2015 AIME I — Problems & Online Timed Exam

Timed

3:00:00

1.

The expressions $A = 1 \times 2$ $+ 3 \times 4$ $+ 5 \times 6$ $+ \cdots$ $+ 37 \times 38 + 39$ and $B = 1$ $+ 2 \times 3$ $+ 4 \times 5$ $+ \cdots$ $+ 36 \times 37$ $+ 38 \times 39$ are obtained by writing multiplication and addition operators in an alternating pattern between successive integers. Find the positive difference between integers $A$ and $B.$

Answer

Answer: 722

Concepts:summation arithmetic sequence pairing and grouping

Difficulty rating: 1890

Small Hint:

Compute $B - A$ by pairing each product in $B$ with the product in $A$ that shares its even factor

Big Hint:

Each pair contributes $(2k+1)(2k)$ $- (2k-1)(2k) = 4k,$ and the leftover terms are $1$ and $-39$

Solution:

Subtract term by term: $\begin{aligned} &B - A = (1 - 39) \\ &\quad {}+ (2 \times 3 - 1 \times 2) \\ &\quad {}+ (4 \times 5 - 3 \times 4) + \cdots \\ &\quad {}+ (38 \times 39 - 37 \times 38). \end{aligned}$ Each parenthesized difference has the form $(2k+1)(2k)$ $- (2k-1)(2k) = 4k$ for $k = 1, 2, \ldots, 19.$

Therefore $\begin{aligned} B - A &= -38 \\ &\quad {}+ 4(1 + 2 + \cdots + 19) \\ &= -38 + 4 \cdot 190 = 722. \end{aligned}$

2.

The nine delegates to the Economic Cooperation Conference include $2$ officials from Mexico, $3$ officials from Canada, and $4$ officials from the United States. During the opening session, three of the delegates fall asleep. Assuming that the three sleepers were determined randomly, the probability that exactly two of the sleepers are from the same country is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Answer

Answer: 139

Concepts:basic probability combinations casework

Difficulty rating: 2110

Small Hint:

All $\binom{9}{3}$ sleeper triples are equally likely; count the favorable ones by which country supplies the pair

Big Hint:

Exactly two from a country of size $s$ means $\binom{s}{2}$ choices for the pair times $9 - s$ choices for the third sleeper

Solution:

There are $\binom{9}{3} = 84$ equally likely sets of three sleepers. Exactly two sleepers come from the same country when one country supplies exactly two of them and the third sleeper comes from a different country: $\binom{4}{2}(2 + 3) = 30$ ways with the pair from the United States, $\binom{3}{2}(2 + 4) = 18$ with the pair from Canada, and $\binom{2}{2}(3 + 4) = 7$ with the pair from Mexico.

The probability is $\frac{30 + 18 + 7}{84} = \frac{55}{84},$ already in lowest terms, so $m + n = 55 + 84 = 139.$

3.

There is a prime number $p$ such that $16p + 1$ is the cube of a positive integer. Find $p.$

Answer

Answer: 307

Concepts:sum and difference of cubes prime parity

Difficulty rating: 2010

Small Hint:

Write $16p + 1 = n^3$ and factor $n^3 - 1$

Big Hint:

$n^2 + n + 1$ is always odd, so the factor $16$ must come entirely from $n - 1;$ primality then forces $n - 1 = 16$

Solution:

Write $16p + 1 = n^3,$ so $16p = n^3 - 1$ $= (n - 1)(n^2 + n + 1).$ Since $16p + 1$ is odd, $n$ is odd, and $n^2 + n + 1$ is odd as well. Therefore all four factors of $2$ must divide $n - 1:$ write $n - 1 = 16k,$ which gives $p = k(n^2 + n + 1).$ For $p$ to be prime we need $k = 1,$ so $n = 17.$

Then $p = 17^2 + 17 + 1 = 307,$ which is indeed prime, and $16 \cdot 307 + 1 = 4913 = 17^3.$

4.

Point $B$ lies on line segment $\overline{AC}$ with $AB = 16$ and $BC = 4.$ Points $D$ and $E$ lie on the same side of line $AC$ forming equilateral triangles $\triangle ABD$ and $\triangle BCE.$ Let $M$ be the midpoint of $\overline{AE},$ and $N$ be the midpoint of $\overline{CD}.$ The area of $\triangle BMN$ is $x.$ Find $x^2.$

Answer

Answer: 507

Concepts:coordinate geometry equilateral triangle distance formula

Difficulty rating: 2390

Small Hint:

Put $B$ at the origin with $A = (-16, 0)$ and $C = (4, 0),$ and write $D$ and $E$ using equilateral-triangle altitudes

Big Hint:

Compute the midpoints $M$ and $N$ and the three distances $BM,$ $MN,$ $NB;$ triangle $BMN$ turns out equilateral

Solution:

Place $B = (0, 0),$ $A = (-16, 0),$ and $C = (4, 0).$ Each equilateral triangle has its apex above the midpoint of its base at height $\frac{\sqrt{3}}{2}$ times the side, so $D = (-8, 8\sqrt{3})$ and $E = (2, 2\sqrt{3}).$ The midpoints are $M = (-7, \sqrt{3})$ and $N = (-2, 4\sqrt{3}).$

Now $BM^2 = 49 + 3 = 52,$ $BN^2 = 4 + 48 = 52,$ and $MN^2 = 25 + 27 = 52,$ so $\triangle BMN$ is equilateral with side $\sqrt{52}.$ Its area is $x = \frac{\sqrt{3}}{4} \cdot 52 = 13\sqrt{3},$ so $x^2 = 169 \cdot 3 = 507.$

5.

In a drawer Sandy has $5$ pairs of socks, each pair a different color. On Monday Sandy selects two individual socks at random from the $10$ socks in the drawer. On Tuesday Sandy selects $2$ of the remaining $8$ socks at random and on Wednesday two of the remaining $6$ socks at random. The probability that Wednesday is the first day Sandy selects matching socks is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Answer

Answer: 341

Concepts:basic probability sampling without replacement symmetry

Difficulty rating: 2510

Small Hint:

Splitting the $10$ socks into a sequence of daily pairs is symmetric: any reordering of the days is equally likely

Big Hint:

So compute instead the probability of a match on Monday and then no match on Tuesday or Wednesday, tracking how many complete pairs remain each day

Solution:

Imagine dealing all ten socks out two per day for five days; every assignment of unordered pairs to days is equally likely, and permuting the days does not change this distribution. Swapping Monday and Wednesday therefore shows that the desired probability (mismatch, mismatch, match) equals the probability of a match on Monday followed by mismatches on Tuesday and Wednesday.

That pattern is easy to compute in order. Monday matches with probability $\frac{1}{9}$ (the second sock must be the first sock’s mate). The remaining $8$ socks then form $4$ complete pairs, so Tuesday mismatches with probability $1 - \frac{4}{\binom{8}{2}} = \frac{6}{7}.$ Tuesday’s mismatch breaks two pairs, leaving $2$ complete pairs among the $6$ remaining socks, so Wednesday mismatches with probability $1 - \frac{2}{\binom{6}{2}} = \frac{13}{15}.$

The probability is $\frac{1}{9} \cdot \frac{6}{7} \cdot \frac{13}{15} = \frac{26}{315},$ so $m + n = 26 + 315 = 341.$

6.

Points $A,$ $B,$ $C,$ $D,$ and $E$ are equally spaced on a minor arc of a circle. Points $E,$ $F,$ $G,$ $H,$ $I,$ and $A$ are equally spaced on a minor arc of a second circle with center $C$ as shown in the figure below. The angle $\angle ABD$ exceeds $\angle AHG$ by $12^\circ.$ Find the degree measure of $\angle BAG.$

Answer

Answer: 58

Concepts:inscribed angle arc angle chasing

Difficulty rating: 2720

Small Hint:

Let $\alpha$ be the common central angle at $C$ in the second circle; then $\angle ACE = 5\alpha$ is simultaneously an inscribed angle of the first circle

Big Hint:

Express every arc of both circles in terms of $\alpha;$ the condition $\angle ABD - \angle AHG = 12^\circ$ pins down $\alpha$

Solution:

Let $\alpha = \angle ECF$ $= \angle FCG$ $= \angle GCH$ $= \angle HCI$ $= \angle ICA,$ the common central angle of the second circle, so $\angle ACE = 5\alpha.$ Since $C$ also lies on the first circle, $\angle ACE$ is an inscribed angle there, so the arc $AE$ not containing $C$ measures $10\alpha,$ and each of the four equal arcs $AB,$ $BC,$ $CD,$ $DE$ measures $\frac{360^\circ - 10\alpha}{4} = 90^\circ - \frac{5\alpha}{2}.$

Angle $ABD$ subtends the arc $AD$ not containing $B,$ which is $360^\circ - 3\left(90^\circ - \frac{5\alpha}{2}\right),$ so $\angle ABD = 45^\circ + \frac{15\alpha}{4}.$ Angle $AHG$ subtends the second circle’s arc $AG$ not containing $H,$ which is $360^\circ - 3\alpha,$ so $\angle AHG = 180^\circ - \frac{3\alpha}{2}.$ The given condition reads $\begin{aligned} &\left(45^\circ + \frac{15\alpha}{4}\right) - \left(180^\circ - \frac{3\alpha}{2}\right) \\ &= \frac{21\alpha}{4} - 135^\circ = 12^\circ, \end{aligned}$ so $\alpha = 28^\circ.$

Finally, $\angle BAE$ subtends the first circle’s arc $BCDE = 3\left(90^\circ - \frac{5\alpha}{2}\right) = 60^\circ,$ giving $\angle BAE = 30^\circ,$ and $\angle EAG$ subtends the second circle’s arc $EFG = 2\alpha,$ giving $\angle EAG = 28^\circ.$ Hence $\angle BAG = \angle BAE + \angle EAG$ $= 30^\circ + 28^\circ = 58^\circ.$

7.

In the diagram below, $ABCD$ is a square. Point $E$ is the midpoint of $\overline{AD}.$ Points $F$ and $G$ lie on $\overline{CE},$ and $H$ and $J$ lie on $\overline{AB}$ and $\overline{BC},$ respectively, so that $FGHJ$ is a square. Points $K$ and $L$ lie on $\overline{GH},$ and $M$ and $N$ lie on $\overline{AD}$ and $\overline{AB},$ respectively, so that $KLMN$ is a square. The area of $KLMN$ is $99.$ Find the area of $FGHJ.$

Answer

Answer: 539

Difficulty rating: 2710

Small Hint:

Every right triangle cut off by the tilted squares ( $HBJ,$ $JFC,$ $MAN,$ $NKH$ ) is similar to $\triangle CDE,$ with legs in ratio $1 : 2$

Big Hint:

Write $BC = BJ + JC$ to get $FGHJ$ ’s side from $AE;$ then $AH = AN + NH$ gives $KLMN$ ’s side in exactly the same way

Solution:

Let $AE = s,$ so the big square has side $2s$ and $CE = s\sqrt{5}.$ The right triangles $CDE,$ $JFC,$ $HBJ,$ $NKH,$ and $MAN$ are all similar, with legs in ratio $1 : 2.$ Let $x$ be the side of $FGHJ.$ In $\triangle HBJ$ the hypotenuse is $HJ = x,$ so $BJ = \frac{x}{\sqrt{5}}$ and $HB = \frac{2x}{\sqrt{5}};$ in $\triangle JFC$ the longer leg is $JF = x,$ so the hypotenuse is $JC = \frac{x\sqrt{5}}{2}.$ Then $\begin{aligned} 2s = BC &= BJ + JC \\ &= x\left(\frac{1}{\sqrt{5}} + \frac{\sqrt{5}}{2}\right) \\ &= \frac{7x}{2\sqrt{5}}, \end{aligned}$ so $x = \frac{4\sqrt{5}\,s}{7}.$

Next, $AH = 2s - HB = 2s - \frac{8s}{7}$ $= \frac{6s}{7}.$ The identical decomposition along $\overline{AB}$ for the square $KLMN$ of side $y$ gives $\frac{6s}{7} = AH = AN + NH$ $= y\left(\frac{1}{\sqrt{5}} + \frac{\sqrt{5}}{2}\right).$ Dividing the two equations, $\frac{x}{y} = \frac{2s}{\frac{6s}{7}} = \frac{7}{3}.$

The areas are therefore in ratio $\left(\frac{7}{3}\right)^2 = \frac{49}{9},$ so the area of $FGHJ$ is $99 \cdot \frac{49}{9} = 539.$

8.

For positive integer $n,$ let $s(n)$ denote the sum of the digits of $n.$ Find the smallest positive integer $n$ satisfying $s(n) = s(n + 864) = 20.$

Answer

Answer: 695

Concepts:digits place value casework

Difficulty rating: 2760

Small Hint:

Each carry in an addition lowers the digit sum by $9,$ so $s(n + 864) = s(n) + 18 - 9c$ where $c$ counts the carries

Big Hint:

Exactly two carries must occur, and the hundreds place always carries, so try each choice of the second carrying place and minimize the hundreds digit

Solution:

Each carry in an addition replaces $10$ in one place by $1$ in the next, lowering the digit sum by $9.$ Hence $s(n + 864) = s(n) + s(864) - 9c$ $= 20 + 18 - 9c,$ where $c$ is the number of carries, and $s(n + 864) = 20$ forces $c = 2.$ For a three-digit candidate $n$ with digits $t,$ $u,$ $v$ summing to $20:$ since $u + v \le 18,$ we have $t \ge 2,$ so the hundreds place always carries ( $t + 8 \ge 10$ ), and exactly one of the units and tens places carries.

If the units carry and the tens do not, the tens computation $u + 6 + 1$ must stay below $10,$ so $u \le 2;$ then $t = 20 - u - v$ $\ge 20 - 2 - 9 = 9,$ forcing $n = 929.$ If the tens carry and the units do not, then $v + 4 \le 9$ gives $v \le 5,$ so $t = 20 - u - v$ $\ge 20 - 9 - 5 = 6,$ and $t = 6,$ $u = 9,$ $v = 5$ works: $n = 695.$

Indeed $s(695) = 20$ and $695 + 864 = 1559$ with $s(1559) = 20,$ so the smallest such $n$ is $695.$

9.

Let $S$ be the set of all ordered triples of integers $(a_1, a_2, a_3)$ with $1 \le a_1,$ $a_2,$ $a_3 \le 10.$ Each ordered triple in $S$ generates a sequence according to the rule $a_n = a_{n-1} \cdot |a_{n-2} - a_{n-3}|$ for $n \ge 4.$ Find the number of such sequences for which $a_n = 0$ for some $n.$

Answer

Answer: 494

Concepts:recursion casework inclusion-exclusion

Difficulty rating: 2990

Small Hint:

If two consecutive terms are ever equal, the sequence hits $0$ two steps later; if they differ by $1,$ it hits $0$ within four steps

Big Hint:

Count triples whose consecutive entries are equal or differ by $1,$ correcting for overlaps, then check what happens when a difference of $2$ meets a $1$

Solution:

If $a_{k-1} = a_k$ then $a_{k+2} = a_{k+1}|a_k - a_{k-1}| = 0,$ and if $|a_k - a_{k-1}| = 1$ then $a_{k+2} = a_{k+1},$ so $a_{k+4} = 0.$ Hence every triple of one of the forms $(j,j,k),$ $(j,k,k),$ $(j,j\pm1,k),$ $(j,k,k\pm1)$ produces a $0.$ These forms contain $100 + 100 + 4 \cdot 90 = 560$ triples, but triples fitting two forms are counted twice: the $10$ of the form $(j,j,j),$ the $9$ in each of the six families $(j,j,j\pm1),$ $(j,j\pm1,j),$ $(j,j\pm1,j\pm1)$ (matching signs), and the $8$ in each of $(j,j+1,j+2)$ and $(j,j-1,j-2).$ That leaves $560 - 10 - 54 - 16 = 480$ triples.

A few other triples also work: if $(a_1, a_2, a_3) = (j, j\pm2, 1),$ then $a_4 = 2$ and $|a_4 - a_3| = 1,$ so $a_8 = 0.$ These $16$ triples include $(3,1,1)$ and $(4,2,1),$ which were already counted, so they add $14$ new ones, for $480 + 14 = 494.$

No other triple reaches $0:$ if both consecutive differences are at least $2$ and $a_3 \ge 2,$ then $a_4 = a_3|a_2 - a_1| \ge 2a_3 \gt a_3$ and $|a_4 - a_3| \ge a_3 \ge 2,$ so inductively the terms grow forever and no factor ever vanishes. If instead $a_3 = 1$ with $|a_2 - a_1| \ge 3,$ then $a_4 \ge 3$ and $|a_4 - a_3| \ge 2,$ and the same growth takes over. The count is $494.$

10.

Let $f(x)$ be a third-degree polynomial with real coefficients satisfying $\begin{aligned} |f(1)| = |f(2)| &= |f(3)| \\ &= |f(5)| = |f(6)| \\ &= |f(7)| = 12. \end{aligned}$ Find $|f(0)|.$

Answer

Answer: 72

Concepts:polynomial Vieta’s Formulas system of equations

Difficulty rating: 2930

Small Hint:

Each of $f(x) - 12$ and $f(x) + 12$ is a cubic, so each has exactly three of $1,$ $2,$ $3,$ $5,$ $6,$ $7$ as roots

Big Hint:

The two cubics differ by a constant, so the two root triples have equal sums and equal pairwise-product sums; only one split works

Solution:

Each of $f(x) - 12$ and $f(x) + 12$ is a cubic, so each vanishes at exactly three of $1,$ $2,$ $3,$ $5,$ $6,$ $7.$ Writing them as $c(x - r_1)(x - r_2)(x - r_3)$ and $c(x - s_1)(x - s_2)(x - s_3),$ the two cubics differ by the constant $24,$ so their $x^2$ and $x$ coefficients agree: the root triples have equal sums and equal sums of pairwise products. The only partition of $\{1,2,3,5,6,7\}$ into two triples of equal sum is $\{2,3,7\}$ and $\{1,5,6\}$ (each summing to $12$ ), and indeed both have pairwise-product sum $41.$

Replacing $f$ by $-f$ if necessary (which does not change $|f(0)|$ ), we have $f(x) = c(x-2)(x-3)(x-7)$ $+ 12$ $= c(x-1)(x-5)(x-6) - 12.$ Setting $x = 0$ gives $-42c + 12 = -30c - 12,$ so $c = 2$ and $f(0) = -42 \cdot 2 + 12 = -72.$ Thus $|f(0)| = 72.$

11.

Triangle $ABC$ has positive integer side lengths with $AB = AC.$ Let $I$ be the intersection of the bisectors of $\angle B$ and $\angle C.$ Suppose $BI = 8.$ Find the smallest possible perimeter of $\triangle ABC.$

Answer

Answer: 108

Concepts:incircle, incenter, and inradius isosceles triangle trigonometric identity divisibility

Difficulty rating: 3160

Small Hint:

Let $M$ be the midpoint of $BC:$ right triangles at $M$ give $\cos\angle ABM = \frac{BM}{AB}$ and $\cos\angle IBM = \frac{BM}{8}$

Big Hint:

Since $\angle IBM$ is half of $\angle ABM,$ the double-angle formula relates $AB$ to $BC;$ integrality and $BM \lt 8$ leave only a few cases to test

Solution:

Let $M$ be the midpoint of $\overline{BC};$ by symmetry $A,$ $I,$ and $M$ are collinear with $AM \perp BC.$ With $a = AB$ and $b = BM,$ right triangles $ABM$ and $IBM$ give $\cos\angle ABM = \frac{b}{a}$ and $\cos\angle IBM = \frac{b}{8}.$ Since $BI$ bisects $\angle ABM,$ the double-angle formula yields $\frac{b}{a} = 2\left(\frac{b}{8}\right)^2 - 1,$ so $a = \frac{32b}{b^2 - 32}.$

Writing $c = BC = 2b,$ this becomes $a = \frac{64c}{c^2 - 128}.$ We need $c^2 \gt 128,$ so $c \ge 12,$ while $\cos\angle IBM = \frac{b}{8} \lt 1$ forces $c \lt 16.$ Testing $c = 12, 13, 14, 15,$ only $c = 12$ makes $a$ an integer, namely $a = \frac{768}{16} = 48.$

The triangle with sides $48,$ $48,$ $12$ satisfies all the conditions, and its perimeter is $48 + 48 + 12 = 108.$

12.

Consider all $1000$ -element subsets of the set $\{1, 2, 3, \ldots, 2015\}.$ From each such subset choose the least element. The arithmetic mean of all of these least elements is $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p + q.$

Answer

Answer: 431

Concepts:combinations expected value bijection

Difficulty rating: 3270

Small Hint:

Exactly $\binom{2015-j}{999}$ of the subsets have least element $j$

Big Hint:

Interpret $\sum j\binom{2015-j}{999}$ as counting $1001$ -element subsets of a $2016$ -element set by their second-smallest element

Solution:

A $1000$ -element subset has least element $j$ exactly when it contains $j$ together with $999$ larger elements, so $\binom{2015-j}{999}$ of the subsets have least element $j.$ The mean is therefore $\frac{\sum_{j} j\binom{2015-j}{999}}{\binom{2015}{1000}}.$

The numerator counts something concrete: to build a $1001$ -element subset of $\{0, 1, \ldots, 2015\}$ whose second-smallest element is $j,$ choose its smallest element from $\{0, \ldots, j-1\}$ ( $j$ ways) and its top $999$ elements from $\{j+1, \ldots, 2015\}.$ Summing over $j$ produces every $1001$ -element subset exactly once, so $\sum_j j\binom{2015-j}{999} = \binom{2016}{1001}.$

Hence the mean is $\frac{\binom{2016}{1001}}{\binom{2015}{1000}} = \frac{2016}{1001} = \frac{288}{143},$ which is in lowest terms, and $p + q = 288 + 143 = 431.$

13.

With all angles measured in degrees, the product $\prod_{k=1}^{45} \csc^2(2k-1)^\circ = m^n,$ where $m$ and $n$ are integers greater than $1.$ Find $m + n.$

Answer

Answer: 91

Concepts:trigonometric identity pairing and grouping

Difficulty rating: 3370

Small Hint:

Let $P$ and $Q$ be the products of the sines of the odd and even degree angles up to $89^\circ;$ pair $\sin k^\circ$ with $\sin(90-k)^\circ = \cos k^\circ$

Big Hint:

Then $P^2Q^2 = \prod \sin k^\circ \cos k^\circ;$ doubling every factor turns the product into $Q^2$ again, leaving only a power of $2$

Solution:

Let $P = \sin 1^\circ \sin 3^\circ \cdots \sin 89^\circ$ and $Q = \sin 2^\circ \sin 4^\circ \cdots \sin 88^\circ,$ so the desired product is $\frac{1}{P^2}.$ Then $PQ = \prod_{k=1}^{89} \sin k^\circ,$ and multiplying this by itself in reverse order, using $\sin(90 - k)^\circ = \cos k^\circ,$ gives $P^2Q^2 = \prod_{k=1}^{89} \sin k^\circ \cos k^\circ.$

Multiply by $2^{89}$ and use $2\sin k^\circ \cos k^\circ = \sin 2k^\circ:$ $\begin{aligned} 2^{89} P^2 Q^2 &= \prod_{k=1}^{89} \sin 2k^\circ \\ &= \left(\prod_{k=1}^{44} \sin 2k^\circ\right) \\ &\quad {}\cdot \left(\prod_{k=46}^{89} \sin 2k^\circ\right) \\ &= Q \cdot Q, \end{aligned}$ since $\sin 90^\circ = 1$ and $\sin(180 - x)^\circ = \sin x^\circ$ turns the second half into $Q$ as well.

Because $Q \ne 0,$ it follows that $P^2 = 2^{-89},$ so $\prod_{k=1}^{45} \csc^2(2k-1)^\circ = 2^{89}.$ Since $89$ is prime, the only representation $m^n$ with $m, n \gt 1$ is $m = 2,$ $n = 89,$ and $m + n = 91.$

14.

For each integer $n \ge 2,$ let $A(n)$ be the area of the region in the coordinate plane defined by the inequalities $1 \le x \lt n$ and $0 \le y \le x\lfloor\sqrt{x}\rfloor,$ where $\lfloor\sqrt{x}\rfloor$ is the greatest integer not exceeding $\sqrt{x}.$ Find the number of values of $n$ with $2 \le n \le 1000$ for which $A(n)$ is an integer.

Answer

Answer: 483

Concepts:floor and ceiling functions trapezoid casework pattern recognition

Difficulty rating: 3500

Small Hint:

Over $m \le x \lt m+1$ with $k = \lfloor\sqrt{m}\rfloor,$ the strip is a trapezoid of area $\frac{(2m+1)k}{2}$ — an integer exactly when $k$ is even

Big Hint:

So integrality of $A(n)$ persists while $k$ is even and alternates while $k$ is odd; track the blocks $k^2 \lt n \le (k+1)^2$ according to $k \bmod 4$

Solution:

On the strip $m \le x \lt m + 1$ we have $\lfloor\sqrt{x}\rfloor = k = \lfloor\sqrt{m}\rfloor,$ so the region above it is a trapezoid under $y = kx$ with area $A(m+1) - A(m) = \frac{(2m+1)k}{2}:$ an integer when $k$ is even, a half-integer when $k$ is odd. Hence as $n$ grows by $1,$ the integrality of $A(n)$ is unchanged while $k$ is even and flips at every step while $k$ is odd.

Consider the block of $2k + 1$ values $k^2 \lt n \le (k+1)^2.$ Starting from $A(1) = 0,$ the statuses of $A(k^2)$ cycle with period $4:$ integer for $k \equiv 0, 1$ and non-integer for $k \equiv 2, 3 \pmod 4$ (an odd block flips the status an odd number of times, an even block preserves it). Counting integer values of $A(n)$ inside each block: for $k = 4j - 3$ the block alternates, beginning and ending with non-integers, giving $4j - 3;$ for $k = 4j - 2$ every value is a non-integer, giving $0;$ for $k = 4j - 1$ it alternates, beginning and ending with integers, giving $4j;$ for $k = 4j$ all $8j + 1$ values are integers.

For $j = 1, \ldots, 7,$ covering $2 \le n \le 29^2,$ the four blocks contribute $(4j - 3) + 0 + 4j$ $+ (8j + 1) = 16j - 2$ integers, totaling $\sum_{j=1}^{7}(16j - 2) = 434.$ Then the block $k = 29$ contributes $29$ integers for $841 \lt n \le 900,$ the block $k = 30$ contributes none, and for $k = 31$ the alternation over $961 \lt n \le 1000$ begins with an integer at $n = 962$ and gives $20$ more. The total is $434 + 29 + 20 = 483.$

15.

A block of wood has the shape of a right circular cylinder with radius $6$ and height $8,$ and its entire surface has been painted blue. Points $A$ and $B$ are chosen on the edge of one of the circular faces of the cylinder so that arc $\overset{\frown}{AB}$ on that face measures $120^\circ.$ The block is then sliced in half along the plane that passes through point $A,$ point $B,$ and the center of the cylinder, revealing a flat, unpainted face on each half. The area of one of these unpainted faces is $a\cdot\pi + b\sqrt{c},$ where $a,$ $b,$ and $c$ are integers and $c$ is not divisible by the square of any prime. Find $a + b + c.$

Answer

Answer: 53

Concepts:cylinder 3D geometry sector trigonometry

Difficulty rating: 3700

Small Hint:

Project the cut face straight down onto the circular face containing $A$ and $B:$ the image is the disk minus two $120^\circ$ circular segments

Big Hint:

With $O$ the cylinder’s center, $O'$ the face center, and $M$ the midpoint of $AB,$ the tilt satisfies $\cos\theta = \frac{O'M}{OM};$ divide the projected area by $\cos\theta$

Solution:

Stand the block on the face containing $A$ and $B,$ and let $O'$ be the center of that face, $M$ the midpoint of $\overline{AB},$ and $O$ the center of the cylinder. The cutting plane meets the bottom face in chord $\overline{AB}$ and, by symmetry through $O,$ meets the top face in the reflected chord, so the cut face projects vertically onto the region $R'$ between chord $\overline{AB}$ and its mirror image through $O'$ (shaded below). Each $120^\circ$ circular segment cut off has area $\frac{1}{3}\pi \cdot 6^2 - \frac{1}{2} \cdot 6 \cdot 6 \sin 120^\circ$ $= 12\pi - 9\sqrt{3},$ so $R'$ has area $36\pi - 2\left(12\pi - 9\sqrt{3}\right)$ $= 12\pi + 18\sqrt{3}.$

Since $\overset{\frown}{AB} = 120^\circ,$ triangle $AO'B$ gives $O'M = 6\cos 60^\circ = 3,$ and $OO' = 4,$ so $OM = 5.$ The cut face is planar and tilted from the horizontal only in the direction of $\overline{O'M},$ at the angle $\theta$ with $\cos\theta = \frac{O'M}{OM} = \frac{3}{5}.$ Undoing the projection therefore multiplies areas by $\frac{5}{3},$ so the unpainted face has area $\frac{5}{3}\left(12\pi + 18\sqrt{3}\right) = 20\pi + 30\sqrt{3}.$ Thus $a + b + c = 20 + 30 + 3 = 53.$