac{5}{2}, \frac{3}{2}\right),$ $\left(\frac{1}{2}, \frac{5}{2}\right),$ and $\left(\frac{3}{2}, \frac{5}{2}\right).$ The pair of circles tangent at $A = \left(1, \frac{1}{2}\right)$ is symmetric about $A,$ so any line through $A$ bisects that pair's area; similarly for the pair tangent at $B = \left(\frac{3}{2}, 2\right).$ The line $AB$ has slope $\frac{2 - 1/2}{3/2 - 1} = 3.$

Line $AB$ misses the remaining four circles entirely, and exactly two of them lie on each side of it, so it divides $\mathcal{R}$ into two regions of equal area. Sliding a slope-$3$ line strictly shifts area from one side to the other, so $\ell$ must be this line.

Its equation is $y - \frac{1}{2} = 3(x - 1),$ that is, $6x = 2y + 5.$ With $\gcd(6, 2, 5) = 1,$ the answer is $a^2 + b^2 + c^2 = 36 + 4 + 25 = 65.$

Solution:

Let $S(n)$ be the number of legal towers using the cubes of edge-lengths $1, \ldots, n.$ Given a legal tower of $n \ge 2$ cubes, cube $n + 1$ can be inserted in exactly three places: at the bottom, immediately on top of cube $n,$ or immediately on top of cube $n - 1$ (anywhere else it would rest on a cube of edge-length at most $n - 2,$ violating the rule). Each insertion stays legal, because the cube that ends up on top of cube $n + 1$ has edge-length at most $n \lt (n+1) + 2.$ Conversely, deleting cube $n + 1$ from a legal tower of $n + 1$ cubes leaves a legal tower: the cube that was above it, of edge-length at most $n,$ lands on a cube of edge-length at least $n - 1.$

Hence $S(n + 1) = 3 S(n)$ for $n \ge 2.$ Since $S(2) = 2$ (either of the two cubes may be on top), we get $T = S(8) = 2 \cdot 3^6 = 1458,$ and the remainder is $458.$

Solution:

By the product-to-sum identity, $2 \cos 4x \cos x = \cos 5x + \cos 3x,$ so the right side is $(\cos 5x + \cos 3x)^3.$ Setting $y = \cos 3x$ and $z = \cos 5x,$ the equation becomes $y^3 + z^3 = (y + z)^3,$ and since $(y+z)^3 - y^3 - z^3 = 3yz(y + z),$ it holds exactly when $$\cos 3x = 0, \qquad \cos 5x = 0, \qquad \text{or} \qquad \cos 4x \cos x = 0.$$

For $100 \lt x \lt 200$ in degrees: $\cos 3x = 0$ gives $x = 150;$ $\cos 5x = 0$ gives $x = 126,$ $162,$ $198;$ $\cos 4x = 0$ gives $x = 112.5,$ $157.5;$ and $\cos x = 0$ gives no solutions in the interval.

The sum is $150 + 126 + 162 + 198 + 112.5 + 157.5 = 906.$

Solution:

$S_n$ is the sum of $g$ over the even numbers $2, 4, \ldots, 2^n.$ Among these $2^{n-1}$ numbers, exactly $2^{n-1-i}$ are divisible by $2^i$ but not $2^{i+1}$ (so have $g = 2^i$) for each $1 \le i \le n - 1,$ and the single number $2^n$ has $g = 2^n.$ Hence $$S_n = \sum_{i=1}^{n-1} 2^i \cdot 2^{n-1-i} + 2^n = (n-1) 2^{n-1} + 2^n = (n+1) 2^{n-1}.$$

If $n$ is even, then $n + 1$ is odd and the exponent $n - 1$ is odd, so $S_n$ has an odd number of factors of $2$ and cannot be a perfect square. If $n$ is odd, then $2^{n-1}$ is already a perfect square, so $S_n$ is a perfect square exactly when $n + 1$ is.

For odd $n,$ the number $n + 1$ is even, and the greatest even perfect square at most $1000$ is $900 = 30^2.$ Thus the greatest valid $n \lt 1000$ is $n = 899.$

Solution:

Place the top at $T = (0, 0, 4).$ Each leg has length $5,$ so each foot is $3$ from the origin: $A = (3, 0, 0),$ $B = \left(-\frac{3}{2}, \frac{3\sqrt{3}}{2}, 0\right),$ $C = \left(-\frac{3}{2}, -\frac{3\sqrt{3}}{2}, 0\right).$ The broken leg has length $4,$ so its tip is $A' = \frac{4}{5}(3, 0, 0) + \frac{1}{5}(0, 0, 4) = \left(\frac{12}{5}, 0, \frac{4}{5}\right),$ and the tripod now stands on the plane $A'BC,$ with $h$ equal to the distance from $T$ to that plane.

The plane $A'BC$ contains $\overline{BC},$ which is parallel to the $y$-axis and passes through the midpoint $M = \left(-\frac{3}{2}, 0, 0\right),$ so its distance from $T$ can be measured in the $xz$-plane: it is the distance from $(0, 4)$ to the line through $\left(-\frac{3}{2}, 0\right)$ and $\left(\frac{12}{5}, \frac{4}{5}\right),$ whose equation is $8x - 39z + 12 = 0.$ Therefore $$h = \frac{|8 \cdot 0 - 39 \cdot 4 + 12|}{\sqrt{8^2 + 39^2}} = \frac{144}{\sqrt{1585}}.$$

Here $m = 144$ and $n = 1585 = 5 \cdot 317$ is squarefree. Since $39^2 = 1521 \lt 1585 \lt 1600,$ we get $\lfloor 144 + \sqrt{1585} \rfloor = 144 + 39 = 183.$

Solution:

Squaring the recurrence gives $x_k^2 = (x_{k-1} + 3)^2 = x_{k-1}^2 + 6 x_{k-1} + 9.$ Summing for $k = 1$ to $2007$ telescopes: $$x_{2007}^2 = x_0^2 + 6 \sum_{k=0}^{2006} x_k + 9 \cdot 2007,$$ so with $x_0 = 0,$ $$x_1 + x_2 + \cdots + x_{2006} = \frac{x_{2007}^2 - 18063}{6}.$$

Induction shows each $x_k$ is a multiple of $3$ whose parity matches that of $k,$ so $x_{2007}$ is an odd multiple of $3.$ To minimize $\left|x_{2007}^2 - 18063\right|,$ take the odd multiple of $3$ whose square is nearest $18063:$ that is $\pm 135,$ with $135^2 = 18225,$ giving $\frac{18225 - 18063}{6} = 27$ (the neighbors $129$ and $141$ give $237$ and $303$).

This value is attained: take $x_k = 3k$ for $k \le 45,$ and thereafter alternate $x_k = -138$ for even $k$ and $x_k = 135$ for odd $k;$ then $x_{2007} = 135$ and the sum is $27.$ So the minimum is $27.$