Small Hint:

Multiply consecutive terms using the recurrence

Big Hint:

Group the requested product as $(x_1x_2)(x_3x_4)(x_5x_6)(x_7x_8)$

Solution:

The recurrence gives $x_{n-1}x_n=n.$ Therefore $$\begin{aligned} x_1x_2\cdots x_8 &=(x_1x_2)(x_3x_4)\\ &\quad{}\cdot(x_5x_6)(x_7x_8)\\ &=2\cdot4\cdot6\cdot8\\ &=384. \end{aligned}$$

Small Hint:

Write the two cone volumes in terms of the triangle’s legs $a$ and $b$

Big Hint:

Dividing the volume equations gives the ratio of the two legs

Solution:

Let the legs be $a$ and $b.$ In a suitable order, $$\begin{aligned} \frac13\pi b^2a&=800\pi,\\ \frac13\pi a^2b&=1920\pi. \end{aligned}$$ Their ratio gives $\frac{a}{b}=\frac{12}{5},$ so write $a=12k$ and $b=5k.$ The first equation becomes $100k^3=800,$ hence $k=2.$ The legs are $24$ and $10,$ so the hypotenuse is $\sqrt{24^2+10^2}=26.$

Small Hint:

Expand $(a+bi)^3$ and set its imaginary part equal to $107$

Big Hint:

The resulting equation shows that the positive integer $b$ divides $107$

Solution:

Expanding and using that $c$ is real gives $$b(3a^2-b^2)=107.$$ Since $107$ is prime, $b=1$ or $107.$ The latter would require $3a^2=11450,$ which is impossible. Thus $b=1,$ and $3a^2-1=107,$ so $a=6.$ The real part is $$c=a^3-3ab^2=216-18=198.$$

Small Hint:

Place the unit square on a coordinate plane and write equations for two parallel construction lines

Big Hint:

The distance between the parallel lines is the side length of the small square

Solution:

Put the outer square at $(0,0),$ $(1,0),$ $(1,1),$ $(0,1).$ One pair of construction lines has equations $$\begin{aligned} nx-(n-1)y&=0,\\ nx-(n-1)y&=1. \end{aligned}$$ The other pair is perpendicular to this pair, and the two pairs have the same separation. Thus the inner square has side length $$\frac1{\sqrt{n^2+(n-1)^2}}$$ and area $\frac{1}{n^2+(n-1)^2}.$ Hence $$n^2+(n-1)^2=1985,$$ or $n^2-n-992=0.$ Its positive root is $n=\frac{1+63}{2}=32.$

Small Hint:

Write the first six terms in terms of $a_1$ and $a_2$

Big Hint:

The sequence repeats every six terms, and each six-term block has sum $0$

Solution:

Put $a_1=x$ and $a_2=y.$ The first six terms are $$x,\ y,\ y-x,\ -x,\ -y,\ x-y,$$ after which the sequence repeats; these six terms sum to $0.$ Since $1492\equiv4\pmod6$ and $1985\equiv5\pmod6,$ the given equations are $$\begin{aligned} 2y-x&=1985,\\ y-x&=1492. \end{aligned}$$ Thus $y=493.$ Since $2001\equiv3\pmod6,$ the requested sum is $x+y+(y-x)=2y=986.$

Small Hint:

Call the upper-right unlabeled area $x$ and the upper-left unlabeled area $y$

Big Hint:

Use equal-altitude area ratios along the sides, together with Ceva’s theorem

Solution:

Let the unlabeled upper-right and upper-left areas be $x$ and $y,$ respectively. The three side-division ratios and Ceva’s theorem give $$\frac43\cdot\frac{35}{x}\cdot\frac{84}{y}=1,$$ so $xy=3920.$

The cevian from $A$ meets $BC.$ The ratio of the two segments of $BC$ is $\frac{35}{x}.$ Computing the same ratio from the two large triangles with vertex $A$ gives $$\frac{35}{x}=\frac{40+30+35}{x+y+84},$$ so $y+84=2x.$ Solving with $xy=3920$ yields $x=70$ and $y=56.$ Therefore $$\begin{aligned} [ABC]&=84+70+35\\ &\quad{}+30+40+56=315. \end{aligned}$$

Small Hint:

Parametrize the solutions of $a^5=b^4$ and $c^3=d^2$

Big Hint:

Factor the resulting difference $s^2-t^4$

Solution:

Comparing prime exponents, write $$\begin{aligned} a&=t^4,\quad b=t^5,\\ c&=s^2,\quad d=s^3 \end{aligned}$$ for positive integers $s,$ $t.$ Then $$(s-t^2)(s+t^2)=s^2-t^4=19.$$ Since $19$ is prime, the factors are $1$ and $19,$ giving $s=10$ and $t^2=9.$ Thus $t=3,$ and $$\begin{aligned} d-b&=10^3-3^5\\ &=1000-243=757. \end{aligned}$$

Small Hint:

Starting from seven $3$’s, the integer sum must be reduced by $2$

Big Hint:

To minimize the worst error, round the two smallest $a_i$’s down to $2$

Solution:

Choose $A_1=A_2=2$ and $A_3=\cdots=A_7=3.$ The sum is $19,$ and the largest error is $|A_2-a_2|=0.61.$

If $M\lt0.61,$ then $A_2,\ldots,A_7$ must all equal $3,$ and $A_1$ can only be $2$ or $3.$ Their sum would therefore be at least $20,$ a contradiction. Thus the minimum is $M=0.61,$ and $100M=61.$

Small Hint:

A chord subtending angle $\theta$ has length $2R\sin(\frac{\theta}{2})$

Big Hint:

Let $x=\cos(\frac{\alpha}{2})$ and $y=\cos(\frac{\beta}{2}),$ then eliminate the common radius

Solution:

Put $k=\frac{1}{2R},$ $x=\cos(\frac{\alpha}{2}),$ and $y=\cos(\frac{\beta}{2}).$ The chord data give $$\begin{aligned} \sin\frac\alpha2&=2k,\\ \sin\frac\beta2&=3k,\\ \sin\frac{\alpha+\beta}{2}&=4k. \end{aligned}$$ The addition formula yields $2y+3x=4.$ Also $$\frac{1-x^2}{4}=\frac{1-y^2}{9},$$ so $9x^2-4y^2=5.$ Substituting $y=\frac{4-3x}{2}$ gives $x=\frac{7}{8}.$ Hence $$\cos\alpha=2x^2-1=\frac{17}{32},$$ and the requested sum is $17+32=49.$

Small Hint:

Substitute $y=2x$ and separate $y$ into its integer and fractional parts

Big Hint:

Determine the values of $\lfloor r\rfloor+\lfloor2r\rfloor+\lfloor3r\rfloor+\lfloor4r\rfloor$ for $0\leq r\lt1$

Solution:

Let $y=2x=m+r,$ where $m$ is an integer and $0\leq r\lt1.$ The expression is $$10m+\lfloor r\rfloor+\lfloor2r\rfloor+\lfloor3r\rfloor+\lfloor4r\rfloor.$$ Checking the breakpoints $r=\frac{1}{4},$ $\frac{1}{3},$ $\frac{1}{2},$ $\frac{2}{3},$ $\frac{3}{4}$ shows that the fractional-part contribution takes exactly the values $0,$ $1,$ $2,$ $4,$ $5,$ and $6.$ Thus precisely six residue classes modulo $10$ are attainable. Among the first $1000$ positive integers, this gives $100\cdot6=600.$

Small Hint:

At tangency, the constant sum of distances is the minimum such sum for a point on the $x$-axis

Big Hint:

Reflect one focus across the $x$-axis to turn the broken path into a straight segment

Solution:

Reflect $(49,55)$ across the $x$-axis to $(49,-55).$ For a point $P$ on the $x$-axis, the sum of its distances to the original foci equals the length of a broken path from $(9,20)$ through $P$ to $(49,-55).$ Its minimum is the straight-line distance $$\begin{aligned} &\sqrt{(49-9)^2+(-55-20)^2}\\ &\qquad{}=\sqrt{40^2+75^2}=85. \end{aligned}$$ Tangency means that the ellipse’s constant distance sum equals this minimum. That sum is the major-axis length, so the answer is $85.$

Small Hint:

Let $p_k$ be the probability of being at $A$ after $k$ steps

Big Hint:

From any vertex other than $A,$ the probability of moving to $A$ is $\frac{1}{3}$

Solution:

Let $p_k$ be the probability that the bug is at $A$ after $k$ steps. It cannot stay at $A,$ while from any other vertex it moves to $A$ with probability $\frac{1}{3}.$ Hence $$p_{k+1}=\frac{1-p_k}{3},\qquad p_0=1.$$ Solving this recurrence gives $$p_k=\frac14+\frac34\left(-\frac13\right)^k.$$ Thus $p_7=\frac14-\frac{3}{4\cdot3^7}=\frac{182}{729},$ so $n=182.$

Small Hint:

A common divisor of consecutive terms also divides their difference $2n+1$

Big Hint:

Combine $n^2+100$ and $2n+1$ to show that the gcd divides a fixed prime

Solution:

A common divisor $d_n$ divides $$a_{n+1}-a_n=2n+1.$$ It therefore also divides $$\begin{aligned} &4(n^2+100)-(2n+1)^2\\ &\qquad{}+2(2n+1)=401. \end{aligned}$$ Since $401$ is prime, $d_n\leq401.$ Equality occurs at $n=200,$ because $2n+1=401$ and $n^2+100=40100=100\cdot401.$ Hence the maximum is $401.$

Small Hint:

Sum the scores of the ten lowest-scoring players and count their internal games

Big Hint:

Let $m$ be the number of other players and double-count points from games across the two groups

Solution:

The games among the ten lowest players contribute $\binom{10}{2}=45$ total points. These are half of those ten players’ combined score, so their combined score is $90.$ Hence they earned $45$ points in games against the other $m$ players.

The other players therefore earned $10m-45$ points against the lowest ten. By the condition, this is half their combined score, which is $\binom{m+10}{2}-90.$ Thus $$2(10m-45)=\binom{m+10}{2}-90,$$ giving $(m-6)(m-15)=0.$ If $m=6,$ the lowest ten average $9$ points while the other six average only $5,$ impossible for the latter group to rank above them. Hence $m=15,$ and the total number of players is $10+15=25.$

Small Hint:

Recognize each $A$ as a square face with one corner cut off and each $B$ as that corner triangle

Big Hint:

A plane through six edge midpoints of a cube cuts a regular hexagon and divides the cube into two congruent parts

Solution:

Consider a $12\times12\times12$ cube and the plane through the midpoints of the six edges that join opposite groups of three vertices. In coordinates $0\leq x,y,z\leq12,$ this is the plane $x+y+z=18.$ Its cross-section is a regular hexagon with side $6\sqrt2,$ the same as the cut edge joining adjacent side midpoints.

On three faces of the cube, the plane leaves a square with a corner triangle removed, exactly piece $A.$ On the other three faces, it leaves the complementary right-isosceles corner triangle, exactly piece $B.$ Thus the pictured net is one of the two pieces into which this plane cuts the cube. Central symmetry interchanges the two pieces, so each has half the cube’s volume: $$V=\frac{12^3}{2}=864.$$