2004 AMC 10B — Problems & Online Timed Exam

Timed

1:15:00

1.

Each row of the Misty Moon Amphitheater has $33$ seats. Rows $12$ through $22$ are reserved for a youth club. How many seats are reserved for this club?

$297$

$330$

$363$

$396$

$726$

Answer: C

Concepts:counting integers in a range basic counting

Difficulty rating: 770

Small Hint:

Count how many rows run from $12$ through $22$ inclusive

Big Hint:

Multiply the number of rows by $33$

Solution:

Rows $12$ through $22$ inclusive make up $22 - 12 + 1 = 11$ rows.

Each row has $33$ seats, so the total is $33 \times 11 = 363.$

Thus, the correct answer is C.

2.

How many two-digit positive integers have at least one $7$ as a digit?

$10$

$18$

$19$

$20$

$30$

Answer: B

Concepts:digits inclusion-exclusion

Difficulty rating: 920

Small Hint:

Count the numbers with $7$ in the tens place and those with $7$ in the units place separately

Big Hint:

The number $77$ belongs to both groups

Solution:

The numbers $70$ through $79$ give $10$ with a $7$ in the tens place.

The numbers $17, 27, \ldots, 97$ give $9$ with a $7$ in the units place.

Since $77$ is counted twice, the total is $10 + 9 - 1 = 18.$

Thus, the correct answer is B.

3.

At each basketball practice last week, Jenny made twice as many free throws as she made at the previous practice. At her fifth practice she made $48$ free throws. How many free throws did she make at the first practice?

$3$

$6$

$9$

$12$

$15$

Answer: A

Concepts:geometric sequence work backwards

Difficulty rating: 820

Small Hint:

Each practice count is twice the previous one, so work backward by halving

Big Hint:

Halve $48$ repeatedly to reach the first practice

Solution:

Working backward from the fifth practice, the counts are $48, 24, 12, 6,$ and $3$ at the fourth, third, second, and first practices.

Thus, the correct answer is A.

4.

A standard six-sided die is rolled, and $P$ is the product of the five numbers that are visible. What is the largest number that is certain to divide $P?$

$6$

$12$

$24$

$144$

$720$

Answer: B

Concepts:prime factorization divisibility bounding to limit cases

Difficulty rating: 1190

Small Hint:

The product of all six faces is $6! = 720$ ; exactly one face is hidden

Big Hint:

For each prime, find how many copies must remain no matter which face is hidden

Solution:

Since $6! = 720 = 2^4 \cdot 3^2 \cdot 5,$ the visible product uses only the primes $2,$ $3,$ and $5.$

Hiding $4$ leaves the fewest $2$ ’s, namely $2^2.$ Hiding $3$ or $6$ leaves the fewest $3$ ’s, namely one. Hiding $5$ leaves no factor of $5.$

Therefore $P$ is always divisible by $2^2 \cdot 3 = 12,$ but not necessarily by any larger number.

Thus, the correct answer is B.

5.

In the expression $c \cdot a^b - d,$ the values of $a,$ $b,$ $c,$ and $d$ are $0,$ $1,$ $2,$ and $3,$ although not necessarily in that order. What is the maximum possible value of the result?

$5$

$6$

$8$

$9$

$10$

Answer: D

Concepts:exponent optimization casework

Difficulty rating: 1120

Small Hint:

To maximize the result, make the subtracted value $d = 0$

Big Hint:

With $1,$ $2,$ $3$ remaining, compare $c \cdot a^b$ for the sensible assignments

Solution:

Setting $d = 0$ removes the subtraction, so we maximize $c \cdot a^b$ using $1,$ $2,$ $3.$

Taking $c = 1,$ $a = 3,$ $b = 2$ gives $3^2 = 9.$ The alternative $2^3 = 8$ is smaller, and any assignment with $c \gt 1$ forces a smaller power. The maximum is $9.$

Thus, the correct answer is D.

6.

Which of the following numbers is a perfect square?

$98! \cdot 99!$

$98! \cdot 100!$

$99! \cdot 100!$

$99! \cdot 101!$

$100! \cdot 101!$

Answer: C

Concepts:factorial perfect square

Difficulty rating: 1290

Small Hint:

For $m \lt n,$ write $m! \cdot n!$ as $(m!)^2 \cdot (m+1)(m+2)\cdots n$

Big Hint:

The product is a perfect square exactly when $(m+1)(m+2)\cdots n$ is

Solution:

For $m \lt n,$ $m! \cdot n!$ equals $(m!)^2 \cdot (m+1)(m+2)\cdots n,$ which is a perfect square precisely when $(m+1)\cdots n$ is a perfect square.

For the five choices this leftover factor is $99,$ $99 \cdot 100,$ $100,$ $100 \cdot 101,$ and $101.$ Only $100 = 10^2$ is a perfect square.

Therefore $99! \cdot 100! = (99! \cdot 10)^2$ is the perfect square.

Thus, the correct answer is C.

7.

On a trip from the United States to Canada, Isabella took $d$ U.S. dollars. At the border she exchanged them all, receiving $10$ Canadian dollars for every $7$ U.S. dollars. After spending $60$ Canadian dollars, she had $d$ Canadian dollars left. What is the sum of the digits of $d?$

$5$

$6$

$7$

$8$

$9$

Answer: A

Concepts:linear equation ratio and proportion

Difficulty rating: 1100

Small Hint:

Exchanging $d$ U.S. dollars gives $\dfrac{10}{7}d$ Canadian dollars

Big Hint:

Set $\dfrac{10}{7}d - 60 = d$ and solve for $d$

Solution:

Isabella received $\dfrac{10}{7}d$ Canadian dollars and spent $60,$ leaving $d.$ So $\dfrac{10}{7}d - 60 = d.$

Then $\dfrac{3}{7}d = 60,$ so $d = 140.$ The sum of its digits is $1 + 4 + 0 = 5.$

Thus, the correct answer is A.

8.

Minneapolis-St. Paul International Airport is $8$ miles southwest of downtown St. Paul and $10$ miles southeast of downtown Minneapolis. Which of the following is closest to the number of miles between downtown St. Paul and downtown Minneapolis?

$13$

$14$

$15$

$16$

$17$

Answer: A

Concepts:Pythagorean Theorem estimation

Difficulty rating: 1030

Small Hint:

Southwest and southeast are perpendicular directions

Big Hint:

The distance is $\sqrt{8^2 + 10^2}$ ; estimate the nearest whole number

Solution:

The two given directions are perpendicular, so the airport sits at the right angle of a right triangle with legs $8$ and $10.$

The distance between the downtowns is $\sqrt{8^2 + 10^2} = \sqrt{164} \approx 12.8,$ which is closest to $13.$

Thus, the correct answer is A.

9.

A square has sides of length $10,$ and a circle centered at one of its vertices has radius $10.$ What is the area of the union of the regions enclosed by the square and the circle?

$200 + 25\pi$

$100 + 75\pi$

$75 + 100\pi$

$100 + 100\pi$

$100 + 125\pi$

Answer: B

Concepts:circle area sector inclusion-exclusion

Difficulty rating: 1270

Small Hint:

Add the areas of the square and the circle, then subtract the overlap

Big Hint:

The overlap is the quarter of the circle that lies inside the square

Solution:

The square has area $10^2 = 100$ and the circle has area $\pi(10)^2 = 100\pi.$

Since the circle is centered at a vertex of the square, exactly one quarter of the circle, area $25\pi,$ lies inside the square.

The union has area $100 + 100\pi - 25\pi = 100 + 75\pi.$

Thus, the correct answer is B.

10.

A grocer makes a display of cans in which the top row has one can and each lower row has two more cans than the row above it. If the display contains $100$ cans, how many rows does it contain?

$5$

$8$

$9$

$10$

$11$

Answer: D

Concepts:sum of first n odd numbers

Difficulty rating: 1170

Small Hint:

The rows contain $1, 3, 5, \ldots$ cans

Big Hint:

The sum of the first $n$ odd numbers equals $n^2$

Solution:

The rows hold $1, 3, 5, \ldots$ cans, and the sum of the first $n$ odd numbers is $n^2.$

Setting $n^2 = 100$ gives $n = 10.$

Thus, the correct answer is D.

11.

Two eight-sided dice each have faces numbered $1$ through $8.$ When the dice are rolled, each face has an equal probability of appearing on the top. What is the probability that the product of the two top numbers is greater than their sum?

$\dfrac{1}{2}$

$\dfrac{47}{64}$

$\dfrac{3}{4}$

$\dfrac{55}{64}$

$\dfrac{7}{8}$

Answer: C

Concepts:dice (probability)Simon’s Favorite Factoring Trick complementary counting

Difficulty rating: 1430

Small Hint:

The condition $mn \gt m + n$ rearranges to $(m-1)(n-1) \gt 1$

Big Hint:

Count the ordered pairs where $(m-1)(n-1) \le 1$ instead

Solution:

There are $8 \cdot 8 = 64$ ordered pairs. The inequality $mn \gt m + n$ is equivalent to $(m-1)(n-1) \gt 1.$

This fails only when $m = 1,$ $n = 1,$ or $m = n = 2,$ which account for $8 + 8 - 1 + 1 = 16$ pairs.

The probability is $\dfrac{64 - 16}{64} = \dfrac{48}{64} = \dfrac{3}{4}.$

Thus, the correct answer is C.

12.

An annulus is the region between two concentric circles. The concentric circles in the figure have radii $b$ and $c,$ with $b \gt c.$ Let $\overline{OX}$ be a radius of the larger circle, let $\overline{XZ}$ be tangent to the smaller circle at $Z,$ and let $\overline{OY}$ be the radius of the larger circle that contains $Z.$ Let $a = XZ,$ $d = YZ,$ and $e = XY.$ What is the area of the annulus?

$\pi a^2$

$\pi b^2$

$\pi c^2$

$\pi d^2$

$\pi e^2$

Answer: A

Concepts:annulus circle area tangent line Pythagorean Theorem

Difficulty rating: 1390

Small Hint:

The annulus area is $\pi b^2 - \pi c^2$

Big Hint:

Since $\overline{XZ}$ is tangent at $Z,$ triangle $OZX$ is right-angled, giving $b^2 - c^2 = a^2$

Solution:

The annulus is the difference of the two circular areas, $\pi b^2 - \pi c^2.$

Because $\overline{XZ}$ is tangent to the small circle at $Z,$ it is perpendicular to the radius $\overline{OZ}.$ In right triangle $OZX$ with $OX = b,$ $OZ = c,$ and $XZ = a,$ we get $b^2 - c^2 = a^2.$

Therefore the area of the annulus is $\pi a^2.$

Thus, the correct answer is A.

13.

In the United States, coins have the following thicknesses: penny, $1.55$ mm; nickel, $1.95$ mm; dime, $1.35$ mm; quarter, $1.75$ mm. If a stack of these coins is exactly $14$ mm high, how many coins are in the stack?

$7$

$8$

$9$

$10$

$11$

Answer: B

Concepts:place value decimal bounding to limit cases

Difficulty rating: 1530

Small Hint:

Every coin thickness ends in a $5$ in the hundredths place, so consider the height’s last two decimal digits

Big Hint:

For the height to be a whole number of mm, the number of coins must be a multiple of $4$

Solution:

Measure every thickness in hundredths of a millimeter. The four possible thicknesses are $135,$ $155,$ $175,$ $195,$ all congruent to $15\pmod{20}.$ If the stack has $k$ coins and an integer height, its total in hundredths is divisible by $20,$ so $15k$ is divisible by $20.$ Therefore $k$ must be a multiple of $4.$

A stack of $4$ coins is at most $4(1.95) = 7.8$ mm, and a stack of $12$ coins is at least $12(1.35) = 16.2$ mm, so only $8$ coins can total $14$ mm.

Indeed, $8$ quarters give $8(1.75) = 14$ mm.

Thus, the correct answer is B.

14.

A bag initially contains red marbles and blue marbles only, with more blue than red. Red marbles are added to the bag until only $\tfrac{1}{3}$ of the marbles in the bag are blue. Then yellow marbles are added to the bag until only $\tfrac{1}{5}$ of the marbles in the bag are blue. Finally, the number of blue marbles in the bag is doubled. What fraction of the marbles now in the bag are blue?

$\dfrac{1}{5}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{2}{5}$

$\dfrac{1}{2}$

Answer: C

Concepts:fraction ratio and proportion

Difficulty rating: 1240

Small Hint:

Let $B$ be the number of blue marbles and track the total at each stage

Big Hint:

Just before doubling there are $B$ blue out of $5B$ total

Solution:

Let there be $B$ blue marbles. After adding red marbles the total is $3B;$ after adding yellow marbles the total is $5B,$ still with $B$ blue.

Doubling the blue marbles gives $2B$ blue out of $6B$ total, which is $\dfrac{2B}{6B} = \dfrac{1}{3}.$

Thus, the correct answer is C.

15.

Patty has $20$ coins consisting of nickels and dimes. If her nickels were dimes and her dimes were nickels, she would have $70$ cents more. How much are her coins worth?

$\$1.15$

$\$1.20$

$\$1.25$

$\$1.30$

$\$1.35$

Answer: A

Concepts:system of equations money

Difficulty rating: 1240

Small Hint:

Swapping raises the value, so she has more nickels than dimes

Big Hint:

Each nickel-for-dime swap changes the total by $5$ cents

Solution:

Swapping increases the value, so Patty has more nickels than dimes. Each swapped coin changes the total by $5$ cents, so she has $\frac{70}{5} = 14$ more nickels than dimes.

With $n + d = 20$ and $n - d = 14,$ she has $17$ nickels and $3$ dimes.

Her coins are worth $17 \cdot 5 + 3 \cdot 10 = 115$ cents, or $\$1.15.$

Thus, the correct answer is A.

16.

Three circles of radius $1$ are externally tangent to each other and internally tangent to a larger circle. What is the radius of the large circle?

$\dfrac{2 + \sqrt{6}}{3}$

$2$

$\dfrac{2 + 3\sqrt{2}}{3}$

$\dfrac{3 + 2\sqrt{3}}{3}$

$\dfrac{3 + \sqrt{3}}{2}$

Answer: D

Concepts:tangent circles equilateral triangle special right triangle

Difficulty rating: 1640

Small Hint:

The three small centers form an equilateral triangle of side $2$

Big Hint:

The large radius is $1$ plus the distance from that triangle’s center to a vertex

Solution:

The centers of the three unit circles form an equilateral triangle with side $2.$ Its center is the center of the large circle.

For an equilateral triangle of side length $2,$ the distance from its center to a vertex is $\dfrac{2}{\sqrt3} = \dfrac{2\sqrt3}{3}.$

Adding the unit radius, the large radius is $1 + \dfrac{2\sqrt3}{3} = \dfrac{3 + 2\sqrt3}{3}.$

Thus, the correct answer is D.

17.

The two digits in Jack’s age are the same as the digits in Bill’s age, but in reverse order. In five years Jack will be twice as old as Bill will be then. What is the difference in their current ages?

$9$

$18$

$27$

$36$

$45$

Answer: B

Concepts:ages digits Diophantine Equation

Difficulty rating: 1410

Small Hint:

Write Jack’s age as $10x + y$ and Bill’s as $10y + x$

Big Hint:

The condition $10x + y + 5 = 2(10y + x + 5)$ reduces to $8x = 19y + 5$

Solution:

Let Jack’s age be $10x + y$ and Bill’s be $10y + x.$ In five years $10x + y + 5 = 2(10y + x + 5),$ which simplifies to $8x = 19y + 5.$

Since $x$ and $y$ are digits, the only solution is $y = 1,$ $x = 3.$

So Jack is $31$ and Bill is $13,$ a difference of $18.$

Thus, the correct answer is B.

18.

In right triangle $\triangle ACE,$ we have $AC = 12,$ $CE = 16,$ and $EA = 20.$ Points $B,$ $D,$ and $F$ are located on $AC,$ $CE,$ and $EA,$ respectively, so that $AB = 3,$ $CD = 4,$ and $EF = 5.$ What is the ratio of the area of $\triangle BDF$ to that of $\triangle ACE?$

$\dfrac{1}{4}$

$\dfrac{9}{25}$

$\dfrac{3}{8}$

$\dfrac{11}{25}$

$\dfrac{7}{16}$

Answer: E

Concepts:area ratio area decomposition triangle area

Difficulty rating: 1630

Small Hint:

The area of $\triangle ACE$ is $\tfrac12(12)(16) = 96$

Big Hint:

Subtract the three corner triangles; each has a base and height that are known fractions of the big triangle’s

Solution:

The area of $\triangle ACE$ is $\tfrac12(12)(16) = 96.$

Each corner triangle $\triangle ABF,$ $\triangle BCD,$ and $\triangle DEF$ has a base and an altitude that are $\tfrac34$ and $\tfrac14$ of a corresponding base and altitude of $\triangle ACE.$ So each has area $\tfrac14 \cdot \tfrac34 = \tfrac{3}{16}$ of $\triangle ACE.$

Hence $\begin{aligned} \dfrac{[BDF]}{[ACE]} &= 1 - 3 \cdot \dfrac{3}{16} \\ &= 1 - \dfrac{9}{16} = \dfrac{7}{16}. \end{aligned}$

Thus, the correct answer is E.

19.

In the sequence $2001,$ $2002,$ $2003,$ $\ldots,$ each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is $2001 + 2002 - 2003 = 2000.$ What is the $2004$ th term in this sequence?

$-2004$

$-2$

$0$

$4003$

$6007$

Answer: C

Concepts:recursion arithmetic sequence pattern recognition

Difficulty rating: 1460

Small Hint:

Write out several terms: $2001,$ $2002,$ $2003,$ $2000,$ $2005,$ $1998,$ $\ldots$

Big Hint:

The even-position terms decrease by $2$ each time

Solution:

The recurrence $a_{k+1} = a_{k-2} + a_{k-1} - a_k$ gives $a_{k+1} - a_{k-1} = -(a_k - a_{k-2}).$ The sequence begins $2001,$ $2002,$ $2003,$ $2000,$ $2005,$ $1998,$ $\ldots$

So the even-position terms form the arithmetic sequence $2002, 2000, 1998, \ldots$ with common difference $-2.$ The $2004$ th term is its $1002$ nd term, $2002 + 1001(-2) = 0.$

Thus, the correct answer is C.

20.

In $\triangle ABC$ points $D$ and $E$ lie on $\overline{BC}$ and $\overline{AC},$ respectively. If $\overline{AD}$ and $\overline{BE}$ intersect at $T$ so that $\frac{AT}{DT} = 3$ and $\frac{BT}{ET} = 4,$ what is $\frac{CD}{BD}?$

$\dfrac{1}{8}$

$\dfrac{2}{9}$

$\dfrac{3}{10}$

$\dfrac{4}{11}$

$\dfrac{5}{12}$

Answer: D

Difficulty rating: 1840

Small Hint:

Draw the line through $D$ parallel to $\overline{BE}$ meeting $\overline{AC}$ at a point $F$

Big Hint:

Two pairs of similar triangles give $\frac{CD}{BC},$ which converts to $\frac{CD}{BD}$

Solution:

Let $F$ be on $\overline{AC}$ with $DF \parallel BE,$ and write $ET = x,$ $BT = 4x.$

From $\triangle ATE \sim \triangle ADF,$ $\dfrac{DF}{x} = \dfrac{AD}{AT} = \dfrac{4}{3},$ so $DF = \dfrac{4x}{3}.$

From $\triangle BEC \sim \triangle DFC,$ $\dfrac{CD}{BC} = \dfrac{DF}{BE} = \dfrac{\frac{4x}{3}}{5x} = \dfrac{4}{15}.$

Therefore $\begin{aligned} \dfrac{CD}{BD} &= \dfrac{\frac{CD}{BC}}{1 - \frac{CD}{BC}} \\ &= \dfrac{\frac{4}{15}}{\frac{11}{15}} = \dfrac{4}{11}. \end{aligned}$

Thus, the correct answer is D.

21.

Let $1,$ $4,$ $\ldots$ and $9,$ $16,$ $\ldots$ be two arithmetic progressions. The set $S$ is the union of the first $2004$ terms of each sequence. How many distinct numbers are in $S?$

$3722$

$3732$

$3914$

$3924$

$4007$

Answer: A

Concepts:arithmetic sequence least common multiple inclusion-exclusion

Difficulty rating: 1740

Small Hint:

The first sequence has common difference $3$ and last term $1 + 3 \cdot 2003 = 6010$

Big Hint:

Common terms start at $16$ and are spaced by $\mathrm{lcm}(3, 7) = 21;$ count the terms not exceeding $6010$

Solution:

The first sequence is $1 + 3k$ with largest term $6010,$ and the second is $9 + 7j$ with a much larger last term, so the binding limit is $6010.$

A common value has the form $16 + 21m$ (the first shared term is $16,$ spaced by $\mathrm{lcm}(3, 7) = 21$ ). Requiring $16 + 21m \le 6010$ gives $0 \le m \le 285,$ that is $286$ common numbers.

The number of distinct values is $2004 + 2004 - 286 = 3722.$

Thus, the correct answer is A.

22.

A triangle with sides of $5,$ $12,$ and $13$ has both an inscribed and a circumscribed circle. What is the distance between the centers of those circles?

$\dfrac{3\sqrt{5}}{2}$

$\dfrac{7}{2}$

$\sqrt{15}$

$\dfrac{\sqrt{65}}{2}$

$\dfrac{9}{2}$

Answer: D

Concepts:incircle, incenter, and inradius circumcircle, circumcenter, and circumradius right triangle coordinate geometry

Difficulty rating: 1770

Small Hint:

Place the right triangle with vertices $(0, 0),$ $(5, 0),$ and $(0, 12)$

Big Hint:

The circumcenter is the midpoint of the hypotenuse; a right triangle’s inradius is $r = \tfrac{a + b - c}{2}$

Solution:

Since $5^2 + 12^2 = 13^2,$ the triangle is right. Place it at $(0, 0),$ $(5, 0),$ $(0, 12).$ The circumcenter is the midpoint of the hypotenuse, $\left(\tfrac52, 6\right).$

The inradius satisfies $(12 - r) + (5 - r) = 13,$ so $r = 2$ and the incenter is $(2, 2).$

The distance is $\begin{gathered} \sqrt{\left(\tfrac52 - 2\right)^2 + (6 - 2)^2} \\ = \sqrt{\tfrac14 + 16} \\ = \dfrac{\sqrt{65}}{2}. \end{gathered}$

Thus, the correct answer is D.

23.

Each face of a cube is painted either red or blue, each with probability $\tfrac12.$ The color of each face is determined independently. What is the probability that the painted cube can be placed on a horizontal surface so that the four vertical faces are all the same color?

$\dfrac{1}{4}$

$\dfrac{5}{16}$

$\dfrac{3}{8}$

$\dfrac{7}{16}$

$\dfrac{1}{2}$

Answer: B

Concepts:basic probability casework cube geometry

Difficulty rating: 1990

Small Hint:

Fixing the cube’s orientation, there are $2^6 = 64$ equally likely colorings

Big Hint:

The working colorings are: all one color, exactly five one color, or four of one color with the opposite pair the other

Solution:

Fixing the orientation, there are $2^6 = 64$ colorings.

A coloring works if all six faces match ( $2$ ways), exactly five match ( $\binom{6}{5} \cdot 2 = 12$ ways), or four faces share a color with the remaining pair being opposite faces of the other color ( $3$ opposite pairs, $2$ colors, giving $6$ ways).

The total is $2 + 12 + 6 = 20,$ so the probability is $\dfrac{20}{64} = \dfrac{5}{16}.$

Thus, the correct answer is B.

24.

In $\triangle ABC$ we have $AB = 7,$ $AC = 8,$ and $BC = 9.$ Point $D$ is on the circumscribed circle of the triangle so that $\overline{AD}$ bisects $\angle BAC.$ What is the value of $\frac{AD}{CD}?$

$\dfrac{9}{8}$

$\dfrac{5}{3}$

$2$

$\dfrac{17}{7}$

$\dfrac{5}{2}$

Answer: B

Concepts:inscribed angle similarity angle bisector theorem

Difficulty rating: 2030

Small Hint:

Let $\overline{AD}$ meet $\overline{BC}$ at $E;$ inscribed angles give $\angle ABE = \angle ADC$

Big Hint:

Triangles $ABE$ and $ADC$ are similar, so $\frac{AD}{CD} = \frac{AB}{BE};$ find $BE$ from the Angle Bisector Theorem

Solution:

Let $\overline{AD}$ meet $\overline{BC}$ at $E.$ Since $\angle ABC$ and $\angle ADC$ subtend the same arc, they are equal, and $\angle EAB = \angle CAD,$ so $\triangle ABE \sim \triangle ADC.$

Hence $\dfrac{AD}{CD} = \dfrac{AB}{BE}.$

By the Angle Bisector Theorem, $\dfrac{BE}{EC} = \dfrac{AB}{AC},$ so $BE = \dfrac{AB \cdot BC}{AB + AC} = \dfrac{7 \cdot 9}{15}.$

Therefore $\begin{aligned} \dfrac{AD}{CD} &= \dfrac{AB}{BE} = \dfrac{AB + AC}{BC} \\ &= \dfrac{15}{9} = \dfrac{5}{3}. \end{aligned}$

Thus, the correct answer is B.

25.

A circle of radius $1$ is internally tangent to two circles of radius $2$ at points $A$ and $B,$ where $AB$ is a diameter of the smaller circle. What is the area of the region, shaded in the figure, that is outside the smaller circle and inside each of the two larger circles?

$\dfrac{5}{3}\pi - 3\sqrt{2}$

$\dfrac{5}{3}\pi - 2\sqrt{3}$

$\dfrac{8}{3}\pi - 3\sqrt{3}$

$\dfrac{8}{3}\pi - 3\sqrt{2}$

$\dfrac{8}{3}\pi - 2\sqrt{3}$

Answer: B

Concepts:sector area decomposition special right triangle symmetry

Difficulty rating: 2270

Small Hint:

By symmetry the shaded region splits into four congruent pieces; compute one quarter

Big Hint:

One quarter is a $60^\circ$ sector of a large circle minus a right triangle minus a quarter of the small circle

Solution:

Let the large circles have centers $A$ and $B,$ let $C$ be the center of the small circle, and let $D$ be a point where the two large circles meet.

Then $\triangle ACD$ is right with $AC = 1$ and $AD = 2,$ so $CD = \sqrt3,$ $\angle CAD = 60^\circ,$ and its area is $\dfrac{\sqrt3}{2}.$

One quarter of the shaded region equals the $60^\circ$ sector of the radius- $2$ circle (area $\dfrac{2\pi}{3}$ ) minus $\triangle ACD$ (area $\dfrac{\sqrt3}{2}$ ) minus a quarter of the small circle (area $\dfrac{\pi}{4}$ ), giving $\dfrac{2\pi}{3} - \dfrac{\sqrt3}{2} - \dfrac{\pi}{4} = \dfrac{5\pi}{12} - \dfrac{\sqrt3}{2}.$

Multiplying by $4,$ the shaded area is $\dfrac{5\pi}{3} - 2\sqrt3.$

Thus, the correct answer is B.