2009 AMC 12B Solutions | LIVE by Po-Shen Loh

1.

Each morning of her five-day workweek, Jane bought either a $50$ -cent muffin or a $75$ -cent bagel. Her total cost for the week was a whole number of dollars. How many bagels did she buy?

$1$

$2$

$3$

$4$

$5$

Concepts:parity money casework

Difficulty rating: 820

Small Hint:

She buys $5$ items total, so muffins and bagels together number $5$

Big Hint:

The total $50m + 75b$ cents must be a multiple of $100;$ reduce the cost modulo $100$

Solution:

With $b$ bagels she buys $5 - b$ muffins, costing $50(5-b) + 75b = 250 + 25b$ cents. For a whole number of dollars this must be a multiple of $100.$ Thus $25b\equiv50\pmod{100},$ or $b\equiv2\pmod4.$ Since $0\le b\le5,$ only $b=2$ works, giving $300$ cents $= \$3.$

Thus, the correct answer is B.

2.

Paula the painter had just enough paint for $30$ identically sized rooms. Unfortunately, on the way to work, three cans of paint fell off her truck, so she had only enough paint for $25$ rooms. How many cans of paint did she use for the $25$ rooms?

$10$

$12$

$15$

$18$

$25$

Concepts:ratio and proportion

Difficulty rating: 900

Small Hint:

The $3$ lost cans account for the $5$ rooms she can no longer paint

Big Hint:

The ratio of cans to rooms is $3 : 5$

Solution:

Losing $3$ cans cost her $5$ rooms, so $3$ cans paint $5$ rooms and each room needs $\dfrac{3}{5}$ of a can.

For $25$ rooms she used $25 \cdot \dfrac{3}{5} = 15$ cans.

Thus, the correct answer is C.

3.

Twenty percent less than $60$ is one-third more than what number?

$16$

$30$

$32$

$36$

$48$

Concepts:percentage fraction linear equation

Difficulty rating: 1000

Small Hint:

Twenty percent less than $60$ is $0.8 \cdot 60$

Big Hint:

One-third more than a number $n$ is $\dfrac{4}{3}n$

Solution:

Twenty percent less than $60$ is $0.8 \cdot 60 = 48.$

One-third more than $n$ is $\dfrac{4}{3}n,$ so $\dfrac{4}{3}n = 48$ gives $n = 36.$

Thus, the correct answer is D.

4.

A rectangular yard contains two flower beds in the shape of congruent isosceles right triangles. The remainder of the yard has a trapezoidal shape, as shown. The parallel sides of the trapezoid have lengths $15$ and $25$ meters. What fraction of the yard is occupied by the flower beds?

$\dfrac{1}{8}$

$\dfrac{1}{6}$

$\dfrac{1}{5}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

Concepts:area ratio triangle area trapezoid

Difficulty rating: 1100

Small Hint:

The two triangle legs equal half the difference of the parallel sides

Big Hint:

Each isosceles right triangle has legs $\dfrac{25 - 15}{2} = 5$

Solution:

The parallel sides differ by $25 - 15 = 10,$ so each triangle has legs $\dfrac{10}{2} = 5$ and area $\dfrac{1}{2} \cdot 5^2 = \dfrac{25}{2}.$ The two beds total $25.$

The rectangle measures $25$ by $5,$ so its area is $125,$ and the fraction occupied is $\dfrac{25}{125} = \dfrac{1}{5}.$

Thus, the correct answer is C.

5.

Kiana has two older twin brothers. The product of their three ages is $128.$ What is the sum of their three ages?

$10$

$12$

$16$

$18$

$24$

Concepts:prime factorization power of 2

Difficulty rating: 1080

Small Hint:

$128 = 2^7,$ so every age is a power of $2$

Big Hint:

If each twin is $t$ years old, then $t^2$ times Kiana’s age is $128,$ and Kiana is younger

Solution:

Since $128 = 2^7,$ each age is a power of $2.$ The twins share an age $t,$ so Kiana’s age is $\dfrac{128}{t^2}.$

Taking $t = 8$ gives Kiana $\dfrac{128}{64} = 2,$ who is younger than the twins. (Smaller twins would make Kiana older, which is not allowed.) The sum is $8 + 8 + 2 = 18.$

Thus, the correct answer is D.

6.

By inserting parentheses, it is possible to give the expression $2 \times 3 + 4 \times 5$ several values. How many different values can be obtained?

$2$

$3$

$4$

$5$

$6$

Concepts:order of operations systematic listing

Difficulty rating: 1250

Small Hint:

The result depends on when the addition is performed relative to the two multiplications

Big Hint:

Enumerate the distinct groupings; several parenthesizations collapse to the same value

Solution:

The genuinely different groupings give

$(2 \times 3) + (4 \times 5) = 26,$ $\ ((2 \times 3) + 4) \times 5 = 50,$ $\ 2 \times (3 + (4 \times 5)) = 46,$ and $\ 2 \times (3 + 4) \times 5 = 70.$

These are all distinct, so $4$ values can be obtained.

Thus, the correct answer is C.

7.

In a certain year the price of gasoline rose by $20\%$ during January, fell by $20\%$ during February, rose by $25\%$ during March, and fell by $x\%$ during April. The price of gasoline at the end of April was the same as it had been at the beginning of January. To the nearest integer, what is $x?$

$12$

$17$

$20$

$25$

$35$

Concepts:percentage

Difficulty rating: 1330

Small Hint:

Multiply the successive factors $1.2,$ $0.8,$ and $1.25$

Big Hint:

After three months the price is $1.2$ times the start, so April must undo that factor

Solution:

After January through March the price is $1.2 \cdot 0.8 \cdot 1.25 = 1.2$ times the original.

To return to the original, April must multiply by $\dfrac{1}{1.2},$ a decrease of $1 - \dfrac{1}{1.2} = \dfrac{1}{6} \approx 16.7\%.$ To the nearest integer, $x = 17.$

Thus, the correct answer is B.

8.

When a bucket is two-thirds full of water, the bucket and water weigh $a$ kilograms. When the bucket is one-half full of water the total weight is $b$ kilograms. In terms of $a$ and $b,$ what is the total weight in kilograms when the bucket is full of water?

$\dfrac{2}{3}a + \dfrac{1}{3}b$

$\dfrac{3}{2}a - \dfrac{1}{2}b$

$\dfrac{3}{2}a + b$

$\dfrac{3}{2}a + 2b$

$3a - 2b$

Concepts:system of equations substitution

Difficulty rating: 1370

Small Hint:

Let the empty bucket weigh $x$ and a full load of water weigh $y,$ so $x + \tfrac{2}{3}y = a$ and $x + \tfrac{1}{2}y = b$

Big Hint:

Subtract the two equations to find $y,$ then compute $x + y$

Solution:

Let $x$ be the bucket’s weight and $y$ the weight of a full bucket of water. Then $x + \dfrac{2}{3}y = a$ and $x + \dfrac{1}{2}y = b.$

Subtracting gives $\dfrac{1}{6}y = a - b,$ so $y = 6a - 6b$ and $x = b - \dfrac{1}{2}y = 4b - 3a.$ The full weight is $x + y = 3a - 2b.$

Thus, the correct answer is E.

9.

Triangle $ABC$ has vertices $A = (3, 0),$ $B = (0, 3),$ and $C,$ where $C$ is on the line $x + y = 7.$ What is the area of $\triangle ABC?$

$6$

$8$

$10$

$12$

$14$

Concepts:coordinate geometry triangle area parallel lines

Difficulty rating: 1390

Small Hint:

Line $AB$ is $x + y = 3,$ which is parallel to $x + y = 7$

Big Hint:

Since $C$ slides along a line parallel to $AB,$ the area is fixed; pick a convenient $C$

Solution:

Line $AB$ has equation $x + y = 3,$ which is parallel to $x + y = 7,$ so the area is independent of where $C$ lies on that line.

Take $C = (7, 0).$ Then the base $AC = 4$ lies on the $x$ -axis with height $3,$ giving area $\dfrac{1}{2} \cdot 4 \cdot 3 = 6.$

Thus, the correct answer is A.

10.

A particular $12$ -hour digital clock displays the hour and minute of a day. Unfortunately, whenever it is supposed to display a $1,$ it mistakenly displays a $9.$ For example, when it is $1{:}16$ pm the clock incorrectly shows $9{:}96$ pm. What fraction of the day will the clock show the correct time?

$\dfrac{1}{2}$

$\dfrac{5}{8}$

$\dfrac{3}{4}$

$\dfrac{5}{6}$

$\dfrac{9}{10}$

Concepts:digits complementary counting fraction

Difficulty rating: 1480

Small Hint:

The hour is shown correctly unless it contains a $1,$ namely hours $1, 10, 11, 12$

Big Hint:

A minute is correct only when neither of its digits is a $1$ ; count the bad minutes out of $60$

Solution:

The hours containing a $1$ are $1, 10, 11, 12,$ so $8$ of the $12$ hours display correctly, a fraction $\dfrac{2}{3}.$

A minute is wrong if either digit is $1$ : the tens digit gives $10\text{-}19$ ( $10$ minutes), and the ones digit adds $01, 21, 31, 41, 51$ ( $5$ more), $15$ in all. So $\dfrac{45}{60} = \dfrac{3}{4}$ of minutes are correct.

The fraction of the day is $\dfrac{2}{3} \cdot \dfrac{3}{4} = \dfrac{1}{2}.$

Thus, the correct answer is A.

11.

On Monday, Millie puts a quart of seeds, $25\%$ of which are millet, into a bird feeder. On each successive day she adds another quart of the same mix of seeds without removing any seeds that are left. Each day the birds eat only $25\%$ of the millet in the feeder, but they eat all of the other seeds. On which day, just after Millie has placed the seeds, will the birds find that more than half the seeds in the feeder are millet?

Tuesday

Wednesday

Thursday

Friday

Saturday

Concepts:geometric sequence recursion inequality

Difficulty rating: 1610

Small Hint:

Each morning $\tfrac{3}{4}$ of the previous millet remains and $\tfrac{1}{4}$ quart of new millet is added, while the other seeds always total $\tfrac{3}{4}$ quart

Big Hint:

After $n$ days the millet is $1 - \left(\tfrac{3}{4}\right)^n$ quart; require this to exceed $\tfrac{3}{4}$

Solution:

Each day the birds leave $\dfrac{3}{4}$ of the millet and Millie adds $\dfrac{1}{4}$ quart of new millet, so after $n$ days the millet is $\begin{aligned} &\dfrac{1}{4}\left(1 + \dfrac{3}{4} + \cdots + \left(\dfrac{3}{4}\right)^{n-1}\right) \\ &= 1 - \left(\dfrac{3}{4}\right)^n \end{aligned}$ quart.

The non-millet seeds always total $\dfrac{3}{4}$ quart, so millet exceeds half when $1 - \left(\dfrac{3}{4}\right)^n \gt \dfrac{3}{4},$ that is, $\left(\dfrac{3}{4}\right)^n \lt \dfrac{1}{4}.$

Since $\left(\dfrac{3}{4}\right)^4 = \dfrac{81}{256} \gt \dfrac{1}{4}$ and $\left(\dfrac{3}{4}\right)^5 = \dfrac{243}{1024} \lt \dfrac{1}{4},$ this first happens on day $5,$ which is Friday.

Thus, the correct answer is D.

12.

The fifth and eighth terms of a geometric sequence of real numbers are $7!$ and $8!$ respectively. What is the first term?

$60$

$75$

$120$

$225$

$315$

Concepts:geometric sequence factorial

Difficulty rating: 1500

Small Hint:

The ratio of the eighth term to the fifth term is $r^3$

Big Hint:

$\dfrac{8!}{7!} = 8 = r^3,$ so $r = 2$ ; then divide $7!$ by $r^4$

Solution:

The eighth term divided by the fifth term is $r^3 = \dfrac{8!}{7!} = 8,$ so $r = 2.$

The fifth term is $a r^4 = 7!,$ so $a = \dfrac{7!}{16} = \dfrac{5040}{16} = 315.$

Thus, the correct answer is E.

13.

Triangle $ABC$ has $AB = 13$ and $AC = 15,$ and the altitude to $BC$ has length $12.$ What is the sum of the two possible values of $BC?$

$15$

$16$

$17$

$18$

$19$

Concepts:Pythagorean Theorem altitude casework

Difficulty rating: 1560

Small Hint:

The foot of the altitude splits $BC$ ; use the Pythagorean theorem in the two right triangles

Big Hint:

The foot distances are $\sqrt{13^2 - 12^2} = 5$ and $\sqrt{15^2 - 12^2} = 9$ ; $BC$ is their sum or difference

Solution:

Let $D$ be the foot of the altitude from $A.$ Then $BD = \sqrt{13^2 - 12^2} = 5$ and $DC = \sqrt{15^2 - 12^2} = 9.$

If $D$ lies between $B$ and $C,$ then $BC = 5 + 9 = 14$ ; if the triangle is obtuse, $BC = 9 - 5 = 4.$ The sum is $14 + 4 = 18.$

Thus, the correct answer is D.

14.

Five unit squares are arranged in the coordinate plane as shown, with the lower left corner at the origin. The slanted line, extending from $(a, 0)$ to $(3, 3),$ divides the entire region into two regions of equal area. What is $a?$

$\dfrac{1}{2}$

$\dfrac{3}{5}$

$\dfrac{2}{3}$

$\dfrac{3}{4}$

$\dfrac{4}{5}$

Concepts:triangle area area decomposition coordinate geometry

Difficulty rating: 1610

Small Hint:

The total area is $5,$ so each region must be $\dfrac{5}{2}$

Big Hint:

The region on the lower-right side of the line is a triangle of base $3 - a$ and height $3,$ minus one unit square

Solution:

The five squares have total area $5,$ so each region must have area $\dfrac{5}{2}.$

The line from $(a, 0)$ to $(3, 3)$ together with the axes bounds a triangle of base $3 - a$ and height $3$ ; the region on the lower-right side of the line is this triangle with one unit square removed. Setting $\dfrac{3(3 - a)}{2} - 1 = \dfrac{5}{2}$ gives $3(3 - a) = 7,$ so $a = \dfrac{2}{3}.$

Thus, the correct answer is C.

15.

Assume $0 \lt r \lt 3.$ Below are five equations for $x.$ Which equation has the largest solution $x?$

$3(1 + r)^x = 7$

$3(1 + \frac{r}{10})^x = 7$

$3(1 + 2r)^x = 7$

$3(1 + \sqrt{r})^x = 7$

$3(1 + \frac{1}{r})^x = 7$

Concepts:logarithm inequality

Difficulty rating: 1710

Small Hint:

Solving each gives $x = \dfrac{\log(\frac{7}{3})}{\log(1 + f(r))}$ ; a smaller base makes $x$ larger

Big Hint:

Compare the values $r,\ \frac{r}{10},\ 2r,\ \sqrt{r},\ \frac{1}{r}$ for $0 \lt r \lt 3$ and find the smallest

Solution:

Each equation gives $x = \dfrac{\log(\frac{7}{3})}{\log(1 + f(r))},$ which is largest when the positive quantity $f(r)$ is smallest.

For $0 \lt r \lt 3,$ we have $\dfrac r{10}\lt r\lt2r$ and $\dfrac r{10}\lt\sqrt r$ because $r\lt100.$ Also $r^2\lt9\lt10,$ so $\dfrac r{10}\lt\dfrac1r.$ Thus $\frac{r}{10}$ is the smallest of the five quantities, and equation (B) has the largest solution.

Thus, the correct answer is B.

16.

Trapezoid $ABCD$ has $AD \parallel BC,$ $BD = 1,$ $\angle DBA = 23^\circ,$ and $\angle BDC = 46^\circ.$ The ratio $BC : AD$ is $9 : 5.$ What is $CD?$

$\dfrac{7}{9}$

$\dfrac{4}{5}$

$\dfrac{13}{15}$

$\dfrac{8}{9}$

$\dfrac{14}{15}$

Concepts:angle bisector theorem trapezoid parallel lines

Difficulty rating: 1800

Small Hint:

Draw the line through $D$ parallel to $AB,$ meeting $BC$ at $E$ ; then $BE = AD$

Big Hint:

Show $DE$ bisects $\angle BDC,$ then apply the angle bisector theorem in $\triangle BDC$

Solution:

Draw the line through $D$ parallel to $AB,$ meeting $BC$ at $E,$ so $ABED$ is a parallelogram with $BE = AD.$ Then $\angle BDE = \angle DBA = 23^\circ,$ and since $\angle BDC = 46^\circ,$ segment $DE$ bisects $\angle BDC.$

By the angle bisector theorem in $\triangle BDC,$ $\dfrac{EC}{BE} = \dfrac{DC}{DB},$ so $\begin{aligned} CD &= DB \cdot \dfrac{BC - AD}{AD} \\ &= 1 \cdot \left(\dfrac{9}{5} - 1\right) \\ &= \dfrac{4}{5}. \end{aligned}$

Thus, the correct answer is B.

17.

Each face of a cube is given a single narrow stripe painted from the center of one edge to the center of its opposite edge. The choice of the edge pairing is made at random and independently for each face. What is the probability that there is a continuous stripe encircling the cube?

$\dfrac{1}{8}$

$\dfrac{3}{16}$

$\dfrac{1}{4}$

$\dfrac{3}{8}$

$\dfrac{1}{2}$

Concepts:basic probability cube geometry casework

Difficulty rating: 1890

Small Hint:

Each face’s stripe has $2$ orientations, giving $2^6 = 64$ equally likely configurations

Big Hint:

An encircling stripe runs around one of $3$ pairs of opposite faces; count the configurations forming a loop

Solution:

Each of the $6$ faces has $2$ equally likely stripe orientations, for $2^6 = 64$ configurations.

An encircling stripe runs around one of the $3$ pairs of opposite faces. Fixing such a band, the four faces it passes through must be aligned, with probability $\left(\dfrac{1}{2}\right)^4 = \dfrac{1}{16},$ while the two remaining faces are free. The $3$ possible bands are disjoint events, so the probability is $3 \cdot \dfrac{1}{16} = \dfrac{3}{16}.$

Thus, the correct answer is B.

18.

Rachel and Robert run on a circular track. Rachel runs counterclockwise and completes a lap every $90$ seconds, and Robert runs clockwise and completes a lap every $80$ seconds. Both start from the start line at the same time. At some random time between $10$ minutes and $11$ minutes after they begin to run, a photographer standing inside the track takes a picture that shows one-fourth of the track, centered on the starting line. What is the probability that both Rachel and Robert are in the picture?

$\dfrac{1}{16}$

$\dfrac{1}{8}$

$\dfrac{3}{16}$

$\dfrac{1}{4}$

$\dfrac{5}{16}$

Concepts:geometric probability distance rate and time

Difficulty rating: 1930

Small Hint:

The picture spans $\tfrac{1}{8}$ lap on each side of the start; find when each runner is within that arc during the $10$ th minute

Big Hint:

Rachel is in view for $18.75\text{-}41.25$ s and Robert for $30\text{-}50$ s; take the overlap over $60$ s

Solution:

The picture covers the arc within $\dfrac{1}{8}$ lap of the start on each side. At $600$ s Rachel has run $6\tfrac{2}{3}$ laps, $30$ s short of the line; a quarter lap takes her $22.5$ s, so she is in view between $30 - 11.25 = 18.75$ s and $30 + 11.25 = 41.25$ s of the $10$ th minute.

At $600$ s Robert is $40$ s from the line; a quarter lap takes $20$ s, so he is in view between $30$ and $50$ s. Both appear between $30$ and $41.25$ s, a window of $11.25$ s out of $60,$ giving probability $\dfrac{11.25}{60} = \dfrac{3}{16}.$

Thus, the correct answer is C.

19.

For each positive integer $n,$ let $f(n) = n^4 - 360n^2 + 400.$ What is the sum of all values of $f(n)$ that are prime numbers?

$794$

$796$

$798$

$800$

$802$

Concepts:difference of squares factoring prime

Difficulty rating: 2000

Small Hint:

Rewrite as a difference of squares: $n^4 - 360n^2 + 400$ $= (n^2 + 20)^2 - (20n)^2$

Big Hint:

Factor it; for the product to be prime, the smaller factor must equal $1$

Solution:

Write $\begin{aligned} f(n) &= n^4 + 40n^2 + 400 - 400n^2 \\ &= (n^2 + 20)^2 - (20n)^2 \\ &= (n^2 + 20n + 20) \\ &\quad {}\cdot (n^2 - 20n + 20). \end{aligned}$

For $f(n)$ to be prime the smaller factor must be $1$ : solving $n^2 - 20n + 20 = 1$ gives $(n - 1)(n - 19) = 0,$ so $n = 1$ or $n = 19.$

Then $f(1) = 41$ and $f(19) = 761$ are both prime, summing to $802.$

Thus, the correct answer is E.

20.

A convex polyhedron $Q$ has vertices $V_1,$ $V_2,$ $\ldots,$ $V_n,$ and $100$ edges. The polyhedron is cut by planes $P_1,$ $P_2,$ $\ldots,$ $P_n$ in such a way that plane $P_k$ cuts only those edges that meet at vertex $V_k.$ In addition, no two planes intersect inside or on $Q.$ The cuts produce $n$ pyramids and a new polyhedron $R.$ How many edges does $R$ have?

$200$

$2n$

$300$

$400$

$4n$

Concepts:polyhedron double counting

Difficulty rating: 2040

Small Hint:

Each original edge is cut once near each of its two endpoints, creating new vertices

Big Hint:

Count the new edges (one per edge-endpoint) plus the surviving middle portion of each original edge

Solution:

Each of the $100$ edges is cut once near each endpoint, so $R$ has $2 \cdot 100 = 200$ vertices.

The cut at vertex $V_k$ creates a small polygon whose number of edges equals the degree of $V_k$ ; summed over all vertices this is $200,$ the total number of edge-endpoints. The middle portion of each original edge also survives, adding $100$ edges. So $R$ has $200 + 100 = 300$ edges.

Thus, the correct answer is C.

21.

Ten women sit in $10$ seats in a line. All of the $10$ get up and then reseat themselves using all $10$ seats, each sitting in the seat she was in before or a seat next to the one she occupied before. In how many ways can the women be reseated?

$89$

$90$

$120$

$210$

$2^{38}$

Concepts:recursion Fibonacci

Difficulty rating: 2160

Small Hint:

Let $S_n$ count the valid reseatings of $n$ women; consider the rightmost woman’s choice

Big Hint:

If she stays she leaves $S_{n-1}$ ways; if she moves left a swap is forced, leaving $S_{n-2}$ — so $S_n = S_{n-1} + S_{n-2}$

Solution:

Let $S_n$ be the number of valid reseatings of $n$ women. The rightmost woman either keeps her seat, leaving $S_{n-1}$ ways for the rest, or swaps with her left neighbor — the only other way to fill the end seat — leaving $S_{n-2}$ ways.

Thus $S_n = S_{n-1} + S_{n-2}$ with $S_1 = 1$ and $S_2 = 2,$ giving the Fibonacci values $1, 2, 3, 5, 8, 13, 21, 34, 55, 89.$ So $S_{10} = 89.$

Thus, the correct answer is A.

22.

Parallelogram $ABCD$ has area $1{,}000{,}000.$ Vertex $A$ is at $(0, 0)$ and all other vertices are in the first quadrant. Vertices $B$ and $D$ are lattice points on the lines $y = x$ and $y = kx$ for some integer $k \gt 1,$ respectively. How many such parallelograms are there?

$49$

$720$

$784$

$2009$

$2048$

Concepts:lattice point stars and bars prime factorization

Difficulty rating: 2340

Small Hint:

With $B = (b, b)$ and $D = (d, kd),$ the area of the parallelogram is $(k - 1)bd$

Big Hint:

You need $(k - 1)bd = 10^6 = 2^6 \cdot 5^6$ ; count ordered factorizations into three positive integers

Solution:

Let $B = (b, b)$ and $D = (d, kd)$ with $b, d, k$ positive integers and $k \gt 1.$ The area is $(k - 1)bd = 1{,}000{,}000 = 2^6 \cdot 5^6.$

Each parallelogram corresponds to an ordered triple $(k - 1, b, d)$ of positive integers with product $2^6 \cdot 5^6.$ The six $2$ ’s distribute among the three factors in $\binom{6 + 2}{2} = 28$ ways, and likewise the six $5$ ’s in $28$ ways, giving $28^2 = 784.$

Thus, the correct answer is C.

23.

A region $S$ in the complex plane is defined by $\scriptsize S = \{x + iy : -1 \le x \le 1,\ -1 \le y \le 1\}.$

A complex number $z = x + iy$ is chosen uniformly at random from $S.$ What is the probability that $\left(\dfrac{3}{4} + \dfrac{3}{4}i\right)z$ is also in $S?$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

$\dfrac{3}{4}$

$\dfrac{7}{9}$

$\dfrac{7}{8}$

Concepts:complex number geometric probability area

Difficulty rating: 2340

Small Hint:

Multiplying gives real part $\tfrac{3}{4}(x - y)$ and imaginary part $\tfrac{3}{4}(x + y)$ ; both must lie in $[-1, 1]$

Big Hint:

These conditions are $|x - y| \le \tfrac{4}{3}$ and $|x + y| \le \tfrac{4}{3}$ ; find the area of $S$ satisfying them

Solution:

Expanding, $\left(\dfrac{3}{4} + \dfrac{3}{4}i\right)(x + iy)$ $= \dfrac{3}{4}(x - y)$ $+ \dfrac{3}{4}(x + y)i.$ Both parts lie in $[-1, 1]$ iff $|x - y| \le \dfrac{4}{3}$ and $|x + y| \le \dfrac{4}{3}.$

Within the square $S$ (area $4$ ) these fail only in four corner triangles. Near $(1, 1),$ the line $x + y = \dfrac{4}{3}$ cuts off a right triangle with legs $\dfrac{2}{3},$ area $\dfrac{1}{2} \cdot \dfrac{2}{3} \cdot \dfrac{2}{3} = \dfrac{2}{9}.$

The four corners remove $4 \cdot \dfrac{2}{9} = \dfrac{8}{9},$ leaving $4 - \dfrac{8}{9} = \dfrac{28}{9}.$ The probability is $\dfrac{\frac{28}{9}}{4} = \dfrac{7}{9}.$

Thus, the correct answer is D.

24.

For how many values of $x$ in $[0, \pi]$ is $\sin^{-1}(\sin 6x) = \cos^{-1}(\cos x)?$

Note: The functions $\sin^{-1} = \arcsin$ and $\cos^{-1} = \arccos$ denote inverse trigonometric functions.

$3$

$4$

$5$

$6$

$7$

Concepts:trigonometry casework

Difficulty rating: 2460

Small Hint:

On $[0, \pi],\ \cos^{-1}(\cos x) = x,$ and the left side lies in $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}],$ so any solution needs $x \in [0, \tfrac{\pi}{2}]$

Big Hint:

Since $\sin$ is one-to-one on $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}],$ the equation reduces to $\sin 6x = \sin x$ on $[0, \tfrac{\pi}{2}]$

Solution:

On $[0, \pi],\ \cos^{-1}(\cos x) = x.$ Since $\sin^{-1}$ takes values in $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}],$ any solution requires $x \in [0, \tfrac{\pi}{2}],$ where the equation becomes $\sin 6x = \sin x.$

The equation $\sin6x=\sin x$ holds exactly when $\begin{aligned} 6x&=x+2k\pi, \\ \text{or}\qquad 6x&=\pi-x+2k\pi \end{aligned}$ for some integer $k.$

The first family gives $x=\tfrac{2k\pi}{5},$ contributing $0$ and $\tfrac{2\pi}{5}.$ The second gives $x=\tfrac{(2k+1)\pi}{7},$ contributing $\tfrac\pi7$ and $\tfrac{3\pi}{7}.$ Hence there are $4$ solutions in $[0,\tfrac\pi2].$

Thus, the correct answer is B.

25.

The set $G$ is defined by the points $(x, y)$ with integer coordinates, $3 \le |x| \le 7,$ and $3 \le |y| \le 7.$ How many squares of side at least $6$ have their four vertices in $G?$

$125$

$150$

$175$

$200$

$225$

Concepts:lattice point bijection counting shapes in figures

Difficulty rating: 2650

Small Hint:

$G$ splits into four $5 \times 5$ blocks, one per quadrant; a side- $\ge 6$ square takes exactly one vertex from each block

Big Hint:

Slide the four blocks together onto a single $5 \times 5$ grid $G';$ each valid square maps to a point of $G'$ or a square in $G'$

Solution:

$G$ consists of four $5 \times 5$ blocks $G_1, \ldots, G_4,$ one in each quadrant. Any square of side $\ge 6$ uses exactly one vertex in each block, since two points in one block are less than $6$ apart while points in different blocks are at least $6$ apart.

Sliding each block inward by $(\pm 5, \pm 5)$ superimposes them on one $5 \times 5$ grid $G'$ (points with $|x|, |y| \le 2$ ). Each such square maps to either a single point of $G'$ or a square in $G'.$ So the count equals the number of points of $G'$ plus $4$ times the number of squares with vertices in $G'.$

The grid has $4^2+3^2+2^2+1^2=30$ axis-parallel squares. For a tilted square, let one side move $a$ units horizontally and $b$ units vertically, where $a,b\gt0$ and $a+b\le4.$ For each ordered pair $(a,b),$ there are $(5-a-b)^2$ placements. The totals for $a+b=2,3,4$ are $9,8,3,$ respectively, giving $20$ tilted squares and $50$ squares altogether. Therefore the required count is $25+4\cdot50=225.$

Thus, the correct answer is E.