Small Hint:

Each square has one fourth the area of the preceding square

Big Hint:

Adjacent squares overlap in one fourth of the smaller square, and nonadjacent interiors do not overlap

Solution:

The sum of the five square areas is $$\begin{aligned}1+\frac14+\frac1{16}&+\frac1{64}+\frac1{256}\\&=\frac{1364}{1024}.\end{aligned}$$ The perpendicular-bisector placement makes the overlap of each adjacent pair one fourth of the smaller square. These four overlaps are disjoint and have total area $$\begin{aligned}\frac1{16}+\frac1{64}&+\frac1{256}+\frac1{1024}\\&=\frac{85}{1024}.\end{aligned}$$ Thus the union has area $\frac{1364-85}{1024}=\frac{1279}{1024},$ and $m-n=1279-1024=255.$

Small Hint:

Set $y=\log_{1995}x,$ so $x=1995^y$

Big Hint:

Compare exponents of $1995$, then use the sum of the two values of $y$

Solution:

Put $y=\log_{1995}x,$ so $x=1995^y.$ The equation becomes $$1995^{\frac{1}{2}+y^2}=1995^{2y},$$ and hence $(y-1)^2=\frac{1}{2}.$ The two values of $y$ have sum $2,$ so the product of the corresponding positive roots is $1995^2.$ Since $1995\equiv-5\pmod {1000},$ its last three digits are $025.$ The requested AIME answer is $25.$

Small Hint:

The target can first be reached only after $4$ or $6$ steps

Big Hint:

From the six-step paths ending at the target, subtract those that already arrived at step $4$

Solution:

There are $\binom42=6$ four-step paths to $(2,2).$ There are $60+60=120$ six-step paths ending there: the extra opposite pair is either left-right or down-up. Of these, $6\cdot4=24$ first reach the target at step $4$ and then make a two-step return. Therefore $$p=\frac6{4^4}+\frac{120-24}{4^6}=\frac3{64}.$$ Thus $m+n=3+64=67.$

Small Hint:

The three circle centers are collinear because their pairwise distances are $6,$ $3,$ and $9$

Big Hint:

Use signed distances from the two smaller centers to the common tangent to find its distance from the large center

Solution:

Put the radius-$9$ circle at the origin and the smaller centers at $(-6,0)$ and $(3,0).$ Write the common external tangent as $\mathbf n\cdot(x,y)=c,$ where $\mathbf n$ is a unit normal. Its signed distances from the two centers differ by $6-3=3,$ so $9n_x=3$ and $n_x=\frac{1}{3}.$ Using the distance $3$ from $(-6,0)$ gives $|c|=5.$ Thus the chord lies $5$ units from the large center, and its squared length is $$4(9^2-5^2)=4\cdot56=224.$$

Small Hint:

Because the polynomial has real coefficients, its non-real roots occur in conjugate pairs

Big Hint:

Group the six pairwise products into the products within the two groups and the four cross-products

Solution:

Let the first two roots be $\alpha$ and $\beta.$ Since $\alpha\beta=13+i$ is not real, they are not conjugates, so the other roots are $\overline\alpha$ and $\overline\beta.$ Hence $\alpha+\beta=3-4i$ and $\overline\alpha\,\overline\beta=13-i.$ By Vieta’s formulas, $$\begin{aligned}b&=\alpha\beta+\overline\alpha\,\overline\beta\\&\quad+(3-4i)(3+4i)\\&=26+25=51.\end{aligned}$$

Small Hint:

Pair each divisor $d$ of $n^2$ with $\frac{n^2}{d}$

Big Hint:

Among the divisors below $n$, subtract those that already divide $n$

Solution:

The number $n^2=2^{62}3^{38}$ has $63\cdot39=2457$ divisors. Pairing $d$ with $\frac{n^2}{d}$ leaves only $n$ unpaired, so $\frac{2457-1}{2}=1228$ divisors lie below $n.$ The number $n$ has $32\cdot20=640$ divisors, of which $639$ are below $n.$ Therefore the requested count is $1228-639=589.$

Small Hint:

Set $u=\sin t+\cos t$ and express $\sin t\cos t$ in terms of $u$

Big Hint:

Both given products become half of a perfect square in $u$

Solution:

Let $u=\sin t+\cos t.$ Since $\sin t\cos t=\frac{u^2-1}{2},$ $$\begin{aligned}(1+\sin t)&(1+\cos t)\\&=\frac{(u+1)^2}{2}=\frac54.\end{aligned}$$ The feasible sign gives $u+1=\sqrt{\frac{5}{2}}.$ Therefore $$\begin{aligned}(1-\sin t)&(1-\cos t)\\&=\frac{(1-u)^2}{2}\\&=\frac{13}{4}-\sqrt{10}.\end{aligned}$$ Thus $k+m+n=10+13+4=27.$

Small Hint:

Write $x=ay$ and reduce $x+1$ modulo $y+1$

Big Hint:

The quotient $a$ must have the form $1+t(y+1)$; count the possible positive values of $t$

Solution:

Write $x=ay.$ Modulo $y+1,$ we have $y\equiv-1,$ so $x+1$ is divisible by $y+1$ exactly when $a\equiv1\pmod {y+1}.$ Since $x>y,$ write $a=1+t(y+1)$ with $t\geq1.$ The bound $x\leq100$ becomes $$t\leq\left\lfloor\frac{100-y}{y(y+1)}\right\rfloor.$$ Only $1\leq y\leq9$ contribute, giving $$\begin{aligned}49+16+8&+4+3\\&+2+1+1+1\\&=85.\end{aligned}$$

Small Hint:

Let $BM=x$ and let half the apex angle be $\alpha$

Big Hint:

Use $\tan\alpha=\frac{x}{11},$ $DM=1,$ and the triple-angle formula for tangent

Solution:

Let $BM=x$ and $\alpha=\angle BAM,$ so $\tan\alpha=\frac{x}{11}.$ Because $DM=AM-AD=1,$ symmetry gives $\angle BDC=2\arctan x,$ while $\angle BAC=2\alpha.$ Hence $\arctan x=3\alpha.$ Put $t=\tan\alpha=\frac{x}{11}.$ Then $$\frac{3t-t^3}{1-3t^2}=11t,$$ so $t^2=\frac{1}{4}$ and $x=\frac{11}{2}.$ Thus $BC=11$ and $AB=\frac{11\sqrt5}{2},$ making the perimeter $$11+11\sqrt5=11+\sqrt{605}.$$ Therefore $a+b=11+605=616.$

Small Hint:

For each residue modulo $42$, find the smallest positive composite integer in that residue

Big Hint:

Once one number in a residue class is representable, every number $42$ larger in that class is representable

Solution:

A number in residue class $r\pmod {42}$ is representable once it exceeds the first positive composite in that class by a positive multiple of $42.$ Composite residues $r$ themselves supply that first value. For the remaining residues, suitable first composites are $$\begin{aligned}&0:42,\ 1:85,\ 2:44,\\&3:45,\ 5:215,\ 7:49,\\&11:95,\ 13:55,\ 17:143,\\&19:145,\ 23:65,\ 29:155,\\&31:115,\ 37:121,\ 41:125.\end{aligned}$$ The largest entry is $215,$ and $215-42,$ $215-84,$ $215-126,$ $215-168,$ and $215-210$ are all prime. Hence $215$ is not representable, while every larger integer is.

Small Hint:

Sort the three side lengths of the smaller prism and compare them in order with $a,$ $b,$ and $c$

Big Hint:

The two unchanged dimensions force $\frac{a}{1995}=\frac{1995}{c}$

Solution:

Let the similar smaller prism have sorted sides $x\leq y\leq z.$ It shares two side lengths with $P,$ and all three of its sorted sides are smaller than the corresponding sides of $P.$ The only possible matching is $y=a$ and $z=b=1995.$ Similarity then gives $$\frac{x}{a}=\frac{a}{1995}=\frac{1995}{c},$$ so $ac=1995^2.$ Conversely every factor pair $a<c$ gives a nondegenerate cut. Since $1995=3\cdot5\cdot7\cdot19,$ its square has $3^4=81$ divisors. Excluding the central pair $a=c=1995$ and taking one divisor from each remaining pair gives $\frac{81-1}{2}=40.$

Small Hint:

Place the square’s vertices at $(\pm1,\pm1,0)$ and the apex at $(0,0,h)$

Big Hint:

Find $h^2$ from $\angle AOB,$ then take the supplement of the angle between suitable face normals

Solution:

Take adjacent base vertices $A=(1,1,0),$ $B=(-1,1,0),$ $C=(-1,-1,0)$ and $O=(0,0,h).$ From $\angle AOB=45^\circ,$ $$\begin{aligned}\frac{h^2}{h^2+2}&=\frac1{\sqrt2},\\h^2&=2+2\sqrt2.\end{aligned}$$ Normals to the two faces may be taken as $(0,2h,2)$ and $(-2h,0,2).$ Their acute angle has cosine $\frac{1}{h^2+1}=3-2\sqrt2.$ The interior dihedral angle is its supplement, so $$\cos\theta=2\sqrt2-3=-3+\sqrt8.$$ Thus $m+n=-3+8=5.$

Small Hint:

Count the integers $n$ for which $j-\tfrac12<\sqrt[4]n<j+\tfrac12$

Big Hint:

The number of occurrences of $f(n)=j$ simplifies to $4j^3+j$

Solution:

For $j\geq1,$ the number of positive integers $n$ for which $f(n)=j$ is $$\begin{aligned}\left(j+\frac12\right)^4&-\left(j-\frac12\right)^4\\&=4j^3+j.\end{aligned}$$ For $j=1,\ldots,6,$ these account for $$\sum_{j=1}^6(4j^3+j)=1785$$ values, and their contribution to the requested sum is $$\begin{aligned}\sum_{j=1}^6\frac{4j^3+j}{j}&=4\sum_{j=1}^6j^2+6\\&=370.\end{aligned}$$ The remaining $1995-1785=210$ values have $f(n)=7,$ contributing $30.$ The total is $400.$

Small Hint:

Each chord is $9\sqrt3$ from the center, so determine the two possible line directions through the intersection point

Big Hint:

The unequal chord segments have lengths $30$ and $48,$ and their endpoints subtend $60^\circ$ at the center

Solution:

Put the center at $O=(0,0)$ and the intersection at $P=(18,0).$ A length-$78$ chord is $9\sqrt3$ from $O,$ so a line through $P$ containing such a chord makes angle $60^\circ$ or $120^\circ$ with $OP.$ Solving along either line gives segment lengths $30$ and $48.$

For either region bordered by unequal segments, the two arc endpoints subtend $60^\circ$ at $O.$ Its area is the sector minus $\triangle OAB$ plus $\triangle PAB:$ $$\begin{aligned}\frac{60}{360}\pi(42)^2&-\frac12(42)^2\sin60^\circ\\&+\frac12(30)(48)\sin60^\circ\\&=294\pi-81\sqrt3.\end{aligned}$$ Hence $m+n+d=378.$

Small Hint:

Use states for $0,$ $1,$ $2,$ $3,$ and $4$ consecutive heads and one separate state for a single trailing tail

Big Hint:

Express every head-run state’s success probability in terms of the trailing-tail state

Solution:

Let $q_i$ be the success probability with $i$ consecutive heads and no trailing tail, and let $t$ be the probability after one tail. Then $$\begin{aligned}q_i&=\frac{q_{i+1}+t}{2}\quad(0\leq i<4),\\q_4&=\frac{1+t}{2},\\t&=\frac{q_1}{2}.\end{aligned}$$ Working backward gives $q_1=\frac{1+15t}{16}.$ Since $q_1=2t,$ we get $t=\frac{1}{17}$ and $q_1=\frac{2}{17}.$ Therefore $$p=q_0=\frac{q_1+t}{2}=\frac3{34},$$ so $m+n=3+34=37.$