Answer: 275

Small Hint:

Both clocks measure the same fraction of a full day, and a metric day has $1000$ metric minutes

Big Hint:

The former $6{:}36$ AM comes $396$ of the $1440$ ordinary minutes into the day; the alarm setting is that same fraction of $1000$ metric minutes

Solution:

An ordinary day has $60 \cdot 24 = 1440$ minutes, and $6{:}36$ AM comes $6 \cdot 60 + 36 = 396$ minutes after midnight. A metric day has $10 \cdot 100 = 1000$ metric minutes, so the equivalent metric time is $$\frac{396}{1440} \cdot 1000 = 275$$ metric minutes after midnight, which the new clock displays as $2{:}75.$

Therefore $100A + 10B + C = 275.$

Answer: 881

Small Hint:

Peel the logarithms from the outside in: the outer equation says the next expression equals $1,$ then the next equals $2^a$

Big Hint:

Everything reduces to $b \cdot 2^a = 1000 = 2^3 \cdot 125,$ so $2^a$ can only be $2,$ $4,$ or $8$

Solution:

Working from the outside in, $\log_2(\cdot) = 0$ forces $\log_{2^a}(\log_{2^b}(2^{1000})) = 1,$ so $\log_{2^b}(2^{1000}) = 2^a,$ so $2^{1000} = (2^b)^{2^a} = 2^{b \cdot 2^a}.$ Hence $$b \cdot 2^a = 1000 = 2^3 \cdot 125.$$

Since $a \ge 1$ and $b$ is a positive integer, $2^a$ must be one of $2,$ $4,$ $8,$ giving $(a, b) = (1, 500),$ $(2, 250),$ and $(3, 125).$ The sum of all possible values of $a + b$ is $501 + 252 + 128 = 881.$

Answer: 350

Small Hint:

Burning the first $x$ centimeters takes $10(1 + 2 + \cdots + x) = 5x(x+1)$ seconds

Big Hint:

Set $5x(x+1)$ equal to $\frac{T}{2} = 5 \cdot 60 \cdot 119$ and write $7140$ as a product of two consecutive integers

Solution:

Burning the first $x$ centimeters takes $10(1 + 2 + \cdots + x) = 5x(x+1)$ seconds, so $T = 5 \cdot 119 \cdot 120 = 71400$ and $\frac{T}{2} = 35700.$

Setting $5x(x+1) = 35700$ gives $x(x+1) = 7140 = 84 \cdot 85,$ so at time $\frac{T}{2}$ the candle has burned down exactly $84$ centimeters. Its height is $h = 119 - 84 = 35,$ and $10h = 350.$

Answer: 40

Small Hint:

$C$ is the midpoint of $\overline{AB}$ shifted perpendicularly by $\frac{\sqrt{3}}{2}$ times the length of $\overline{AB},$ and $P$ is the centroid

Big Hint:

A unit vector perpendicular to $\overrightarrow{AB} = (1, 2\sqrt{3})$ is $\frac{1}{\sqrt{13}}(2\sqrt{3}, -1);$ average the three vertices to get $P$

Solution:

The midpoint of $\overline{AB}$ is $M = \left(\frac{3}{2}, \sqrt{3}\right),$ and $AB = \sqrt{1 + 12} = \sqrt{13}.$ The third vertex lies at distance $\frac{\sqrt{3}}{2}\sqrt{13}$ from $M$ along a direction perpendicular to $\overrightarrow{AB} = (1, 2\sqrt{3});$ a unit perpendicular is $\frac{1}{\sqrt{13}}(2\sqrt{3}, -1).$ Taking the sign that lands in the first quadrant, $$\begin{aligned} C &= M + \frac{\sqrt{3}}{2}\,(2\sqrt{3}, -1) \\ &= \left(\frac{9}{2}, \frac{\sqrt{3}}{2}\right) \end{aligned}$$ (the other choice has negative $x$-coordinate).

The center of an equilateral triangle is its centroid, the average of the vertices: $$\begin{aligned} P &= \small \left(\frac{1 + 2 + \frac{9}{2}}{3},\ \frac{0 + 2\sqrt{3} + \frac{\sqrt{3}}{2}}{3} \right) \\ &= \left(\frac{5}{2}, \frac{5\sqrt{3}}{6}\right). \end{aligned}$$

Then $x \cdot y = \frac{5}{2} \cdot \frac{5\sqrt{3}}{6} = \frac{25\sqrt{3}}{12},$ so $p + q + r = 25 + 3 + 12 = 40.$

Answer: 20

Small Hint:

Take side length $6,$ so $BD = DE = EC = 2,$ and find $AD = AE$ with the law of cosines

Big Hint:

Triangle $ADE$ has one third the area of $ABC;$ compare that with $\frac{1}{2} \cdot AD \cdot AE \cdot \sin(\angle DAE)$

Solution:

Scale so the side length is $6,$ with $BD = DE = EC = 2.$ In triangle $AEC,$ the law of cosines gives $$\begin{aligned} AE^2 &= 6^2 + 2^2 - 2 \cdot 6 \cdot 2 \cos 60^\circ \\ &= 28, \end{aligned}$$ so $AE = 2\sqrt{7},$ and $AD = 2\sqrt{7}$ by symmetry.

Since $DE$ is one third of $BC$ and triangles $ADE$ and $ABC$ share the apex $A,$ we get $[ADE] = \frac{1}{3}[ABC]$ $= \frac{1}{3} \cdot \frac{36\sqrt{3}}{4}$ $= 3\sqrt{3}.$ On the other hand $[ADE] = \frac{1}{2} \cdot AD \cdot AE$ $\cdot \sin(\angle DAE)$ $= 14\sin(\angle DAE).$

Therefore $\sin(\angle DAE) = \frac{3\sqrt{3}}{14},$ and $a + b + c = 3 + 3 + 14 = 20.$

Answer: 282

Small Hint:

Consecutive squares differ by $2x + 1,$ so a block of $1000$ integers can be skipped only once $x \ge 500$

Big Hint:

Write $x = 500 + a,$ so $x^2 = 250000 + 1000a + a^2.$ Find the least $a$ with $a^2 + 2a + 1001 \ge 2000.$

Solution:

The block $\{1000N, \ldots, 1000N + 999\}$ misses all squares exactly when some consecutive squares $x^2$ and $(x+1)^2$ jump over it, which requires $(x+1)^2 - x^2 = 2x + 1 \gt 1000,$ so $x \ge 500.$ In particular every block below $500^2 = 250000$ contains a square, so we search from there.

Write $x = 500 + a$ with $a \ge 0.$ Then $x^2 = 1000(250 + a) + a^2,$ so as long as $a^2 \lt 1000$ (that is, $a \le 31$), the square $x^2$ lies in block $250 + a;$ these cover blocks $250$ through $281.$ Block $251 + a$ is skipped exactly when $$\begin{aligned} (x+1)^2 &= 1000(250 + a) + a^2 \\ &\quad {}+ 2a + 1001 \\ &\ge 1000(252 + a), \end{aligned}$$ that is, $a^2 + 2a \ge 999.$ For $a \le 30$ this fails (so $(x+1)^2$ lands in block $251 + a$), and it first holds at $a = 31,$ since $961 + 62 = 1023.$

Indeed $531^2 = 281961$ and $532^2 = 283024$ straddle the block starting at $282000.$ The least such $N$ is $251 + 31 = 282.$

Answer: 945

Small Hint:

With $n$ clerks at the start and $k$ removed at the end of each hour, add up the files sorted in each hour plus the final $10$ minutes

Big Hint:

The total reduces to $19n - 21k = 355;$ solve in positive integers by working modulo $19,$ keeping $n - 3k \gt 0$

Solution:

Let $n$ clerks start and $k$ be reassigned at the end of each hour. In the final $10$ minutes each remaining clerk sorts $5$ files, so $$\begin{aligned} &30n + 30(n - k) + 30(n - 2k) \\ &\quad {}+ 5(n - 3k) = 1775, \end{aligned}$$ which simplifies to $95n - 105k = 1775,$ or $19n - 21k = 355.$

Modulo $19$ this reads $-2k \equiv 13,$ so $2k \equiv 6$ and $k \equiv 3 \pmod{19}.$ Taking $k = 3$ gives $n = \frac{355 + 63}{19} = 22;$ the next candidate, $k = 22,$ gives $n = 43,$ for which $n - 3k \lt 0.$ So $n = 22$ and $k = 3.$

In the first hour $22$ clerks sort $30 \cdot 22 = 660$ files, and in the next half hour $19$ clerks sort $15 \cdot 19 = 285,$ for a total of $660 + 285 = 945.$

Answer: 272

Small Hint:

If each $22$-chord subtends central angle $\alpha$ and each $20$-chord subtends $\beta,$ then $4\alpha + 2\beta = 360^\circ,$ so $\sin\frac{\beta}{2} = \cos\alpha$

Big Hint:

The law of cosines gives $\cos\alpha = 1 - \frac{242}{r^2},$ while $\sin\frac{\beta}{2} = \frac{10}{r};$ equating them yields a quadratic in $r$

Solution:

Half a $20$-chord gives $\sin\frac{\beta}{2} = \frac{10}{r},$ and the law of cosines on the isosceles triangle with legs $r$ and base $22$ gives $22^2 = 2r^2(1 - \cos\alpha),$ so $\cos\alpha = 1 - \frac{242}{r^2}.$ Equating, $$\begin{aligned} 1 - \frac{242}{r^2} &= \frac{10}{r} \\ &\quad\Longrightarrow\quad r^2 - 10r - 242 \\ &= 0, \end{aligned}$$ so $r = 5 + \sqrt{267}$ (taking the positive root).

Therefore $p + q = 5 + 267 = 272.$

Answer: 106

Small Hint:

First count all tilings when $k$ colors are allowed: scan left to right, and note each new square either extends the current tile or starts a fresh tile

Big Hint:

That count is $k(k+1)^6.$ Then use inclusion-exclusion over which of the three colors go unused.

Solution:

First count colored tilings when $k$ colors are available. The first square’s tile can be colored in $k$ ways, and each of the remaining $6$ squares either extends the current tile or starts a new tile in one of the $k$ colors, giving $k + 1$ choices per square. So there are $k(k+1)^6$ tilings.

With three colors that is $3 \cdot 4^6 = 12288$ tilings. By inclusion-exclusion over the unused colors, the number using all three colors is $$\begin{aligned} N &= 3 \cdot 4^6 - 3 \cdot (2 \cdot 3^6) \\ &\quad {}+ 3 \cdot (1 \cdot 2^6) \\ &= 12288 - 4374 + 192 \\ &= 8106. \end{aligned}$$

The remainder when $N$ is divided by $1000$ is $106.$

Answer: 146

Small Hint:

$B$ lies on segment $OA$ with $AB = 4.$ For any line through $A,$ the distances from $O$ and from $B$ to that line are in the ratio $OA : AB$

Big Hint:

So $[BKL] = \frac{4}{4 + \sqrt{13}}[OKL],$ and $[OKL] = \frac{13}{2}\sin(\angle KOL)$ is largest when $\angle KOL = 90^\circ$

Solution:

The nearest point $B$ lies on segment $OA$ with $OB = \sqrt{13}$ and $AB = 4.$ Triangles $OKL$ and $BKL$ share the base $KL,$ and their heights are the distances from $O$ and $B$ to the line through $A.$ For any point $P$ on line $OA,$ that distance is $PA\sin\varphi,$ where $\varphi$ is the angle between the two lines, so $$\frac{[BKL]}{[OKL]} = \frac{AB}{AO} = \frac{4}{4 + \sqrt{13}}.$$

Since $OK = OL = \sqrt{13},$ we have $[OKL] = \frac{13}{2}\sin(\angle KOL) \le \frac{13}{2},$ with equality when $\angle KOL = 90^\circ.$ Such a chord lies at distance $\sqrt{\frac{13}{2}}$ from $O,$ which is less than $OA,$ so a line through $A$ can achieve it.

The maximum area is $$\begin{aligned} [BKL] &= \frac{13}{2} \cdot \frac{4}{4 + \sqrt{13}} \\ &= \frac{26}{4 + \sqrt{13}} = \frac{26(4 - \sqrt{13})}{3} \\ &= \frac{104 - 26\sqrt{13}}{3}, \end{aligned}$$ so $a + b + c + d$ $= 104 + 26 + 13 + 3$ $= 146.$

Answer: 399

Small Hint:

If $f(f(x)) = a$ always, let $S$ be the set of elements that $f$ sends to $a.$ Show $f(a) = a$ and every element outside $S$ maps into $S \setminus \{a\}$

Big Hint:

With $|S| = k:$ choose $a,$ choose the rest of $S,$ then send each of the $7 - k$ outside elements anywhere in $S \setminus \{a\},$ and sum over $k$

Solution:

Say $f(f(x)) = a$ for all $x,$ and let $S = \{x : f(x) = a\}.$ Picking any $t \in S,$ we get $a = f(f(t)) = f(a),$ so $a \in S.$ Every $x$ satisfies $f(x) \in S$ (because $f(f(x)) = a$), and if $x \notin S$ then $f(x) \ne a,$ so $f$ maps the complement of $S$ into $S \setminus \{a\}.$ Conversely, any $f$ built this way works.

If $|S| = k,$ we choose the constant $a$ in $7$ ways, the remaining $k - 1$ elements of $S$ in $\binom{6}{k-1}$ ways, and an image in $S \setminus \{a\}$ for each of the $7 - k$ other elements in $(k-1)^{7-k}$ ways. Hence $$\begin{aligned} N &= \scriptsize 7\sum_{k=1}^{7} \binom{6}{k-1}(k-1)^{7-k} \\ &\scriptsize = 7\,(0 + 6 + 240 + 540 + 240 + 30 + 1) \\ &= 7 \cdot 1057 = 7399. \end{aligned}$$

The remainder when $N$ is divided by $1000$ is $399.$

Answer: 540

Small Hint:

Nonreal roots come in conjugate pairs, so either all three roots are among $\pm 20$ and $\pm 13,$ or one real root is joined by a conjugate pair on one of the two circles

Big Hint:

For roots $r \pm s\mathrm{i}$ and real $k,$ expanding shows integer coefficients hold exactly when $2r$ is an integer; count the half-integers $r$ with $|r|$ less than the modulus

Solution:

A cubic with real coefficients has either three real roots or one real root and a conjugate pair. The only real numbers with modulus $20$ or $13$ are $\pm 20$ and $\pm 13,$ so in the all-real case the roots form a multiset of size $3$ from those $4$ values: $\binom{6}{3} = 20$ polynomials.

Otherwise the roots are $k \in \{\pm 20, \pm 13\}$ and a conjugate pair $r \pm s\mathrm{i}$ with $s \ne 0$ and $r^2 + s^2 = 400$ or $169.$ Expanding $$(z - k)\bigl(z^2 - 2rz + (r^2 + s^2)\bigr)$$ shows the coefficients are $-(2r + k),$ $r^2 + s^2 + 2rk,$ and $-(r^2 + s^2)k,$ which are all integers exactly when $2r$ is an integer. On the circle of radius $20$ we need $|r| \lt 20,$ allowing $2r \in \{-39, \ldots, 39\}:$ $79$ choices; on the circle of radius $13,$ $2r \in \{-25, \ldots, 25\}:$ $51$ choices. With $4$ choices of $k,$ that gives $4(79 + 51) = 520$ polynomials, each distinct since the roots determine the polynomial.

In total there are $20 + 520 = 540$ such polynomials.

Answer: 10

Small Hint:

Set $AB = 2x$ and $AC = BC = y;$ then $\cos B = \frac{x}{y},$ and the law of cosines in triangle $ABD$ gives $AD^2$ in terms of $x$ and $y$

Big Hint:

$CE$ and $BE$ are medians of triangles $ACD$ and $ABD$ to side $AD;$ the median formula $4m_a^2 = 2b^2 + 2c^2 - a^2$ gives two equations in $x^2$ and $y^2$

Solution:

Let $AB = 2x$ and $AC = BC = y,$ so $BD = \frac{y}{4},$ $CD = \frac{3y}{4},$ and $\cos B = \frac{x}{y}$ (drop the altitude from $C$ to the midpoint of $\overline{AB}$). The law of cosines in triangle $ABD$ gives $$\begin{aligned} AD^2 &= 4x^2 + \frac{y^2}{16} \\ &\quad {}- 2 \cdot 2x \cdot \frac{y}{4} \cdot \frac{x}{y} \\ &= 3x^2 + \frac{y^2}{16}. \end{aligned}$$

Both $CE$ and $BE$ are medians to $\overline{AD},$ in triangles $ACD$ and $ABD$ respectively. The median formula $4m^2 = 2b^2 + 2c^2 - a^2$ gives $$\begin{aligned} 28 &= 2y^2 + \frac{18y^2}{16} - AD^2 \\ &= \frac{49y^2}{16} - 3x^2, \end{aligned}$$ $$\begin{aligned} 36 &= 8x^2 + \frac{2y^2}{16} - AD^2 \\ &= 5x^2 + \frac{y^2}{16}. \end{aligned}$$ From the second equation $\frac{y^2}{16} = 36 - 5x^2;$ substituting into the first gives $49(36 - 5x^2) - 3x^2 = 28,$ so $248x^2 = 1736,$ $x^2 = 7,$ and then $y^2 = 16.$