2015 AMC 12A — Problems & Online Timed Exam

Timed

1:15:00

1.

What is the value of $(2^0-1+5^2+0)^{-1}\times 5?$

$-125$

$-120$

$\dfrac{1}{5}$

$\dfrac{5}{24}$

$25$

Answer: C

Concepts:order of operations exponent

Difficulty rating: 890

Small Hint:

Evaluate inside the parentheses first, using $2^0=1$

Big Hint:

A $-1$ exponent means take the reciprocal before multiplying by $5$

Solution:

Inside the parentheses, $\begin{aligned} &2^0-1+5^2+0 \\ &\quad = 1-1+25+0 = 25. \end{aligned}$

Then $(25)^{-1}\times 5 = \dfrac{5}{25} = \dfrac{1}{5}.$

Thus, the correct answer is C.

2.

Two of the three sides of a triangle are $20$ and $15.$ Which of the following numbers is not a possible perimeter of the triangle?

$52$

$57$

$62$

$67$

$72$

Answer: E

Concepts:triangle inequality bounding to limit cases

Difficulty rating: 1020

Small Hint:

The third side lies strictly between $20-15$ and $20+15$

Big Hint:

Add $20+15$ to that range to bound the perimeter

Solution:

By the Triangle Inequality, the third side $s$ satisfies $20-15 \lt s \lt 20+15,$ that is $5 \lt s \lt 35.$

The perimeter is $35 + s,$ so it lies strictly between $40$ and $70.$ Among the choices, only $72$ falls outside this range.

Thus, the correct answer is E.

3.

Mr. Patrick teaches math to $15$ students. He was grading tests and found that when he graded everyone’s test except Payton’s, the average grade for the class was $80.$ After he graded Payton’s test, the class average became $81.$ What was Payton’s score on the test?

$81$

$85$

$91$

$94$

$95$

Answer: E

Concepts:mean

Difficulty rating: 1130

Small Hint:

The other $14$ students together scored $14\times 80$

Big Hint:

All $15$ scores together total $15\times 81$ ; subtract

Solution:

The sum of the $14$ other scores was $14\cdot 80 = 1120.$ The sum of all $15$ scores was $15\cdot 81 = 1215.$

Therefore Payton’s score was $1215 - 1120 = 95.$

Thus, the correct answer is E.

4.

The sum of two positive numbers is $5$ times their difference. What is the ratio of the larger number to the smaller number?

$\dfrac{5}{4}$

$\dfrac{3}{2}$

$\dfrac{9}{5}$

$2$

$\dfrac{5}{2}$

Answer: B

Concepts:ratio and proportion linear equation

Difficulty rating: 1200

Small Hint:

Let the numbers be $x \gt y$ and write $x+y=5(x-y)$

Big Hint:

Rearrange to relate $x$ and $y$ , then form $\dfrac{x}{y}$

Solution:

Let $x$ and $y$ be the numbers with $x \gt y \gt 0.$ Then $x+y = 5(x-y).$

Expanding gives $x+y = 5x-5y,$ so $6y = 4x$ and $\dfrac{x}{y} = \dfrac{3}{2}.$

Thus, the correct answer is B.

5.

Amelia needs to estimate the quantity $\dfrac{a}{b}-c,$ where $a,$ $b,$ and $c$ are large positive integers. She rounds each of the integers so that the calculation will be easier to do mentally. In which of these situations will her answer necessarily be greater than the exact value of $\dfrac{a}{b}-c?$

She rounds all three numbers up.

She rounds $a$ and $b$ up, and she rounds $c$ down.

She rounds $a$ and $c$ up, and she rounds $b$ down.

She rounds $a$ up, and she rounds $b$ and $c$ down.

She rounds $c$ up, and she rounds $a$ and $b$ down.

Answer: D

Concepts:estimation inequality

Difficulty rating: 1270

Small Hint:

Increasing $a$ or decreasing $b$ makes $\dfrac{a}{b}$ larger

Big Hint:

Decreasing $c$ makes $-c$ larger; find the choice where every change raises the result

Solution:

To make $\dfrac{a}{b}$ larger, round the numerator $a$ up and the denominator $b$ down. To make $-c$ larger, round $c$ down.

Only choice $\text{(D)}$ does all three: it rounds $a$ up while rounding $b$ and $c$ down, so every change pushes the estimate above the exact value. In the other choices at least one change works the wrong way, so the estimate is not guaranteed to be larger.

Thus, the correct answer is D.

6.

Two years ago Pete was three times as old as his cousin Claire. Two years before that, Pete was four times as old as Claire. In how many years will the ratio of their ages be $2 : 1?$

$2$

$4$

$5$

$6$

$8$

Answer: B

Concepts:ages system of equations

Difficulty rating: 1350

Small Hint:

Let $p,c$ be their current ages: $p-2=3(c-2)$ and $p-4=4(c-4)$

Big Hint:

Solve for $p$ and $c,$ then find $n$ with $p+n = 2(c+n)$

Solution:

Let $p$ and $c$ be Pete’s and Claire’s current ages. Then $p-2 = 3(c-2)$ and $p-4 = 4(c-4).$

Solving these gives $p = 20$ and $c = 8,$ so Pete is $12$ years older than Claire.

The ratio is $2 : 1$ when Claire is $12,$ which is $12 - 8 = 4$ years from now.

Thus, the correct answer is B.

7.

Two right circular cylinders have the same volume. The radius of the second cylinder is $10\%$ more than the radius of the first. What is the relationship between the heights of the two cylinders?

The second height is $10\%$ less than the first.

The first height is $10\%$ more than the second.

The second height is $21\%$ less than the first.

The first height is $21\%$ more than the second.

The second height is $80\%$ of the first.

Answer: D

Concepts:cylinder volume percentage

Difficulty rating: 1380

Small Hint:

Equal volumes give $\pi r^2 h = \pi R^2 H$ with $R = 1.1r$

Big Hint:

Divide by $\pi r^2$ to compare $h$ and $H;$ note $1.1^2 = 1.21$

Solution:

Let $r,h$ and $R,H$ be the radii and heights of the first and second cylinders. The volumes are equal, so $\pi r^2 h = \pi R^2 H,$ and $R = 1.1r.$

Then $\pi r^2 h = \pi(1.1r)^2 H$ $= \pi(1.21 r^2) H.$ Dividing by $\pi r^2$ yields $h = 1.21H,$ so the first height is $21\%$ more than the second.

Thus, the correct answer is D.

8.

The ratio of the length to the width of a rectangle is $4 : 3.$ If the rectangle has diagonal of length $d,$ then the area may be expressed as $kd^2$ for some constant $k.$ What is $k?$

$\dfrac{2}{7}$

$\dfrac{3}{7}$

$\dfrac{12}{25}$

$\dfrac{16}{25}$

$\dfrac{3}{4}$

Answer: C

Concepts:Pythagorean Theorem ratio and proportion area

Difficulty rating: 1440

Small Hint:

Let the sides be $4a$ and $3a,$ so the diagonal is $5a$

Big Hint:

Write the area $12a^2$ in terms of $d = 5a$

Solution:

Let the sides of the rectangle be $4a$ and $3a.$ By the Pythagorean Theorem the diagonal is $5a = d,$ so $a = \dfrac{d}{5}.$

The area is $4a\cdot 3a = 12a^2$ $= 12\left(\dfrac{d}{5}\right)^2$ $= \dfrac{12}{25}d^2,$ so $k = \dfrac{12}{25}.$

Thus, the correct answer is C.

9.

A box contains $2$ red marbles, $2$ green marbles, and $2$ yellow marbles. Carol takes $2$ marbles from the box at random; then Claudia takes $2$ of the remaining marbles at random; and then Cheryl takes the last $2$ marbles. What is the probability that Cheryl gets $2$ marbles of the same color?

$\dfrac{1}{10}$

$\dfrac{1}{6}$

$\dfrac{1}{5}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

Answer: C

Concepts:basic probability symmetry

Difficulty rating: 1500

Small Hint:

Because all draws are random, Cheryl’s marbles are just a random pair of the six

Big Hint:

Fix her first marble; her second is equally likely to be any of the other $5$

Solution:

Because the marbles left for Cheryl are determined at random, her two marbles are equally likely to be any pair. Fixing her first marble, the second is equally likely to be any of the $5$ remaining marbles.

Exactly one of those $5$ matches the first marble in color, so the probability is $\dfrac{1}{5}.$

Thus, the correct answer is C.

10.

Integers $x$ and $y$ with $x \gt y \gt 0$ satisfy $x+y+xy = 80.$ What is $x?$

$8$

$10$

$15$

$18$

$26$

Answer: E

Concepts:Simon’s Favorite Factoring Trick Diophantine Equation

Difficulty rating: 1530

Small Hint:

Add $1$ to both sides so the left side factors

Big Hint:

$(x+1)(y+1) = 81;$ use $x \gt y \gt 0$ with $81 = 3^4$

Solution:

Adding $1$ to both sides and factoring gives $(x+1)(y+1) = 81 = 3^4.$

Because $x$ and $y$ are distinct positive integers with $x \gt y,$ the only possibility is $x+1 = 3^3 = 27$ and $y+1 = 3.$ Therefore $x = 26.$

Thus, the correct answer is E.

11.

On a sheet of paper, Isabella draws a circle of radius $2,$ a circle of radius $3,$ and all possible lines simultaneously tangent to both circles. Isabella notices that she has drawn exactly $k \ge 0$ lines. How many different values of $k$ are possible?

$2$

$3$

$4$

$5$

$6$

Answer: D

Concepts:tangent line circle casework

Difficulty rating: 1570

Small Hint:

The number of common tangents depends on how the two circles are positioned

Big Hint:

Move the circles from one inside the other out to fully separated and count at each stage

Solution:

The number of common tangent lines depends on the relative position of the two circles:

If the smaller circle is inside the larger, there are $0$ tangents. If it is internally tangent, there is $1.$ If the circles intersect at two points, there are $2.$ If they are externally tangent, there are $3.$ If they are separated, there are $4.$

Thus $k$ can be any of $0, 1, 2, 3, 4,$ which gives $5$ possible values.

Thus, the correct answer is D.

12.

The parabolas $y = ax^2 - 2$ and $y = 4 - bx^2$ intersect the coordinate axes in exactly four points, and these four points are the vertices of a kite of area $12.$ What is $a + b?$

$1$

$1.5$

$2$

$2.5$

$3$

Answer: B

Concepts:parabola kite coordinate geometry area

Difficulty rating: 1630

Small Hint:

The two $y$ -intercepts are $-2$ and $4;$ the $x$ -intercepts are symmetric about the $y$ -axis

Big Hint:

The kite’s diagonals are the vertical distance $6$ and the horizontal spread of the $x$ -intercepts

Solution:

The $y$ -intercepts of the two parabolas are $-2$ and $4.$ To intersect the $x$ -axis, the first parabola opens upward and the second opens downward, so their $x$ -intercepts are $\pm t$ for some $t \gt 0.$

The kite has one diagonal of length $4 - (-2) = 6$ along the $y$ -axis and the other of length $2t.$ Its area is $\dfrac{1}{2}\cdot 6\cdot 2t = 6t = 12,$ so $t = 2.$

Thus the $x$ -intercepts are $\pm 2.$ For the first parabola, $0 = a(2)^2 - 2$ gives $a = \dfrac{1}{2};$ for the second, $0 = 4 - b(2)^2$ gives $b = 1.$ Therefore $a + b = 1.5.$

Thus, the correct answer is B.

13.

A league with $12$ teams holds a round-robin tournament, with each team playing every other team exactly once. Games either end with one team victorious or else end in a draw. A team scores $2$ points for every game it wins and $1$ point for every game it draws. Which of the following is not a true statement about the list of $12$ scores?

There must be an even number of odd scores.

There must be an even number of even scores.

There cannot be two scores of $0.$

The sum of the scores must be at least $100.$

The highest score must be at least $12.$

Answer: E

Concepts:parity invariant counterexample

Difficulty rating: 1660

Small Hint:

Every game contributes exactly $2$ points in total, so the sum of all scores is fixed

Big Hint:

Consider what happens if every single game ends in a draw

Solution:

Each of the $12$ teams plays $11$ games, so $\dfrac{12\cdot 11}{2} = 66$ games are played, and each game adds $2$ points to the list. The total of all scores is $66\cdot 2 = 132.$

If every game is a draw, each team scores $11,$ so the highest score need not reach $12;$ thus statement $\text{(E)}$ can fail. The other statements always hold: the sum $132 \ge 100,$ the sum being even forces an even number of odd scores and hence an even number of even scores, and two teams cannot both score $0$ because their mutual game gives at least one of them a point.

Thus, the correct answer is E.

14.

What is the value of $a$ for which $\dfrac{1}{\log_2 a} + \dfrac{1}{\log_3 a} + \dfrac{1}{\log_4 a} = 1?$

$9$

$12$

$18$

$24$

$36$

Answer: D

Concepts:logarithm

Difficulty rating: 1730

Small Hint:

$\dfrac{1}{\log_b a} = \log_a b$

Big Hint:

The sum becomes $\log_a(2\cdot 3\cdot 4)$

Solution:

By the change-of-base formula, $\dfrac{1}{\log_b a} = \log_a b.$ Therefore $\begin{aligned} &1 = \log_a 2 + \log_a 3 \\ &\quad {}+ \log_a 4 = \log_a 24. \end{aligned}$

It follows that $a = 24.$

Thus, the correct answer is D.

15.

What is the minimum number of digits to the right of the decimal point needed to express the fraction $\dfrac{123456789}{2^{26}\cdot 5^4}$ as a decimal?

$4$

$22$

$26$

$30$

$104$

Answer: C

Concepts:decimal prime factorization

Difficulty rating: 1800

Small Hint:

The numerator shares no factor of $2$ or $5$ with the denominator

Big Hint:

Multiply the top and bottom to turn the denominator into a power of $10$

Solution:

The numerator and denominator share no common factors. To write the fraction as a decimal, rewrite it with a power of $10$ in the denominator; the smallest that works is $10^{26}:$ $\dfrac{123456789}{2^{26}\cdot 5^4} = \dfrac{123456789\cdot 5^{22}}{10^{26}}.$

Since the numerator $123456789\cdot 5^{22}$ is not divisible by $10,$ the decimal has exactly $26$ places after the point.

Thus, the correct answer is C.

16.

Tetrahedron $ABCD$ has $AB = 5,$ $AC = 3,$ $BC = 4,$ $BD = 4,$ $AD = 3,$ and $CD = \dfrac{12}{5}\sqrt{2}.$ What is the volume of the tetrahedron?

$3\sqrt{2}$

$2\sqrt{5}$

$\dfrac{24}{5}$

$3\sqrt{3}$

$\dfrac{24}{5}\sqrt{2}$

Answer: C

Concepts:3D geometry volume right triangle

Difficulty rating: 1840

Small Hint:

Triangles $ABC$ and $ABD$ are both $3$ - $4$ - $5$ right triangles sharing the edge $AB$

Big Hint:

Drop altitudes from $C$ and $D$ to $AB;$ their common foot shows $CD$ is perpendicular to base $ABC$

Solution:

Triangles $ABC$ and $ABD$ are $3$ - $4$ - $5$ right triangles with area $6$ and common hypotenuse $AB.$ Let $E$ be the foot of the altitude from $C$ to $AB;$ then $CE = \dfrac{3\cdot 4}{5} = \dfrac{12}{5}.$ Likewise the altitude from $D$ meets $AB$ at the same point $E$ with $DE = \dfrac{12}{5}.$

Triangle $CDE$ has sides $\dfrac{12}{5},$ $\dfrac{12}{5},$ and $CD = \dfrac{12}{5}\sqrt{2},$ so it is an isosceles right triangle with the right angle at $E.$ Thus $DE \perp CE$ and $DE \perp AB,$ making $DE$ perpendicular to the plane of $ABC.$

The tetrahedron’s volume is $\dfrac{1}{3}\cdot [ABC]\cdot DE = \dfrac{1}{3}\cdot 6\cdot \dfrac{12}{5}$ $= \dfrac{24}{5}.$

Thus, the correct answer is C.

17.

Eight people are sitting around a circular table, each holding a fair coin. All eight people flip their coins and those who flip heads stand while those who flip tails remain seated. What is the probability that no two adjacent people will stand?

$\dfrac{47}{256}$

$\dfrac{3}{16}$

$\dfrac{49}{256}$

$\dfrac{25}{128}$

$\dfrac{51}{256}$

Answer: A

Concepts:circular arrangements basic probability casework

Difficulty rating: 1910

Small Hint:

There are $2^8 = 256$ equally likely outcomes; count those with no two adjacent heads

Big Hint:

Group by the number of people standing $(0,1,2,3,4);$ more than $4$ forces an adjacency

Solution:

There are $2^8 = 256$ equally likely outcomes. Count the arrangements of standers (heads) with no two adjacent around the circle of $8$ seats, grouped by how many people stand.

The number of ways to choose $k$ non-adjacent seats from a circle of $n$ is $\dfrac{n}{n-k}\dbinom{n-k}{k}.$ For $n = 8$ this gives $1,\; 8,\; 20,\; 16,\; 2$ for $k = 0,1,2,3,4,$ and more than $4$ standers is impossible without an adjacency.

The total is $1 + 8 + 20 + 16 + 2 = 47,$ so the probability is $\dfrac{47}{256}.$

Thus, the correct answer is A.

18.

The zeros of the function $f(x) = x^2 - ax + 2a$ are integers. What is the sum of the possible values of $a?$

$7$

$8$

$16$

$17$

$18$

Answer: C

Concepts:Vieta’s Formulas Simon’s Favorite Factoring Trick

Difficulty rating: 1990

Small Hint:

If the integer roots are $p,q$ then $p+q = a$ and $pq = 2a$

Big Hint:

Eliminate $a$ to get $pq = 2(p+q),$ i.e. $(p-2)(q-2) = 4$

Solution:

Let the integer zeros be $p$ and $q.$ By Vieta’s formulas $p + q = a$ and $pq = 2a,$ so $pq = 2(p+q).$ Rearranging gives $(p-2)(q-2) = 4.$

The integer factor pairs of $4$ are $(1,4),$ $(2,2),$ $(4,1),$ $(-1,-4),$ $(-2,-2),$ $(-4,-1),$ which yield $(p,q)$ pairs summing to $a = 9, 8, -1,$ and $0.$

The distinct possible values of $a$ are $9, 8, 0, -1,$ whose sum is $16.$

Thus, the correct answer is C.

19.

For some positive integers $p,$ there is a quadrilateral $ABCD$ with positive integer side lengths, perimeter $p,$ right angles at $B$ and $C,$ $AB = 2,$ and $CD = AD.$ How many different values of $p \lt 2015$ are possible?

$30$

$31$

$61$

$62$

$63$

Answer: B

Concepts:Pythagorean Theorem counting integers in a range

Difficulty rating: 2010

Small Hint:

Drop a perpendicular from $A$ to $\overline{CD};$ with right angles at $B,C,$ that creates a rectangle

Big Hint:

If $AE = x$ and $DE = y$ then $x^2 + y^2 = (2+y)^2,$ which forces $x$ to be even

Solution:

In every such quadrilateral $CD \ge AB.$ Let $E$ be the foot of the perpendicular from $A$ to $\overline{CD};$ then $CE = 2$ and $AE = BC.$ Let $x = AE$ and $y = DE,$ so $AD = 2 + y.$

By the Pythagorean Theorem $x^2 + y^2 = (2+y)^2,$ so $x^2 = 4 + 4y$ and $x$ is even. Writing $x = 2z$ gives $y = z^2 - 1,$ and the perimeter is $x + 2y + 6 = 2z^2 + 2z + 4.$

Increasing values $z = 1, 2, 3, \dots$ give the required quadrilaterals with increasing perimeter. For $z = 31$ the perimeter is $1988,$ and for $z = 32$ it is $2116.$ Therefore there are $31$ possible values of $p \lt 2015.$

Thus, the correct answer is B.

20.

Isosceles triangles $T$ and $T'$ are not congruent but have the same area and the same perimeter. The sides of $T$ have lengths $5,$ $5,$ and $8,$ while those of $T'$ have lengths $a,$ $a,$ and $b.$ Which of the following numbers is closest to $b?$

$3$

$4$

$5$

$6$

$8$

Answer: A

Concepts:isosceles triangle system of equations polynomial

Difficulty rating: 2110

Small Hint:

$T$ has perimeter $18$ and area $12;$ write those same two conditions for $T'$

Big Hint:

Eliminate $a$ using $2a + b = 18;$ you reach a cubic with the extraneous root $b = 8$ to factor out

Solution:

The altitude of $T$ to its base of length $8$ is $\sqrt{5^2 - 4^2} = 3,$ so $T$ has area $\dfrac{1}{2}\cdot 8\cdot 3 = 12$ and perimeter $18.$

For $T'$ we need $2a + b = 18$ and area $\dfrac{1}{4}b\sqrt{4a^2 - b^2} = 12.$ Substituting $a = \dfrac{18 - b}{2}$ and squaring leads to $(b - 8)(b^2 - b - 8) = 0.$

Since $T$ and $T'$ are not congruent, $b \ne 8,$ so $b^2 - b - 8 = 0$ and $b = \dfrac{1 + \sqrt{33}}{2}.$ Because $25 \lt 33 \lt 36,$ this is between $3$ and $3.5,$ so the closest integer is $3.$

Thus, the correct answer is A.

21.

A circle of radius $r$ passes through both foci of, and exactly four points on, the ellipse with equation $x^2 + 16y^2 = 16.$ The set of all possible values of $r$ is an interval $[a, b).$ What is $a + b?$

$5\sqrt{2} + 4$

$\sqrt{17} + 7$

$6\sqrt{2} + 3$

$\sqrt{15} + 8$

$12$

Answer: D

Concepts:ellipse circle coordinate geometry

Difficulty rating: 2170

Small Hint:

The ellipse is $\dfrac{x^2}{16} + y^2 = 1,$ with foci at $(\pm\sqrt{15}, 0)$

Big Hint:

A circle through both foci is centered at $(0, k);$ require its bottom point to lie outside the ellipse

Solution:

The ellipse $\dfrac{x^2}{16} + y^2 = 1$ has semi-axes $4$ and $1,$ so $c^2 = 16 - 1 = 15$ and the foci are $(\pm\sqrt{15}, 0).$

A circle through both foci has its center on the $y$ -axis, say $(0, k),$ with radius $\sqrt{k^2 + 15}.$ Its top point always lies outside the ellipse. For four intersection points, its bottom point $(0, k - \sqrt{k^2 + 15})$ must be below $y = -1,$ which happens exactly when $0 \le k \lt 7.$

As $k$ ranges over $[0, 7),$ the radius $\sqrt{k^2 + 15}$ ranges over $[\sqrt{15}, 8),$ so $a + b = \sqrt{15} + 8.$

Thus, the correct answer is D.

22.

For each positive integer $n,$ let $S(n)$ be the number of sequences of length $n$ consisting solely of the letters $A$ and $B,$ with no more than three $A$ s in a row and no more than three $B$ s in a row. What is the remainder when $S(2015)$ is divided by $12?$

$0$

$4$

$6$

$8$

$10$

Answer: D

Concepts:recursive counting modular arithmetic Chinese Remainder Theorem

Difficulty rating: 2270

Small Hint:

A valid sequence ends in a run of $1,$ $2,$ or $3$ equal letters, giving $S(n) = S(n-1)$ $+ S(n-2)$ $+ S(n-3)$

Big Hint:

Reduce the recurrence modulo $3$ and modulo $4$ separately to find its period

Solution:

Note $S(1) = 2,$ $S(2) = 4,$ $S(3) = 8.$ Every valid sequence ends in a run of one, two, or three equal letters; removing that run leaves a valid sequence of length $n-1,$ $n-2,$ or $n-3.$ Thus $\begin{aligned} &S(n) = S(n-1) + S(n-2) \\ &\quad {}+ S(n-3). \end{aligned}$

Modulo $3,$ the first $13$ terms are $2,1,2,2,2,0,1,0,1,2,0,0,2.$ The next three terms are $2,1,2,$ which reproduce the initial state, so the recurrence repeats with period $13.$ Since $2015 = 13\cdot 155,$ it follows that $S(2015) \equiv S(13) \equiv 2 \pmod 3.$

Modulo $4,$ the terms repeat as $2,0,0,2,$ because the next three-term state after these four terms is again $(2,0,0).$ Thus the period is $4.$ As $2015 = 4\cdot 503 + 3,$ we have $S(2015) \equiv S(3) \equiv 0 \pmod 4.$

Writing $S(2015) = 4k,$ the condition $4k \equiv 2 \pmod 3$ gives $k \equiv 2 \pmod 3,$ so $S(2015) \equiv 8 \pmod{12}.$

Thus, the correct answer is D.

23.

Let $S$ be a square of side length $1.$ Two points are chosen independently at random on the sides of $S.$ The probability that the straight-line distance between the points is at least $\dfrac12$ is $\dfrac{a - b\pi}{c},$ where $a,$ $b,$ and $c$ are positive integers and $\gcd(a, b, c) = 1.$ What is $a + b + c?$

$59$

$60$

$61$

$62$

$63$

Answer: A

Concepts:geometric probability casework

Difficulty rating: 2380

Small Hint:

Condition on whether the two points lie on the same side, opposite sides, or adjacent sides

Big Hint:

The adjacent-sides case produces the $\pi$ term via a quarter-circle of radius $\dfrac12$

Solution:

The second point is on the same side as the first with probability $\dfrac14,$ on the opposite side with probability $\dfrac14,$ and on an adjacent side with probability $\dfrac12.$

Opposite sides: the distance is at least $1 \ge \dfrac12$ always, probability $1.$

Same side: for points $(a, 0)$ and $(b, 0),$ the condition $|a - b| \ge \dfrac12$ has probability $\dfrac14.$

Adjacent sides: for points $(a, 0)$ and $(0, b),$ the condition $\sqrt{a^2 + b^2} \ge \dfrac12$ is the region outside a quarter-circle of radius $\dfrac12,$ with probability $1 - \dfrac14\pi\left(\dfrac12\right)^2 = 1 - \dfrac{\pi}{16}.$

The total probability is $\begin{aligned} &\dfrac14\cdot 1 + \dfrac14\cdot\dfrac14 \\ &\quad {}+ \dfrac12\left(1 - \dfrac{\pi}{16}\right) \\ &\quad = \dfrac{26 - \pi}{32}. \end{aligned}$ Thus $a + b + c = 26 + 1 + 32 = 59.$

Thus, the correct answer is A.

24.

Rational numbers $a$ and $b$ are chosen at random among all rational numbers in the interval $[0, 2)$ that can be written as fractions $\dfrac{n}{d}$ where $n$ and $d$ are integers with $1 \le d \le 5.$ What is the probability that $(\cos(a\pi) + i\sin(b\pi))^4$ is a real number?

$\dfrac{3}{50}$

$\dfrac{4}{25}$

$\dfrac{41}{200}$

$\dfrac{6}{25}$

$\dfrac{13}{50}$

Answer: D

Concepts:complex number De Moivre’s Theorem basic counting

Difficulty rating: 2520

Small Hint:

There are $20$ allowed values for each of $a$ and $b;$ $(x + iy)^4$ is real iff $x = 0,$ $y = 0,$ or $x = \pm y$

Big Hint:

Translate those into $\cos(a\pi) = 0,$ $\sin(b\pi) = 0,$ or $\cos(a\pi) = \pm\sin(b\pi)$

Solution:

There are $20$ possible values for each of $a$ and $b.$ In reduced form, denominators $1,2,3,4,5$ contribute respectively $2,2,4,4,8$ values in $[0,2),$ for a total of $20.$

Writing $x = \cos(a\pi)$ and $y = \sin(b\pi),$ the fourth power $(x + iy)^4$ is real if and only if $x = 0,$ $y = 0,$ or $x = \pm y.$

The case $x = 0$ means $a \in \left\{\dfrac12, \dfrac32\right\},$ giving $2\cdot 20 = 40$ pairs; the case $y = 0$ means $b \in \{0, 1\},$ giving another $40$ pairs, of which $4$ were already counted.

For the remaining condition $\cos(a\pi) = \pm\sin(b\pi),$ with neither side zero, the allowed values of $b$ are $\dfrac14,$ $\dfrac12,$ $\dfrac34,$ $\dfrac54,$ $\dfrac32,$ and $\dfrac74.$ The corresponding numbers of allowed values of $a$ are respectively $4,2,4,4,2,4.$ For example, when $b = \dfrac14,$ the values are $a = \dfrac14,$ $\dfrac34,$ $\dfrac54,$ and $\dfrac74;$ the other rows follow from the same quarter-turn identities. Hence this condition contributes $4+2+4+4+2+4=20$ more pairs.

In all there are $40 + 40 - 4 + 20 = 96$ valid pairs out of $400,$ so the probability is $\dfrac{96}{400} = \dfrac{6}{25}.$

Thus, the correct answer is D.

25.

A collection of circles in the upper half-plane, all tangent to the $x$ -axis, is constructed in layers as follows. Layer $L_0$ consists of two circles of radii $70^2$ and $73^2$ that are externally tangent. For $k \ge 1,$ the circles in $\bigcup_{j=0}^{k-1} L_j$ are ordered according to their points of tangency with the $x$ -axis. For every pair of consecutive circles in this order, a new circle is constructed externally tangent to each of the two circles in the pair. Layer $L_k$ consists of the $2^{k-1}$ circles constructed in this way. Let $S = \bigcup_{j=0}^{6} L_j,$ and for every circle $C$ denote by $r(C)$ its radius. What is $\sum_{C \in S} \dfrac{1}{\sqrt{r(C)}}?$

$\dfrac{286}{35}$

$\dfrac{583}{70}$

$\dfrac{715}{73}$

$\dfrac{143}{14}$

$\dfrac{1573}{146}$

Answer: D

Concepts:tangent circles geometric sequence induction

Difficulty rating: 2650

Small Hint:

For a circle nestled between two circles tangent to the same line, $\dfrac{1}{\sqrt{r}} = \dfrac{1}{\sqrt{r_1}} + \dfrac{1}{\sqrt{r_2}}$

Big Hint:

Let $x = \dfrac{1}{70} + \dfrac{1}{73};$ the layer sums are $x,$ $x,$ $3x,$ $9x, \dots$

Solution:

If a circle of radius $r$ is tangent to the $x$ -axis and nestled in the crevice between two circles of radii $r_1$ and $r_2$ that are also tangent to the axis and to each other, then $\dfrac{1}{\sqrt{r}} = \dfrac{1}{\sqrt{r_1}} + \dfrac{1}{\sqrt{r_2}}.$

Let $x = \dfrac{1}{\sqrt{70^2}} + \dfrac{1}{\sqrt{73^2}}$ $= \dfrac{1}{70} + \dfrac{1}{73},$ which is the sum over $L_0.$ The single circle of $L_1$ also contributes $x.$ For $k \ge 2,$ each new circle contributes the sum of its two neighbors, and every earlier circle is counted twice except the two circles of $L_0;$ this yields a sum of $3^{k-1}x$ over $L_k.$

Therefore $\begin{gathered} \sum_{C \in S} \dfrac{1}{\sqrt{r(C)}} = x \\ {}+ \sum_{k=1}^{6} 3^{k-1}x \\ = x\left(1 + \dfrac{3^6 - 1}{2}\right) \\ = x\cdot\dfrac{3^6 + 1}{2} \\ = 365x. \end{gathered}$

Since $x = \dfrac{1}{70} + \dfrac{1}{73}$ $= \dfrac{143}{70\cdot 73}$ $= \dfrac{143}{5110},$ the sum is $365\cdot\dfrac{143}{5110} = \dfrac{143}{14}.$

Thus, the correct answer is D.