problem stringlengths 16 4.31k | answer stringlengths 29 6.77k |
|---|---|
The sides of an isosceles triangle are $\cos x,$ $\cos x,$ and $\cos 7x,$ and its vertex angle is $2x.$ (All angle measurements are in degrees.) Enter all possible values of $x,$ separated by commas. | Note that angle $x$ must be acute.
If we drop an altitude from the vertex of the isosceles triangle, then we obtain two right triangles, where one of the angles is $x,$ the opposite side is $\frac{\cos 7x}{2},$ and the hypotenuse is $\cos x.$ Hence,
\[\sin x = \frac{\frac{\cos 7x}{2}}{\cos x} = \frac{\cos 7x}{2 \cos ... |
Find $\cos \frac{5 \pi}{4}.$ | Converting to degrees,
\[\frac{5 \pi}{4} = \frac{180^\circ}{\pi} \cdot \frac{5 \pi}{4} = 225^\circ.\]Then $\cos 225^\circ = -\cos (225^\circ - 180^\circ) = -\cos 45^\circ = \boxed{-\frac{1}{\sqrt{2}}}.$ |
Let
\[\mathbf{A} = \begin{pmatrix} 4 & 1 \\ -9 & -2 \end{pmatrix}.\]Compute $\mathbf{A}^{100}.$ | Note that
\begin{align*}
\mathbf{A}^2 &= \begin{pmatrix} 4 & 1 \\ -9 & -2 \end{pmatrix} \begin{pmatrix} 4 & 1 \\ -9 & -2 \end{pmatrix} \\
&= \begin{pmatrix} 7 & 2 \\ -18 & -5 \end{pmatrix} \\
&= 2 \begin{pmatrix} 4 & 1 \\ -9 & -2 \end{pmatrix} - \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \\
&= 2 \mathbf{A} - \mathbf{... |
If
\[\frac{\sin^4 \theta}{a} + \frac{\cos^4 \theta}{b} = \frac{1}{a + b},\]then find the value of
\[\frac{\sin^8 \theta}{a^3} + \frac{\cos^8 \theta}{b^3}\]in terms of $a$ and $b.$ | Let $x = \sin^2 \theta$ and $y = \cos^2 \theta,$ so $x + y = 1.$ Also,
\[\frac{x^2}{a} + \frac{y^2}{b} = \frac{1}{a + b}.\]Substituting $y = 1 - x,$ we get
\[\frac{x^2}{a} + \frac{(1 - x)^2}{b} = \frac{1}{a + b}.\]This simplifies to
\[(a^2 + 2ab + b^2) x^2 - (2a^2 + 2ab) x + a^2 = 0,\]which nicely factors as $((a + b)... |
A plane is expressed parametrically by
\[\mathbf{v} = \begin{pmatrix} 1 + s - t \\ 2 - s \\ 3 - 2s + 2t \end{pmatrix}.\]Find the equation of the plane. Enter your answer in the form
\[Ax + By + Cz + D = 0,\]where $A,$ $B,$ $C,$ $D$ are integers such that $A > 0$ and $\gcd(|A|,|B|,|C|,|D|) = 1.$ | We can express the vector as
\[\mathbf{v} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + s \begin{pmatrix} 1 \\ -1 \\ -2 \end{pmatrix} + t \begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix}.\]Thus, the plane is generated by $\begin{pmatrix} 1 \\ -1 \\ -2 \end{pmatrix}$ and $\begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix},$ so we can... |
Find the equation of the plane passing through the point $(0,7,-7)$ and containing the line
\[\frac{x + 1}{-3} = \frac{y - 3}{2} = \frac{z + 2}{1}.\]Enter your answer in the form
\[Ax + By + Cz + D = 0,\]where $A,$ $B,$ $C,$ $D$ are integers such that $A > 0$ and $\gcd(|A|,|B|,|C|,|D|) = 1.$ | From the equation, $\frac{x + 1}{-3} = \frac{y - 3}{2},$
\[2x + 3y - 7 = 0.\]From the equation $\frac{y - 3}{2} = \frac{z + 2}{1},$
\[y - 2z - 7 = 0.\]So, any point on the line given in the problem will satisfy $2x + 3y - 7 = 0$ and $y - 2z - 7 = 0,$ which means it will also satisfy any equation of the form
\[a(2x + 3y... |
Compute $\tan \left (\operatorname{arccot} \frac{4}{7} \right).$ | Consider a right triangle where the adjacent side is 4 and the opposite side is 7.
[asy]
unitsize (0.5 cm);
draw((0,0)--(4,0)--(4,7)--cycle);
label("$4$", (2,0), S);
label("$7$", (4,7/2), E);
label("$\theta$", (0.8,0.5));
[/asy]
Then $\cot \theta = \frac{4}{7},$ so $\theta = \operatorname{arccot} \frac{4}{7}.$ Hen... |
Let $\mathbf{A} = \begin{pmatrix} 2 & 3 \\ 0 & 1 \end{pmatrix}.$ Find $\mathbf{A}^{20} - 2 \mathbf{A}^{19}.$ | First, we can write $\mathbf{A}^{20} - 2 \mathbf{A}^{19} = \mathbf{A}^{19} (\mathbf{A} - 2 \mathbf{I}).$ We can compute that
\[\mathbf{A} - 2 \mathbf{I} =
\begin{pmatrix} 2 & 3 \\ 0 & 1 \end{pmatrix}
- 2
\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
=
\begin{pmatrix} 0 & 3 \\ 0 & -1 \end{pmatrix}
.\]Then
\[\mathbf{A} (... |
The solutions to $z^4 = -16i$ can be expressed in the form
\begin{align*}
z_1 &= r_1 (\cos \theta_1 + i \sin \theta_1), \\
z_2 &= r_2 (\cos \theta_2 + i \sin \theta_2), \\
z_3 &= r_3 (\cos \theta_3 + i \sin \theta_3), \\
z_4 &= r_4 (\cos \theta_4 + i \sin \theta_4),
\end{align*}where $r_k > 0$ and $0^\circ \le \theta_k... | First, we can write $z^4 = -16i = 16 \operatorname{cis} 270^\circ.$ Therefore, the four roots are
\begin{align*}
&2 \operatorname{cis} 67.5^\circ, \\
&2 \operatorname{cis} (67.5^\circ + 90^\circ) = 2 \operatorname{cis} 157.5^\circ, \\
&2 \operatorname{cis} (67.5^\circ + 180^\circ) = 2 \operatorname{cis} 247.5^\circ, \... |
If $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = 4,$ then find
\[\begin{vmatrix} a & 7a + 3b \\ c & 7c +3d \end{vmatrix}.\] | Since $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = 4,$ $ad - bc = 4.$ Then
\[\begin{vmatrix} a & 7a + 3b \\ c & 7c +3d \end{vmatrix} = a(7c + 3d) - (7a + 3b)c = 3ad - 3bc = 3(ad - bc) = \boxed{12}.\] |
The foot of the perpendicular from the origin to a plane is $(12,-4,3).$ Find the equation of the plane. Enter your answer in the form
\[Ax + By + Cz + D = 0,\]where $A,$ $B,$ $C,$ $D$ are integers such that $A > 0$ and $\gcd(|A|,|B|,|C|,|D|) = 1.$ | We can take $\begin{pmatrix} 12 \\ -4 \\ 3 \end{pmatrix}$ as the normal vector of the plane. Then the equation of the plane is of the form
\[12x - 4y + 3z + D = 0.\]Substituting in the coordinates of $(12,-4,3),$ we find that the equation of the plane is $\boxed{12x - 4y + 3z - 169 = 0}.$ |
There exist vectors $\mathbf{a}$ and $\mathbf{b}$ such that
\[\mathbf{a} + \mathbf{b} = \begin{pmatrix} 6 \\ -3 \\ -6 \end{pmatrix},\]where $\mathbf{a}$ is parallel to $\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix},$ and $\mathbf{b}$ is orthogonal to $\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}.$ Find $\mathbf{b}.$ | Since $\mathbf{a}$ is parallel to $\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix},$
\[\mathbf{a} = t \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} t \\ t \\ t \end{pmatrix}\]for some scalar $t.$ Then
\[\mathbf{b} = \begin{pmatrix} 6 \\ -3 \\ -6 \end{pmatrix} - \begin{pmatrix} t \\ t \\ t \end{pmatrix} = \begi... |
The matrix for reflecting over a certain line $\ell,$ which passes through the origin, is given by
\[\begin{pmatrix} \frac{7}{25} & -\frac{24}{25} \\ -\frac{24}{25} & -\frac{7}{25} \end{pmatrix}.\]Find the direction vector of line $\ell.$ Enter your answer in the form $\begin{pmatrix} a \\ b \end{pmatrix},$ where $a,$... | Since $\begin{pmatrix} a \\ b \end{pmatrix}$ actually lies on $\ell,$ the reflection takes this vector to itself.
[asy]
unitsize(1.5 cm);
pair D = (4,-3), V = (2,1), P = (V + reflect((0,0),D)*(V))/2;
draw((4,-3)/2--(-4,3)/2,dashed);
draw((-2,0)--(2,0));
draw((0,-2)--(0,2));
draw((0,0)--P,Arrow(6));
label("$\ell$", ... |
Let $\mathbf{a},$ $\mathbf{b},$ and $\mathbf{c}$ be unit vectors such that $\mathbf{a} \cdot \mathbf{b} = \mathbf{a} \cdot \mathbf{c} = 0,$ and the angle between $\mathbf{b}$ and $\mathbf{c}$ is $\frac{\pi}{4}.$ Then
\[\mathbf{a} = k (\mathbf{b} \times \mathbf{c})\]for some constant $k.$ Enter all the possible values... | First, note that since $\mathbf{a}$ is orthogonal to both $\mathbf{b}$ and $\mathbf{c},$ $\mathbf{a}$ is a scalar multiple of their cross product $\mathbf{b} \times \mathbf{c}.$ Furthermore,
\[\|\mathbf{b} \times \mathbf{c}\| = \|\mathbf{b}\| \|\mathbf{c}\| \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}.\]Hence,
\[\|\mathbf{... |
Find the matrix $\mathbf{M}$ such that
\[\mathbf{M} \begin{pmatrix} 1 & -4 \\ 3 & -2 \end{pmatrix} = \begin{pmatrix} -16 & -6 \\ 7 & 2 \end{pmatrix}.\] | The inverse of $\begin{pmatrix} 1 & -4 \\ 3 & -2 \end{pmatrix}$ is
\[\frac{1}{(1)(-2) - (-4)(3)} \begin{pmatrix} -2 & 4 \\ -3 & 1 \end{pmatrix} = \frac{1}{10} \begin{pmatrix} -2 & 4 \\ -3 & 1 \end{pmatrix}.\]So, multiplying by this inverse on the right, we get
\[\mathbf{M} = \begin{pmatrix} -16 & -6 \\ 7 & 2 \end{pmatr... |
For real numbers $t,$ the point of intersection of the lines $tx - 2y - 3t = 0$ and $x - 2ty + 3 = 0$ is plotted. All the plotted points lie on what kind of curve?
(A) Line
(B) Circle
(C) Parabola
(D) Ellipse
(E) Hyperbola
Enter the letter of the correct option. | Solving for $x$ and $y$ in the equations $tx - 2y - 3t = 0$ and $x - 2ty + 3 = 0,$ we find
\[x = \frac{3t^2 + 3}{t^2 - 1}, \quad y = \frac{3t}{t^2 - 1}.\]Then
\[x^2 = \frac{(3t^2 + 3)^2}{(t^2 - 1)^2} = \frac{9t^4 + 18t^2 + 9}{t^4 - 2t^2 + 1},\]and
\[y^2 = \frac{9t^2}{(t^2 - 1)^2} = \frac{9t^2}{t^4 - 2t^2 + 1}.\]Thus,
\... |
Let $\mathbf{R}$ be the matrix for rotating about the origin counter-clockwise by an angle of $58^\circ.$ Find $\det \mathbf{R}.$ | The matrix corresponding to rotating about the origin counter-clockwise by an angle of $\theta$ is given by
\[\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}.\]The determinant of this matrix is then
\[\cos^2 \theta - (-\sin \theta)(\sin \theta) = \cos^2 \theta + \sin^2 \theta = \bo... |
Compute $\arcsin 0.$ Express your answer in radians. | Since $\sin 0 = 0,$ $\arcsin 0 = \boxed{0}.$ |
Compute
\[\cos^2 0^\circ + \cos^2 1^\circ + \cos^2 2^\circ + \dots + \cos^2 90^\circ.\] | Let $S = \cos^2 0^\circ + \cos^2 1^\circ + \cos^2 2^\circ + \dots + \cos^2 90^\circ.$ Then
\begin{align*}
S &= \cos^2 0^\circ + \cos^2 1^\circ + \cos^2 2^\circ + \dots + \cos^2 90^\circ \\
&= \cos^2 90^\circ + \cos^2 89^\circ + \cos^2 88^\circ + \dots + \cos^2 0^\circ \\
&= \sin^2 0^\circ + \sin^2 1^\circ + \sin^2 2^\... |
Let $\mathbf{p}$ and $\mathbf{q}$ be two three-dimensional unit vectors such that the angle between them is $30^\circ.$ Find the area of the parallelogram whose diagonals correspond to $\mathbf{p} + 2 \mathbf{q}$ and $2 \mathbf{p} + \mathbf{q}.$ | Suppose that vectors $\mathbf{a}$ and $\mathbf{b}$ generate the parallelogram. Then the vectors corresponding to the diagonals are $\mathbf{a} + \mathbf{b}$ and $\mathbf{b} - \mathbf{a}.$
[asy]
unitsize(0.4 cm);
pair A, B, C, D, trans;
A = (0,0);
B = (7,2);
C = (1,3);
D = B + C;
trans = (10,0);
draw(B--D--C);
draw... |
Simplify
\[\frac{\sin x + \sin 2x}{1 + \cos x + \cos 2x}.\] | We can write
\begin{align*}
\frac{\sin x + \sin 2x}{1 + \cos x + \cos 2x} &= \frac{\sin x + 2 \sin x \cos x}{1 + \cos x + 2 \cos^2 x - 1} \\
&= \frac{\sin x + 2 \sin x \cos x}{\cos x + 2 \cos^2 x} \\
&= \frac{\sin x (1 + 2 \cos x)}{\cos x (1 + 2 \cos x)} \\
&= \frac{\sin x}{\cos x} = \boxed{\tan x}.
\end{align*} |
For real numbers $t \neq 0,$ the point
\[(x,y) = \left( \frac{t + 1}{t}, \frac{t - 1}{t} \right)\]is plotted. All the plotted points lie on what kind of curve?
(A) Line
(B) Circle
(C) Parabola
(D) Ellipse
(E) Hyperbola
Enter the letter of the correct option. | For $x = \frac{t + 1}{t}$ and $y = \frac{t - 1}{t},$
\[x + y = \frac{t + 1}{t} + \frac{t - 1}{t} = \frac{2t}{t} = 2.\]Thus, all the plotted points lie on a line. The answer is $\boxed{\text{(A)}}.$ |
There exists a scalar $c$ so that
\[\mathbf{i} \times (\mathbf{v} \times \mathbf{i}) + \mathbf{j} \times (\mathbf{v} \times \mathbf{j}) + \mathbf{k} \times (\mathbf{v} \times \mathbf{k}) = c \mathbf{v}\]for all vectors $\mathbf{v}.$ Find $c.$ | In general, the vector triple product states that for any vectors $\mathbf{a},$ $\mathbf{b},$ and $\mathbf{c},$
\[\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c}.\]So
\begin{align*}
\mathbf{i} \times (\mathbf{v} \times \mathbf{i}) &=... |
Simplify
\[\frac{\sin{10^\circ}+\sin{20^\circ}}{\cos{10^\circ}+\cos{20^\circ}}.\]Enter your answer is a trigonometric function evaluated at an integer, such as "sin 7". (The angle should be positive and as small as possible.) | From the product-to-sum identities,
\[\frac{\sin{10^\circ}+\sin{20^\circ}}{\cos{10^\circ}+\cos{20^\circ}} = \frac{2 \sin 15^\circ \cos (-5^\circ)}{2 \cos 15^\circ \cos(-5^\circ)} = \frac{\sin 15^\circ}{\cos 15^\circ} = \boxed{\tan 15^\circ}.\] |
Convert the point $(\rho,\theta,\phi) = \left( 3, \frac{5 \pi}{12}, 0 \right)$ in spherical coordinates to rectangular coordinates. | We have that $\rho = 3,$ $\theta = \frac{5 \pi}{12},$ and $\phi = 0,$ so
\begin{align*}
x &= \rho \sin \phi \cos \theta = 3 \sin 0 \cos \frac{5 \pi}{12} = 0, \\
y &= \rho \sin \phi \sin \theta = 3 \sin 0 \sin \frac{5 \pi}{12} = 0, \\
z &= \rho \cos \phi = 3 \cos 0 = 3.
\end{align*}Therefore, the rectangular coordinates... |
Find the reflection of $\begin{pmatrix} 0 \\ 4 \end{pmatrix}$ over the vector $\begin{pmatrix} 1 \\ 3 \end{pmatrix}.$ | Let $\mathbf{r}$ be the reflection of $\begin{pmatrix} 0 \\ 4 \end{pmatrix}$ over the vector $\begin{pmatrix} 1 \\ 3 \end{pmatrix},$ and let $\mathbf{p}$ be the projection of $\begin{pmatrix} 0 \\ 4 \end{pmatrix}$ onto $\begin{pmatrix} 1 \\ 3 \end{pmatrix}.$
[asy]
usepackage("amsmath");
unitsize(1 cm);
pair D, P, R,... |
Find $x.$
[asy]
unitsize(0.7 cm);
pair A, B, C, D, O;
O = (0,0);
A = 4*dir(160);
B = 5*dir(160 + 180);
C = 8*dir(20);
D = 4*dir(20 + 180);
draw(A--B);
draw(C--D);
draw(A--C);
draw(B--D);
label("$4$", (A + O)/2, SW);
label("$10$", (C + O)/2, SE);
label("$4$", (D + O)/2, NW);
label("$5$", (B + O)/2, NE);
label("$8$"... | Let $\theta = \angle AOC = \angle BOD.$ Then by the Law of Cosines on triangle $BOD,$
\[\cos \theta = \frac{4^2 + 5^2 - 8^2}{2 \cdot 4 \cdot 5} = -\frac{23}{40}.\]Then by the Law of Cosines on triangle $AOC,$
\begin{align*}
x^2 &= 4^2 + 10^2 - 2 \cdot 4 \cdot 10 \cos \theta \\
&= 4^2 + 10^2 - 2 \cdot 4 \cdot 10 \cdot ... |
In triangle $ABC,$ $M$ is the midpoint of $\overline{BC},$ $AB = 12,$ and $AC = 16.$ Let $E$ be on $\overline{AC},$ and $F$ be on $\overline{AB},$ and let $G$ be the intersection of $\overline{EF}$ and $\overline{AM}.$ If $AE = 2AF,$ then find $\frac{EG}{GF}.$
[asy]
unitsize(0.3 cm);
pair A, B, C, E, F, G, M;
real ... | Let $x = AF,$ so $AE = 2x.$ Then $BF = 12 - x$ and $CE = 16 - 2x.$
[asy]
unitsize(0.3 cm);
pair A, B, C, E, F, G, M;
real x = 4;
B = (0,0);
C = (18,0);
A = intersectionpoint(arc(B,12,0,180),arc(C,16,0,180));
M = (B + C)/2;
F = interp(A,B,x/12);
E = interp(A,C,2*x/16);
G = extension(E,F,A,M);
draw(A--B--C--cycle);
... |
Find all real numbers $k$ for which there exists a nonzero, 2-dimensional vector $\mathbf{v}$ such that
\[\begin{pmatrix} 1 & 8 \\ 2 & 1 \end{pmatrix} \mathbf{v} = k \mathbf{v}.\]Enter all the solutions, separated by commas. | Let $\mathbf{v} = \begin{pmatrix} x \\ y \end{pmatrix}$. Then
\[\begin{pmatrix} 1 & 8 \\ 2 & 1 \end{pmatrix} \mathbf{v} = \begin{pmatrix} 1 & 8 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x + 8y \\ 2x + y \end{pmatrix},\]and
\[k \mathbf{v} = k \begin{pmatrix} x \\ y \end{pmatrix} = \b... |
A translation of the plane takes $-3 + 2i$ to $-7 - i.$ Find the complex number that the translation takes $-4 + 5i$ to. | This translation takes $z$ to $z + w,$ where $w$ is a fixed complex number. Thus,
\[-7 - i = (-3 + 2i) + w.\]Hence, $w = -4 - 3i.$ Then the translation takes $-4 + 5i$ to $(-4 + 5i) + (-4 - 3i) = \boxed{-8 + 2i}.$ |
Find the range of the function
\[f(x) = \frac{\sin^3 x + 6 \sin^2 x + \sin x + 2 \cos^2 x - 8}{\sin x - 1},\]as $x$ ranges over all real numbers such that $\sin x \neq 1.$ Enter your answer using interval notation. | Since $\cos^2 x = 1 - \sin^2 x,$ we can write
\begin{align*}
f(x) &= \frac{\sin^3 x + 6 \sin^2 x + \sin x + 2(1 - \sin^2 x) - 8}{\sin x - 1} \\
&= \frac{\sin^3 x + 4 \sin^2 x + \sin x - 6}{\sin x - 1} \\
&= \frac{(\sin x - 1)(\sin x + 2)(\sin x + 3)}{\sin x - 1} \\
&= (\sin x + 2)(\sin x + 3) \\
&= \sin^2 x + 5 \sin x ... |
Find all angles $\theta,$ $0 \le \theta \le 2 \pi,$ with the following property: For all real numbers $x,$ $0 \le x \le 1,$
\[x^2 \cos \theta - x(1 - x) + (1 - x)^2 \sin \theta > 0.\] | Taking $x = 0,$ we get $\sin \theta > 0.$ Taking $x = 1,$ we get $\cos \theta > 0.$ Hence, $0 < \theta < \frac{\pi}{2}.$
Then we can write
\begin{align*}
&x^2 \cos \theta - x(1 - x) + (1 - x)^2 \sin \theta \\
&= x^2 \cos \theta - 2x (1 - x) \sqrt{\cos \theta \sin \theta} + (1 - x)^2 \sin \theta + 2x (1 - x) \sqrt{\c... |
Find the point on the line defined by
\[\begin{pmatrix} 4 \\ 0 \\ 1 \end{pmatrix} + t \begin{pmatrix} -2 \\ 6 \\ -3 \end{pmatrix}\]that is closest to the point $(2,3,4).$ | A point on the line is given by
\[\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 4 \\ 0 \\ 1 \end{pmatrix} + t \begin{pmatrix} -2 \\ 6 \\ -3 \end{pmatrix} = \begin{pmatrix} 4 - 2t \\ 6t \\ 1 - 3t \end{pmatrix}.\][asy]
unitsize (0.6 cm);
pair A, B, C, D, E, F, H;
A = (2,5);
B = (0,0);
C = (8,0);
D = (A + ... |
Find the point in the $xz$-plane that is equidistant from the points $(1,-1,0),$ $(2,1,2),$ and $(3,2,-1).$ | Since the point lies in the $xz$-plane, it is of the form $(x,0,z).$ We want this point to be equidistant to the points $(1,-1,0),$ $(2,1,2),$ and $(3,2,-1),$ which gives us the equations
\begin{align*}
(x - 1)^2 + 1^2 + z^2 &= (x - 2)^2 + 1^2 + (z - 2)^2, \\
(x - 1)^2 + 1^2 + z^2 &= (x - 3)^2 + 2^2 + (z + 1)^2.
\end{... |
The projection of $\begin{pmatrix} 0 \\ 1 \\ 4 \end{pmatrix}$ onto a certain vector $\mathbf{w}$ is $\begin{pmatrix} 1 \\ -1/2 \\ 1/2 \end{pmatrix}.$ Find the projection of $\begin{pmatrix} 3 \\ 3 \\ -2 \end{pmatrix}$ onto $\mathbf{w}.$ | Since the projection of $\begin{pmatrix} 0 \\ 1 \\ 4 \end{pmatrix}$ onto $\mathbf{w}$ is $\begin{pmatrix} 1 \\ -1/2 \\ 1/2 \end{pmatrix},$ $\mathbf{w}$ must be a scalar multiple of $\begin{pmatrix} 1 \\ -1/2 \\ 1/2 \end{pmatrix}.$ Furthermore, the projection of a vector onto $\mathbf{w}$ is the same as the projection ... |
Compute the number of real solutions $(x,y,z,w)$ to the system of equations:
\begin{align*}
x &= z+w+zwx, \\
y &= w+x+wxy, \\
z &= x+y+xyz, \\
w &= y+z+yzw.
\end{align*} | We can re-write the first equation as
\[x = \frac{w+z}{1-wz}.\]which is an indication to consider trigonometric substitution.
Let $x = \tan a,$ $y = \tan b,$ $z = \tan c,$ and $w = \tan d,$ where $-90^{\circ} < a,$ $b,$ $c,$ $d < 90^{\circ}$. Then
\[\tan a = \frac{\tan d + \tan c}{1 - \tan d \tan c} = \tan (c + d).\]... |
Let $O$ be the origin, and let $(a,b,c)$ be a fixed point. A plane passes through $(a,b,c)$ and intersects the $x$-axis, $y$-axis, and $z$-axis at $A,$ $B,$ and $C,$ respectively, all distinct from $O.$ Let $(p,q,r)$ be the center of the sphere passing through $A,$ $B,$ $C,$ and $O.$ Find
\[\frac{a}{p} + \frac{b}{q}... | Let $A = (\alpha,0,0),$ $B = (0,\beta,0),$ and $C = (0,0,\gamma).$ Since $(p,q,r)$ is equidistant from $O,$ $A,$ $B,$ and $C,$
\begin{align*}
p^2 + q^2 + r^2 &= (p - \alpha)^2 + q^2 + r^2, \\
p^2 + q^2 + r^2 &= p^2 + (q - \beta)^2 + r^2, \\
p^2 + q^2 + r^2 &= p^2 + q^2 + (r - \gamma)^2.
\end{align*}The first equation ... |
The perpendicular bisectors of the sides of triangle $ABC$ meet its circumcircle at points $A',$ $B',$ and $C',$ as shown. If the perimeter of triangle $ABC$ is 35 and the radius of the circumcircle is 8, then find the area of hexagon $AB'CA'BC'.$
[asy]
unitsize(2 cm);
pair A, B, C, Ap, Bp, Cp, O;
O = (0,0);
A = di... | Note that the perpendicular bisectors meet at $O,$ the circumcenter of triangle $ABC.$
[asy]
unitsize(2 cm);
pair A, B, C, Ap, Bp, Cp, O;
O = (0,0);
A = dir(210);
B = dir(60);
C = dir(330);
Ap = dir(15);
Bp = dir(270);
Cp = dir(135);
draw(Circle(O,1));
draw(A--B--C--cycle);
draw(O--Ap);
draw(O--Bp);
draw(O--Cp);
dr... |
Find the number of $x$-intercepts on the graph of $y = \sin \frac{1}{x}$ (evaluated in terms of radians) in the interval $(0.0001, 0.001).$ | The intercepts occur where $\sin \frac{1}{x}= 0$, that is, where $x = \frac{1}{k\pi}$ and $k$ is a nonzero integer. Solving
\[0.0001 < \frac{1}{k\pi} < 0.001\]yields
\[\frac{1000}{\pi} < k < \frac{10{,}000}{\pi}.\]Thus the number of $x$ intercepts in $(0.0001, 0.001)$ is
\[\left\lfloor\frac{10{,}000}{\pi}\right\rfloor ... |
Find the area of the triangle with vertices $(6,5,3),$ $(3,3,1),$ and $(15,11,9).$ | Let $\mathbf{u} = \begin{pmatrix} 6 \\ 5 \\ 3 \end{pmatrix},$ $\mathbf{v} = \begin{pmatrix} 3 \\ 3 \\ 1 \end{pmatrix},$ and $\mathbf{w} = \begin{pmatrix} 15 \\ 11 \\ 9 \end{pmatrix}.$
Then
\[\mathbf{v} - \mathbf{u} = \begin{pmatrix} 3 \\ 2 \\ 2 \end{pmatrix}\]and
\[\mathbf{w} - \mathbf{u} = \begin{pmatrix} 9 \\ 6 \\ 6... |
The matrix
\[\begin{pmatrix} a & 3 \\ -8 & d \end{pmatrix}\]is its own inverse, for some real numbers $a$ and $d.$ Find the number of possible pairs $(a,d).$ | Since $\begin{pmatrix} a & 3 \\ -8 & d \end{pmatrix}$ is its own inverse,
\[\begin{pmatrix} a & 3 \\ -8 & d \end{pmatrix}^2 = \begin{pmatrix} a & 3 \\ -8 & d \end{pmatrix} \begin{pmatrix} a & 3 \\ -8 & d \end{pmatrix} = \mathbf{I}.\]This gives us
\[\begin{pmatrix} a^2 - 24 & 3a + 3d \\ -8a - 8d & d^2 - 24 \end{pmatrix}... |
Let $a$ and $b$ be angles such that
\[\cos (a + b) = \cos a + \cos b.\]Find the maximum value of $\cos a.$ | From $\cos (a + b) = \cos a + \cos b,$ $\cos a = \cos (a + b) - \cos b.$ Then from sum-to-product,
\[\cos (a + b) - \cos b = -2 \sin \frac{a + 2b}{2} \sin \frac{a}{2}.\]Let $k = \sin \frac{a + 2b}{2},$ so
\[\cos a = -2k \sin \frac{a}{2}.\]Then
\[\cos^2 a = 4k^2 \sin^2 \frac{a}{2} = 4k^2 \cdot \frac{1}{2} (1 - \cos a) ... |
In triangle $ABC,$ $AB = 9,$ $BC = 10,$ and $AC = 11.$ If $D$ and $E$ are chosen on $\overline{AB}$ and $\overline{AC}$ so that $AD = 4$ and $AE = 7,$ then find the area of triangle $ADE.$
[asy]
unitsize (1 cm);
pair A, B, C, D, E;
A = (2,3);
B = (0,0);
C = (6,0);
D = interp(A,B,0.4);
E = interp(A,C,3/5);
draw(A--... | By Heron's formula, the area of triangle $ABC$ is $30 \sqrt{2}.$ Then
\[\frac{1}{2} \cdot 10 \cdot 11 \sin A = 30 \sqrt{2},\]so $\sin A = \frac{20 \sqrt{2}}{33}.$ Therefore,
\[[ADE] = \frac{1}{2} \cdot 4 \cdot 7 \cdot \frac{20 \sqrt{2}}{33} = \boxed{\frac{280 \sqrt{2}}{33}}.\] |
In triangle $ABC$, $AB = BC$, and $\overline{BD}$ is an altitude. Point $E$ is on the extension of $\overline{AC}$ such that $BE =
10$. The values of $\tan \angle CBE$, $\tan \angle DBE$, and $\tan \angle ABE$ form a geometric progression, and the values of $\cot \angle DBE$, $\cot \angle CBE$, $\cot \angle DBC$ form ... | Let $\angle DBE = \alpha$ and $\angle DBC = \beta$. Then $\angle CBE = \alpha - \beta$ and $\angle ABE = \alpha +
\beta$, so $\tan(\alpha - \beta)\tan(\alpha + \beta) = \tan^2
\alpha$. Thus \[\frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}\cdot \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan\beta} = \... |
Compute $\begin{pmatrix} 2 & 0 \\ 5 & -3 \end{pmatrix} \begin{pmatrix} 8 & -2 \\ 1 & 1 \end{pmatrix}.$ | We have that
\[\begin{pmatrix} 2 & 0 \\ 5 & -3 \end{pmatrix} \begin{pmatrix} 8 & -2 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} (2)(8) + (0)(1) & (2)(-2) + (0)(1) \\ (5)(8) + (-3)(1) & (5)(-2) + (-3)(1) \end{pmatrix} = \boxed{\begin{pmatrix} 16 & -4 \\ 37 & -13 \end{pmatrix}}.\] |
The sum $10 e^{2 \pi i/11} + 10 e^{15 \pi i/22}$ is expressed as $re^{i \theta}.$ Enter the ordered pair $(r, \theta).$ | The average of $\frac{2 \pi}{11}$ and $\frac{15 \pi}{22}$ is $\frac{19 \pi}{44}.$ We can then write
\begin{align*}
10 e^{2 \pi i/11} + 10 e^{15 \pi i/22} &= 10 e^{19 \pi i/44} (e^{-\pi i/4} + e^{\pi i/4}) \\
&= 10 e^{19 \pi i/44} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} + \cos \frac{\pi}{4} - i \sin \frac{\pi}... |
Compute $\tan\left(\frac{\pi}{7}\right)\tan\left(\frac{2\pi}{7}\right)\tan\left(\frac{3\pi}{7}\right)$. | In general, By DeMoivre's Theorem,
\begin{align*}
\operatorname{cis} n \theta &= (\operatorname{cis} \theta)^n \\
&= (\cos \theta + i \sin \theta)^n \\
&= \cos^n \theta + \binom{n}{1} i \cos^{n - 1} \theta \sin \theta - \binom{n}{2} \cos^{n - 2} \theta \sin^2 \theta - \binom{n}{3} i \cos^{n - 3} \theta \sin^3 \theta + ... |
The dilation, centered at $-1 + 4i,$ with scale factor $-2,$ takes $2i$ to which complex number? | Let $z$ be the image of $2i$ under the dilation.
[asy]
unitsize(0.5 cm);
pair C, P, Q;
C = (-1,4);
P = (0,2);
Q = (-3,8);
draw((-5,0)--(5,0));
draw((0,-1)--(0,10));
draw(P--Q,dashed);
dot("$-1 + 4i$", C, SW);
dot("$2i$", P, E);
dot("$-3 + 8i$", Q, NW);
[/asy]
Since the dilation is centered at $-1 + 4i,$ with scale... |
Find the phase shift of the graph of $y = \sin (3x - \pi).$ | Since the graph of $y = \sin (3x - \pi)$ is the same as the graph of $y = \sin 3x$ shifted $\frac{\pi}{3}$ units to the right, the phase shift is $\boxed{\frac{\pi}{3}}.$
[asy]import TrigMacros;
size(400);
real g(real x)
{
return sin(3*x - pi);
}
real f(real x)
{
return sin(3*x);
}
draw(graph(g,-2*pi,2*pi,n=700,... |
Let triangle $ABC$ be a right triangle with right angle at $C.$ Let $D$ and $E$ be points on $\overline{AB}$ with $D$ between $A$ and $E$ such that $\overline{CD}$ and $\overline{CE}$ trisect $\angle C.$ If $\frac{DE}{BE} = \frac{8}{15},$ then find $\tan B.$ | Without loss of generality, set $CB = 1$. Then, by the Angle Bisector Theorem on triangle $DCB$, we have $CD = \frac{8}{15}$.
[asy]
unitsize(0.5 cm);
pair A, B, C, D, E;
A = (0,4*sqrt(3));
B = (11,0);
C = (0,0);
D = extension(C, C + dir(60), A, B);
E = extension(C, C + dir(30), A, B);
draw(A--B--C--cycle);
draw(C--... |
Equilateral triangle $ABC$ has been creased and folded so that vertex $A$ now rests at $A'$ on $\overline{BC}$ as shown. If $BA' = 1$ and $A'C = 2,$ then find the length of crease $\overline{PQ}.$
[asy]
unitsize(1 cm);
pair A, Ap, B, C, P, Q;
A = 3*dir(60);
B = (0,0);
C = (3,0);
Ap = (1,0);
P = 8/5*dir(60);
Q = C +... | The side length of equilateral triangle $ABC$ is 3.
Let $x = BP.$ Then $AP = A'P = 3 - x,$ so by the Law of Cosines on triangle $PBA',$
\[(3 - x)^2 = x^2 + 3^2 - 2 \cdot x \cdot 3 \cdot \cos 60^\circ = x^2 - 3x + 9.\]Solving, we find $x = \frac{8}{5}.$
Let $y = CQ.$ Then $AQ = A'Q = 3 - y,$ so by the Law of Cosines... |
The graph of
\[r = -2 \cos \theta + 6 \sin \theta\]is a circle. Find the area of the circle. | From the equation $r = -2 \cos \theta + 6 \sin \theta,$
\[r^2 = -2r \cos \theta + 6r \sin \theta.\]Then $x^2 + y^2 = -2x + 6y.$ Completing the square in $x$ and $y,$ we get
\[(x + 1)^2 + (y - 3)^2 = 10.\]Thus, the graph is the circle centered at $(-1,3)$ with radius $\sqrt{10}.$ Its area is $\boxed{10 \pi}.$
[asy]
... |
Find $\begin{pmatrix} 3 \\ -7 \end{pmatrix} + \begin{pmatrix} -6 \\ 11 \end{pmatrix}.$ | We have that
\[\begin{pmatrix} 3 \\ -7 \end{pmatrix} + \begin{pmatrix} -6 \\ 11 \end{pmatrix} = \begin{pmatrix} 3 + (-6) \\ (-7) + 11 \end{pmatrix} = \boxed{\begin{pmatrix} -3 \\ 4 \end{pmatrix}}.\] |
Let $O$ be the origin. There exists a scalar $k$ so that for any points $A,$ $B,$ $C,$ and $D$ such that
\[3 \overrightarrow{OA} - 2 \overrightarrow{OB} + 5 \overrightarrow{OC} + k \overrightarrow{OD} = \mathbf{0},\]the four points $A,$ $B,$ $C,$ and $D$ are coplanar. Find $k.$ | From the given equation,
\[3 \overrightarrow{OA} - 2 \overrightarrow{OB} = -5 \overrightarrow{OC} - k \overrightarrow{OD}.\]Let $P$ be the point such that
\[\overrightarrow{OP} = 3 \overrightarrow{OA} - 2 \overrightarrow{OB} = -5 \overrightarrow{OC} - k \overrightarrow{OD}.\]Since $3 + (-2) = 1,$ $P$ lies on line $AB.$... |
Find the dot product of $\begin{pmatrix} 3 \\ -4 \\ -3 \end{pmatrix}$ and $\begin{pmatrix} -5 \\ 2 \\ 1 \end{pmatrix}.$ | The dot product of $\begin{pmatrix} 3 \\ -4 \\ -3 \end{pmatrix}$ and $\begin{pmatrix} -5 \\ 2 \\ 1 \end{pmatrix}$ is
\[(3)(-5) + (-4)(2) + (-3)(1) = \boxed{-26}.\] |
If triangle $ABC$ has sides of length $AB = 6,$ $AC = 5,$ and $BC = 4,$ then calculate
\[\frac{\cos \frac{A - B}{2}}{\sin \frac{C}{2}} - \frac{\sin \frac{A - B}{2}}{\cos \frac{C}{2}}.\] | We can write the expression as
\[\frac{\cos \frac{A - B}{2} \cos \frac{C}{2} - \sin \frac{A - B}{2} \sin \frac{C}{2}}{\sin \frac{C}{2} \cos \frac{C}{2}}.\]The numerator is
\[\cos \left (\frac{A - B}{2} + \frac{C}{2} \right) = \cos \frac{A - B + C}{2} = \cos \frac{(180^\circ - B) - B}{2} = \cos (90^\circ - B) = \sin B,\... |
Solve $\arcsin x + \arcsin (1 - x) = \arccos x.$ | Taking the sine of both sides, we get
\[\sin (\arcsin x + \arcsin (1 - x)) = \sin (\arccos x).\]Then from the angle addition formula,
\[\sin (\arcsin x) \cos (\arcsin (1 - x)) + \cos (\arcsin x) \sin (\arcsin (1 - x)) = \sin (\arccos x),\]or
\[x \sqrt{1 - (1 - x)^2} + \sqrt{1 - x^2} (1 - x) = \sqrt{1 - x^2}.\]Then
\[x ... |
If $\|\mathbf{v}\| = 4,$ then find $\mathbf{v} \cdot \mathbf{v}.$ | We have that $\mathbf{v} \cdot \mathbf{v} = \|\mathbf{v}\|^2 = \boxed{16}.$ |
The distance between two vectors is the magnitude of their difference. Find the value of $t$ for which the vector
\[\bold{v} = \begin{pmatrix} 2 \\ -3 \\ -3 \end{pmatrix} + t \begin{pmatrix} 7 \\ 5 \\ -1 \end{pmatrix}\]is closest to
\[\bold{a} = \begin{pmatrix} 4 \\ 4 \\ 5 \end{pmatrix}.\] | The equation
\[\bold{v} = \begin{pmatrix} 2 \\ -3 \\ -3 \end{pmatrix} + \begin{pmatrix} 7 \\ 5 \\ -1 \end{pmatrix} t = \begin{pmatrix} 2 + 7t \\ -3 + 5t \\ -3 - t \end{pmatrix}\]describes a line, so if $\bold{v}$ is the vector that is closest to $\bold{a}$, then the vector joining $\bold{v}$ and $\bold{a}$ is orthogona... |
Find the smallest positive integer $k$ such that $
z^{10} + z^9 + z^6+z^5+z^4+z+1
$ divides $z^k-1$. | First, we factor the given polynomial. The polynomial has almost all the powers of $z$ from 1 to $z^6,$ which we can fill in by adding and subtracting $z^2$ and $z^3.$ This allows us to factor as follows:
\begin{align*}
z^{10} + z^9 + z^6 + z^5 + z^4 + z + 1 &= (z^{10} - z^3) + (z^9 - z^2) + (z^6 + z^5 + z^4 + z^3 + ... |
The solid $S$ consists of the set of all points $(x,y,z)$ such that $|x| + |y| \le 1,$ $|x| + |z| \le 1,$ and $|y| + |z| \le 1.$ Find the volume of $S.$ | By symmetry, we can focus on the octant where $x,$ $y,$ $z$ are all positive. In this octant, the condition $|x| + |y| = 1$ becomes $x + y = 1,$ which is the equation of a plane. Hence, the set of points in this octant such that $|x| + |y| \le 1$ is the set of points bound by the plane $x + y = 1,$ $x = 0,$ and $y = ... |
Find the inverse of the matrix
\[\begin{pmatrix} 6 & -4 \\ -3 & 2 \end{pmatrix}.\]If the inverse does not exist, then enter the zero matrix. | Since the determinant is $(6)(2) - (-4)(-3) = 0,$ the inverse does not exist, so the answer is the zero matrix $\boxed{\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}}.$ |
Let $P$ be the plane passing through the origin with normal vector $\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}.$ Find the matrix $\mathbf{R}$ such that for any vector $\mathbf{v},$ $\mathbf{R} \mathbf{v}$ is the reflection of $\mathbf{v}$ through plane $P.$ | Let $\mathbf{v} = \begin{pmatrix} x \\ y \\ z \end{pmatrix},$ and let $\mathbf{p}$ be the projection of $\mathbf{p}$ onto plane $P.$ Then $\mathbf{v} - \mathbf{p}$ is the projection of $\mathbf{v}$ onto the normal vector $\mathbf{n} = \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}.$
[asy]
import three;
size(160);
curren... |
For a constant $c,$ in spherical coordinates $(\rho,\theta,\phi),$ find the shape described by the equation
\[\theta = c.\](A) Line
(B) Circle
(C) Plane
(D) Sphere
(E) Cylinder
(F) Cone
Enter the letter of the correct option. | In spherical coordinates, $\theta$ denotes the angle a point makes with the positive $x$-axis. Thus, for a fixed angle $\theta = c,$ all the points lie on a plane. The answer is $\boxed{\text{(C)}}.$ Note that we can obtain all points in this plane by taking $\rho$ negative.
[asy]
import three;
import solids;
size... |
Compute the least positive value of $t$ such that
\[\arcsin (\sin \alpha), \ \arcsin (\sin 2 \alpha), \ \arcsin (\sin 7 \alpha), \ \arcsin (\sin t \alpha)\]is a geometric progression for some $\alpha$ with $0 < \alpha < \frac{\pi}{2}.$ | Let $r$ be the common ratio. Since $0 < \alpha < \frac{\pi}{2},$ both $\arcsin (\sin \alpha)$ and $\arcsin (\sin 2 \alpha)$ are positive, so $r$ is positive. The positive portions of the graphs of $y = \arcsin (\sin x),$ $y = \arcsin (2 \sin x),$ and $y = \arcsin (7 \sin x)$ are shown below. (Note that each graph is... |
Find the area of the triangle with vertices $(-1,4),$ $(7,0),$ and $(11,5).$ | Let $A = (-1,4),$ $B = (7,0),$ and $C = (11,5).$ Let $\mathbf{v} = \overrightarrow{CA} = \begin{pmatrix} -1 - 11 \\ 4 - 5 \end{pmatrix} = \begin{pmatrix} -12 \\ -1 \end{pmatrix}$ and $\mathbf{w} = \overrightarrow{CB} = \begin{pmatrix} 7 - 11 \\ 0 - 5 \end{pmatrix} = \begin{pmatrix} -4 \\ -5 \end{pmatrix}.$ The area o... |
Let $\mathbf{M} = \begin{pmatrix} 2 & 7 \\ -3 & -1 \end{pmatrix}.$ There exist scalars $p$ and $q$ such that
\[\mathbf{M}^2 = p \mathbf{M} + q \mathbf{I}.\]Enter the ordered pair $(p,q).$ | Since $\mathbf{M}^2 = \begin{pmatrix} 2 & 7 \\ -3 & -1 \end{pmatrix} \begin{pmatrix} 2 & 7 \\ -3 & -1 \end{pmatrix} = \begin{pmatrix} -17 & 7 \\ -3 & -20 \end{pmatrix},$ we seek $p$ and $q$ such that
\[\begin{pmatrix} -17 & 7 \\ -3 & -20 \end{pmatrix} = p \begin{pmatrix} 2 & 7 \\ -3 & -1 \end{pmatrix} + q \begin{pmatri... |
Compute $\tan 60^\circ$. | Let $P$ be the point on the unit circle that is $60^\circ$ counterclockwise from $(1,0)$, and let $D$ be the foot of the altitude from $P$ to the $x$-axis, as shown below.
[asy]
pair A,C,P,O,D;
draw((0,-1.2)--(0,1.2),p=black+1.2bp,Arrows(0.15cm));
draw((-1.2,0)--(1.2,0),p=black+1.2bp,Arrows(0.15cm));
A = (1,0);
O... |
Let $x = \cos \frac{2 \pi}{7} + i \sin \frac{2 \pi}{7}.$ Compute the value of
\[(2x + x^2)(2x^2 + x^4)(2x^3 + x^6)(2x^4 + x^8)(2x^5 + x^{10})(2x^6 + x^{12}).\] | Note that $x^7 = \cos 2 \pi + i \sin 2 \pi = 1,$ so $x^7 - 1 = 0,$ which factors as
\[(x - 1)(x^6 + x^5 + x^4 + x^3 + x^2 + x + 1) = 0.\]Since $x \neq 1,$
\[x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 = 0.\]Then
\begin{align*}
(2x + x^2)(2x^6 + x^{12}) &= 4x^7 + 2x^8 + 2x^{13} + x^{14} = 4 + 2x + 2x^6 + 1 = 5 + 2x + 2x^6, \\
(... |
There exist constants $a$ and $b$ so that
\[\cos^3 \theta = a \cos 3 \theta + b \cos \theta\]for all angles $\theta.$ Enter the ordered pair $(a,b).$ | From the triple angle formulas, $\cos 3 \theta = 4 \cos^3 \theta - 3 \cos \theta.$ Hence,
\[\cos^3 \theta = \frac{1}{4} \cos 3 \theta + \frac{3}{4} \cos \theta,\]so $(a,b) = \boxed{\left( \frac{1}{4}, \frac{3}{4} \right)}.$ |
The matrix $\mathbf{A} = \begin{pmatrix} 2 & 3 \\ 5 & d \end{pmatrix}$ satisfies
\[\mathbf{A}^{-1} = k \mathbf{A}\]for some constant $k.$ Enter the ordered pair $(d,k).$ | For $\mathbf{A} = \begin{pmatrix} 2 & 3 \\ 5 & d \end{pmatrix},$
\[\mathbf{A}^{-1} = \frac{1}{2d - 15} \begin{pmatrix} d & -3 \\ -5 & 2 \end{pmatrix}\]Comparing entries to $k \mathbf{A},$ we get
\begin{align*}
\frac{d}{2d - 15} &= 2k, \\
\frac{-3}{2d - 15} &= 3k, \\
\frac{-5}{2d - 15} &= 5k, \\
\frac{2}{2d - 15} &= dk.... |
Convert the point $(0, -3 \sqrt{3}, 3)$ in rectangular coordinates to spherical coordinates. Enter your answer in the form $(\rho,\theta,\phi),$ where $\rho > 0,$ $0 \le \theta < 2 \pi,$ and $0 \le \phi \le \pi.$ | We have that $\rho = \sqrt{0^2 + (-3 \sqrt{3})^2 + 3^2} = 6.$ We want $\phi$ to satisfy
\[3 = 6 \cos \phi,\]so $\phi = \frac{\pi}{3}.$
We want $\theta$ to satisfy
\begin{align*}
0 &= 6 \sin \frac{\pi}{3} \cos \theta, \\
-3 \sqrt{3} &= 6 \sin \frac{\pi}{3} \sin \theta.
\end{align*}Thus, $\theta = \frac{3 \pi}{2},$ so ... |
If $\tan x = 2,$ then find $\tan \left( x + \frac{\pi}{4} \right).$ | From the angle addition formula,
\begin{align*}
\tan \left( x + \frac{\pi}{4} \right) &= \frac{\tan x + \tan \frac{\pi}{4}}{1 - \tan x \tan \frac{\pi}{4}} \\
&= \frac{1 + 2}{1 - 2 \cdot 1} \\
&= \boxed{-3}.
\end{align*} |
Find the curve defined by the equation
\[r = \frac{1}{1 - \cos \theta}.\](A) Line
(B) Circle
(C) Parabola
(D) Ellipse
(E) Hyperbola
Enter the letter of the correct option. | From $r = \frac{1}{1 - \cos \theta},$
\[r - r \cos \theta = 1.\]Then $r = 1 + r \cos \theta = x + 1,$ so
\[r^2 = (x + 1)^2 = x^2 + 2x + 1.\]Hence, $x^2 + y^2 = x^2 + 2x + 1,$ so
\[y^2 = 2x + 1.\]This represents the graph of a parabola, so the answer is $\boxed{\text{(C)}}.$
[asy]
unitsize(0.5 cm);
pair moo (real t) {... |
If the six solutions of $x^6=-64$ are written in the form $a+bi$, where $a$ and $b$ are real, find the product of those solutions with $a>0.$ | We write $-64 = 2^6 \operatorname{cis} 180^\circ,$ so $x^6 = 2^6 \operatorname{cis} 180^\circ.$ The solutions are of the form
\[x = 2 \operatorname{cis} (30^\circ + 60^\circ k),\]where $0 \le k \le 5.$
[asy]
unitsize(1 cm);
int i;
draw(Circle((0,0),2));
draw((-2.2,0)--(2.2,0));
draw((0,-2.2)--(0,2.2));
dot("$30^\c... |
Given that $x+\sin y=2008$ and $x+2008 \cos y=2007$, where $0 \leq y \leq \frac{\pi}{2}$, find the value of $x+y$. | Subtracting the two equations gives $\sin y - 2008 \cos y = 1$. But since $0 \leq y \leq \frac{\pi}{2}$, the maximum of $\sin y$ is 1 and the minimum of $\cos y$ is 0, so we must have $\sin y = 1$, so $y = \frac{\pi}{2}$ and $x = 2007,$ so $x+y = \boxed{2007 + \frac\pi 2}$. |
Find the maximum value of
\[\sin \frac{\theta}{2} \cdot (1 + \cos \theta)\]for $0 < \theta < \pi.$ | From the double angle formula,
\[\sin \frac{\theta}{2} \cdot (1 + \cos \theta) = \sin \frac{\theta}{2} \left( 2 \cos^2 \frac{\theta}{2} \right) = 2 \sin \frac{\theta}{2} \left( 1 - \sin^2 \frac{\theta}{2} \right).\]Let $x = \sin \frac{\theta}{2}.$ We want to maximize
\[y = 2x (1 - x^2).\]Note that
\[y^2 = 4x^2 (1 - x^... |
Let $P$ be a point in coordinate space, where all the coordinates of $P$ are positive. The line between the origin and $P$ is drawn. The angle between this line and the $x$-, $y$-, and $z$-axis are $\alpha,$ $\beta,$ and $\gamma,$ respectively. If $\cos \alpha = \frac{1}{3}$ and $\cos \beta = \frac{1}{5},$ then dete... | Let $O$ be the origin, and let $P = (x,y,z).$ Let $X$ be the foot of the perpendicular from $P$ to the $x$-axis. Then $\angle POX = \alpha,$ $OP = \sqrt{x^2 + y^2 + z^2},$ and $OX = x,$ so
\[\cos \alpha = \frac{x}{\sqrt{x^2 + y^2 + z^2}}.\][asy]
unitsize(1 cm);
draw((0,0)--(3,0)--(3,2)--cycle);
label("$P = (x,y,z)$... |
Let $z = \cos \frac{4 \pi}{7} + i \sin \frac{4 \pi}{7}.$ Compute
\[\frac{z}{1 + z^2} + \frac{z^2}{1 + z^4} + \frac{z^3}{1 + z^6}.\] | Note $z^7 - 1 = \cos 4 \pi + i \sin 4 \pi - 1 = 0,$ so
\[(z - 1)(z^6 + z^5 + z^4 + z^3 + z^2 + z + 1) = 0.\]Since $z \neq 1,$ $z^6 + z^5 + z^4 + z^3 + z^2 + z + 1 = 0.$
Then
\begin{align*}
\frac{z}{1 + z^2} + \frac{z^2}{1 + z^4} + \frac{z^3}{1 + z^6} &= \frac{z}{1 + z^2} + \frac{z^2}{1 + z^4} + \frac{z^3}{(1 + z^2)(1 ... |
For a positive constant $c,$ in spherical coordinates $(\rho,\theta,\phi),$ find the shape described by the equation
\[\rho = c.\](A) Line
(B) Circle
(C) Plane
(D) Sphere
(E) Cylinder
(F) Cone
Enter the letter of the correct option. | In spherical coordinates, $\rho$ is the distance from a point to the origin. So if this distance is fixed, then we obtain a sphere. The answer is $\boxed{\text{(D)}}.$
[asy]
import three;
import solids;
size(180);
currentprojection = perspective(6,3,2);
currentlight = (1,0,1);
draw((-1,0,0)--(-2,0,0));
draw((0,-1... |
If $\sin x = 3 \cos x,$ then what is $\sin x \cos x$? | We know that $\sin^2 x + \cos^2 x = 1.$ Substituting $\sin x = 3 \cos x,$ we get
\[9 \cos^2 x + \cos^2 x = 1,\]so $10 \cos^2 x = 1,$ or $\cos^2 x = \frac{1}{10}.$ Then
\[\sin x \cos x = (3 \cos x)(\cos x) = 3 \cos^2 x = \boxed{\frac{3}{10}}.\] |
If
\[\sin x + \cos x + \tan x + \cot x + \sec x + \csc x = 7,\]then find $\sin 2x.$ | Expressing everything in terms of $\sin x$ and $\cos x,$ we get
\[\sin x + \cos x + \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} + \frac{1}{\sin x} + \frac{1}{\cos x} = 7.\]Then
\[\sin x + \cos x + \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} + \frac{\sin x + \cos x}{\sin x \cos x} = 7,\]which becomes
\[\sin x + \cos x +... |
In triangle $ABC,$ $D$ is on $\overline{AB}$ such that $AD:DB = 3:2,$ and $E$ is on $\overline{BC}$ such that $BE:EC = 3:2.$ If lines $DE$ and $AC$ intersect at $F,$ then find $\frac{DE}{EF}.$ | Let $\mathbf{a}$ denote $\overrightarrow{A},$ etc. Then from the given information
\[\mathbf{d} = \frac{2}{5} \mathbf{a} + \frac{3}{5} \mathbf{b}\]and
\[\mathbf{e} = \frac{2}{5} \mathbf{b} + \frac{3}{5} \mathbf{c}.\][asy]
unitsize(0.6 cm);
pair A, B, C, D, E, F;
A = (2,5);
B = (0,0);
C = (6,0);
D = interp(A,B,3/5);
... |
Let $A$ and $B$ be the endpoints of a semicircular arc of radius $2$. The arc is divided into seven congruent arcs by six equally spaced points $C_1$, $C_2$, $\dots$, $C_6$. All chords of the form $\overline {AC_i}$ or $\overline {BC_i}$ are drawn. Find the product of the lengths of these twelve chords. | Let $\omega = e^{2 \pi i/14}.$ We can identify $A$ with $2,$ $B$ with $-2,$ and $C_k$ with the complex number $2 \omega^k.$
[asy]
unitsize (3 cm);
int i;
pair A, B;
pair[] C;
A = (1,0);
B = (-1,0);
C[1] = dir(1*180/7);
C[2] = dir(2*180/7);
C[3] = dir(3*180/7);
C[4] = dir(4*180/7);
C[5] = dir(5*180/7);
C[6] = dir(6*... |
Let $a$ and $b$ be acute angles such that
\begin{align*}
3 \sin^2 a + 2 \sin^2 b &= 1, \\
3 \sin 2a - 2 \sin 2b &= 0.
\end{align*}Find $a + 2b,$ as measured in radians. | From the first equation, using the double angle formula,
\[3 \sin^2 a = 1 - 2 \sin^2 b = \cos 2b.\]From the second equation, again using the double angle formula,
\[\sin 2b = \frac{3}{2} \sin 2a = 3 \cos a \sin a.\]Since $\cos^2 2b + \sin^2 2b = 1,$
\[9 \sin^4 a + 9 \cos^2 a \sin^2 a = 1.\]Then $9 \sin^2 a (\sin^2 a + ... |
Compute $(\cos 185^\circ + i \sin 185^\circ)^{54}.$ | By DeMoivre's Theorem,
\begin{align*}
(\cos 185^\circ + i \sin 185^\circ)^{54} &= \cos 9990^\circ + i \sin 9990^\circ \\
&= \cos 270^\circ + i \sin 270^\circ \\
&= \boxed{-i}.
\end{align*} |
Find $\sec 135^\circ.$ | We have that
\[\sec 135^\circ = \frac{1}{\cos 135^\circ}.\]Then $\cos 135^\circ = -\cos (135^\circ - 180^\circ) = -\cos (-45^\circ) = -\cos 45^\circ = -\frac{1}{\sqrt{2}},$ so
\[\frac{1}{\cos 135^\circ} = \boxed{-\sqrt{2}}.\] |
Convert the point $\left( 2 \sqrt{3}, \frac{2 \pi}{3} \right)$ in polar coordinates to rectangular coordinates. | In rectangular coordinates, $\left( 2 \sqrt{3}, \frac{2 \pi}{3} \right)$ becomes
\[\left( 2 \sqrt{3} \cos \frac{2 \pi}{3}, 2 \sqrt{3} \sin \frac{2 \pi}{3} \right) = \boxed{(-\sqrt{3}, 3)}.\] |
Lines $l_1^{}$ and $l_2^{}$ both pass through the origin and make first-quadrant angles of $\frac{\pi}{70}$ and $\frac{\pi}{54}$ radians, respectively, with the positive $x$-axis. For any line $l$, the transformation $R(l)$ produces another line as follows: $l$ is reflected in $l_1$, and the resulting line is reflecte... | More generally, suppose we have a line $l$ that is reflect across line $l_1$ to obtain line $l'.$
[asy]
unitsize(3 cm);
draw(-0.2*dir(35)--dir(35));
draw(-0.2*dir(60)--dir(60));
draw(-0.2*dir(10)--dir(10));
draw((-0.2,0)--(1,0));
draw((0,-0.2)--(0,1));
label("$l$", dir(60), NE);
label("$l_1$", dir(35), NE);
label("$... |
Given vectors $\mathbf{a}$ and $\mathbf{b},$ let $\mathbf{p}$ be a vector such that
\[\|\mathbf{p} - \mathbf{b}\| = 2 \|\mathbf{p} - \mathbf{a}\|.\]Among all such vectors $\mathbf{p},$ there exists constants $t$ and $u$ such that $\mathbf{p}$ is at a fixed distance from $t \mathbf{a} + u \mathbf{b}.$ Enter the ordered... | From $\|\mathbf{p} - \mathbf{b}\| = 2 \|\mathbf{p} - \mathbf{a}\|,$
\[\|\mathbf{p} - \mathbf{b}\|^2 = 4 \|\mathbf{p} - \mathbf{a}\|^2.\]This expands as
\[\|\mathbf{p}\|^2 - 2 \mathbf{b} \cdot \mathbf{p} + \|\mathbf{b}\|^2 = 4 \|\mathbf{p}\|^2 - 8 \mathbf{a} \cdot \mathbf{p} + 4 \|\mathbf{a}\|^2,\]which simplifies to $3... |
The vertices of a cube have coordinates $(0,0,0),$ $(0,0,4),$ $(0,4,0),$ $(0,4,4),$ $(4,0,0),$ $(4,0,4),$ $(4,4,0),$ and $(4,4,4).$ A plane cuts the edges of this cube at the points $P = (0,2,0),$ $Q = (1,0,0),$ $R = (1,4,4),$ and two other points. Find the distance between these two points. | Let $\mathbf{p} = \begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix},$ $\mathbf{q} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix},$ and $\mathbf{r} = \begin{pmatrix} 1 \\ 4 \\ 4 \end{pmatrix}.$ Then the normal vector to the plane passing through $P,$ $Q,$ and $R$ is
\[(\mathbf{p} - \mathbf{q}) \times (\mathbf{p} - \mathbf{r}) = \... |
Find the cross product of $\begin{pmatrix} 2 \\ 0 \\ 3 \end{pmatrix}$ and $\begin{pmatrix} 5 \\ -1 \\ 7 \end{pmatrix}.$ | The cross product of $\begin{pmatrix} 2 \\ 0 \\ 3 \end{pmatrix}$ and $\begin{pmatrix} 5 \\ -1 \\ 7 \end{pmatrix}$ is
\[\begin{pmatrix} (0)(7) - (-1)(3) \\ (3)(5) - (7)(2) \\ (2)(-1) - (5)(0) \end{pmatrix} = \boxed{\begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}}.\] |
If
\[\begin{pmatrix} 1 & 2 & a \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{pmatrix}^n = \begin{pmatrix} 1 & 18 & 2007 \\ 0 & 1 & 36 \\ 0 & 0 & 1 \end{pmatrix},\]then find $a + n.$ | Let $\mathbf{A} = \begin{pmatrix} 1 & 2 & a \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{pmatrix}.$ Then we can write $\mathbf{A} = \mathbf{I} + \mathbf{B},$ where
\[\mathbf{B} = \begin{pmatrix} 0 & 2 & a \\ 0 & 0 & 4 \\ 0 & 0 & 0 \end{pmatrix}.\]Note that
\[\mathbf{B}^2 = \begin{pmatrix} 0 & 2 & a \\ 0 & 0 & 4 \\ 0 & 0 & 0 \end{pm... |
Simplify $\cot 10 + \tan 5.$
Enter your answer as a trigonometric function evaluated at an integer, such as "sin 7". | We can write
\[\cot 10 + \tan 5 = \frac{\cos 10}{\sin 10} + \frac{\sin 5}{\cos 5} = \frac{\cos 10 \cos 5 + \sin 5 \sin 10}{\sin 10 \cos 5}.\]From the angle subtraction formula, the numerator is equal to $\cos (10 - 5) = \cos 5,$ so
\[\frac{\cos 10 \cos 5 + \sin 5 \sin 10}{\sin 10 \cos 5} = \frac{\cos 5}{\sin 10 \cos 5}... |
In a polar coordinate system, the midpoint of the line segment whose endpoints are $\left( 8, \frac{5 \pi}{12} \right)$ and $\left( 8, -\frac{3 \pi}{12} \right)$ is the point $(r, \theta).$ Enter $(r, \theta),$ where $r > 0$ and $0 \le \theta < 2 \pi.$ | Let $A = \left( 8, \frac{5 \pi}{12} \right)$ and $B = \left( 8, -\frac{3 \pi}{12}\right).$ Note that both $A$ and $B$ lie on the circle with radius 8. Also, $\angle AOB = \frac{2 \pi}{3},$ where $O$ is the origin.
[asy]
unitsize (0.3 cm);
pair A, B, M, O;
A = 8*dir(75);
B = 8*dir(-45);
O = (0,0);
M = (A + B)/2;
d... |
Let $x$ and $y$ be distinct real numbers such that
\[
\begin{vmatrix} 1 & 4 & 9 \\ 3 & x & y \\ 3 & y & x \end{vmatrix}
= 0.\]Find $x + y.$ | Expanding the determinant, we obtain
\begin{align*}
\begin{vmatrix} 1 & 4 & 9 \\ 3 & x & y \\ 3 & y & x \end{vmatrix} &= \begin{vmatrix} x & y \\ y & x \end{vmatrix} - 4 \begin{vmatrix} 3 & y \\ 3 & x \end{vmatrix} + 9 \begin{vmatrix} 3 & x \\ 3 & y \end{vmatrix} \\
&= (x^2 - y^2) - 4(3x - 3y) + 9(3y - 3x) \\
&= x^2 - ... |
A $180^\circ$ rotation around the origin in the counter-clockwise direction is applied to $-6 - 3i.$ What is the resulting complex number? | A $180^\circ$ rotation in the counter-clockwise direction corresponds to multiplication by $\operatorname{cis} 180^\circ = -1.$
[asy]
unitsize(0.5 cm);
pair A = (-6,-3), B = (6,3);
draw((-8,0)--(8,0));
draw((0,-4)--(0,4));
draw((0,0)--A,dashed);
draw((0,0)--B,dashed);
dot("$-6 - 3i$", A, SW);
dot("$6 + 3i$", B, NE... |
The following line is parameterized, so that its direction vector is of the form $\begin{pmatrix} a \\ -1 \end{pmatrix}.$ Find $a.$
[asy]
unitsize(0.4 cm);
pair A, B, L, R;
int i, n;
for (i = -8; i <= 8; ++i) {
draw((i,-8)--(i,8),gray(0.7));
draw((-8,i)--(8,i),gray(0.7));
}
draw((-8,0)--(8,0),Arrows(6));
draw(... | The line passes through $\begin{pmatrix} -2 \\ 5 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 0 \end{pmatrix},$ so its direction vector is proportional to
\[\begin{pmatrix} 1 \\ 0 \end{pmatrix} - \begin{pmatrix} -2 \\ 5 \end{pmatrix} = \begin{pmatrix} 3 \\ -5 \end{pmatrix}.\]To get a $y$-coordinate of $-1,$ we can multiply... |
Compute
\[\begin{pmatrix} 1 & 1 & -2 \\ 0 & 4 & -3 \\ -1 & 4 & 3 \end{pmatrix} \begin{pmatrix} 2 & -2 & 0 \\ 1 & 0 & -3 \\ 4 & 0 & 0 \end{pmatrix}.\] | We have that
\[\begin{pmatrix} 1 & 1 & -2 \\ 0 & 4 & -3 \\ -1 & 4 & 3 \end{pmatrix} \begin{pmatrix} 2 & -2 & 0 \\ 1 & 0 & -3 \\ 4 & 0 & 0 \end{pmatrix} = \boxed{\begin{pmatrix} -5 & -2 & -3 \\ -8 & 0 & -12 \\ 14 & 2 & -12 \end{pmatrix}}.\] |
Let $\mathbf{A} = \begin{pmatrix} a & 1 \\ -2 & d \end{pmatrix}$ for some real numbers $a$ and $d.$ If
\[\mathbf{A} + \mathbf{A}^{-1} = \mathbf{0},\]then find $\det \mathbf{A}.$ | From the formula for the inverse,
\[\mathbf{A}^{-1} = \frac{1}{ad + 2} \begin{pmatrix} d & -1 \\ 2 & a \end{pmatrix} = \begin{pmatrix} \frac{d}{ad + 2} & -\frac{1}{ad + 2} \\ \frac{2}{ad + 2} & \frac{a}{ad + 2} \end{pmatrix},\]so we want
\[\begin{pmatrix} a & 1 \\ -2 & d \end{pmatrix} + \begin{pmatrix} \frac{d}{ad + 2}... |
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