Small Hint:

Rewrite each given logarithm with base $w$

Big Hint:

Expand $\log_w(xyz)$ as a sum of three logarithms

Solution:

Taking reciprocals of the given logarithms gives $$\begin{aligned} \log_w x&=\frac1{24},\\ \log_w y&=\frac1{40},\\ \log_w(xyz)&=\frac1{12}. \end{aligned}$$ Therefore $$\log_w z=\frac1{12}-\frac1{24}-\frac1{40} =\frac1{60}.$$ Taking the reciprocal yields $\log_z w=60.$

Small Hint:

Determine the sign of each expression inside an absolute value on the given interval

Big Hint:

On $p\leq x\leq15,$ simplify $f(x)$ to a decreasing linear function

Solution:

Since $p\leq x\leq15,$ we have $$\begin{aligned} |x-p|&=x-p,\\ |x-15|&=15-x,\\ |x-p-15|&=p+15-x. \end{aligned}$$ Hence $f(x)=30-x.$ This is minimized at the right endpoint $x=15,$ where its value is $15.$

Small Hint:

The expression outside the radical is $15$ less than the radicand

Big Hint:

Set $t=\sqrt{x^2+18x+45}$ and solve $t^2-15=2t$

Solution:

Set $t=\sqrt{x^2+18x+45},$ so $t\geq0.$ The equation becomes $$t^2-15=2t,$$ or $(t-5)(t+3)=0.$ Thus $t=5,$ and $$x^2+18x+45=25,$$ so the real roots satisfy $x^2+18x+20=0.$ By Vieta’s formulas, their product is $20.$

Small Hint:

Put $B=(0,0),$ $A=(0,6),$ and $C=(2,0)$

Big Hint:

The center lies on the perpendicular bisector of $AC$ at distance $\sqrt{40}$ from its midpoint

Solution:

Put $B=(0,0),$ $A=(0,6),$ and $C=(2,0).$ The midpoint of $AC$ is $M=(1,3),$ and $AC=\sqrt{40}.$ If $O$ is the center, then $$\begin{aligned} OM&=\sqrt{OA^2-AM^2}\\ &=\sqrt{50-10}=\sqrt{40}. \end{aligned}$$ A vector perpendicular to $AC=(2,-6)$ is $(6,2),$ which already has length $\sqrt{40}.$ Thus the two possible centers are $$\begin{aligned} M+(6,2)&=(7,5),\\ M-(6,2)&=(-5,1). \end{aligned}$$ The pictured notched circle has its center on the side opposite the notch, so $O=(-5,1).$ Therefore $BO^2=(-5)^2+1^2=26.$

Small Hint:

Let $s=x+y$ and $p=xy$

Big Hint:

Use $x^2+y^2=s^2-2p$ to eliminate $p$ from $x^3+y^3=s^3-3ps$

Solution:

Let $s=x+y$ and $p=xy.$ From $x^2+y^2=s^2-2p=7,$ we get $p=\frac{s^2-7}{2}.$ Also, $$\begin{aligned} 10&=x^3+y^3\\ &=s^3-3ps\\ &=s^3-\frac{3s(s^2-7)}2. \end{aligned}$$ Hence $$\begin{aligned} s^3-21s+20&=0,\\ (s-1)(s-4)(s+5)&=0. \end{aligned}$$ Each root gives a possible pair of complex roots of $t^2-st+p=0,$ so the largest real possible value of $s=x+y$ is $4.$

Small Hint:

Write $6=7-1$ and $8=7+1$

Big Hint:

Modulo $49,$ only the constant and linear terms of each binomial expansion survive

Solution:

Modulo $49,$ every binomial term containing $7^2$ vanishes. Since $83$ is odd, $$6^{83}=(7-1)^{83}\equiv -1+83\cdot7$$ and $$8^{83}=(7+1)^{83}\equiv 1+83\cdot7.$$ Their sum is congruent to $166\cdot7=1162\equiv35\pmod{49}.$

Small Hint:

Count the complementary selections in which no two chosen knights are adjacent

Big Hint:

Split into selections containing a fixed knight and selections not containing that knight

Solution:

There are $\binom{25}{3}=2300$ selections. Fix one seat. If it is not selected, choosing three nonadjacent seats among the remaining $24$ seats is equivalent to choosing three nonconsecutive positions from a row of $24,$ giving $\binom{22}{3}.$ If the fixed seat is selected, its two neighbors are forbidden, and the other two selected seats must be nonconsecutive among the remaining row of $22,$ giving $\binom{21}{2}.$

Thus the number with no adjacent selected seats is $$\begin{aligned} \binom{22}{3}+\binom{21}{2} &=1540+210\\ &=1750. \end{aligned}$$ Therefore $$P=1-\frac{1750}{2300}=\frac{11}{46},$$ and the requested sum is $11+46=57.$

Small Hint:

For a prime $p>50,$ compare the exponent of $p$ in $200!$ with twice its exponent in $100!$

Big Hint:

Primes from $67$ through $99$ cancel; primes from $51$ through $66$ do not

Solution:

For a prime $p>50,$ we have $p^2>200,$ so $$\begin{aligned} v_p\left(\binom{200}{100}\right) &=\left\lfloor\frac{200}{p}\right\rfloor\\ &\quad-2\left\lfloor\frac{100}{p}\right\rfloor. \end{aligned}$$ If $67\leq p<100,$ this is $2-2=0.$ If $50<p\leq66,$ it is $3-2=1.$ Hence no prime greater than $66$ divides the binomial coefficient, while every prime between $50$ and $66$ does. The largest such prime is $61.$

Small Hint:

Set $t=x\sin x,$ which is positive on the given interval

Big Hint:

Apply AM-GM to $9t+\frac4t$ and verify that equality is attainable

Solution:

Let $t=x\sin x>0.$ The expression becomes $$9t+\frac4t\geq2\sqrt{9t\cdot\frac4t}=12,$$ with equality when $9t=\frac{4}{t},$ or $t=\frac{2}{3}.$ The continuous function $x\sin x$ tends to $0$ as $x$ tends to $0$ and equals $\frac{\pi}{2}>\frac{2}{3}$ at $x=\frac{\pi}{2}.$ Thus it takes the value $\frac{2}{3}$ somewhere in $(0,\frac{\pi}{2}),$ so the lower bound $12$ is attained.

Small Hint:

Separate the case where the repeated digit is $1$ from the case where it is not

Big Hint:

In each case, choose the repeated digit’s positions before choosing the remaining distinct digit

Solution:

If $1$ is the repeated digit, exactly one of the last three positions contains $1.$ There are $3$ choices for that position, $9$ choices for the next digit, and $8$ for the last digit, since those two digits must differ from each other and from $1.$ This gives $3\cdot9\cdot8=216$ numbers.

If a digit other than $1$ is repeated, there are $9$ choices for that digit, $3$ ways to choose its two positions among the last three, and $8$ choices for the remaining digit. This gives another $9\cdot3\cdot8=216$ numbers. The total is $216+216=432.$

Small Hint:

Find the height by projecting either endpoint of the upper edge onto the base

Big Hint:

At a fraction $t$ of the height, the cross-section is a rectangle with sides $s(1-t)$ and $s(1+t)$

Solution:

Each endpoint of the upper edge is joined to the endpoints of one side of the square base. Its distance to that side’s midpoint is therefore $\frac{\sqrt3s}{2}.$ The projection of each upper endpoint lies $\frac{s}{2}$ beyond that midpoint, because the two projected endpoints are $2s$ apart while the two opposite-side midpoints are $s$ apart. Thus the height $h$ satisfies $$h^2=\left(\frac{\sqrt3s}{2}\right)^2-\left(\frac{s}{2}\right)^2 =\frac{s^2}{2},$$ so $h=\frac{s}{\sqrt2}.$

At a fraction $t$ of the height above the base, the cross-section is a rectangle whose dimensions are $s(1-t)$ and $s(1+t).$ Its area is $s^2(1-t^2).$ By Cavalieri’s principle, the volume is the volume $s^2h$ of a prism minus the volume $\frac{s^2h}{3}$ of a pyramid: $$V=\frac23s^2h=\frac{\sqrt2}{3}s^3.$$ Substituting $s=6\sqrt2$ gives $V=288.$

Small Hint:

Write the diameter as $10a+b$ and the chord as $10b+a$

Big Hint:

The squared distance from the center to the chord is one fourth the difference of their squares

Solution:

Let the diameter be $10a+b$ and the chord be $10b+a.$ Since a perpendicular from the center bisects a chord, $$\begin{aligned} 4OH^2&=(10a+b)^2\\ &\quad-(10b+a)^2\\ &=99(a^2-b^2). \end{aligned}$$ For $OH$ to be rational, $11(a^2-b^2)$ must be a square. Thus it equals $121c^2,$ so $$(a-b)(a+b)=11c^2.$$ Because $a$ and $b$ are digits with $a>b,$ the left side is at most $81,$ so $c=1$ or $2.$ If $c=1,$ the factors are $1$ and $11,$ giving $a=6$ and $b=5.$ If $c=2,$ the factor pairs of $44$ either have different parity or give $a>9.$ Hence the diameter is $65.$

Small Hint:

Add the contribution of each number separately over all subsets

Big Hint:

The sign of $k$ depends only on whether the subset contains an even or odd number of elements greater than $k$

Solution:

Fix $k.$ Once $k$ is included, the $k-1$ smaller elements may be chosen arbitrarily, contributing a factor of $2^{k-1}.$ If exactly $j$ of the $7-k$ larger elements are chosen, the sign of $k$ is $(-1)^j.$ Therefore the coefficient of $k$ in the total is $$\begin{aligned} &2^{k-1}\sum_{j=0}^{7-k} (-1)^j\binom{7-k}{j}\\ &\qquad=2^{k-1}(1-1)^{7-k}. \end{aligned}$$ This is $0$ for $k<7$ and $2^6$ for $k=7.$ Hence the total is $7\cdot2^6=448.$

Small Hint:

Put the centers at $(-6,0)$ and $(6,0)$ and find the coordinates of $P$

Big Hint:

If $Q=P-\ell u$ and $R=P+\ell u,$ subtract the two equal-radius equations

Solution:

Put the center of the radius-$8$ circle at $O_1=(-6,0)$ and the other center at $O_2=(6,0).$ Subtracting the two circle equations gives $$P=\left(\frac76,\frac{\sqrt{455}}6\right),$$ where the positive $y$-coordinate selects the pictured intersection.

Let the common chord length be $\ell,$ and let $u$ be the unit vector from $Q$ toward $R.$ Then $Q=P-\ell u$ and $R=P+\ell u.$ Comparing $Q$ with $P$ in the first circle and $R$ with $P$ in the second gives $$\begin{aligned} u\mathbin{\cdot}(P-O_1)&=\frac{\ell}{2},\\ u\mathbin{\cdot}(P-O_2)&=-\frac{\ell}{2}. \end{aligned}$$ Adding shows that $u\perp P,$ so $$u=\frac{(\sqrt{455},-7)}{\sqrt{504}}.$$ Subtracting the two dot-product equations gives $$\ell=u\mathbin{\cdot}(O_2-O_1) =\frac{12\sqrt{455}}{\sqrt{504}}.$$ Therefore $\ell^2=\frac{144\cdot455}{504}=130.$

Small Hint:

The midpoints of all chords from $A$ lie on the circle with diameter joining $A$ to the center

Big Hint:

Uniqueness makes $BC$ tangent to that midpoint circle at the midpoint of $AD$

Solution:

Let $H$ be the midpoint of $AD.$ Put $H=(0,0),$ $A=(-a,0),$ $D=(a,0),$ and let the circle’s center be $O=(0,k).$ Then $a^2+k^2=25.$ The midpoints of all chords starting at $A$ form the circle with center $\frac{A+O}{2}$ and radius $\frac{5}{2}.$ Because $AD$ is the only such chord bisected by the line $BC,$ that line is tangent to the midpoint circle at $H.$

Hence a unit direction vector for $BC$ may be taken as $u=\frac{(k,a)}{5},$ perpendicular to $A+O=(-a,k).$ Write $B=-pu$ and $C=qu,$ where $p,q>0.$ Then $p+q=6,$ and intersecting chords give $pq=a^2.$ Substituting the line into the original circle also gives $$q-p=2u\mathbin{\cdot}O=\frac{2ak}{5}.$$ Therefore $$\begin{aligned} 36-4a^2&=(q-p)^2\\ &=\frac{4a^2(25-a^2)}{25}, \end{aligned}$$ so $a^4-50a^2+225=0.$ Its roots are $a^2=5$ and $45,$ but $a^2=pq\leq\frac{(p+q)^2}{4}=9.$ Thus $a^2=5,$ and $\{p,q\}=\{1,5\}.$

Choose $a=\sqrt5,$ $k=2\sqrt5,$ and $p=1$ for the endpoint $B$ on minor arc $AD.$ Then $$\begin{aligned} \overrightarrow{OA} &=(-\sqrt5,-2\sqrt5),\\ \overrightarrow{OB} &=\left(-\frac2{\sqrt5}, -\frac{11}{\sqrt5}\right). \end{aligned}$$ The sine of the central angle subtending minor arc $AB$ is the absolute determinant of these radius vectors divided by $5^2,$ namely $$\sin\overset{\frown}{AB}=\frac{|11-4|}{25}=\frac7{25}.$$ Hence $mn=7\cdot25=175.$