2006 AMC 12B Solutions | LIVE by Po-Shen Loh

1.

What is $(-1)^1 + (-1)^2 + \cdots + (-1)^{2006}?$

$-2006$

$-1$

$0$

$1$

$2006$

Concepts:exponent pairing and grouping

Difficulty rating: 840

Small Hint:

$(-1)^k$ equals $1$ when $k$ is even and $-1$ when $k$ is odd

Big Hint:

Group the $2006$ terms into consecutive pairs $(-1)+1$

Solution:

Since $(-1)^k = -1$ for odd $k$ and $(-1)^k = 1$ for even $k,$ the terms alternate $-1, 1, -1, 1, \ldots$

There are $2006$ terms, forming $1003$ pairs, each equal to $(-1) + 1 = 0.$ The total is $0.$

Thus, the correct answer is C.

2.

For real numbers $x$ and $y,$ define $x \spadesuit y = (x + y)(x - y).$ What is $3 \spadesuit (4 \spadesuit 5)?$

$-72$

$-27$

$-24$

$24$

$72$

Concepts:custom operation difference of squares

Difficulty rating: 990

Small Hint:

Note that $x \spadesuit y = x^2 - y^2$

Big Hint:

Evaluate the inner $4 \spadesuit 5$ first, then apply the operation again

Solution:

Since $x \spadesuit y = x^2 - y^2,$ the inner value is $4 \spadesuit 5 = 16 - 25 = -9.$

Then $3 \spadesuit (-9) = 3^2 - (-9)^2$ $= 9 - 81 = -72.$

Thus, the correct answer is A.

3.

A football game was played between two teams, the Cougars and the Panthers. The two teams scored a total of $34$ points, and the Cougars won by a margin of $14$ points. How many points did the Panthers score?

$10$

$14$

$17$

$20$

$24$

Concepts:system of equations

Difficulty rating: 940

Small Hint:

Let $c$ and $p$ be the two scores, with $c + p = 34$ and $c - p = 14$

Big Hint:

Subtracting the two equations isolates $2p$

Solution:

Let $c$ and $p$ be the Cougars’ and Panthers’ scores. Then $c + p = 34$ and $c - p = 14.$

Subtracting gives $2p = 20,$ so $p = 10.$

Thus, the correct answer is A.

4.

Mary is about to pay for five items at the grocery store. The prices of the items are $\$7.99,$ $\$4.99,$ $\$2.99,$ $\$1.99,$ and $\$0.99.$ Mary will pay with a twenty-dollar bill. Which of the following is closest to the percentage of the $\$20.00$ that she will receive in change?

$5$

$10$

$15$

$20$

$25$

Concepts:estimation percentage

Difficulty rating: 1080

Small Hint:

Round each price to the nearest dollar to estimate the total quickly

Big Hint:

The change is $\$20$ minus the total; compare it to $\$20$ as a fraction

Solution:

The prices total $\$18.95,$ so the change is $\$1.05.$ As a percentage of $\$20.00,$ this is $\frac{1.05}{20}\cdot100\% = 5.25\%.$ The closest listed percentage is $5\%.$

Thus, the correct answer is A.

5.

John is walking east at a speed of $3$ miles per hour, while Bob is also walking east, but at a speed of $5$ miles per hour. If Bob is now $1$ mile west of John, how many minutes will it take for Bob to catch up to John?

$30$

$50$

$60$

$90$

$120$

Concepts:relative speed distance rate and time

Difficulty rating: 1100

Small Hint:

Bob gains ground at the difference of the two speeds

Big Hint:

He must close a $1$ -mile gap at $5 - 3$ miles per hour

Solution:

Bob closes the gap at a relative speed of $5 - 3 = 2$ miles per hour. To cover the $1$ -mile gap takes $\frac{1}{2}$ hour, or $30$ minutes.

Thus, the correct answer is A.

6.

Francesca uses $100$ grams of lemon juice, $100$ grams of sugar, and $400$ grams of water to make lemonade. There are $25$ calories in $100$ grams of lemon juice and $386$ calories in $100$ grams of sugar. Water contains no calories. How many calories are in $200$ grams of her lemonade?

$129$

$137$

$174$

$223$

$411$

Concepts:ratio and proportion

Difficulty rating: 1070

Small Hint:

Find the total calories and total grams of the whole batch first

Big Hint:

$200$ grams is one third of the full $600$ -gram batch

Solution:

The full batch weighs $100 + 100 + 400 = 600$ grams and contains $25 + 386 = 411$ calories.

Since $200$ grams is one third of the batch, it has $\frac{411}{3} = 137$ calories.

Thus, the correct answer is B.

7.

Mr. and Mrs. Lopez have two children. When they get into their family car, two people sit in the front, and the other two sit in the back. Either Mr. Lopez or Mrs. Lopez must sit in the driver’s seat. How many seating arrangements are possible?

$4$

$12$

$16$

$24$

$48$

Concepts:multiplication principle arrangements with restrictions

Difficulty rating: 1180

Small Hint:

Fill the seats in order: driver, then front passenger, then the two back seats

Big Hint:

There are $2$ choices for the driver and $3$ for the front passenger, then $2$ orders in back

Solution:

The driver is one of the two parents: $2$ choices.

Any of the remaining $3$ people can sit in the front passenger seat, and the last $2$ people fill the back in $2$ orders.

The total is $2 \cdot 3 \cdot 2 = 12.$

Thus, the correct answer is B.

8.

The lines $x = \tfrac14 y + a, \qquad y = \tfrac14 x + b$ intersect at the point $(1, 2).$ What is $a + b?$

$0$

$\dfrac{3}{4}$

$1$

$2$

$\dfrac{9}{4}$

Concepts:system of equations substitution

Difficulty rating: 1250

Small Hint:

Substitute $x = 1$ and $y = 2$ into both equations

Big Hint:

Solve each equation for $a$ and $b$ separately, then add

Solution:

Substituting $(1, 2)$ gives $1 = \frac{2}{4} + a \quad\Rightarrow\quad a = \frac{1}{2},$ and $2 = \frac{1}{4} + b \quad\Rightarrow\quad b = \frac{7}{4}.$

Therefore $a + b = \frac{1}{2} + \frac{7}{4} = \frac{9}{4}.$

Thus, the correct answer is E.

9.

How many even three-digit integers have the property that their digits, read left to right, are in strictly increasing order?

$21$

$34$

$51$

$72$

$150$

Concepts:digits combinations casework

Difficulty rating: 1390

Small Hint:

The units digit is even; the two smaller digits must both be less than it

Big Hint:

For each even units digit $c,$ count the ways to choose two increasing digits below $c$

Solution:

Let the digits be $a \lt b \lt c$ with $c$ even. Since $a \geq 1,$ no digit is zero, and $c \neq 2$ (there is no room for two smaller nonzero digits).

Once the units digit $c$ is fixed, any two distinct digits below it can be arranged in increasing order in exactly one way. So the count for each $c$ is $\binom{c-1}{2}.$

For $c = 4, 6, 8$ this gives $\begin{aligned} &\binom{3}{2} + \binom{5}{2} \\ &\quad {}+ \binom{7}{2} = 3 + 10 + 21 \\ &\quad = 34. \end{aligned}$

Thus, the correct answer is B.

10.

In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is $15.$ What is the greatest possible perimeter of the triangle?

$43$

$44$

$45$

$46$

$47$

Concepts:triangle inequality optimization

Difficulty rating: 1450

Small Hint:

Let the sides be $x,$ $3x,$ and $15,$ then apply the triangle inequality

Big Hint:

The binding condition is $x + 15 \gt 3x,$ which caps how large $x$ can be

Solution:

Let the sides be $x,$ $3x,$ and $15.$ The triangle inequality requires $x + 3x \gt 15,$ so $x \geq 4,$ and $x + 15 \gt 3x,$ so $x \leq 7.$

The perimeter $4x + 15$ is largest when $x = 7,$ giving $7 + 21 + 15 = 43.$

Thus, the correct answer is A.

11.

Joe and JoAnn each bought $12$ ounces of coffee in a $16$ -ounce cup. Joe drank $2$ ounces of his coffee and then added $2$ ounces of cream. JoAnn added $2$ ounces of cream, stirred the coffee well, and then drank $2$ ounces. What is the resulting ratio of the amount of cream in Joe’s coffee to that in JoAnn’s coffee?

$\dfrac{6}{7}$

$\dfrac{13}{14}$

$1$

$\dfrac{14}{13}$

$\dfrac{7}{6}$

Concepts:mixture ratio and proportion

Difficulty rating: 1510

Small Hint:

Joe adds cream after drinking, so all $2$ ounces of his cream remain

Big Hint:

JoAnn drinks from a well-mixed $14$ -ounce cup, removing cream in proportion

Solution:

Joe adds the cream last, so his cup holds all $2$ ounces of cream.

JoAnn’s cup has $14$ ounces of mixture containing $2$ ounces of cream. Drinking $2$ ounces removes a fraction $\tfrac{2}{14}$ of everything, leaving $2 \cdot \frac{12}{14} = \frac{12}{7}$ ounces of cream.

The ratio is $\frac{2}{\,\frac{12}{7}\,} = \frac{14}{12} = \frac{7}{6}.$

Thus, the correct answer is E.

12.

The parabola $y = ax^2 + bx + c$ has vertex $(p, p)$ and $y$ -intercept $(0, -p),$ where $p \neq 0.$ What is $b?$

$-p$

$0$

$2$

$4$

$p$

Concepts:parabola quadratic

Difficulty rating: 1530

Small Hint:

Write the parabola in vertex form $y = a(x - p)^2 + p$

Big Hint:

Use the $y$ -intercept to find $a$ in terms of $p,$ then read off the coefficient of $x$

Solution:

The vertex form is $y = a(x - p)^2 + p.$

At $x = 0,$ $y = ap^2 + p = -p,$ so $ap^2 = -2p$ and $a = -\dfrac{2}{p}.$

Expanding, $y = a x^2 - 2ap\, x + ap^2 + p,$ so $b = -2ap = -2\left(-\dfrac{2}{p}\right)p = 4.$

Thus, the correct answer is D.

13.

Rhombus $ABCD$ is similar to rhombus $BFDE.$ The area of rhombus $ABCD$ is $24,$ and $\angle BAD = 60^\circ.$ What is the area of rhombus $BFDE?$

$6$

$4\sqrt{3}$

$8$

$9$

$6\sqrt{3}$

Concepts:rhombus equilateral triangle area decomposition

Difficulty rating: 1660

Small Hint:

Since $\angle BAD = 60^\circ,$ triangle $ABD$ is equilateral, so $BD$ equals a side

Big Hint:

If the diagonals meet at $O,$ then $\triangle ABO$ is a $30^\circ$ – $60^\circ$ – $90^\circ$ triangle; compare the long and short diagonals

Solution:

Let the diagonals of $ABCD$ meet at $O.$ They bisect each other at right angles, and since $\angle BAD=60^\circ,$ triangle $ABO$ is a $30^\circ$ – $60^\circ$ – $90^\circ$ triangle. Hence the half-diagonals satisfy $AO=\sqrt3\,BO.$

The segment $BD$ is the short diagonal of $ABCD$ and the long diagonal of the similar rhombus $BFDE.$ Thus the smaller-to-larger length ratio is $\frac{BD}{AC}=\frac{BO}{AO}=\frac{1}{\sqrt3}.$ Areas scale by the square of this ratio, so the area of $BFDE$ is $24\cdot\dfrac13=8.$

Thus, the correct answer is C.

14.

Elmo makes $N$ sandwiches for a fundraiser. For each sandwich he uses $B$ globs of peanut butter at $4$ ¢ per glob and $J$ blobs of jam at $5$ ¢ per blob. The cost of the peanut butter and jam to make all the sandwiches is $\$2.53.$ Assume that $B,$ $J,$ and $N$ are positive integers with $N \gt 1.$ What is the cost of the jam Elmo uses to make the sandwiches?

$\$1.05$

$\$1.25$

$\$1.45$

$\$1.65$

$\$1.85$

Concepts:Diophantine Equation prime factorization casework

Difficulty rating: 1680

Small Hint:

The total cost is $N(4B + 5J) = 253$ cents; factor $253$

Big Hint:

Test $N = 11, 23, 253,$ then determine the numbers of peanut-butter globs $B$ and jam blobs $J$

Solution:

The total cost in cents is $N(4B + 5J) = 253 = 11 \cdot 23.$ Since $N \gt 1,$ the value of $N$ is $11,$ $23,$ or $253.$

If $N = 253$ then $4B + 5J = 1,$ and if $N = 23$ then $4B + 5J = 11;$ neither has a positive integer solution.

So $N = 11$ and $4B + 5J = 23,$ whose only positive solution is $B = 2,$ $J = 3.$

The jam costs $N \cdot J \cdot 5$ ¢ $= 11 \cdot 3 \cdot 5 = 165$ cents, or $\$1.65.$

Thus, the correct answer is D.

15.

Circles with centers $O$ and $P$ have radii $2$ and $4,$ respectively, and are externally tangent. Points $A$ and $B$ are on the circle centered at $O,$ and points $C$ and $D$ are on the circle centered at $P,$ such that $AD$ and $BC$ are common external tangents to the circles. What is the area of hexagon $AOBCPD?$

$18\sqrt{3}$

$24\sqrt{2}$

$36$

$24\sqrt{3}$

$32\sqrt{2}$

Concepts:tangent circles trapezoid Pythagorean Theorem

Difficulty rating: 1680

Small Hint:

The centers are $OP = 2 + 4 = 6$ apart, and the radii to the tangent points are perpendicular to the tangent

Big Hint:

$AOPD$ is a right trapezoid; find its slant side with the Pythagorean theorem, then double the area

Solution:

The circles are externally tangent, so $OP = 2 + 4 = 6.$ In quadrilateral $AOPD,$ both $OA = 2$ and $PD = 4$ are perpendicular to the tangent line $AD,$ making it a right trapezoid.

Drawing the line through $O$ parallel to $AD$ creates a right triangle with hypotenuse $OP = 6$ and one leg $PD - OA = 2,$ so $AD = \sqrt{6^2 - 2^2} = \sqrt{32} = 4\sqrt2.$

The trapezoid $AOPD$ has area $\frac{1}{2}(2 + 4)(4\sqrt2) = 12\sqrt2.$

By symmetry the hexagon $AOBCPD$ is made of two such trapezoids, so its area is $2 \cdot 12\sqrt2 = 24\sqrt2.$

Thus, the correct answer is B.

16.

Regular hexagon $ABCDEF$ has vertices $A$ and $C$ at $(0, 0)$ and $(7, 1),$ respectively. What is its area?

$20\sqrt{3}$

$22\sqrt{3}$

$25\sqrt{3}$

$27\sqrt{3}$

$50$

Concepts:regular polygon distance formula area

Difficulty rating: 1740

Small Hint:

$A$ and $C$ are two vertices apart, so $AC$ is a short diagonal of the hexagon

Big Hint:

For side length $s,$ the short diagonal is $s\sqrt3;$ the area is $\dfrac{3\sqrt3}{2}s^2$

Solution:

The distance is $AC = \sqrt{7^2 + 1^2} = \sqrt{50}.$ In a regular hexagon with side $s,$ the distance between vertices two apart is $s\sqrt3,$ so $s^2 \cdot 3 = 50,$ giving $s^2 = \dfrac{50}{3}.$

The hexagon’s area is $\frac{3\sqrt3}{2}s^2 = \frac{3\sqrt3}{2} \cdot \frac{50}{3} = 25\sqrt3.$

Thus, the correct answer is C.

17.

For a particular peculiar pair of dice, the probabilities of rolling $1,$ $2,$ $3,$ $4,$ $5,$ and $6$ on each die are in the ratio $1 : 2 : 3 : 4 : 5 : 6.$ What is the probability of rolling a total of $7$ on the two dice?

$\dfrac{4}{63}$

$\dfrac{1}{8}$

$\dfrac{8}{63}$

$\dfrac{1}{6}$

$\dfrac{2}{7}$

Concepts:dice (probability)basic probability

Difficulty rating: 1740

Small Hint:

The probability of rolling $k$ is $\dfrac{k}{1+2+\cdots+6} = \dfrac{k}{21}$

Big Hint:

Sum $\dfrac{k}{21} \cdot \dfrac{7-k}{21}$ over the pairs that total $7$

Solution:

Since the weights sum to $21,$ the probability of rolling $k$ is $\dfrac{k}{21}.$

A total of $7$ comes from $(1,6), (2,5), \ldots, (6,1),$ so the probability is $\begin{aligned} &\scriptsize \frac{1 \cdot 6 + 2 \cdot 5 + 3 \cdot 4 + 4 \cdot 3 + 5 \cdot 2 + 6 \cdot 1}{21^2} \\ &= \frac{56}{441} = \frac{8}{63}. \end{aligned}$

Thus, the correct answer is C.

18.

An object in the plane moves from one lattice point to another. At each step, the object may move one unit to the right, one unit to the left, one unit up, or one unit down. If the object starts at the origin and takes a ten-step path, how many different points could be the final point?

$120$

$121$

$221$

$230$

$231$

Concepts:lattice point parity basic counting

Difficulty rating: 1820

Small Hint:

After $10$ steps the endpoint $(a, b)$ satisfies $|a| + |b| \le 10$ with $a + b$ even

Big Hint:

Count lattice points with even coordinate sum inside the diamond $|a| + |b| \le 10$

Solution:

Each step changes the coordinate sum by $1,$ so after $10$ steps the endpoint $(a, b)$ has $a + b$ even, and $|a| + |b| \le 10.$ Any such point is reachable: walk $|a| + |b|$ steps to it, then use the remaining even number of steps going out and back.

The reachable points lie on the lines $a + b = 2k$ for $-5 \le k \le 5.$ Each such line meets the diamond in exactly $11$ lattice points.

With $11$ lines and $11$ points each, there are $121$ points.

Thus, the correct answer is B.

19.

Mr. Jones has eight children of different ages. On a family trip his oldest child, who is $9,$ spots a license plate with a $4$ -digit number in which each of two digits appears two times. “Look, daddy!” she exclaims. “That number is evenly divisible by the age of each of us kids!” “That’s right,” replies Mr. Jones, “and the last two digits just happen to be my age.” Which of the following is not the age of one of Mr. Jones’s children?

$4$

$5$

$6$

$7$

$8$

Concepts:divisibility digits casework

Difficulty rating: 1920

Small Hint:

The oldest child is $9,$ so the number is divisible by $9,$ forcing the two digits to sum to $9$

Big Hint:

A number ending in a two-digit age cannot end in $0,$ so think about which small ages it must avoid

Solution:

The number has the form $aabb,$ $abab,$ or $baab.$ Divisibility by $9$ means $2(a + b)$ is a multiple of $9,$ so $a + b = 9.$

The eight distinct ages are eight of the nine integers from $1$ through $9,$ so at least one of ages $4$ and $8$ occurs. Therefore the number is divisible by $4.$ The possibilities become $1188, 2772, 3636,$ $5544, 6336, 7272, 9900.$

Since the last two digits are Mr. Jones’s age, $9900$ is impossible, and none of the others is a multiple of $5.$ So the children’s ages cannot include $5.$ Indeed $5544$ is divisible by $1, 2, 3, 4, 6, 7, 8, 9.$

Thus, the correct answer is B.

20.

Let $x$ be chosen at random from the interval $(0, 1).$ What is the probability that $\lfloor \log_{10} 4x \rfloor - \lfloor \log_{10} x \rfloor = 0?$ Here $\lfloor x \rfloor$ denotes the greatest integer that is less than or equal to $x.$

$\dfrac{1}{8}$

$\dfrac{3}{20}$

$\dfrac{1}{6}$

$\dfrac{1}{5}$

$\dfrac{1}{4}$

Concepts:floor and ceiling functions logarithm geometric probability

Difficulty rating: 2090

Small Hint:

The condition says $x$ and $4x$ have the same power-of-ten magnitude

Big Hint:

On each interval $[10^n, 10^{n+1}),$ the condition holds for $x \lt \dfrac{10^{n+1}}{4}$

Solution:

The equation says $\lfloor \log_{10} x \rfloor = \lfloor \log_{10} 4x \rfloor,$ i.e. $x$ and $4x$ lie in the same interval $[10^n, 10^{n+1}).$

This holds exactly when $10^n \le x$ and $4x \lt 10^{n+1},$ that is $10^n \le x \lt \dfrac{10^{n+1}}{4}.$

Within $[10^n, 10^{n+1}),$ the favorable fraction is $\begin{aligned} &\frac{\frac{10^{n+1}}{4} - 10^n}{10^{n+1} - 10^n} \\ &= \frac{\frac{10}{4} - 1}{10 - 1} = \frac{1}{6}. \end{aligned}$

Since this fraction is the same on every such interval, the overall probability is $\dfrac{1}{6}.$

Thus, the correct answer is C.

21.

Rectangle $ABCD$ has area $2006.$ An ellipse with area $2006\pi$ passes through $A$ and $C$ and has foci at $B$ and $D.$ What is the perimeter of the rectangle? (The area of an ellipse is $\pi ab,$ where $2a$ and $2b$ are the lengths of its axes.)

$\dfrac{16\sqrt{2006}}{\pi}$

$\dfrac{1003}{4}$

$8\sqrt{1003}$

$6\sqrt{2006}$

$\dfrac{32\sqrt{1003}}{\pi}$

Concepts:ellipse algebraic manipulation

Difficulty rating: 2150

Small Hint:

For the ellipse, the sum of distances from $A$ to the foci $B, D$ equals a rectangle diagonal plus a side

Big Hint:

Let the sides be $x, y;$ then $x + y = 2a$ and the diagonal $\sqrt{x^2 + y^2} = 2\sqrt{a^2 - b^2}$

Solution:

Let the rectangle’s sides be $x$ and $y.$ Point $A$ is on the ellipse with foci $B$ and $D,$ so $x + y = AB + AD = 2a.$ The distance between the foci is the diagonal, so $\sqrt{x^2 + y^2} = 2\sqrt{a^2 - b^2}.$

Then $2xy = (x + y)^2 - (x^2 + y^2)$ $= 4a^2 - (4a^2 - 4b^2)$ $= 4b^2,$ so $xy = 2b^2.$ The area gives $2b^2 = 2006,$ hence $b^2 = 1003.$

The ellipse area gives $\pi ab = 2006\pi,$ so $ab = 2006$ and $a = \dfrac{2006}{\sqrt{1003}} = 2\sqrt{1003}.$

The perimeter is $2(x + y) = 4a = 8\sqrt{1003}.$

Thus, the correct answer is C.

22.

Suppose $a,$ $b,$ and $c$ are positive integers with $a + b + c = 2006,$ and $a!\,b!\,c! = m \cdot 10^n,$ where $m$ and $n$ are integers and $m$ is not divisible by $10.$ What is the smallest possible value of $n?$

$489$

$492$

$495$

$498$

$501$

Concepts:Legendre’s Formula trailing zeros optimization

Difficulty rating: 2300

Small Hint:

Factors of $5$ are scarcer than factors of $2,$ so $n$ counts the factors of $5$ in $a!\,b!\,c!$

Big Hint:

Use $\lfloor x \rfloor + \lfloor y \rfloor$ $+ \lfloor z \rfloor \ge \lfloor x + y + z \rfloor - 2$ at each power of $5$

Solution:

Since factors of $2$ are more plentiful than factors of $5,$ $n$ equals the number of factors of $5$ in $a!\,b!\,c!,$ namely $n = \sum_{k \ge 1}\left(\left\lfloor \tfrac{a}{5^k}\right\rfloor + \left\lfloor \tfrac{b}{5^k}\right\rfloor + \left\lfloor \tfrac{c}{5^k}\right\rfloor\right).$

For each $k,$ $\lfloor \frac{a}{5^k} \rfloor + \lfloor \frac{b}{5^k} \rfloor$ $+ \lfloor \frac{c}{5^k} \rfloor$ $\ge \lfloor \frac{2006}{5^k} \rfloor - 2.$ Summing over $k = 1, 2, 3, 4$ (as $2006 \lt 5^5$ ) gives $\begin{aligned} &n \ge (401 + 80 + 16 + 3) \\ &\quad {}- 4 \cdot 2 = 492. \end{aligned}$

Equality is attainable, for example with $a = b = 624$ and $c = 758.$ So the minimum is $492.$

Thus, the correct answer is B.

23.

Isosceles $\triangle ABC$ has a right angle at $C.$ Point $P$ is inside $\triangle ABC,$ such that $PA = 11,$ $PB = 7,$ and $PC = 6.$ Legs $\overline{AC}$ and $\overline{BC}$ have length $s = \sqrt{a + b\sqrt{2}},$ where $a$ and $b$ are positive integers. What is $a + b?$

$85$

$91$

$108$

$121$

$127$

Concepts:transformation law of cosines Pythagorean Theorem

Difficulty rating: 2390

Small Hint:

Rotate $\triangle ABC$ by $90^\circ$ about $C$ so that $A$ maps onto $B$

Big Hint:

The image of $P$ forms an isosceles right triangle with $P,$ which pins down $\angle BPC$

Solution:

Rotate $\triangle ABC$ by $90^\circ$ about $C,$ sending $A$ to $B$ and $P$ to $P'.$ Then $CP' = CP = 6$ and $\angle PCP' = 90^\circ,$ so $\triangle PCP'$ is an isosceles right triangle with $PP' = 6\sqrt2.$

Also $BP' = AP = 11.$ Since $(6\sqrt2)^2 + 7^2 = 72 + 49$ $= 121 = 11^2,$ triangle $BPP'$ has a right angle at $P.$ Hence $\angle BPC = \angle BPP'$ $+ \angle P'PC = 90^\circ$ $+ 45^\circ = 135^\circ.$

By the Law of Cosines in $\triangle BPC,$ $\begin{aligned} &BC^2 = 6^2 + 7^2 \\ &\quad {}- 2 \cdot 6 \cdot 7 \cos 135^\circ \\ &\quad = 85 + 42\sqrt2. \end{aligned}$

So $s^2 = 85 + 42\sqrt2,$ giving $a = 85,$ $b = 42,$ and $a + b = 127.$

Thus, the correct answer is E.

24.

Let $S$ be the set of all points $(x, y)$ in the coordinate plane such that $0 \le x \le \dfrac{\pi}{2}$ and $0 \le y \le \dfrac{\pi}{2}.$ What is the area of the subset of $S$ for which $\sin^2 x - \sin x \sin y + \sin^2 y \le \frac{3}{4}?$

$\dfrac{\pi^2}{9}$

$\dfrac{\pi^2}{8}$

$\dfrac{\pi^2}{6}$

$\dfrac{3\pi^2}{16}$

$\dfrac{2\pi^2}{9}$

Concepts:trigonometric identity quadratic area

Difficulty rating: 2480

Small Hint:

Treat the expression as a quadratic in $\sin x$ and find where it equals $\dfrac34$

Big Hint:

The boundary reduces to $\sin x = \sin\!\left(y \pm \dfrac{\pi}{3}\right),$ giving straight lines in $S$

Solution:

Fixing $y,$ solve $\sin^2 x - \sin x \sin y + \sin^2 y = \dfrac34$ as a quadratic in $\sin x:$ $\begin{aligned} &\sin x = \frac{1}{2}\sin y \\ &\quad {}\pm \frac{\sqrt3}{2}\cos y \\ &\quad = \sin\!\left(y \pm \frac{\pi}{3}\right). \end{aligned}$

Within $S,$ $\sin x = \sin\!\left(y - \tfrac{\pi}{3}\right)$ gives the line $x = y - \tfrac{\pi}{3},$ while $\sin x = \sin\!\left(y + \tfrac{\pi}{3}\right)$ gives $x = y + \tfrac{\pi}{3}$ for $y \le \tfrac{\pi}{6}$ and $x = -y + \tfrac{2\pi}{3}$ for $y \ge \tfrac{\pi}{6}.$

These lines split $S$ into regions; testing the corners shows the inequality holds only in the middle band. Its area is $\begin{aligned} &\left(\frac{\pi}{2}\right)^2 - \frac{1}{2}\left(\frac{\pi}{3}\right)^2 \\ &\quad {}- 2 \cdot \frac{1}{2}\left(\frac{\pi}{6}\right)^2 \\ &\quad = \frac{\pi^2}{6}. \end{aligned}$

Thus, the correct answer is C.

25.

A sequence $a_1,$ $a_2,$ $\ldots$ of non-negative integers is defined by the rule $a_{n+2} = |a_{n+1} - a_n|$ for $n \ge 1.$ If $a_1 = 999,$ $a_2 \lt 999,$ and $a_{2006} = 1,$ how many different values of $a_2$ are possible?

$165$

$324$

$495$

$499$

$660$

Concepts:greatest common divisor parity inclusion-exclusion

Difficulty rating: 2520

Small Hint:

The terms $a_n$ and $a_{n+3}$ always share the same parity; use this with $a_{2006} = 1$

Big Hint:

Every term is a multiple of $\gcd(a_1, a_2),$ so $\gcd(999, a_2) = 1;$ note $999 = 3^3 \cdot 37$

Solution:

The rule gives $a_n \equiv a_{n+3} \pmod 2,$ so $a_2$ has the same parity as $a_{2006} = 1;$ thus $a_2$ is odd.

Every term is a multiple of $\gcd(a_1, a_2),$ and $a_{2006} = 1$ forces $\gcd(999, a_2) = 1.$ Since $999 = 3^3 \cdot 37,$ we need $a_2$ not divisible by $3$ or $37.$

Among the odd integers in $[1, 998]$ there are $499;$ removing the $166$ multiples of $3$ and $13$ multiples of $37,$ then adding back the $4$ multiples of $111,$ leaves $499 - 166 - 13 + 4 = 324.$

Each such $a_2$ works. For consecutive positive terms $u,v,$ the update $(u,v)\mapsto(v,|v-u|)$ reduces their maximum within at most two steps. Since both initial terms are at most $999,$ some $a_N=0$ occurs by $N\le1999.$

The gcd of each consecutive pair is invariant, so the equal terms immediately before that zero both equal $\gcd(999,a_2)=1.$ The sequence then cycles through $1,1,0.$ Finally, $2006\equiv2\pmod3,$ so $a_{2006}$ has the same odd parity as $a_2;$ in this cycle it must therefore be $1.$

Thus, the correct answer is B.