Answer: 630

Small Hint:

If the number is $10a + b,$ divisibility by the tens digit $a$ forces $b$ to be a multiple of $a$

Big Hint:

Write $b = ka.$ Then $10a + b$ divisible by $b$ forces $k = 1,$ $2,$ or $5$

Solution:

Let the number be $10a + b$ with tens digit $a$ and units digit $b.$ Since $10a + b$ is divisible by $a,$ we need $b$ to be divisible by $a,$ so $b = ka$ for some positive integer $k.$ Since $10a + b$ is divisible by $b,$ we need $10a$ to be divisible by $b,$ that is $10a$ is divisible by $ka,$ so $10$ is divisible by $k.$ Because $b = ka \le 9,$ only $k = 1,$ $2,$ and $5$ are possible.

For $k = 1$ the numbers are $11, 22, \ldots, 99,$ with sum $11 \cdot 45 = 495.$ For $k = 2$ they are $12, 24, 36, 48,$ with sum $120.$ For $k = 5$ the only one is $15.$

The total is $495 + 120 + 15 = 630.$

Answer: 651

Small Hint:

Let $\mathcal{S}$ have $n$ elements with mean $x,$ and translate each condition into an equation

Big Hint:

The conditions give $x - 13(n+1) = 1$ and $x + 27(n+1) = 2001.$ Subtract them.

Solution:

Let $\mathcal{S}$ have $n$ elements with mean $x,$ so the elements sum to $nx.$ The two conditions say $$\frac{nx + 1}{n + 1} = x - 13$$ and $$\frac{nx + 2001}{n + 1} = x + 27.$$

Clearing denominators, $nx + 1 = (n+1)x - 13(n+1)$ gives $x - 13(n+1) = 1,$ and $nx + 2001$ $= (n+1)x + 27(n+1)$ gives $x + 27(n+1) = 2001.$ Subtracting the first equation from the second yields $40(n+1) = 2000,$ so $n + 1 = 50.$

Then $x = 1 + 13 \cdot 50 = 651.$

Answer: 500

Small Hint:

Expanding $\left(\frac{1}{2} - x\right)^{2001}$ cancels the $x^{2001}$ term, leaving a polynomial of degree $2000$

Big Hint:

The sum of the roots is minus the coefficient of $x^{1999}$ divided by the coefficient of $x^{2000}$

Solution:

Expand $\left(\frac{1}{2} - x\right)^{2001}$ by the binomial theorem. Its leading term $(-x)^{2001} = -x^{2001}$ cancels the $x^{2001}$ in the equation, so what remains is a polynomial of degree $2000:$ $$\begin{aligned} &2001 \cdot \frac{1}{2}\,x^{2000} \\ &\quad {}- \binom{2001}{2}\frac{1}{4}\,x^{1999} + \cdots = 0. \end{aligned}$$

By Vieta’s formulas, the sum of the $2000$ roots is $$\begin{aligned} \frac{\frac{\binom{2001}{2}}{4}}{\frac{2001}{2}} &= \frac{\frac{2001 \cdot 2000}{8}}{\frac{2001}{2}} \\ &= \frac{2000}{4} = 500. \end{aligned}$$

Answer: 291

Small Hint:

$\angle C = 75^\circ$ and $\angle ATC = 75^\circ$ too, so $AC = AT = 24$

Big Hint:

Drop the altitude from $C$ to $\overline{AB}:$ it cuts $ABC$ into a $30$-$60$-$90$ triangle and a $45$-$45$-$90$ triangle

Solution:

Since $\angle A = 60^\circ$ and $\angle B = 45^\circ,$ we have $\angle C = 75^\circ.$ In triangle $ATC,$ angle $TAC = 30^\circ$ (half of angle $A$), so $\angle ATC = 180^\circ - 30^\circ - 75^\circ$ $= 75^\circ.$ Thus triangle $ACT$ is isosceles with $AC = AT = 24.$

Drop the altitude $CH$ to $\overline{AB}.$ Triangle $ACH$ is $30$-$60$-$90,$ so $AH = 12$ and $CH = 12\sqrt{3}.$ Triangle $BCH$ is $45$-$45$-$90,$ so $BH = CH = 12\sqrt{3}.$

The area is $\frac{1}{2} \cdot CH \cdot AB$ $= \frac{1}{2} \cdot 12\sqrt{3}\,(12 + 12\sqrt{3})$ $= 216 + 72\sqrt{3}.$ Then $a + b + c = 216 + 72 + 3 = 291.$

Answer: 937

Small Hint:

By symmetry the other two vertices are $(\pm x, y),$ and the side through $(0, 1)$ makes a $120^\circ$ angle with the positive $x$-axis

Big Hint:

Substitute $y = -\sqrt{3}\,x + 1$ into $x^2 + 4y^2 = 4$ and solve for $x;$ the side length is $2x$

Solution:

Since one altitude lies along the $y$-axis, the other two vertices are symmetric: $(x, y)$ and $(-x, y)$ with $x \gt 0.$ The side from $(0,1)$ to $(x,y)$ makes a $120^\circ$ angle with the positive $x$-axis, so it lies on the line $y = -\sqrt{3}\,x + 1.$

Substituting into $x^2 + 4y^2 = 4$ gives $x^2 + 4(1 - \sqrt{3}x)^2 = 4,$ which simplifies to $13x^2 - 8\sqrt{3}\,x = 0,$ so $x = \frac{8\sqrt{3}}{13}.$

The side length is $2x = \frac{16\sqrt{3}}{13},$ whose square is $\frac{768}{169}.$ Since $\gcd(768, 169) = 1,$ the answer is $768 + 169 = 937.$

Answer: 79

Small Hint:

Every collection of four values (with repeats allowed) can be arranged in exactly one non-decreasing order

Big Hint:

Count multisets of size $4$ from $\{1, \ldots, 6\}$ by stars and bars: $\binom{9}{4}$

Solution:

The rolls must form a non-decreasing sequence. Every multiset of four values from $\{1, \ldots, 6\}$ can be arranged in non-decreasing order in exactly one way, so the number of successful outcomes equals the number of such multisets. By stars and bars (4 stars and $5$ dividers), that count is $\binom{9}{4} = 126.$

The probability is $\frac{126}{6^4} = \frac{126}{1296} = \frac{7}{72},$ so $m + n = 7 + 72 = 79.$

Answer: 923

Small Hint:

Triangles $ADE$ and $ABC$ are similar with ratio $\frac{h - r}{h},$ where $h$ is the height from $A$ and $r$ is the inradius

Big Hint:

$r = \frac{K}{s}$ and $h = \frac{2K}{BC}$ where $K$ is the area, so $\frac{r}{h} = \frac{BC}{2s}$ — no need to compute $K$

Solution:

Since $\overline{DE} \parallel \overline{BC},$ triangles $ADE$ and $ABC$ are similar, and the ratio equals the ratio of their heights from $A.$ The line $DE$ passes through the incenter, which sits at height $r$ (the inradius) above $BC,$ so the ratio is $\frac{h - r}{h} = 1 - \frac{r}{h},$ where $h$ is the height from $A$ to $\overline{BC}.$

If $K$ is the area and $s = \frac{21 + 22 + 20}{2} = \frac{63}{2}$ the semiperimeter, then $r = \frac{K}{s}$ and $h = \frac{2K}{20},$ so $$\frac{r}{h} = \frac{20}{2s} = \frac{20}{63}.$$

Therefore $DE = 20\left(1 - \frac{20}{63}\right)$ $= 20 \cdot \frac{43}{63}$ $= \frac{860}{63},$ which is in lowest terms, and $m + n = 860 + 63 = 923.$

Answer: 315

Small Hint:

If $N$ has base-$7$ digits $d_2 d_1 d_0,$ the condition says $100 d_2 + 10 d_1 + d_0$ $= 2(49 d_2 + 7 d_1 + d_0)$

Big Hint:

That simplifies to $2 d_2 = 4 d_1 + d_0.$ A fourth digit is impossible because $10^3 - 2 \cdot 7^3 = 314$ is bigger than the other terms could balance.

Solution:

Suppose $N$ has base-$7$ digits $d_k \ldots d_1 d_0.$ The condition is $\sum d_i \, 10^i = 2 \sum d_i \, 7^i,$ that is $\sum d_i \,(10^i - 2 \cdot 7^i) = 0.$ The coefficients $10^i - 2 \cdot 7^i$ for $i = 0, 1, 2, 3$ are $-1,$ $-4,$ $2,$ $314.$ Every coefficient for $i\ge3$ is positive and at least $314.$ Thus if any digit in a place $i\ge3$ were nonzero, its positive contribution could not be canceled by the two negative terms, whose total magnitude is at most $4\cdot6+6=30.$ So $N$ has at most three base-$7$ digits.

For three digits the condition reads $2 d_2 = 4 d_1 + d_0.$ To maximize $N = 49 d_2 + 7 d_1 + d_0,$ take $d_2 = 6,$ so $4 d_1 + d_0 = 12;$ the largest value of $7 d_1 + d_0$ comes from $d_1 = 3,$ $d_0 = 0.$

Thus $N = 49 \cdot 6 + 7 \cdot 3 = 315,$ whose base-$7$ representation is $630 = 2 \cdot 315.$

Answer: 61

Small Hint:

Subtract the three corner triangles: $\frac{[DEF]}{[ABC]} = 1$ $- p(1-r)$ $- q(1-p)$ $- r(1-q)$

Big Hint:

$pq + qr + rp = \frac{(p+q+r)^2 - (p^2+q^2+r^2)}{2}$

Solution:

Each corner triangle’s area is a product of side fractions: $[ADF] = p(1-r)[ABC],$ $[BED] = q(1-p)[ABC],$ and $[CFE] = r(1-q)[ABC],$ using the formula $\frac{1}{2}xy\sin\theta$ on the shared angles. Subtracting, $$\begin{aligned} \frac{[DEF]}{[ABC]} &= 1 - p(1-r) \\ &\quad {}- q(1-p) - r(1-q) \\ &= 1 - (p+q+r) \\ &\quad {}+ (pq+qr+rp). \end{aligned}$$

From the given values, $pq + qr + rp = \frac{(\frac{2}{3})^2 - \frac{2}{5}}{2}$ $= \frac{\frac{4}{9} - \frac{2}{5}}{2}$ $= \frac{1}{45}.$

Therefore the ratio is $1 - \frac{2}{3} + \frac{1}{45} = \frac{16}{45},$ and $m + n = 16 + 45 = 61.$

Answer: 200

Small Hint:

The midpoint has integer coordinates exactly when the two points have the same parity in each coordinate

Big Hint:

Count ordered same-parity pairs coordinate by coordinate ($5$ ways for $x,$ $8$ for $y,$ $13$ for $z$), then subtract the pairs where both points coincide

Solution:

The midpoint is a lattice point exactly when the two chosen points agree in parity in each coordinate. Count ordered pairs (allowing equality) coordinate by coordinate. For $x \in \{0, 1, 2\}$ there are $2$ even and $1$ odd values, giving $2^2 + 1^2 = 5$ same-parity ordered pairs. For $y \in \{0, \ldots, 3\}:$ $2^2 + 2^2 = 8.$ For $z \in \{0, \ldots, 4\}:$ $3^2 + 2^2 = 13.$

That gives $5 \cdot 8 \cdot 13 = 520$ ordered pairs, including the $60$ pairs where the two points are equal, so $520 - 60 = 460$ ordered pairs of distinct points, or $230$ unordered pairs. The total number of unordered pairs is $\binom{60}{2} = 1770.$

The probability is $\frac{230}{1770} = \frac{23}{177},$ and since $177 = 3 \cdot 59,$ this is in lowest terms. Thus $m + n = 23 + 177 = 200.$

Answer: 149

Small Hint:

If $P_i$ is in column $c_i,$ then $x_i = (i-1)N + c_i$ and $y_i = 5(c_i - 1) + i$

Big Hint:

The first two equations give $N = 24c_2 - 19;$ the other three force a divisibility condition on $c_2.$ Solve it modulo $31.$

Solution:

Let $P_i$ sit in column $c_i.$ Then $x_i = (i-1)N + c_i$ and $y_i = 5(c_i - 1) + i.$ The five conditions become $$\begin{aligned} c_1 &= 5c_2 - 3, \\ N + c_2 &= 5c_1 - 4, \\ 2N + c_3 &= 5c_4 - 1, \end{aligned}$$ $$\begin{aligned} 3N + c_4 &= 5c_5, \\ 4N + c_5 &= 5c_3 - 2. \end{aligned}$$

Substituting $c_1 = 5c_2 - 3$ into the second equation gives $N = 24c_2 - 19.$ Eliminating $c_3$ and $c_4$ from the last three equations yields $124 c_5 = 89N + 7.$ Substituting $N = 24c_2 - 19$ and reducing, $534 c_2 - 421$ is divisible by $31,$ i.e. $7 c_2 \equiv 18 \pmod{31},$ whose smallest positive solution is $c_2 = 7.$

Then $N = 24 \cdot 7 - 19 = 149,$ and back-substituting gives valid columns $(c_1, \ldots, c_5)=$ $(32, 7, 141, 88, 107),$ all at most $149.$ Indeed the corresponding pairs are $(x_1,y_2)=(32,32),$ $(x_2,y_1)=(156,156),$ $(x_3,y_4)=(439,439),$ $(x_4,y_5)=(535,535),$ and $(x_5,y_3)=(703,703).$ So the smallest possible $N$ is $149.$

Answer: 5

Small Hint:

Decompose the tetrahedron into four pyramids with apex at the insphere center: $V = \frac{1}{3} r S$ where $S$ is the total surface area

Big Hint:

$V = \frac{1}{6} \cdot 6 \cdot 4 \cdot 2,$ the three right-angle faces have areas $12,$ $6,$ $4,$ and face $ABC$ has area $14$

Solution:

Connecting the incenter to the four vertices splits the tetrahedron into four pyramids of height $r$ over the faces, so $V = \frac{1}{3} r S,$ i.e. $r = \frac{3V}{S},$ where $S$ is the total surface area.

Here $V = \frac{1}{6} \cdot 6 \cdot 4 \cdot 2 = 8.$ The three faces on the coordinate planes are right triangles with areas $\frac{1}{2} \cdot 6 \cdot 4 = 12,$ $\frac{1}{2} \cdot 6 \cdot 2 = 6,$ and $\frac{1}{2} \cdot 4 \cdot 2 = 4.$ For face $ABC,$ the cross product of $\overrightarrow{AB} = (-6, 4, 0)$ and $\overrightarrow{AC} = (-6, 0, 2)$ is $(8, 12, 24),$ with length $4\sqrt{2^2 + 3^2 + 6^2} = 28,$ so that face has area $14.$

Then $S = 12 + 6 + 4 + 14 = 36$ and $r = \frac{24}{36} = \frac{2}{3},$ giving $m + n = 2 + 3 = 5.$

Answer: 174

Small Hint:

A chord of a $\theta$-degree arc has length $2R\sin\frac{\theta}{2}.$ Set $t = \frac{d}{2}$ and use $\sin 2t = 2\sin t\cos t,$ $\sin 3t = \sin t\,(4\cos^2 t - 1).$

Big Hint:

Divide everything by $2R \sin t = 22$ to get $44\cos^2 t - 22\cos t - 1 = 0,$ and note the $3d$-degree chord equals $44 \cos t - 20$

Solution:

A chord subtending a $\theta$-degree arc in a circle of radius $R$ has length $2R\sin\frac{\theta}{2}.$ Write $t = \frac{d}{2},$ so the three chords are $2R\sin t = 22,$ $2R \sin 2t,$ and $2R \sin 3t.$ Using $\sin 2t = 2 \sin t \cos t$ and $\sin 3t = \sin t\,(4\cos^2 t - 1),$ the chords of the $2d$- and $3d$-degree arcs are $22 \cdot 2\cos t = 44 \cos t$ and $22\,(4\cos^2 t - 1).$