Small Hint:

Pair the outer factors and the inner factors

Big Hint:

The two paired products are consecutive even integers

Solution:

We have $31\cdot28=868$ and $30\cdot29=870.$ These products lie one below and one above $869,$ so their product is $869^2-1.$ After the final $1$ is added, the radicand is $869^2.$ Its positive square root is $869.$

Small Hint:

Each choice of at least three marked points determines one convex polygon

Big Hint:

Count all subsets and remove those of sizes $0,$ $1,$ and $2$

Solution:

Every subset of at least three points determines exactly one convex polygon. There are $2^{10}=1024$ subsets in all. Of these, $\binom{10}{0}=1,$ $\binom{10}{1}=10,$ and $\binom{10}{2}=45$ have fewer than three points. Therefore the required number is $1024-1-10-45=968.$

Small Hint:

Convert the repeating three-digit block into a fraction with denominator $999$

Big Hint:

Use the integrality of $n$ to determine which digit makes $4d+1$ divisible by $37$

Solution:

The repeating decimal is $\frac{100d+25}{999}.$ Thus $$\begin{aligned}n&=\frac{810(100d+25)}{999}\\&=\frac{750(4d+1)}{37}.\end{aligned}$$ Since $0\leq d\leq9,$ the only way $4d+1$ can be divisible by $37$ is $4d+1=37,$ so $d=9$ and $n=750.$

Small Hint:

Express both sums in terms of the middle integer $c$

Big Hint:

Compare the prime exponents in $3c$ and $5c$ modulo $2$ and modulo $3$

Solution:

The two sums are $3c$ and $5c.$ If $c=\prod p^{v_p},$ then $3c$ being a square and $5c$ being a cube impose conditions on every exponent. For $p=3,$ the smallest exponent in $c$ that is odd and divisible by $3$ is $3.$ For $p=5,$ the smallest exponent that is even and congruent to $2\pmod3$ is $2.$ Every other prime exponent can be $0.$ Thus the least possible value is $c=3^3\cdot5^2=675.$

Small Hint:

Let $p$ be the probability of heads and equate the two binomial probabilities

Big Hint:

Cancel the nonzero common factors before solving for $p$

Solution:

Let $p$ be the probability of heads. The condition gives $$5p(1-p)^4=10p^2(1-p)^3.$$ The stated nonzero condition permits cancellation, yielding $1-p=2p,$ so $p=\frac{1}{3}.$ The probability of exactly three heads is $$\binom53\left(\frac13\right)^3\left(\frac23\right)^2=\frac{40}{243}.$$ Therefore $i+j=40+243=283.$

Small Hint:

After $t$ seconds, Allie is $8t$ meters from $A$ and Billie can be $7t$ meters from $B$

Big Hint:

Apply the Law of Cosines to the triangle and select the smaller positive time

Solution:

Suppose the skaters meet after $t$ seconds. Their distances from $A$ and $B$ are $8t$ and $7t,$ and the included angle at $A$ is $60^\circ.$ The Law of Cosines gives $$\begin{aligned}(7t)^2&=(8t)^2+100^2\\&\quad-2(8t)(100)\cos60^\circ.\end{aligned}$$ Hence $3t^2-160t+2000=0,$ whose roots are $20$ and $\frac{100}{3}.$ The earliest meeting occurs at $t=20,$ so Allie skates $8(20)=160$ meters.

Small Hint:

Write the three arithmetic-sequence terms as $x,$ $x+r,$ and $x+2r$

Big Hint:

Subtract adjacent square equations, then subtract those two resulting equations

Solution:

Let the three terms be $x,$ $x+r,$ and $x+2r.$ Subtracting the square equations gives $$r(2x+r)=300-36=264$$ and $$r(2x+3r)=596-300=296.$$ Their difference is $2r^2=32,$ so $r=\pm4.$ Negating all three terms does not change their squares, so take $r=4.$ Then $4(2x+4)=264,$ giving $x=31.$ Therefore $k=x^2-36=961-36=925.$

Small Hint:

Let $S(t)=\sum_{i=1}^7(i+t)^2x_i$

Big Hint:

Because $S(t)$ is quadratic in $t,$ its second finite differences are constant

Solution:

Define $S(t)=\sum_{i=1}^7(i+t)^2x_i.$ This is a quadratic polynomial in $t,$ and the equations say $S(0)=1,$ $S(1)=12,$ and $S(2)=123.$ Its first two differences are $11$ and $111,$ so the constant second difference is $100.$ The next first difference is therefore $211,$ giving $S(3)=123+211=334.$

Small Hint:

Estimate the fifth root to narrow the possible integer values of $n$

Big Hint:

Evaluate the fifth powers by repeated squaring and multiplication, then compare their sum with the nearby candidate

Solution:

Direct integer arithmetic gives $$\begin{aligned}133^5&=41{,}615{,}795{,}893,\\110^5&=16{,}105{,}100{,}000,\\84^5&=4{,}182{,}119{,}424,\\27^5&=14{,}348{,}907.\end{aligned}$$ Their sum is $61{,}917{,}364{,}224.$ Repeated multiplication also gives $144^5=61{,}917{,}364{,}224,$ so the positive integer $n$ is $144.$

Small Hint:

Simplify $\cot\alpha+\cot\beta$ using $\alpha+\beta=\pi-\gamma$

Big Hint:

Use the Law of Sines for the resulting sine factors and the Law of Cosines for $\cos\gamma$

Solution:

First, $$\begin{aligned}\cot\alpha+\cot\beta&=\frac{\sin(\alpha+\beta)}{\sin\alpha\sin\beta}\\&=\frac{\sin\gamma}{\sin\alpha\sin\beta}.\end{aligned}$$ Hence the desired ratio is $$\frac{\cos\gamma\sin\alpha\sin\beta}{\sin^2\gamma}=\frac{ab\cos\gamma}{c^2},$$ where the Law of Sines was used in the final equality. By the Law of Cosines, $$\begin{aligned}2ab\cos\gamma&=a^2+b^2-c^2\\&=1988c^2.\end{aligned}$$ Therefore the ratio is $\frac{1988}{2}=994.$

Small Hint:

By symmetry, place the mode at the low endpoint and push every other entry as high as the frequency restriction allows

Big Hint:

If the mode occurs $f$ times, every other value may occur at most $f-1$ times; optimize separately over $f$

Solution:

By reflecting every value $x$ to $1001-x,$ it suffices to maximize the mean minus the mode. For a fixed modal frequency $f,$ the extremal sample has $f$ copies of $1,$ then fills the largest available integers with at most $f-1$ copies each.

Put $121-f=q(f-1)+r,$ where $0\leq r<f-1.$ The nonmodal entries are $f-1$ copies of each of $1000,$ $999,$ $\ldots,$ $1001-q,$ followed by $r$ copies of $1000-q.$ For $f=2,$ $f=3,$ $f=4,$ $f=5,$ and $f=6,$ this formula gives floors $924,$ $945,$ $947,$ $944,$ and $939,$ respectively. If $f\geq7,$ there are at most $114$ nonmodal terms, so even the weaker bound $D\leq\frac{114(999)}{121}<942$ suffices. Thus the maximum occurs at $f=4.$

The extremal sample contains four $1$’s and three copies of every integer from $962$ through $1000.$ Put $T=962+963+\cdots+1000.$ Since $T=38259,$ $$\begin{aligned}D&=\frac{3T+4}{121}-1\\&=\frac{114660}{121}\\&=947+\frac{73}{121}.\end{aligned}$$ Therefore the largest possible floor is $947.$

Small Hint:

Represent the vertices by vectors and write the vector between the two midpoints

Big Hint:

Expand $\lVert A+B-C-D\rVert^2$ in terms of the six edge lengths

Solution:

Let the vertex names also denote their position vectors. The vector between the midpoints is $\frac{A+B-C-D}{2}.$ Expanding squared lengths gives $$\begin{aligned}4d^2={}&AC^2+AD^2\\&+BC^2+BD^2\\&-AB^2-CD^2.\end{aligned}$$ Therefore $$\begin{aligned}4d^2={}&7^2+18^2+36^2+27^2\\&-41^2-13^2=548,\end{aligned}$$ so $d^2=137.$

Small Hint:

Within any eleven consecutive integers, the forbidden-difference graph is an $11$-cycle

Big Hint:

For a matching construction, look for five allowable residue classes modulo $11$

Solution:

On any eleven consecutive integers, join two numbers when their difference is $4$ or $7.$ Because $7=11-4,$ this graph is an $11$-cycle, whose largest independent set has size $5.$ Similarly, any ten consecutive integers induce a path on ten vertices and contribute at most $5.$ Since $1989=179\cdot11+2\cdot10,$ this gives $|S|\leq179(5)+2(5)=905.$

This bound is attained by taking every integer whose residue modulo $11$ is $1,$ $3,$ $4,$ $6,$ or $9.$ No two selected residues differ by $4$ or $7$ modulo $11,$ and each of these five residues occurs $181$ times from $1$ through $1989.$ Thus the maximum is $5(181)=905.$

Small Hint:

Compute the second and third powers of $-3+i$ and set the imaginary part of the expansion equal to zero

Big Hint:

The digit bounds leave only two possible triples $(a_3,a_2,a_1)$; then let $a_0$ range over all digits

Solution:

Let $b=-3+i.$ Then $b^2=8-6i$ and $b^3=-18+26i.$

The imaginary part of $a_3b^3+a_2b^2+a_1b+a_0$ is $$26a_3-6a_2+a_1.$$ Thus $a_1=6a_2-26a_3.$ With $1\leq a_3\leq9$ and $0\leq a_1,a_2\leq9,$ the only possibilities are $$\begin{gathered}(a_3,a_2,a_1)=(1,5,4),\\(a_3,a_2,a_1)=(2,9,2).\end{gathered}$$

The corresponding real parts are $10+a_0$ and $30+a_0,$ respectively. As $a_0$ ranges from $0$ through $9,$ the required sum is $$\begin{aligned}&(10+11+\cdots+19)\\&\quad+(30+31+\cdots+39)\\&=145+345=490.\end{aligned}$$

Small Hint:

Use the two known cevian ratios to find the barycentric weights of $A$ and $B$ at $P$

Big Hint:

Place $P$ at the origin; the resulting vector relation determines the angle between $PA$ and $PB$

Solution:

Since $AP=PD,$ the barycentric weight of $A$ at $P$ is $\frac{1}{2}.$ Since $BP:PE=3:1,$ the weight of $B$ is $\frac{1}{4},$ so the weight of $C$ is also $\frac{1}{4}.$ Along $CF,$ this means $\frac{PF}{CF}=\frac{1}{4},$ hence $PF=5$ and $CP=15.$

Place $P$ at the origin and denote the position vectors $A,$ $B,$ and $C$ by the same letters. The barycentric relation is $2A+B+C=0.$ Thus $C=-2A-B.$ Using $|A|=6,$ $|B|=9,$ and $|C|=15,$ $$\begin{aligned}225&=|2A+B|^2\\&=4(36)+81+4A\mathbin{\cdot}B,\end{aligned}$$ so $A\mathbin{\cdot}B=0.$ Therefore $PA\perp PB.$ Expanding the cross product gives $(B-A)\mathbin{\times}(C-A)=4(A\mathbin{\times}B).$ Consequently, $$\begin{aligned}{}[ABC]&=2|A\mathbin{\times}B|\\&=2(6)(9)=108.\end{aligned}$$