2011 AMC 12A — Problems & Online Timed Exam

Timed

1:15:00

1.

A cell phone plan costs $\$20$ each month, plus $5$ ¢ per text message sent, plus $10$ ¢ for each minute used over $30$ hours. In January Michelle sent $100$ text messages and talked for $30.5$ hours. How much did she have to pay?

$\$24.00$

$\$24.50$

$\$25.50$

$\$28.00$

$\$30.00$

Answer: D

Concepts:money unit conversion

Difficulty rating: 840

Small Hint:

$30.5$ hours is $30$ minutes over the $30$ -hour allowance

Big Hint:

Add the base fee, $5$ ¢ times the texts, and $10$ ¢ times the overage minutes

Solution:

The text charge is $100 \cdot 5 = 500$ cents $= \$5.$ She talked $30$ minutes past the $30$ -hour allowance, so the overage is $30 \cdot 10 = 300$ cents $= \$3.$

The total is $\$20 + \$5 + \$3 = \$28.$

Thus, the correct answer is D.

2.

There are $5$ coins placed flat on a table according to the figure. What is the order of the coins from top to bottom?

$(C, A, E, D, B)$

$(C, A, D, E, B)$

$(C, D, E, A, B)$

$(C, E, A, D, B)$

$(C, E, D, A, B)$

Answer: E

Concepts:logical deduction

Difficulty rating: 840

Small Hint:

A coin whose outline is a complete, unbroken circle lies on top

Big Hint:

At each overlap, the coin whose arc is drawn without interruption is the higher one

Solution:

Coin $C$ is drawn as a complete, unbroken circle, so nothing covers it and it lies on top.

Reading the remaining overlaps, each coin’s uncovered arc shows it sits above the next: $C$ covers $E,$ $E$ covers $D,$ $D$ covers $B,$ and $A$ covers $B$ while lying under the others. This gives the top-to-bottom order $(C, E, D, A, B).$

Thus, the correct answer is E.

3.

A small bottle of shampoo can hold $35$ milliliters of shampoo, whereas a large bottle can hold $500$ milliliters of shampoo. Jasmine wants to buy the minimum number of small bottles necessary to completely fill a large bottle. How many bottles must she buy?

$11$

$12$

$13$

$14$

$15$

Answer: E

Concepts:floor and ceiling functions estimation

Difficulty rating: 880

Small Hint:

Divide $500$ by $35$

Big Hint:

Since $14$ bottles fall short of $500,$ round the quotient up

Solution:

Fourteen bottles hold $14 \cdot 35 = 490$ milliliters, which is not enough. Fifteen bottles hold $15 \cdot 35 = 525$ milliliters, which suffices.

So Jasmine needs $15$ bottles.

Thus, the correct answer is E.

4.

At an elementary school, the students in third grade, fourth grade, and fifth grade run an average of $12,$ $15,$ and $10$ minutes per day, respectively. There are twice as many third graders as fourth graders, and twice as many fourth graders as fifth graders. What is the average number of minutes run per day by these students?

$12$

$\dfrac{37}{3}$

$\dfrac{88}{7}$

$13$

$14$

Answer: C

Concepts:weighted mean ratio and proportion

Difficulty rating: 1040

Small Hint:

Let the number of fifth graders be $1,$ so fourth graders is $2$ and third graders is $4$

Big Hint:

The average is $\dfrac{4 \cdot 12 + 2 \cdot 15 + 1 \cdot 10}{4 + 2 + 1}$

Solution:

Take the grade sizes in the ratio $4 : 2 : 1$ for third, fourth, and fifth grades. The weighted average is $\begin{aligned} \dfrac{4\cdot12+2\cdot15+10}{7} \\ = \dfrac{48+30+10}{7} \\ = \dfrac{88}{7}. \end{aligned}$

Thus, the correct answer is C.

5.

Last summer $30\%$ of the birds living on Town Lake were geese, $25\%$ were swans, $10\%$ were herons, and $35\%$ were ducks. What percent of the birds that were not swans were geese?

$20$

$30$

$40$

$50$

$60$

Answer: C

Concepts:percentage conditional probability

Difficulty rating: 990

Small Hint:

The non-swan birds make up $75\%$ of the total

Big Hint:

Compute $\dfrac{30}{75}$ as a percent

Solution:

The birds that are not swans make up $100\% - 25\% = 75\%$ of the total, and geese are $30\%$ of the total. The requested fraction is $\dfrac{30}{75} = \dfrac{2}{5} = 40\%.$

Thus, the correct answer is C.

6.

The players on a basketball team made some three-point shots, some two-point shots, and some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team’s total score was $61$ points. How many free throws did they make?

$13$

$14$

$15$

$16$

$17$

Answer: A

Concepts:system of equations substitution

Difficulty rating: 1170

Small Hint:

Let $a$ be the number of two-point shots; the points from two-point and three-point shots are each $2a$

Big Hint:

The free throws number $a + 1,$ so the total is $2a + 2a + (a + 1)$

Solution:

Let $a$ be the number of two-point shots. The two-point shots score $2a$ points, and the three-point shots score the same $2a$ points. The free throws number $a + 1$ and score $a + 1$ points.

The total is $\begin{aligned} 2a + 2a + (a + 1) \\ = 5a + 1 = 61, \end{aligned}$ so $a = 12$ and the free throws number $a + 1 = 13.$

Thus, the correct answer is A.

7.

A majority of the $30$ students in Ms. Demeanor’s class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than $1.$ The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was $\$17.71.$ What was the cost of a pencil in cents?

$7$

$11$

$17$

$23$

$77$

Answer: B

Concepts:prime factorization divisibility bounding to limit cases

Difficulty rating: 1370

Small Hint:

Factor $1771 = 7 \cdot 11 \cdot 23$ and let (students)(pencils)(cost) $= 1771$

Big Hint:

A majority means more than $15$ students, so the number of students is $23$

Solution:

Total cents is $1771 = 7 \cdot 11 \cdot 23.$ Writing (students)(pencils each)(cost per pencil) $= 1771,$ the number of students is a divisor of $1771$ that is a majority of $30,$ hence more than $15.$ The only such divisor is $23.$

Then (pencils)(cost) $= 77 = 7 \cdot 11$ with cost $\gt$ pencils $\gt 1,$ forcing $7$ pencils at $11$ cents each.

Thus, the correct answer is B.

8.

In the eight-term sequence $A,$ $B,$ $C,$ $D,$ $E,$ $F,$ $G,$ $H,$ the value of $C$ is $5$ and the sum of any three consecutive terms is $30.$ What is $A + H?$

$17$

$18$

$25$

$26$

$43$

Answer: C

Concepts:pattern recognition algebraic manipulation

Difficulty rating: 1190

Small Hint:

Consecutive triples having equal sums forces the sequence to repeat every three terms

Big Hint:

So $H$ equals $B,$ which means $A + H$ is the same as $A + B$

Solution:

Since $A + B + C = B + C + D = 30,$ we get $D = A,$ and likewise the sequence repeats with period $3.$ Thus $H,$ the eighth term, equals $B.$

From $A + B + C = 30$ and $C = 5,$ we have $A + H = A + B = 30 - 5 = 25.$

Thus, the correct answer is C.

9.

At a twins and triplets convention, there were $9$ sets of twins and $6$ sets of triplets, all from different families. Each twin shook hands with all the twins except his/her sibling and with half the triplets. Each triplet shook hands with all the triplets except his/her siblings and with half the twins. How many handshakes took place?

$324$

$441$

$630$

$648$

$882$

Answer: B

Concepts:counting pairs casework

Difficulty rating: 1440

Small Hint:

There are $18$ twins and $18$ triplets; count twin-twin, triplet-triplet, and twin-triplet handshakes separately

Big Hint:

Each twin greets $16$ twins and each triplet greets $15$ triplets; divide those two totals by $2$

Solution:

There are $18$ twins and $18$ triplets.

Twin-twin handshakes: each twin shakes $18 - 2 = 16$ other twins, giving $\dfrac{18 \cdot 16}{2} = 144.$

Triplet-triplet handshakes: each triplet shakes $18 - 3 = 15$ other triplets, giving $\dfrac{18 \cdot 15}{2} = 135.$

Twin-triplet handshakes: each twin shakes half the $18$ triplets, giving $18 \cdot 9 = 162$ (each such handshake counted once).

The total is $144 + 135 + 162 = 441.$

Thus, the correct answer is B.

10.

A pair of standard $6$ -sided fair dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle’s circumference?

$\dfrac{1}{36}$

$\dfrac{1}{12}$

$\dfrac{1}{6}$

$\dfrac{1}{4}$

$\dfrac{5}{18}$

Answer: B

Concepts:dice (probability)circle inequality

Difficulty rating: 1370

Small Hint:

With diameter $d,$ the area is $\dfrac{\pi d^2}{4}$ and the circumference is $\pi d$

Big Hint:

The inequality $\dfrac{\pi d^2}{4} \lt \pi d$ reduces to $d \lt 4$

Solution:

For diameter $d,$ area $\lt$ circumference means $\dfrac{\pi d^2}{4} \lt \pi d,$ i.e. $d \lt 4.$ Since $d \ge 2,$ this needs a sum of $2$ or $3.$

A sum of $2$ has probability $\dfrac{1}{36}$ and a sum of $3$ has probability $\dfrac{2}{36},$ totaling $\dfrac{3}{36} = \dfrac{1}{12}.$

Thus, the correct answer is B.

11.

Circles $A,$ $B,$ and $C$ each have radius $1.$ Circles $A$ and $B$ share one point of tangency. Circle $C$ has a point of tangency with the midpoint of $\overline{AB}.$ What is the area inside circle $C$ but outside circle $A$ and circle $B?$

$3 - \dfrac{\pi}{2}$

$\dfrac{\pi}{2}$

$2$

$\dfrac{3\pi}{4}$

$1 + \dfrac{\pi}{2}$

Answer: C

Concepts:circle area tangent circles area decomposition

Difficulty rating: 1540

Small Hint:

Place the centers at $A(-1,0),$ $B(1,0),$ and $C(0,1);$ the distance from $C$ to each of $A, B$ is $\sqrt2$

Big Hint:

The wanted area is the area of $C$ minus the two lens-shaped overlaps of $C$ with $A$ and with $B$

Solution:

Place $A = (-1, 0),$ $B = (1, 0),$ so their tangency point is the origin, the midpoint of $\overline{AB}.$ Then $C = (0, 1),$ since $C$ passes through the origin.

The distance from $C$ to $A$ (and to $B$ ) is $\sqrt2.$ Two unit circles whose centers are $\sqrt2$ apart overlap in a lens of area $\begin{aligned} 2\cos^{-1}\!\left(\tfrac{\sqrt2}{2}\right) \\ {}- \tfrac{\sqrt2}{2}\sqrt{4 - 2} \\ = 2 \cdot \tfrac{\pi}{4} - 1 = \tfrac{\pi}{2} - 1. \end{aligned}$

Circles $A$ and $B$ meet only at the origin, so the two lenses do not overlap. The wanted area is $\pi - 2\left(\tfrac{\pi}{2} - 1\right) = 2.$

Thus, the correct answer is C.

12.

A power boat and a raft both left dock $A$ on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock $B$ downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river $9$ hours after leaving dock $A.$ How many hours did it take the power boat to go from $A$ to $B?$

$3$

$3.5$

$4$

$4.5$

$5$

Answer: D

Concepts:relative speed distance rate and time symmetry

Difficulty rating: 1580

Small Hint:

Work in the frame of the water, in which the raft stays put

Big Hint:

Relative to the water the boat travels at the same speed going out and coming back, so it takes equal times each way

Solution:

Measure everything relative to the water. In that frame the raft is stationary at the point where the boat started, and the boat moves at its constant speed $v$ relative to the water, both downstream and upstream.

The boat leaves the raft, travels away for some time, then returns to it at the same relative speed, so it spends equal times going and returning. Hence the outbound leg to $B$ takes half of $9,$ which is $4.5$ hours.

Thus, the correct answer is D.

13.

Triangle $ABC$ has side-lengths $AB = 12,$ $BC = 24,$ and $AC = 18.$ The line through the incenter of $\triangle ABC$ parallel to $\overline{BC}$ intersects $\overline{AB}$ at $M$ and $\overline{AC}$ at $N.$ What is the perimeter of $\triangle AMN?$

$27$

$30$

$33$

$36$

$42$

Answer: B

Concepts:incircle, incenter, and inradius parallel lines isosceles triangle

Difficulty rating: 1600

Small Hint:

Let $I$ be the incenter. Since $MN \parallel BC,$ the angle $\angle MIB$ equals $\angle IBC$

Big Hint:

That makes $\triangle MBI$ isosceles with $MB = MI,$ and similarly $NC = NI$

Solution:

Let $I$ be the incenter. Because $\overline{BI}$ bisects $\angle B$ and $MN \parallel BC,$ alternate angles give $\angle MIB = \angle IBC = \angle MBI,$ so $\triangle MBI$ is isosceles with $MB = MI.$ Similarly $NC = NI.$

Therefore the perimeter of $\triangle AMN$ is $\begin{gathered} AM + MN + NA \\ = AM + (MI + IN) + NA \\ = AM + MB + NC + NA \\ = AB + AC = 12 + 18 = 30. \end{gathered}$

Thus, the correct answer is B.

14.

Suppose $a$ and $b$ are single-digit positive integers chosen independently and at random. What is the probability that the point $(a, b)$ lies above the parabola $y = ax^2 - bx?$

$\dfrac{11}{81}$

$\dfrac{13}{81}$

$\dfrac{5}{27}$

$\dfrac{17}{81}$

$\dfrac{19}{81}$

Answer: E

Concepts:parabola basic probability casework

Difficulty rating: 1690

Small Hint:

The point is above the parabola when $b \gt a \cdot a^2 - b \cdot a,$ i.e. $b(a + 1) \gt a^3$

Big Hint:

Count valid $b$ in $\{1, \ldots, 9\}$ for each $a;$ only $a = 1, 2, 3$ allow any

Solution:

Substituting $x = a,$ $y = b,$ the point is above the parabola when $b \gt a^3 - ab,$ i.e. $b(a + 1) \gt a^3.$

For $a = 1:$ $b \gt \tfrac12,$ all $9$ values work. For $a = 2:$ $b \gt \tfrac83,$ so $b \ge 3,$ giving $7.$ For $a = 3:$ $b \gt \tfrac{27}{4} = 6.75,$ so $b \ge 7,$ giving $3.$ For $a \ge 4,$ no $b \le 9$ works.

The count is $9 + 7 + 3 = 19$ out of $81,$ so the probability is $\dfrac{19}{81}.$

Thus, the correct answer is E.

15.

The circular base of a hemisphere of radius $2$ rests on the base of a square pyramid of height $6.$ The hemisphere is tangent to the other four faces of the pyramid. What is the edge-length of the base of the pyramid?

$3\sqrt{2}$

$\dfrac{13}{3}$

$4\sqrt{2}$

$6$

$\dfrac{13}{2}$

Answer: A

Concepts:3D geometry pyramid distance formula

Difficulty rating: 1870

Small Hint:

Take the cross-section through the apex and the midpoints of two opposite base edges

Big Hint:

In that cross-section, the slant line runs from $\left(\tfrac{s}{2}, 0\right)$ to $(0, 6),$ and its distance from the center is $2$

Solution:

Let the base have side $s,$ centered at the origin, with apex at height $6.$ Cut with the vertical plane through the apex and the midpoints of two opposite base edges. The slant face appears as the line from $\left(\tfrac{s}{2}, 0\right)$ to $(0, 6).$

This line is $\tfrac{2}{s}x + \tfrac16 y = 1.$ The hemisphere is tangent to the face, so the distance from the origin to this line is the radius $2:$ $\dfrac{1}{\sqrt{\tfrac{4}{s^2} + \tfrac{1}{36}}} = 2.$

Then $\tfrac{4}{s^2} + \tfrac{1}{36} = \tfrac14,$ so $\tfrac{4}{s^2} = \tfrac{2}{9}$ and $s^2 = 18,$ giving $s = 3\sqrt2.$

Thus, the correct answer is A.

16.

Each vertex of convex pentagon $ABCDE$ is to be assigned a color. There are $6$ colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?

$2520$

$2880$

$3120$

$3250$

$3750$

Answer: C

Concepts:graph theory inclusion-exclusion

Difficulty rating: 1820

Small Hint:

The five diagonals $AC, CE, EB, BD, DA$ form a single $5$ -cycle among the vertices

Big Hint:

The number of proper colorings of a cycle of length $n$ with $k$ colors is $(k-1)^n + (-1)^n (k-1)$

Solution:

The diagonals connect the vertices in the order $A - C - E - B - D - A,$ which is a $5$ -cycle. The condition is exactly that this cycle is properly colored.

The number of proper $k$ -colorings of a cycle of length $n$ is $(k-1)^n + (-1)^n (k-1).$ With $n = 5$ and $k = 6,$ $\begin{gathered} 5^5 + (-1)^5 \cdot 5 \\ = 3125 - 5 = 3120. \end{gathered}$

Thus, the correct answer is C.

17.

Circles with radii $1,$ $2,$ and $3$ are mutually externally tangent. What is the area of the triangle determined by the points of tangency?

$\dfrac{3}{5}$

$\dfrac{4}{5}$

$1$

$\dfrac{6}{5}$

$\dfrac{4}{3}$

Answer: D

Concepts:tangent circles coordinate geometry shoelace formula

Difficulty rating: 1920

Small Hint:

The centers form a triangle with sides $1+2,$ $2+3,$ $1+3,$ i.e. a $3$ - $4$ - $5$ right triangle

Big Hint:

Put the right angle at the center of the radius- $1$ circle and locate each tangency point on the connecting segment

Solution:

The centers are separated by the sums of radii: $3,$ $4,$ and $5,$ a right triangle with the right angle at the radius- $1$ center. Place that center at $(0,0),$ the radius- $2$ center at $(3,0),$ and the radius- $3$ center at $(0,4).$

The tangency points lie on the segments at distances equal to the radii: $(1, 0),$ $(0, 1),$ and on the hypotenuse at $(3,0) + 2 \cdot \tfrac{(-3,4)}{5} = \left(\tfrac95, \tfrac85\right).$

By the shoelace formula the area is $\begin{gathered} \tfrac12\left| 1\left(1 - \tfrac85\right) + 0 + \tfrac95(0 - 1) \right| \\ = \tfrac12 \cdot \tfrac{12}{5} = \tfrac65. \end{gathered}$

Thus, the correct answer is D.

18.

Suppose that $|x + y| + |x - y| = 2.$ What is the maximum possible value of $x^2 - 6x + y^2?$

$5$

$6$

$7$

$8$

$9$

Answer: D

Concepts:absolute value optimization square (geometry)

Difficulty rating: 1840

Small Hint:

The equation $|x+y| + |x-y| = 2$ means $\max(|x|, |y|) = 1,$ the boundary of the square $[-1,1]^2$

Big Hint:

On that square, $x^2 - 6x + y^2$ is largest when $x = -1$ and $y = \pm 1$

Solution:

The identity $|x+y| + |x-y|$ $= 2\max(|x|, |y|)$ turns the condition into $\max(|x|, |y|) = 1,$ the boundary of the square with $|x| \le 1$ and $|y| \le 1.$

On this region $x^2 - 6x + y^2$ increases as $x$ decreases and as $y^2$ increases, so the maximum is at $x = -1,$ $y = \pm 1:$ $1 + 6 + 1 = 8.$

Thus, the correct answer is D.

19.

At a competition with $N$ players, the number of players given elite status is equal to $2^{1 + \lfloor \log_2 (N - 1) \rfloor} - N.$ Suppose that $19$ players are given elite status. What is the sum of the two smallest possible values of $N?$

Note: $\lfloor x \rfloor$ is the greatest integer less than or equal to $x.$

$38$

$90$

$154$

$406$

$1024$

Answer: C

Concepts:floor and ceiling functions power of 2 casework

Difficulty rating: 2150

Small Hint:

Let $m = \lfloor \log_2 (N-1) \rfloor,$ so $N = 2^{m+1} - 19$

Big Hint:

For that $m$ to be consistent, $2^m \le N - 1,$ which forces $2^m \ge 20$

Solution:

Let $m = \lfloor \log_2 (N - 1) \rfloor,$ so the elite count is $2^{m+1} - N = 19,$ giving $N = 2^{m+1} - 19.$

Consistency requires $2^m \le N - 1 = 2^{m+1} - 20,$ i.e. $2^m \ge 20,$ so $m \ge 5.$

The two smallest choices are $m = 5$ giving $N = 64 - 19 = 45,$ and $m = 6$ giving $N = 128 - 19 = 109.$ Their sum is $154.$

Thus, the correct answer is C.

20.

Let $f(x) = ax^2 + bx + c,$ where $a,$ $b,$ and $c$ are integers. Suppose that $f(1) = 0,$ $50 \lt f(7) \lt 60,$ $70 \lt f(8) \lt 80,$ and $5000k \lt f(100) \lt 5000(k+1)$ for some integer $k.$ What is $k?$

$1$

$2$

$3$

$4$

$5$

Answer: C

Concepts:quadratic system of equations bounding to limit cases

Difficulty rating: 2030

Small Hint:

From $f(1) = 0,$ $c = -a - b,$ so $f(7) = 6(8a + b)$ and $f(8) = 7(9a + b)$

Big Hint:

The bounds force $8a + b = 9$ and $9a + b = 11$

Solution:

Since $f(1) = a + b + c = 0,$ we have $c = -a - b.$ Then $f(7) = 48a + 6b = 6(8a + b),$ $f(8) = 63a + 7b = 7(9a + b).$

From $50 \lt 6(8a + b) \lt 60$ we get $8a + b = 9,$ and from $70 \lt 7(9a + b) \lt 80$ we get $9a + b = 11.$ Subtracting, $a = 2,$ then $b = -7$ and $c = 5.$

So $f(100) = 20000 - 700 + 5$ $= 19305,$ which lies in $5000 \cdot 3 \lt 19305 \lt 5000 \cdot 4,$ giving $k = 3.$

Thus, the correct answer is C.

21.

Let $f_1(x) = \sqrt{1 - x},$ and for integers $n \ge 2,$ let $f_n(x) = f_{n-1}\!\left(\sqrt{n^2 - x}\right).$ If $N$ is the largest value of $n$ for which the domain of $f_n$ is nonempty, the domain of $f_N$ is $\{c\}.$ What is $N + c?$

$-226$

$-144$

$-20$

$20$

$144$

Answer: A

Concepts:function radical recursion

Difficulty rating: 2270

Small Hint:

Build the domains outward: $f_n$ needs $\sqrt{n^2 - x}$ to lie in the domain of $f_{n-1}$

Big Hint:

The first three domains are $(-\infty,1],$ $[3,4],$ and $[-7,0];$ continue while remembering that a square root is nonnegative

Solution:

Each step requires $\sqrt{n^2 - x}$ to lie in the domain of $f_{n-1}.$ Tracking the domains:

$f_1: (-\infty, 1].$ $f_2: \sqrt{4 - x} \in (-\infty, 1] \Rightarrow [3, 4].$ $f_3: \sqrt{9 - x} \in [3, 4] \Rightarrow [-7, 0].$ $f_4: \sqrt{16 - x} \in [-7, 0] \Rightarrow \{16\}$ (only the value $0$ is possible). $f_5: \sqrt{25 - x} = 16 \Rightarrow \{-231\}.$

For $f_6$ we would need $\sqrt{36 - x} = -231,$ impossible, so the domain is empty. Hence $N = 5,$ $c = -231,$ and $N + c = -226.$

Thus, the correct answer is A.

22.

Let $R$ be a square region and $n \ge 4$ an integer. A point $X$ in the interior of $R$ is called $n$ -ray partitional if there are $n$ rays emanating from $X$ that divide $R$ into $n$ triangles of equal area. How many points are $100$ -ray partitional but not $60$ -ray partitional?

$1500$

$1560$

$2320$

$2480$

$2500$

Answer: C

Concepts:area lattice point complementary counting

Difficulty rating: 2460

Small Hint:

The $n$ -ray partitional points form a grid: for $n = 2m,$ they are the interior points $\left(\tfrac{i}{m}, \tfrac{j}{m}\right),$ an $(m-1) \times (m-1)$ array

Big Hint:

A point is both $100$ - and $60$ -ray partitional exactly when it is $20$ -ray partitional

Solution:

Scale the square to $[0,1]^2$ and write $X=(x,y).$ The rays must include those through the four vertices. Every small triangle has area $\frac{1}{n}.$ The triangles whose bases partition the bottom side together have area $\frac{y}{2},$ so their number is $\frac{ny}{2}.$ Similarly, the numbers along the top, left, and right sides are $\frac{n(1-y)}{2},$ $\frac{nx}{2},$ and $\frac{n(1-x)}{2}.$

These four numbers must be positive integers. Hence $n$ is even and $\begin{gathered} X=\left(\dfrac{2i}{n},\dfrac{2j}{n}\right), \\ 1\le i,j\le\dfrac n2-1. \end{gathered}$ Conversely, partitioning each side into the indicated number of equal segments and joining the division points to $X$ produces $n$ equal-area triangles. Thus these are exactly the partitional points.

For $n=100,$ the points are $(\frac{i}{50},\frac{j}{50})$ with $1\le i,j\le49,$ giving $49^2=2401.$ Such a point is also $60$ -ray partitional exactly when $\frac{i}{50}=\frac{c}{30}$ and $\frac{j}{50}=\frac{d}{30}$ for integers $c,d.$ Thus $i$ and $j$ must both be multiples of $5.$ There are $9$ choices for each, so the overlap has $9^2=81$ points.

So the count is $2401 - 81 = 2320.$

Thus, the correct answer is C.

23.

Let $f(z) = \dfrac{z + a}{z + b}$ and $g(z) = f(f(z)),$ where $a$ and $b$ are complex numbers. Suppose that $|a| = 1$ and $g(g(z)) = z$ for all $z$ for which $g(g(z))$ is defined. What is the difference between the largest and smallest possible values of $|b|?$

$0$

$\sqrt{2} - 1$

$\sqrt{3} - 1$

$1$

$2$

Answer: C

Concepts:complex number matrix roots of unity

Difficulty rating: 2560

Small Hint:

Represent $f$ by the matrix $\begin{pmatrix} 1 & a \\ 1 & b \end{pmatrix};$ then $g(g(z)) = z$ means $f^4$ is the identity

Big Hint:

The order- $4$ case forces $(\operatorname{tr})^2 = 2\det,$ which gives $b^2 = -(1 + 2a)$

Solution:

Direct composition gives $g(z)=\dfrac{Az+B}{Cz+D},$ where $A=1+a,$ $B=a(1+b),$ $C=1+b,$ and $D=a+b^2.$

The matrix $\begin{pmatrix}A&B\\C&D\end{pmatrix}$ represents $g.$ For $g\circ g$ to be the identity, its square must be scalar. Comparing the off-diagonal entries and the two diagonal entries gives two possibilities: either $B=C=0$ and $A=D,$ or $A+D=0.$ The first gives $b=-1$ (with $a\ne-1$ ); the second gives $b^2=-(1+2a).$

In the second case, $|b|^2=|1+2a|.$ As $a$ runs around the unit circle, $|1+2a|$ ranges from $1$ to $3,$ so $1\le|b|\le\sqrt3.$ Both endpoints occur: take $(a,b)=(-1,1)$ and $(a,b)=(1,i\sqrt3).$ The separate case $b=-1$ also has $|b|=1.$ Therefore the requested difference is $\sqrt3-1.$

Thus, the correct answer is C.

24.

Consider all quadrilaterals $ABCD$ such that $AB = 14,$ $BC = 9,$ $CD = 7,$ and $DA = 12.$ What is the radius of the largest possible circle that fits inside or on the boundary of such a quadrilateral?

$\sqrt{15}$

$\sqrt{21}$

$2\sqrt{6}$

$5$

$2\sqrt{7}$

Answer: C

Concepts:incircle, incenter, and inradius Brahmagupta’s Formula cyclic quadrilateral optimization

Difficulty rating: 2460

Small Hint:

If a circle of radius $r$ fits, splitting from its center shows the quadrilateral’s area is at least $21r$

Big Hint:

Bretschneider’s inequality bounds the area by the cyclic case; the equality case is also tangential because opposite side sums are equal

Solution:

Suppose a circle of radius $r$ centered at $X$ fits in one of the quadrilaterals. If $h_1,h_2,h_3,h_4$ are the distances from $X$ to the four side lines, then each $h_i\ge r.$ Splitting the quadrilateral into four triangles gives $\begin{aligned} K&=\dfrac12(14h_1+9h_2 \\ &\qquad+7h_3+12h_4) \\ &\ge21r. \end{aligned}$

Bretschneider’s inequality bounds the area of any quadrilateral with these sides by the cyclic case: $\begin{aligned} K^2&\le(21-14)(21-9) \\ &\qquad\cdot(21-7)(21-12) \\ &=7\cdot12\cdot14\cdot9, \\ K&\le42\sqrt6. \end{aligned}$

Therefore $r\le \frac{K}{21}\le2\sqrt6.$ Equality is attainable: the cyclic quadrilateral with these sides is also tangential because $14+7=9+12,$ and its incircle has radius $\frac{K}{21}=2\sqrt6.$

Thus, the correct answer is C.

25.

Triangle $ABC$ has $\angle BAC = 60^\circ,$ $\angle CBA \le 90^\circ,$ $BC = 1,$ and $AC \ge AB.$ Let $H,$ $I,$ and $O$ be the orthocenter, incenter, and circumcenter of $\triangle ABC,$ respectively. Assume that the area of the pentagon $BCOIH$ is the maximum possible. What is $\angle CBA?$

$60^\circ$

$72^\circ$

$75^\circ$

$80^\circ$

$90^\circ$

Answer: D

Concepts:circumcircle, circumcenter, and circumradius optimization trigonometry

Difficulty rating: 2840

Small Hint:

Because $\angle BAC = 60^\circ,$ the points $B, C, O, I, H$ all lie on one circle

Big Hint:

On that circle, maximize the quadrilateral $BOIH$ by making the three consecutive subarcs from $O$ to $B$ equal

Solution:

Write $B=\angle CBA$ and $C=\angle BCA=120^\circ-B.$ Since $AC\ge AB,$ we have $B\ge C,$ so $60^\circ\le B\le90^\circ.$ The standard angle formulas give $\begin{aligned} \angle BOC&=2\angle A=120^\circ, \\ \angle BHC&=180^\circ-\angle A=120^\circ, \\ \angle BIC&=90^\circ+\dfrac{\angle A}{2}=120^\circ. \end{aligned}$ Hence $B,C,O,I,H$ lie on one circle.

Also $BC=1$ and $A=60^\circ$ fix the circumradius $OB=OC=\frac{1}{\sqrt3},$ so $\triangle BCO$ and the circle through $B,C,O$ are fixed. Angle chasing at $C$ gives $\begin{aligned} \angle OCI&=30^\circ-\dfrac C2, \\ \angle ICH&=30^\circ-\dfrac C2. \end{aligned}$ Thus the corresponding chords satisfy $OI=IH.$

The pentagon’s area is the fixed area $[BCO]$ plus $[BOIH].$ For two points dividing a fixed arc $OB,$ an inscribed quadrilateral has greatest area when its three consecutive subarcs are equal (equivalently, maximize the sum of their sines). Hence at the maximum $OI=IH=HB.$

In $\triangle BOC,$ $\angle OCB=30^\circ.$ Equal chords make $\angle OCI=\angle ICH,$ $\angle ICH=\angle HCB,$ and each of these angles is $10^\circ.$ Therefore $30^\circ-\tfrac C2=10^\circ,$ so $C=40^\circ$ and $B=80^\circ.$

Thus, the correct answer is D.