Answer: 29

Small Hint:

If the common difference is not a multiple of $3,$ one of the first three terms is divisible by $3$

Big Hint:

The difference must be a multiple of $6,$ so try five-term progressions with difference $6$ starting from small primes

Solution:

Let the terms be $p,$ $p + d,$ $\ldots,$ $p + 4d.$ The even prime cannot be the first term: an even difference makes the second term even, while an odd difference makes the third term even. If $d$ were odd, consecutive terms would have opposite parity, so some term other than the first would be even and greater than $2$ — impossible. If $d$ were not a multiple of $3,$ then $p,$ $p + d,$ $p + 2d$ would cover all residues mod $3,$ so some term would be divisible by $3;$ that term would have to be $3$ itself, forcing $p = 3,$ but then $p + 3d = 3(1 + d)$ is composite. Hence $d$ is divisible by $6.$

With $d \ge 6$ the fifth term is at least $p + 24.$ Trying $p = 5$ and $d = 6$ gives $5, 11, 17, 23, 29,$ all prime, and no smaller fifth term is possible since $p \ge 5$ (the starts $p = 2$ and $p = 3$ fail as above). The answer is $29.$

Answer: 118

Small Hint:

A line cuts a parallelogram into two congruent pieces when it passes through the parallelogram’s center of symmetry

Big Hint:

The center is the midpoint of the diagonal from $(10, 45)$ to $(28, 153)$

Solution:

Congruent pieces have equal area. For any fixed direction, there is only one line in that direction that bisects the area of a convex figure; because a parallelogram is centrally symmetric, that line passes through its center. Thus the required line through the origin must also pass through the center. Conversely, the $180^\circ$ rotation about the center swaps the two pieces cut by such a line, so they are congruent. The center is the midpoint of a diagonal: $$\begin{aligned} &\left(\frac{10 + 28}{2}, \frac{45 + 153}{2}\right) \\ &= (19, 99). \end{aligned}$$

The line through the origin and $(19, 99)$ has slope $\frac{99}{19},$ and $\gcd(99, 19) = 1,$ so $m + n = 99 + 19 = 118.$

Answer: 38

Small Hint:

Multiply by $4$ and complete the square: $4(n^2 - 19n + 99)$ $= (2n - 19)^2 + 35$

Big Hint:

If the expression equals $k^2,$ then $(2k)^2 - (2n - 19)^2 = 35;$ factor $35$ as a difference of squares in every way

Solution:

Suppose $n^2 - 19n + 99 = k^2.$ Multiplying by $4$ and completing the square gives $(2n - 19)^2 + 35 = (2k)^2,$ so $$\begin{aligned} &\bigl(2k - (2n - 19)\bigr) \\ &\quad {}\cdot \bigl(2k + (2n - 19)\bigr) \\ &= 35. \end{aligned}$$ The two factors sum to $4k \gt 0,$ so both are positive: the factor pairs are $(1, 35),$ $(5, 7),$ $(7, 5),$ and $(35, 1).$

Subtracting the first factor from the second gives $2(2n - 19) = 34,$ $2,$ $-2,$ or $-34,$ so $n = 18,$ $10,$ $9,$ or $1.$ Each indeed makes the expression a perfect square ($81,$ $9,$ $9,$ $81$), and the sum is $18 + 10 + 9 + 1 = 38.$

Answer: 185

Small Hint:

By the symmetries swapping the two squares and rotating by $90^\circ,$ all eight sides of the octagon are congruent

Big Hint:

Cut the octagon into $8$ triangles with apex $O;$ each side of the octagon lies on a side of a square, at distance $\frac{1}{2}$ from $O$

Solution:

The whole configuration is unchanged by rotating $90^\circ$ about $O,$ which cycles the octagon side $AB$ to $CD,$ $EF,$ $GH,$ and it is also unchanged by the reflection that swaps the two squares, which carries those sides to $BC,$ $DE,$ $FG,$ $HA.$ So all eight sides of the octagon have the same length, $\frac{43}{99}.$

Segments from $O$ to the eight vertices cut the octagon into $8$ triangles. Each has base $\frac{43}{99}$ lying on a side of one of the unit squares, so its height from $O$ is the distance from the center to that side, namely $\frac{1}{2}.$ The area is $$8 \cdot \frac{1}{2} \cdot \frac{43}{99} \cdot \frac{1}{2} = \frac{86}{99}.$$

Since $\gcd(86, 99) = 1,$ the answer is $86 + 99 = 185.$

Answer: 223

Small Hint:

If adding $2$ causes no carrying, $T(x) = 2.$ Carrying happens only when $x$ ends in $8$ or $9.$

Big Hint:

Track how the digit sum changes when $x$ ends in a run of nines; the carrying values form an arithmetic progression with common difference $9$

Solution:

If the last digit of $x$ is at most $7,$ adding $2$ changes no other digit, so $T(x) = 2.$ Otherwise there is carrying. If $x$ ends in the digit $8$ preceded by exactly $m \ge 0$ nines, then $x + 2$ replaces $\ldots a\underbrace{9 \cdots 9}_{m}8$ by $\ldots (a{+}1)\underbrace{0 \cdots 0}_{m}0,$ so $S(x + 2) - S(x) = 1 - 9m - 8$ and $T(x) = 9m + 7.$ If $x$ ends in exactly $m \ge 1$ nines, then $x + 2$ replaces $\ldots a\underbrace{9 \cdots 9}_{m}$ by $\ldots (a{+}1)\underbrace{0 \cdots 0}_{m - 1}1,$ so $S(x + 2) - S(x) = 2 - 9m$ and $T(x) = 9m - 2.$

Both carrying families give exactly the values $7, 16, 25, \ldots,$ that is, $9j + 7$ for $j \ge 0,$ and every such value occurs. So the possible values of $T$ are $2$ together with all $9j + 7.$ Requiring $9j + 7 \le 1999$ gives $j \le 221,$ which is $222$ values, and $T = 2$ adds one more, for a total of $223.$

Answer: 314

Small Hint:

Instead of tracking the vertices, find what the map does to each of the four boundary lines

Big Hint:

$y = 3x$ and $y = \frac{x}{3}$ map to lines through the origin, while $x + y = c$ maps to an arc of $u^2 + v^2 = c$

Solution:

Follow the four edges. Sides $AB$ and $DC$ lie on the lines $y = \frac{x}{3}$ and $y = 3x,$ which map to the lines $v = \frac{u}{\sqrt{3}}$ and $v = \sqrt{3}\,u$ — rays from the origin at angles $30^\circ$ and $60^\circ.$ Sides $AD$ and $BC$ lie on $x + y = 1200$ and $x + y = 2400,$ which map to arcs of the circles $u^2 + v^2 = 1200$ and $u^2 + v^2 = 2400.$

So the image is the part of the annulus between radii $\sqrt{1200}$ and $\sqrt{2400}$ lying between the $30^\circ$ and $60^\circ$ rays, one twelfth of the full annulus: $$\begin{aligned} k &= \frac{30}{360}\,\pi\,(2400 - 1200) \\ &= 100\pi. \end{aligned}$$

Since $3.14 < \pi < 3.15,$ we have $314 < k < 315,$ so the greatest integer not exceeding $k$ is $314.$

Answer: 650

Small Hint:

The switch labeled $d$ is advanced once for every label that is a multiple of $d;$ count those multiples in terms of the exponents

Big Hint:

The label $2^x 3^y 5^z$ advances $(10 - x)(10 - y)(10 - z)$ times; count the triples for which this product is not a multiple of $4$

Solution:

The switch labeled $d$ is advanced exactly once for each step $i$ whose label is a multiple of $d.$ The multiples of $2^x 3^y 5^z$ among the labels are the $2^{x'} 3^{y'} 5^{z'}$ with $x \le x' \le 9,$ etc., so that switch advances $(10 - x)(10 - y)(10 - z)$ times. It returns to position $A$ exactly when this count is a multiple of $4.$

Write $a = 10 - x,$ $b = 10 - y,$ $c = 10 - z,$ each ranging over $1$ through $10.$ We count the triples where $abc$ is not divisible by $4:$ either all three are odd, or exactly one is even but not divisible by $4.$ Among $1, \ldots, 10$ there are $5$ odd values and $3$ values ($2,$ $6,$ $10$) that are twice an odd number. That gives $5^3 = 125$ triples of the first kind and $3 \cdot 3 \cdot 5^2 = 225$ of the second, or $350$ in all.

Therefore $1000 - 350 = 650$ switches end in position $A.$

Answer: 25

Small Hint:

$T$ is a triangle, and each condition like $x \ge \frac{1}{2}$ cuts off a corner with a line parallel to a side

Big Hint:

Since $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 1,$ all three inequalities hold only at one boundary point; apart from that point, each two-inequality region is a triangle similar to $T$

Solution:

$T$ is the triangle with vertices $(1,0,0),$ $(0,1,0),$ $(0,0,1).$ Because $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 1 = x + y + z,$ whenever two of the inequalities $x \ge \frac{1}{2},$ $y \ge \frac{1}{3},$ $z \ge \frac{1}{6}$ hold, the third can hold only at a boundary point of zero area. So $S$ is, up to measure zero, the union of the three regions where a specific pair of inequalities holds.

The region with $x \ge \frac{1}{2}$ and $y \ge \frac{1}{3}$ becomes, after substituting $x = \frac{1}{2} + x'$ and $y = \frac{1}{3} + y',$ a copy of $T$ with coordinate sum $1 - \frac{1}{2} - \frac{1}{3} = \frac{1}{6},$ i.e. a triangle similar to $T$ with ratio $\frac{1}{6}$ and area $\left(\frac{1}{6}\right)^2$ of $T.$ Likewise the pairs $\{x, z\}$ and $\{y, z\}$ give similar triangles with ratios $\frac{1}{3}$ and $\frac{1}{2}.$

The ratio of areas is $$\begin{aligned} &\frac{1}{36} + \frac{1}{9} + \frac{1}{4} = \frac{1 + 4 + 9}{36} \\ &= \frac{7}{18}, \end{aligned}$$ so $m + n = 7 + 18 = 25.$

Answer: 259

Small Hint:

The condition says $|f(z) - z| = |f(z)|$ for every $z;$ write both sides as a modulus times $|z|$

Big Hint:

You need $|a - 1 + bi| = |a + bi|,$ which pins down $a;$ then use $a^2 + b^2 = 64$

Solution:

The condition is $|f(z) - z| = |f(z)|$ for all $z,$ that is, $|(a - 1 + bi)z| = |(a + bi)z|.$ Dividing by $|z|$ (for $z \ne 0$) gives $|a - 1 + bi| = |a + bi|,$ so $$(a - 1)^2 + b^2 = a^2 + b^2,$$ which forces $a = \frac{1}{2}.$

Since $|a + bi| = 8,$ we have $a^2 + b^2 = 64,$ so $b^2 = 64 - \frac{1}{4} = \frac{255}{4}.$ As $\gcd(255, 4) = 1,$ the answer is $255 + 4 = 259.$

Answer: 489

Small Hint:

Four segments can contain at most one triangle, because two different triangles always use at least five segments

Big Hint:

Count favorable choices as: pick a triangle from the $\binom{10}{3}$ possible ones, then pick any fourth segment

Solution:

There are $\binom{10}{2} = 45$ segments, so $\binom{45}{4} = 148995$ equally likely choices. Two distinct triangles share at most one edge, so together they use at least $5$ segments; hence a set of $4$ segments contains at most one triangle, and the favorable sets are counted exactly once by choosing a triangle and then a fourth segment: $$\begin{aligned} &\binom{10}{3} \cdot (45 - 3) = 120 \cdot 42 \\ &= 5040. \end{aligned}$$

The probability is $\frac{5040}{148995} = \frac{16}{473},$ already in lowest terms ($473 = 11 \cdot 43$), so $m + n = 16 + 473 = 489.$

Answer: 177

Small Hint:

Multiply the sum by $2 \sin 2.5^\circ$ and use $2 \sin\alpha \sin\beta$ $= \cos(\alpha - \beta)$ $- \cos(\alpha + \beta)$ to telescope

Big Hint:

The two surviving terms combine because $\cos(180^\circ - \theta) = -\cos\theta;$ finish with $\cot\theta = \tan(90^\circ - \theta)$

Solution:

Multiply the sum by $2 \sin 2.5^\circ$ and apply $2 \sin 5k^\circ \sin 2.5^\circ$ $= \cos(5k - 2.5)^\circ$ $- \cos(5k + 2.5)^\circ,$ so the sum telescopes: $$\begin{aligned} &2 \sin 2.5^\circ \sum_{k=1}^{35} \sin 5k^\circ \\ &= \cos 2.5^\circ - \cos 177.5^\circ \\ &= 2\cos 2.5^\circ. \end{aligned}$$

Hence the sum equals $\frac{\cos 2.5^\circ}{\sin 2.5^\circ}$ $= \cot 2.5^\circ$ $= \tan 87.5^\circ$ $= \tan \frac{175}{2}^\circ.$ Since $\gcd(175, 2) = 1$ and $\frac{175}{2} \lt 90,$ we get $m + n = 175 + 2 = 177.$

Answer: 345

Small Hint:

Tangent segments from each vertex are equal, so the tangent lengths from $A,$ $B,$ $C$ are $23,$ $27,$ and an unknown $z$

Big Hint:

Set the area two ways: $rs$ and Heron’s formula; the resulting equation is linear in $z$

Solution:

Tangent segments from a point are equal, so the tangent lengths from $A,$ $B,$ $C$ are $23,$ $27,$ and some $z.$ Then the sides are $50,$ $23 + z,$ $27 + z,$ the semiperimeter is $s = 50 + z,$ and Heron’s formula gives area $\sqrt{(50 + z) \cdot z \cdot 23 \cdot 27}.$

The area also equals $rs = 21(50 + z).$ Squaring $21(50 + z) = \sqrt{621 z (50 + z)}$ and dividing by $50 + z$ gives $441(50 + z) = 621 z,$ so $180 z = 22050$ and $z = \frac{245}{2}.$

The perimeter is $2s = 2\left(50 + \frac{245}{2}\right) = 345.$

Answer: 742

Small Hint:

Distinct win totals must be exactly $0, 1, \ldots, 39,$ and that ranking determines the winner of every single game

Big Hint:

The probability is $40!$ divided by $2^{\binom{40}{2}};$ reduce by counting the factors of $2$ in $40!$

Solution:

There are $\binom{40}{2} = 780$ games, hence $2^{780}$ equally likely outcomes. If all $40$ win totals are distinct, they must be exactly $0, 1, \ldots, 39.$ In that case the team with $39$ wins beat everyone, the team with $38$ wins beat everyone except that team, and so on: the assignment of totals to teams determines every game. Conversely each of the $40!$ assignments arises from exactly one outcome, so the probability is $\frac{40!}{2^{780}}.$

By Legendre’s formula the power of $2$ dividing $40!$ is $20 + 10 + 5 + 2 + 1 = 38.$ In lowest terms the denominator is therefore $n = 2^{780 - 38} = 2^{742},$ so $\log_2 n = 742.$

Answer: 463

Small Hint:

Call the common angle $\omega$ and compute $BP$ two ways, from triangles $ABP$ and $BCP,$ with the law of sines

Big Hint:

The comparison simplifies to $\cot\omega = \cot A + \cot B + \cot C,$ and each cotangent can be written in terms of the side lengths and the area

Solution:

Let $\omega = \angle PAB = \angle PBC = \angle PCA.$ In triangle $ABP,$ the angles at $A$ and $B$ are $\omega$ and $B - \omega,$ so $\angle APB = 180^\circ - B$ and the law of sines gives $BP = \frac{c \sin\omega}{\sin B}.$ In triangle $BCP,$ the angles at $B$ and $C$ are $\omega$ and $C - \omega,$ so $\angle BPC = 180^\circ - C$ and $BP = \frac{a \sin(C - \omega)}{\sin C}.$

Equating and substituting $a = 2R\sin A,$ $c = 2R\sin C$ yields $\sin^2 C \sin\omega$ $= \sin A \sin B \sin(C - \omega).$ Expanding $\sin(C - \omega)$ and dividing by $\sin A \sin B \sin C \sin\omega,$ $$\frac{\sin C}{\sin A \sin B} = \cot\omega - \cot C,$$ and since $\sin C$ $= \sin(A + B)$ $= \sin A \cos B$ $+ \cos A \sin B,$ the left side is $\cot A + \cot B.$ Hence $\cot\omega = \cot A + \cot B + \cot C.$