2009 AMC 12A — Problems & Online Timed Exam

Timed

1:15:00

1.

Kim’s flight took off from Newark at $10{:}34$ am and landed in Miami at $1{:}18$ pm. Both cities are in the same time zone. If her flight took $h$ hours and $m$ minutes, with $0 \le m \lt 60,$ what is $h + m?$

$46$

$47$

$50$

$53$

$54$

Answer: A

Concepts:date and time clock

Difficulty rating: 730

Small Hint:

Count the minutes from $10{:}34$ am up to $11{:}00$ am, then continue from there

Big Hint:

The flight lasts $2$ hours and $26 + 18$ minutes

Solution:

From $10{:}34$ am to $11{:}00$ am is $26$ minutes, from $11{:}00$ am to $1{:}00$ pm is $2$ hours, and from $1{:}00$ pm to $1{:}18$ pm is $18$ minutes.

So the flight lasted $2$ hours and $26 + 18 = 44$ minutes. Thus $h = 2,$ $m = 44,$ and $h + m = 46.$

Thus, the correct answer is A.

2.

Which of the following is equal to $1 + \dfrac{1}{1 + \dfrac{1}{1 + 1}}?$

$\dfrac{5}{4}$

$\dfrac{3}{2}$

$\dfrac{5}{3}$

$2$

$3$

Answer: C

Concepts:continued fraction fraction

Difficulty rating: 860

Small Hint:

Work outward from the innermost fraction

Big Hint:

First simplify $1 + \dfrac{1}{1+1} = \dfrac{3}{2},$ then take its reciprocal

Solution:

Starting inside, $1 + 1 = 2,$ so $1 + \dfrac{1}{2} = \dfrac{3}{2}.$ Then $\dfrac{1}{\frac{3}{2}} = \dfrac{2}{3},$ and $1 + \dfrac{2}{3} = \dfrac{5}{3}.$

Thus, the correct answer is C.

3.

What number is one third of the way from $\dfrac{1}{4}$ to $\dfrac{3}{4}?$

$\dfrac{1}{3}$

$\dfrac{5}{12}$

$\dfrac{1}{2}$

$\dfrac{7}{12}$

$\dfrac{2}{3}$

Answer: B

Concepts:fraction ratio and proportion

Difficulty rating: 1050

Small Hint:

The distance from $\dfrac{1}{4}$ to $\dfrac{3}{4}$ is $\dfrac{1}{2}$

Big Hint:

Add one third of that distance to $\dfrac{1}{4}$

Solution:

The gap is $\dfrac{3}{4} - \dfrac{1}{4} = \dfrac{1}{2}.$ One third of the way adds $\dfrac{1}{3} \cdot \dfrac{1}{2} = \dfrac{1}{6}.$

So the number is $\dfrac{1}{4} + \dfrac{1}{6} = \dfrac{3}{12} + \dfrac{2}{12} = \dfrac{5}{12}.$

Thus, the correct answer is B.

4.

Four coins are picked out of a piggy bank that contains a collection of pennies, nickels, dimes, and quarters. Which of the following could not be the total value of the four coins, in cents?

$15$

$25$

$35$

$45$

$55$

Answer: A

Concepts:money multiple casework

Difficulty rating: 1200

Small Hint:

A selection that includes at least one penny cannot total a multiple of $5$

Big Hint:

Without a penny, all four coins are worth at least $5$ cents each

Solution:

If the four coins include a penny, the total is not a multiple of $5,$ so it cannot equal any of the five listed multiples of $5.$ If there is no penny, every coin is worth at least $5$ cents, so the total is at least $20$ cents. Either way, $15$ is impossible.

The other amounts are attainable: $25 = 10 + 5 + 5 + 5,$ $35 = 10 + 10 + 10 + 5,$ $45 = 25 + 10 + 5 + 5,$ and $55 = 25 + 10 + 10 + 10.$

Thus, the correct answer is A.

5.

One dimension of a cube is increased by $1,$ another is decreased by $1,$ and the third is left unchanged. The volume of the new rectangular solid is $5$ less than that of the cube. What was the volume of the cube?

$8$

$27$

$64$

$125$

$216$

Answer: D

Concepts:volume difference of squares

Difficulty rating: 1100

Small Hint:

Let the cube have side length $x$ and write the new volume as a product

Big Hint:

The new solid has volume $x(x+1)(x-1) = x^3 - x$

Solution:

Let the cube have side length $x.$ The new solid has dimensions $x + 1,$ $x - 1,$ and $x,$ so its volume is $x(x+1)(x-1) = x^3 - x.$

Setting this equal to $x^3 - 5$ gives $x^3 - x = x^3 - 5,$ so $x = 5.$

The cube’s volume is $5^3 = 125.$

Thus, the correct answer is D.

6.

Suppose that $P = 2^m$ and $Q = 3^n.$ Which of the following is equal to $12^{mn}$ for every pair of integers $(m, n)?$

$P^2 Q$

$P^n Q^m$

$P^n Q^{2m}$

$P^{2m} Q^n$

$P^{2n} Q^m$

Answer: E

Concepts:exponent substitution

Difficulty rating: 1290

Small Hint:

Write $12 = 2^2 \cdot 3$

Big Hint:

Then $12^{mn} = 2^{2mn} \cdot 3^{mn}$ $= (2^m)^{2n}(3^n)^m$

Solution:

Since $12 = 2^2 \cdot 3,$ $\begin{aligned} 12^{mn} &= 2^{2mn} \cdot 3^{mn} \\ &= (2^m)^{2n}(3^n)^m \\ &= P^{2n}Q^m. \end{aligned}$

Thus, the correct answer is E.

7.

The first three terms of an arithmetic sequence are $2x - 3,$ $5x - 11,$ and $3x + 1$ respectively. The $n$ th term of the sequence is $2009.$ What is $n?$

$255$

$502$

$1004$

$1506$

$8037$

Answer: B

Concepts:arithmetic sequence

Difficulty rating: 1250

Small Hint:

Consecutive differences are equal: $(5x - 11) - (2x - 3)$ $= (3x + 1) - (5x - 11)$

Big Hint:

Solve for $x,$ then find the first term and common difference

Solution:

Equal consecutive differences give $(5x - 11) - (2x - 3)$ $= (3x + 1) - (5x - 11),$ that is $3x - 8 = -2x + 12,$ so $x = 4.$

The first three terms are $5, 9, 13,$ with common difference $4.$

The $n$ th term satisfies $2009 = 5 + (n - 1)\cdot 4,$ so $n - 1 = 501$ and $n = 502.$

Thus, the correct answer is B.

8.

Four congruent rectangles are placed as shown. The area of the outer square is $4$ times that of the inner square. What is the ratio of the length of the longer side of each rectangle to the length of its shorter side?

$3$

$\sqrt{10}$

$2 + \sqrt{2}$

$2\sqrt{3}$

$4$

Answer: A

Concepts:area ratio square (geometry)system of equations

Difficulty rating: 1410

Small Hint:

Let each rectangle have shorter side $x$ and longer side $y$

Big Hint:

The outer square has side $x + y,$ the inner square has side $y - x,$ and their side ratio is $\sqrt{4} = 2$

Solution:

Let the rectangles have shorter side $x$ and longer side $y.$ The outer square has side $x + y$ and the inner square has side $y - x.$

Since the outer area is $4$ times the inner area, the side ratio is $\sqrt{4} = 2,$ so $x + y = 2(y - x).$

This gives $y = 3x,$ so the ratio of longer to shorter side is $3.$

Thus, the correct answer is A.

9.

Suppose that $f(x + 3) = 3x^2 + 7x + 4$ and $f(x) = ax^2 + bx + c.$ What is $a + b + c?$

$-1$

$0$

$1$

$2$

$3$

Answer: D

Concepts:function substitution

Difficulty rating: 1350

Small Hint:

$a + b + c = f(1)$

Big Hint:

Write $1 = (-2) + 3$ and use the given formula for $f(x + 3)$

Solution:

Note that $a + b + c = f(1).$

Using $f(x + 3) = 3x^2 + 7x + 4$ with $x = -2$ gives $\begin{aligned} f(1) &= f(-2 + 3) \\ &= 3(-2)^2 + 7(-2) + 4 \\ &= 12 - 14 + 4 = 2. \end{aligned}$

Thus, the correct answer is D.

10.

In quadrilateral $ABCD,$ $AB = 5,$ $BC = 17,$ $CD = 5,$ $DA = 9,$ and $BD$ is an integer. What is $BD?$

$11$

$12$

$13$

$14$

$15$

Answer: C

Concepts:triangle inequality bounding to limit cases

Difficulty rating: 1500

Small Hint:

Apply the triangle inequality to $\triangle BCD$ and $\triangle ABD$

Big Hint:

From $\triangle BCD,$ $BD \gt 12,$ and from $\triangle ABD,$ $BD \lt 14$

Solution:

In $\triangle BCD,$ the triangle inequality gives $BD + CD \gt BC,$ so $BD + 5 \gt 17$ and $BD \gt 12.$

In $\triangle ABD,$ $AB + DA \gt BD,$ so $BD \lt 5 + 9 = 14.$

The only integer with $12 \lt BD \lt 14$ is $13.$

Thus, the correct answer is C.

11.

The figures $F_1,$ $F_2,$ $F_3,$ and $F_4$ shown are the first in a sequence of figures. For $n \ge 3,$ $F_n$ is constructed from $F_{n-1}$ by surrounding it with a square and placing one more diamond on each side of the new square than $F_{n-1}$ had on each side of its outside square. For example, figure $F_3$ has $13$ diamonds. How many diamonds are there in figure $F_{20}?$

$401$

$485$

$585$

$626$

$761$

Answer: E

Concepts:arithmetic sequence summation

Difficulty rating: 1630

Small Hint:

The outside square of $F_n$ has $4(n - 1)$ diamonds

Big Hint:

Summing over all rings, $F_n$ has $1 + 4\big(1 + 2 + \cdots + (n - 1)\big)$ diamonds

Solution:

The outside square of $F_n$ has $4$ more diamonds than that of $F_{n-1},$ and the outside square of $F_2$ has $4,$ so the outside square of $F_n$ has $4(n - 1)$ diamonds.

Adding all the rings, $\begin{gathered} 1 + 4\big(1 + 2 + \cdots + (n - 1)\big) \\ = 1 + 4\cdot\frac{(n - 1)n}{2} \\ = 1 + 2(n - 1)n. \end{gathered}$

For $n = 20,$ this is $1 + 2\cdot 19\cdot 20 = 761.$

Thus, the correct answer is E.

12.

How many positive integers less than $1000$ are $6$ times the sum of their digits?

$0$

$1$

$2$

$4$

$12$

Answer: B

Concepts:digits bounding to limit cases casework

Difficulty rating: 1730

Small Hint:

If $N$ equals $6$ times its digit sum, then $N$ is a multiple of $6$ and $N$ is small

Big Hint:

The digit sum of a number below $1000$ is at most $27,$ so $N \le 162$

Solution:

If $N = 6\cdot(\text{digit sum}),$ then since the digit sum of a number below $1000$ is at most $27,$ we have $N \le 162.$

For a two-digit number $10t + u = 6(t + u)$ gives $4t = 5u,$ forcing $t = 5$ and $u = 4,$ so $N = 54.$ A one-digit number would need $6u = u,$ impossible for $u \gt 0.$ A three-digit number $100h + 10t + u = 6(h + t + u)$ gives $94h + 4t = 5u,$ whose left side is at least $94$ while the right side is at most $45,$ so there is no solution.

Hence exactly one number, $54,$ works.

Thus, the correct answer is B.

13.

A ship sails $10$ miles in a straight line from $A$ to $B,$ turns through an angle between $45^\circ$ and $60^\circ,$ and then sails another $20$ miles to $C.$ Let $AC$ be measured in miles. Which of the following intervals contains $AC^2?$

$[400, 500]$

$[500, 600]$

$[600, 700]$

$[700, 800]$

$[800, 900]$

Answer: D

Concepts:law of cosines bounding to limit cases

Difficulty rating: 1770

Small Hint:

By the Law of Cosines, $AC^2 = 10^2$ $+ 20^2$ $- 2\cdot 10\cdot 20\cos(\angle ABC)$

Big Hint:

The turn is between $45^\circ$ and $60^\circ,$ so $\angle ABC$ is between $120^\circ$ and $135^\circ$

Solution:

By the Law of Cosines, $\begin{aligned} AC^2 &= 10^2 + 20^2 \\ &\quad {}- 2\cdot 10\cdot 20\cos(\angle ABC) \\ &= 500 - 400\cos(\angle ABC). \end{aligned}$

The ship turns through an angle between $45^\circ$ and $60^\circ,$ so the interior angle $\angle ABC$ lies between $120^\circ$ and $135^\circ.$

Since $\cos 120^\circ = -\dfrac{1}{2}$ and $\cos 135^\circ = -\dfrac{\sqrt{2}}{2},$ $\begin{aligned} 700 &= 500 + 200 \\ &\le AC^2 \le 500 + 200\sqrt{2} \\ &\lt 800. \end{aligned}$

So $AC^2$ lies in $[700, 800].$

Thus, the correct answer is D.

14.

A triangle has vertices $(0, 0),$ $(1, 1),$ and $(6m, 0),$ and the line $y = mx$ divides the triangle into two triangles of equal area. What is the sum of all possible values of $m?$

$-\dfrac{1}{3}$

$-\dfrac{1}{6}$

$\dfrac{1}{6}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

Answer: B

Concepts:coordinate geometry median (geometry)Vieta’s Formulas

Difficulty rating: 1820

Small Hint:

The line passes through the vertex $(0, 0),$ so to bisect the area it must hit the midpoint of the opposite side

Big Hint:

Set the midpoint of the segment from $(1, 1)$ to $(6m, 0)$ on the line $y = mx$

Solution:

The line $y = mx$ passes through the vertex $(0, 0),$ so it bisects the triangle’s area exactly when it passes through the midpoint of the opposite side, joining $(1, 1)$ and $(6m, 0).$ That midpoint is $\left(\dfrac{6m + 1}{2}, \dfrac{1}{2}\right).$

Requiring it to satisfy $y = mx$ gives $\frac{1}{2} = m\cdot\frac{6m + 1}{2},$ so $6m^2 + m - 1 = 0,$ that is $(3m - 1)(2m + 1) = 0.$

The possible values are $m = \dfrac{1}{3}$ and $m = -\dfrac{1}{2},$ whose sum is $-\dfrac{1}{6}.$

Thus, the correct answer is B.

15.

For what value of $n$ is $\begin{aligned} &i + 2i^2 + 3i^3 + \cdots + ni^n \\ &= 48 + 49i? \end{aligned}$ Note: here $i = \sqrt{-1}.$

$24$

$48$

$49$

$97$

$98$

Answer: D

Concepts:complex number pairing and grouping

Difficulty rating: 2010

Small Hint:

Group the terms in blocks of four consecutive powers of $i$

Big Hint:

Each block $(k + 1)i^{k+1} + \cdots + (k + 4)i^{k+4}$ with $k$ a multiple of $4$ equals $2 - 2i$

Solution:

For $k$ a multiple of $4,$ $\begin{aligned} &(k + 1)i^{k+1} + (k + 2)i^{k+2} \\ &\quad {}+ (k + 3)i^{k+3} + (k + 4)i^{k+4} \\ &= (k + 1)i - (k + 2) \\ &\quad {}- (k + 3)i + (k + 4) \\ &= 2 - 2i. \end{aligned}$

Summing the first $96$ terms (that is $24$ blocks) gives $24(2 - 2i) = 48 - 48i.$

Adding the next term $97i^{97} = 97i$ yields $48 - 48i + 97i = 48 + 49i.$ So $n = 97.$

Thus, the correct answer is D.

16.

A circle with center $C$ is tangent to the positive $x$ - and $y$ -axes and externally tangent to the circle centered at $(3, 0)$ with radius $1.$ What is the sum of all possible radii of the circle with center $C?$

$3$

$4$

$6$

$8$

$9$

Answer: D

Concepts:tangent circles coordinate geometry Vieta’s Formulas

Difficulty rating: 1910

Small Hint:

A circle tangent to both positive axes with radius $r$ has center $(r, r)$

Big Hint:

External tangency gives distance $r + 1$ between $(r, r)$ and $(3, 0);$ form a quadratic in $r$ and use Vieta

Solution:

A circle tangent to both positive axes with radius $r$ has center $(r, r).$ External tangency to the circle at $(3, 0)$ of radius $1$ means the distance between centers is $r + 1$ : $(r - 3)^2 + r^2 = (r + 1)^2.$

Expanding gives $r^2 - 8r + 8 = 0.$ Both roots $r = 4 \pm 2\sqrt{2}$ are positive, and by Vieta’s formulas their sum is $8.$

Thus, the correct answer is D.

17.

Let $a + ar_1 + ar_1^2 + ar_1^3 + \cdots$ and $a + ar_2 + ar_2^2 + ar_2^3 + \cdots$ be two different infinite geometric series of positive numbers with the same first term. The sum of the first series is $r_1,$ and the sum of the second series is $r_2.$ What is $r_1 + r_2?$

$0$

$\dfrac{1}{2}$

$1$

$\dfrac{1 + \sqrt{5}}{2}$

$2$

Answer: C

Concepts:geometric sequence Vieta’s Formulas

Difficulty rating: 2040

Small Hint:

The sum of an infinite geometric series with first term $a$ and ratio $r$ is $\dfrac{a}{1 - r}$

Big Hint:

Each ratio satisfies $\dfrac{a}{1 - r} = r,$ that is $r^2 - r + a = 0;$ use Vieta on the two roots

Solution:

For a series with first term $a$ and ratio $r,$ the sum is $\dfrac{a}{1 - r} = r,$ so $r^2 - r + a = 0.$

Both $r_1$ and $r_2$ satisfy this same quadratic, and since the two series are different, $r_1 \ne r_2,$ so they are its two distinct roots. By Vieta’s formulas, $r_1 + r_2 = 1.$

Thus, the correct answer is C.

18.

For $k \gt 0,$ let $I_k = 10\ldots064,$ where there are $k$ zeros between the $1$ and the $6.$ Let $N(k)$ be the number of factors of $2$ in the prime factorization of $I_k.$ What is the maximum value of $N(k)?$

$6$

$7$

$8$

$9$

$10$

Answer: B

Concepts:prime factorization sum and difference of cubes

Difficulty rating: 2010

Small Hint:

Write $I_k = 10^{k+2} + 64 = 2^{k+2}5^{k+2} + 2^6$

Big Hint:

Factor out the smaller power of $2;$ the largest count comes at $k = 4,$ where $5^6 + 1$ supplies one extra factor of $2$

Solution:

Note that $I_k = 10^{k+2} + 64$ $= 2^{k+2}5^{k+2} + 2^6.$

For $k \lt 4$ the first term has fewer than $6$ factors of $2,$ so $N(k) \lt 6.$ For $k \gt 4$ the first term is divisible by $2^7$ but the $2^6$ term is not, so $N(k) \lt 7.$

For $k = 4,$ $I_4 = 2^6(5^6 + 1).$ Since $5^6 + 1$ $= (5^2 + 1)\big((5^2)^2 - 5^2 + 1\big)$ $= 26\cdot 601,$ and $26 = 2\cdot 13$ contributes exactly one more factor of $2,$ we get $N(4) = 7.$

So the maximum value is $7.$

Thus, the correct answer is B.

19.

Andrea inscribed a circle inside a regular pentagon, circumscribed a circle around the pentagon, and calculated the area of the region between the two circles. Bethany did the same with a regular heptagon ( $7$ sides). The areas of the two regions were $A$ and $B,$ respectively. Each polygon had a side length of $2.$ Which of the following is true?

$A = \dfrac{25}{49}B$

$A = \dfrac{5}{7}B$

$A = B$

$A = \dfrac{7}{5}B$

$A = \dfrac{49}{25}B$

Answer: C

Concepts:regular polygon Pythagorean Theorem circle area

Difficulty rating: 1910

Small Hint:

For a regular polygon with side $2,$ let $r$ and $R$ be the inradius and circumradius, and look at the center, a side’s midpoint, and its endpoint

Big Hint:

That right triangle has legs $r$ and $1$ and hypotenuse $R,$ so $R^2 - r^2 = 1$ regardless of the number of sides

Solution:

For a regular polygon with side length $2,$ let $O$ be the center, $M$ the midpoint of a side, and $N$ an endpoint of that side. Then $\triangle OMN$ has a right angle at $M,$ with $MN = 1,$ $OM = r$ (inradius), and $ON = R$ (circumradius).

So $R^2 - r^2 = 1,$ and the area between the circles is $\pi(R^2 - r^2) = \pi$ for any number of sides. Hence $A = B.$

Thus, the correct answer is C.

20.

Convex quadrilateral $ABCD$ has $AB = 9$ and $CD = 12.$ Diagonals $AC$ and $BD$ intersect at $E,$ $AC = 14,$ and $\triangle AED$ and $\triangle BEC$ have equal areas. What is $AE?$

$\dfrac{9}{2}$

$\dfrac{50}{11}$

$\dfrac{21}{4}$

$\dfrac{17}{3}$

$6$

Answer: E

Concepts:parallel lines similarity area ratio

Difficulty rating: 1930

Small Hint:

Equal areas of $\triangle AED$ and $\triangle BEC$ imply $\triangle ACD$ and $\triangle BCD$ have equal areas

Big Hint:

That forces $AB \parallel CD,$ making $\triangle ABE \sim \triangle CDE$ with ratio $\dfrac{9}{12}$

Solution:

Adding $\triangle CED$ to each of $\triangle AED$ and $\triangle BEC$ shows $\triangle ACD$ and $\triangle BCD$ have equal areas. They share base $CD,$ so $A$ and $B$ are equidistant from line $CD,$ meaning $AB \parallel CD.$

Then $\triangle ABE \sim \triangle CDE$ with ratio $\dfrac{AB}{CD} = \dfrac{9}{12} = \dfrac{3}{4},$ so $\dfrac{AE}{EC} = \dfrac{3}{4}.$

Writing $AE = 3x$ and $EC = 4x,$ we get $7x = AC = 14,$ so $x = 2$ and $AE = 6.$

Thus, the correct answer is E.

21.

Let $p(x) = x^3 + ax^2 + bx + c,$ where $a,$ $b,$ and $c$ are complex numbers. Suppose that $\begin{gathered} p(2009 + 9002\pi i) \\ = p(2009) \\ = p(9002) = 0. \end{gathered}$ What is the number of nonreal zeros of $x^{12} + ax^8 + bx^4 + c?$

$4$

$6$

$8$

$10$

$12$

Answer: C

Concepts:polynomial complex number roots of unity

Difficulty rating: 2170

Small Hint:

Notice that $x^{12} + ax^8 + bx^4 + c = p(x^4)$

Big Hint:

So the zeros satisfy $x^4 = 2009 + 9002\pi i,$ $x^4 = 2009,$ or $x^4 = 9002;$ count the nonreal fourth roots

Solution:

Since $x^{12} + ax^8 + bx^4 + c = p(x^4),$ a value is a zero exactly when $x^4$ equals one of the roots of $p,$ namely $2009 + 9002\pi i,$ $2009,$ or $9002.$

The equation $x^4 = 2009 + 9002\pi i$ has four distinct nonreal roots. Each of $x^4 = 2009$ and $x^4 = 9002$ has two real roots and two nonreal roots.

So the nonreal zeros number $4 + 2 + 2 = 8.$

Thus, the correct answer is C.

22.

A regular octahedron has side length $1.$ A plane parallel to two of its opposite faces cuts the octahedron into two congruent solids. The polygon formed by the intersection of the plane and the octahedron has area $\dfrac{a\sqrt{b}}{c},$ where $a,$ $b,$ and $c$ are positive integers, $a$ and $c$ are relatively prime, and $b$ is not divisible by the square of any prime. What is $a + b + c?$

$10$

$11$

$12$

$13$

$14$

Answer: E

Concepts:3D geometry regular polygon area

Difficulty rating: 2270

Small Hint:

The cutting plane passes through the midpoints of the six edges that do not belong to the two parallel faces

Big Hint:

Those midpoints form a regular hexagon with side $\dfrac{1}{2};$ find its area

Solution:

Let the two parallel faces be triangles. The plane passes through the midpoints of the six edges not on those faces, forming an equilateral hexagon of side $\dfrac{1}{2},$ which by symmetry is also equiangular and hence regular.

A regular hexagon is six equilateral triangles, so its area is $6\cdot\frac{\sqrt{3}}{4}\left(\frac{1}{2}\right)^2 = \frac{3\sqrt{3}}{8}.$

Thus $a = 3,$ $b = 3,$ $c = 8,$ and $a + b + c = 14.$

Thus, the correct answer is E.

23.

Functions $f$ and $g$ are quadratic, $g(x) = -f(100 - x),$ and the graph of $g$ contains the vertex of the graph of $f.$ The four $x$ -intercepts on the two graphs have $x$ -coordinates $x_1,$ $x_2,$ $x_3,$ and $x_4,$ in increasing order, and $x_3 - x_2 = 150.$ The value of $x_4 - x_1$ is $m + n\sqrt{p},$ where $m,$ $n,$ and $p$ are positive integers, and $p$ is not divisible by the square of any prime. What is $m + n + p?$

$602$

$652$

$702$

$752$

$802$

Answer: D

Concepts:quadratic parabola symmetry

Difficulty rating: 2420

Small Hint:

The map $(x, y) \mapsto (100 - x, -y)$ is a $180^\circ$ rotation about $(50, 0)$ that sends the graph of $f$ to the graph of $g$

Big Hint:

So the roots pair as $x_2 + x_3 = x_1 + x_4 = 100;$ with $x_3 - x_2 = 150,$ find $x_2, x_3$ and use the vertex condition

Solution:

Because $g(x) = -f(100 - x),$ the graphs of $f$ and $g$ are reflections of each other through the point $(50, 0),$ so the four intercepts pair up with $x_2 + x_3 = x_1 + x_4 = 100.$

With $x_3 - x_2 = 150,$ we get $x_2 = -25$ and $x_3 = 125.$

Take $x_1, x_3$ as the roots of $f,$ whose vertex has $x$ -coordinate $h = \dfrac{x_1 + x_3}{2},$ so $x_1 = 2h - 125.$ The condition that the vertex of $f$ lies on the graph of $g$ gives $\begin{aligned} 1 &= \frac{f(h)}{g(h)} \\ &= \frac{(125 - h)(h - 125)}{-(h + 25)(3h - 225)}, \end{aligned}$ which gives $h = -25 \pm 75\sqrt{2}.$ Since $x_1 = 2h-125$ must be less than $x_2=-25,$ we need $h \lt 50,$ so $h = -25-75\sqrt{2}.$

Then $x_4 = 100 - x_1,$ so $\begin{aligned} x_4 - x_1 &= 350 - 4h \\ &= 450 + 300\sqrt{2}. \end{aligned}$ Hence $m + n + p = 450 + 300 + 2$ $= 752.$

Thus, the correct answer is D.

24.

The tower function of twos is defined recursively as follows: $T(1) = 2$ and $T(n + 1) = 2^{T(n)}$ for $n \ge 1.$ Let $A = (T(2009))^{T(2009)}$ and $B = (T(2009))^A.$ What is the largest integer $k$ such that $\underbrace{\log_2 \log_2 \log_2 \ldots \log_2}_{k} B$ is defined?

$2009$

$2010$

$2011$

$2012$

$2013$

Answer: E

Concepts:recursion logarithm exponent

Difficulty rating: 2650

Small Hint:

Repeatedly applying $\log_2$ peels one level off a tower: $\log_2 T(n + 1) = T(n)$

Big Hint:

Track how many logs $B = (T(2009))^A$ can survive before the value drops below $1$

Solution:

Since $\log_2 T(n + 1) = T(n),$ each application of $\log_2$ strips one $2$ off the top of a tower of twos.

Write $T_j=T(j),$ and let $L_j$ be the result of applying $\log_2$ to $B$ exactly $j$ times. The first two results are $\begin{aligned} L_1 &= A\,T_{2008}, \\ L_2 &= T_{2009}T_{2008}+T_{2007}. \end{aligned}$

For the lower bound, $\begin{aligned} L_3 &\gt \log_2(T_{2009}T_{2008}) \\ &= T_{2008}+T_{2007} \\ &\gt T_{2008}. \end{aligned}$ Repeatedly taking logarithms gives $L_{k+3} \gt T_{2008-k}$ for $0\le k\le 2007.$ In particular, $L_{2010}\gt2,$ so $L_{2011}\gt1$ and $L_{2012}\gt0.$ Thus $L_{2013}$ is defined.

For the upper bound, $T_{2007}\lt T_{2008}T_{2009},$ so $\begin{aligned} L_3 &\lt 1+T_{2007}+T_{2008} \\ &\lt 2T_{2008}, \\ L_4 &\lt 1+T_{2007} \lt T_{2008}. \end{aligned}$ Repeating the last comparison gives $L_{k+4}\lt T_{2008-k}$ for $0\le k\le2007.$ Hence $L_{2011}\lt2.$ Together with the lower bound, this yields $0\lt L_{2012}\lt1,$ so $L_{2013}\lt0.$ Therefore $L_{2014}$ is undefined, and the largest possible $k$ is $2013.$

Thus, the correct answer is E.

25.

The first two terms of a sequence are $a_1 = 1$ and $a_2 = \dfrac{1}{\sqrt{3}}.$ For $n \ge 1,$ $a_{n+2} = \dfrac{a_n + a_{n+1}}{1 - a_n a_{n+1}}.$ What is $|a_{2009}|?$

$0$

$2 - \sqrt{3}$

$\dfrac{1}{\sqrt{3}}$

$1$

$2 + \sqrt{3}$

Answer: A

Concepts:trigonometric identity recursion modular arithmetic

Difficulty rating: 2520

Small Hint:

The recursion matches the tangent addition formula $\tan(\alpha + \beta) = \dfrac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}$

Big Hint:

Since $a_1 = \tan\dfrac{\pi}{4}$ and $a_2 = \tan\dfrac{\pi}{6},$ write $a_n = \tan\dfrac{\pi c_n}{12}$ and track $c_n \bmod 12$

Solution:

The recursion is exactly the tangent addition formula, and $a_1 = \tan\dfrac{\pi}{4},$ $a_2 = \tan\dfrac{\pi}{6}.$

Writing $a_n = \tan\dfrac{\pi c_n}{12}$ with $c_1 = 3,$ $c_2 = 2,$ and $c_{n+2} \equiv c_n + c_{n+1} \pmod{12},$ the sequence $c_n$ is $\begin{gathered} 3, 2, 5, 7, 0, 7, 7, 2, \\ 9, 11, 8, 7, 3, 10, 1, 11, \\ 0, 11, 11, 10, 9, 7, 4, 11, \ldots \end{gathered}$ which is periodic with period $24.$

Since $2009 = 24\cdot 83 + 17,$ $c_{2009} = c_{17} = 0,$ so $a_{2009} = \tan 0 = 0$ and $|a_{2009}| = 0.$

Thus, the correct answer is A.