Answer: 200

Small Hint:

Let $e$ be the center entry and $S$ the common sum

Big Hint:

In a $3\times3$ magic square, $S=3e$ and opposite entries sum to $2e$

Solution:

Let the center entry be $e$ and the common sum be $S.$ In any $3\times3$ magic square, $S=3e$ and entries opposite across the center sum to $2e.$ Thus the bottom-left entry is $2e-96.$ The first column and first row give $$\begin{aligned}x+1+(2e-96)&=3e,\\x+19+96&=3e.\end{aligned}$$ The first equation says $e=x-95,$ while the second says $3e=x+115.$ Hence $3x-285=x+115,$ so $x=200.$

Answer: 340

Small Hint:

Translate each possible value of $\lfloor\log_2n\rfloor$ into a power-of-two interval

Big Hint:

The possible even values below $1000$ are $2,$ $4,$ $6,$ and $8$

Solution:

If $\lfloor\log_2n\rfloor=k,$ then $2^k\leq n<2^{k+1}.$ The value of $k$ can be $2,$ $4,$ $6,$ or $8,$ since $2^{10}>1000.$ The corresponding interval sizes are $2^2,$ $2^4,$ $2^6,$ and $2^8.$ Therefore the requested number is $$4+16+64+256=340.$$

Answer: 44

Small Hint:

Factor the expression before raising it to the $n$th power

Big Hint:

Count the distinct choices of the exponents of $x$ and $y$

Solution:

The base factors as $$\begin{gathered}xy-3x+7y-21\\=(x+7)(y-3).\end{gathered}$$ Hence its $n$th power is $(x+7)^n(y-3)^n.$ Each exponent of $x$ from $0$ through $n$ can occur with each exponent of $y$ from $0$ through $n,$ and every resulting coefficient is nonzero. Thus there are $(n+1)^2$ terms. Since $44^2<1996\leq45^2,$ the least possible $n+1$ is $45,$ so $n=44.$

Answer: 166

Small Hint:

Project the cube’s upper face onto the horizontal surface from the light source

Big Hint:

Similar triangles give the projected side length as $\frac{x+1}{x}$

Solution:

The light is $x+1$ centimeters above the surface and $x$ centimeters above the cube’s top face. By similarity, the projection of that unit-square face is a square of side $\frac{x+1}{x}.$ This square contains the unit-square area beneath the cube, so $$\left(\frac{x+1}{x}\right)^2-1=48.$$ Therefore $\frac{x+1}{x}=7,$ giving $x=\frac{1}{6}.$ The greatest integer not exceeding $1000x=166\frac23$ is $166.$

Answer: 23

Small Hint:

Use Vieta’s formulas on the original cubic

Big Hint:

Expand $(a+b)(b+c)(c+a)$ in symmetric sums

Solution:

Vieta’s formulas give $$\begin{aligned}a+b+c&=-3,\\ab+bc+ca&=4,\\abc&=11.\end{aligned}$$ Also, $$\begin{gathered}(a+b)(b+c)(c+a)\\=(a+b+c)(ab+bc+ca)\\\quad-abc=-23.\end{gathered}$$ This is the product of the roots of the second monic cubic, so its constant term is the negative of that product. Hence $t=23.$

Answer: 49

Small Hint:

There are $2^{\binom52}$ equally likely tournament outcomes

Big Hint:

Use inclusion-exclusion on the events that an undefeated or a winless team exists

Solution:

There are $2^{10}=1024$ outcomes. A specified undefeated team forces its four games and leaves the other six arbitrary, so there are $5\cdot2^6=320$ outcomes with an undefeated team. The same count holds for a winless team. If distinct specified teams are undefeated and winless, seven games are forced and the three games among the other teams are arbitrary. Thus the intersection count is $5\cdot4\cdot2^3=160.$ By inclusion-exclusion, the desired count is $$1024-320-320+160=544.$$ The probability is $\frac{544}{1024}=\frac{17}{32},$ so $m+n=49.$

Answer: 300

Small Hint:

Average the numbers of two-square colorings fixed by the four rotations

Big Hint:

Only a half-turn can fix a nontrivial pair of squares

Solution:

Under the identity rotation, all $\binom{49}{2}=1176$ pairs are fixed. A quarter-turn or three-quarter-turn has only orbits of sizes $1$ and $4,$ so it fixes no two-square set. A half-turn fixes exactly the $24$ pairs of squares opposite one another across the center. Burnside’s Lemma therefore gives $$\frac{1176+0+24+0}{4}=300$$ inequivalent colorings.

Answer: 799

Small Hint:

For $N=6^{20},$ rearrange $\frac{2xy}{x+y}=N$ into a product

Big Hint:

Count complementary factor pairs of $N^2$ in which both factors are even

Solution:

Let $N=6^{20}.$ Rearranging the harmonic-mean equation gives $$(2x-N)(2y-N)=N^2.$$ Because $x<N<y$ and $N<2x,$ the two factors are positive, and both must be even. Conversely, each factorization $AB=N^2$ with even $A<B$ gives one valid pair via $x=\frac{A+N}{2}$ and $y=\frac{B+N}{2}.$

Now $N^2=2^{40}3^{40}.$ For both complementary factors to be even, the exponent of $2$ in $A$ can be $1,\ldots,39,$ while the exponent of $3$ can be $0,\ldots,40.$ This gives $39\cdot41=1599$ divisors $A,$ including the central factor $A=N.$ Pairing complementary divisors and excluding that central case gives $\frac{1599-1}{2}=799.$

Answer: 342

Small Hint:

After each trip, the still-closed lockers form an arithmetic sequence

Big Hint:

Record only the first term, common difference, and number of terms after each trip

Solution:

On every trip the student opens the first, third, fifth, and so on among the remaining lockers in his direction of travel. Tracking the closed arithmetic sequence after each trip gives:

Thus only lockers $342$ and $854$ remain after the ninth trip. On the tenth trip, starting from the right, locker $854$ is opened and $342$ remains. Therefore the last locker opened is $342.$

Answer: 159

Small Hint:

Recognize the right-hand side using the tangent addition formula

Big Hint:

Solve the resulting congruence modulo $180$

Solution:

The tangent addition formula gives $$\begin{gathered}\frac{\cos96^\circ+\sin96^\circ}{\cos96^\circ-\sin96^\circ}\\=\tan(45^\circ+96^\circ)\\=\tan141^\circ.\end{gathered}$$ Hence $19x\equiv141\pmod {180}.$ Since $19^2=361\equiv1\pmod {180},$ multiplying by $19$ gives $$x\equiv19\cdot141\equiv159\pmod {180}.$$ The smallest positive solution is $159.$

Answer: 276

Small Hint:

Divide by $z^3$ and set $w=z+z^{-1}$

Big Hint:

Factor the resulting cubic in $w$ and identify each value as $2\cos\phi$

Solution:

No root is zero. Dividing by $z^3$ and setting $w=z+z^{-1}$ gives $$\begin{aligned}w^3-2w+1&=0,\\{}(w-1)(w^2+w-1)&=0.\end{aligned}$$ Its three roots are $$\begin{aligned}1&=2\cos60^\circ,\\\frac{\sqrt5-1}{2}&=2\cos72^\circ,\\-\frac{1+\sqrt5}{2}&=2\cos144^\circ.\end{aligned}$$ For each value $w=2\cos\phi,$ the corresponding roots of the original equation are $e^{i\phi}$ and $e^{-i\phi}.$ Thus the roots with positive imaginary part have arguments $60^\circ,$ $72^\circ,$ and $144^\circ.$ Their product has argument $60+72+144=276^\circ,$ so $\theta=276.$

Answer: 58

Small Hint:

Each of the five absolute differences has the same average

Big Hint:

For a random unordered pair, difference $d$ occurs $10-d$ times

Solution:

Each paired difference has the same expected value as the difference of a uniformly selected unordered pair from $1,\ldots,10.$ Therefore $$\begin{aligned}\mathbb E|a_1-a_2|&=\frac{\sum_{d=1}^9d(10-d)}{\binom{10}{2}}\\&=\frac{10\sum_{d=1}^9d}{45}\\&\quad-\frac{\sum_{d=1}^9d^2}{45}\\&=\frac{450-285}{45}=\frac{11}{3}.\end{aligned}$$ By linearity of expectation, the average of the five-term sum is $\frac{5\cdot11}{3}=\frac{55}{3}.$ Thus $p+q=58.$

Answer: 65

Small Hint:

Let $E$ be the midpoint of $\overline{BC},$ so $A,$ $E,$ and $D$ are collinear

Big Hint:

Find $AE$ with the median formula, then compare two right triangles sharing $BD$

Solution:

Let $E$ be the midpoint of $\overline{BC}.$ Then $A,$ $E,$ and $D$ are collinear. The median formula gives $$\begin{aligned}AE^2&=\frac{2AB^2+2AC^2-BC^2}{4}\\&=\frac{57}{4}.\end{aligned}$$ Since $AB^2>AE^2+BE^2,$ angle $AEB$ is obtuse, so the perpendicular foot $D$ lies beyond $E.$ Both $\triangle ABD$ and $\triangle EBD$ are right at $D.$ Therefore $$\begin{gathered}AB^2-BE^2\\=(AE+DE)^2-DE^2\\=AE^2+2AE\cdot DE.\end{gathered}$$ Substitution gives $30-\frac{15}{4}=\frac{57}{4}+2AE\cdot DE,$ so $\frac{DE}{AE}=\frac{8}{19}.$

Triangles $ABE$ and $DBE$ have bases $AE$ and $DE$ on the same line and share the altitude from $B,$ while $[ABC]=2[ABE].$ Hence $$\begin{aligned}\frac{[ADB]}{[ABC]}&=\frac{[ABE]+[DBE]}{2[ABE]}\\&=\frac12\left(1+\frac8{19}\right)\\&=\frac{27}{38}.\end{aligned}$$ Thus $m+n=65.$

Answer: 768

Small Hint:

Count the coordinate-plane crossings of the space diagonal

Big Hint:

Correct for crossings of two or three grid planes at once using greatest common divisors

Solution:

For an $a\times b\times c$ array, the diagonal crosses $a-1,$ $b-1,$ and $c-1$ internal grid planes of the three orientations. Crossings of two orientations coincide $\gcd(a,b)-1,$ $\gcd(a,c)-1,$ and $\gcd(b,c)-1$ times, and triple crossings occur $\gcd(a,b,c)-1$ times. Adding one for the initial cube and applying inclusion-exclusion gives $$\begin{gathered}a+b+c-\gcd(a,b)\\-\gcd(a,c)-\gcd(b,c)\\+\gcd(a,b,c).\end{gathered}$$ Here the pairwise gcds are $6,$ $75,$ and $3,$ and the triple gcd is $3.$ Thus the number of cube interiors met is $$\begin{gathered}150+324+375\\-6-75-3+3=768.\end{gathered}$$

Answer: 777

Small Hint:

Set $\angle DBA=\alpha$ and compare triangles $ABC$ and $ABD$

Big Hint:

Use the law of sines to obtain an equation involving $\sin5\alpha,$ $\sin2\alpha,$ and $\sin\alpha$

Solution:

Let $\angle DBA=\alpha.$ Then $\angle DBC=\angle CAB=2\alpha,$ so in $\triangle ABC$ the angles are $2\alpha,$ $3\alpha,$ and $180^\circ-5\alpha.$ Because $AD\parallel BC,$ triangle $ABD$ has angles $\alpha,$ $2\alpha,$ and $180^\circ-3\alpha.$

Applying the law of sines in the two triangles and using $AD=BC$ gives $$\frac{AB}{BC}=\frac{\sin5\alpha}{\sin2\alpha}=\frac{\sin2\alpha}{\sin\alpha}.$$ With $u=\cos^2\alpha,$ this becomes $$16u^2-16u+1=0.$$ Since $5\alpha<180^\circ,$ the valid root is $u=\frac{2+\sqrt3}{4}=\cos^2 15^\circ,$ so $\alpha=15^\circ.$ Thus $\angle ACB=105^\circ.$ In $\triangle AOB,$ the angles at $A$ and $B$ are $30^\circ$ and $15^\circ,$ so $\angle AOB=135^\circ.$ Therefore $$\begin{aligned}r&=\frac{105}{135}=\frac79,\\\left\lfloor1000r\right\rfloor&=777.\end{aligned}$$