problem stringlengths 16 4.31k | type stringclasses 7
values | level stringclasses 6
values | solution stringlengths 0 6.9k |
|---|---|---|---|
Find the matrix that corresponds to rotating about the origin by an angle of $120^\circ$ counter-clockwise. | Precalculus | Level 3 | Step 1: The transformation that rotates about the origin by an angle of $120^\circ$ counter-clockwise takes $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ to $\begin{pmatrix} -1/2 \\ \sqrt{3}/2 \end{pmatrix},$ and $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ to $\begin{pmatrix} -\sqrt{3}/2 \\ -1/2 \end{pmatrix},$ so the matrix is
\[... |
A point has rectangular coordinates $(-5,-7,4)$ and spherical coordinates $(\rho, \theta, \phi).$ Find the rectangular coordinates of the point with spherical coordinates $(\rho, \theta, -\phi).$ | Precalculus | Level 4 | Step 1: We have that
\begin{align*}
-5 &= \rho \sin \phi \cos \theta, \\
-7 &= \rho \sin \phi \sin \theta, \\
4 &= \rho \cos \phi
Step 2: \end{align*}Then
\begin{align*}
\rho \sin (-\phi) \cos \theta &= -\rho \sin \phi \cos \theta = 5, \\
\rho \sin (-\phi) \sin \theta &= -\rho \sin \phi \sin \theta = 7, \\
\rho \cos (-... |
Compute $\sin 6^\circ \sin 42^\circ \sin 66^\circ \sin 78^\circ.$ | Precalculus | Level 3 | Step 1: Since $\sin 66^\circ = \cos 24^\circ$ and $\sin 78^\circ = \cos 12^\circ,$ the product is equal to
\[\sin 6^\circ \cos 12^\circ \cos 24^\circ \sin 42^\circ
Step 2: \]Then
\[\sin 6^\circ \cos 12^\circ \cos 24^\circ \sin 42^\circ = \frac{\cos 6^\circ \sin 6^\circ \cos 12^\circ \cos 24^\circ \sin 42^\circ}{\cos 6^... |
Let $H$ be the orthocenter of triangle $ABC.$ For all points $P$ on the circumcircle of triangle $ABC,$
\[PA^2 + PB^2 + PC^2 - PH^2\]is a constant. Express this constant in terms of the side lengths $a,$ $b,$ $c$ and circumradius $R$ of triangle $ABC.$ | Precalculus | Level 5 | Step 1: Let the circumcenter $O$ of triangle $ABC$ be the origin, so $\|\overrightarrow{P}\| = R
Step 2: $ Also, $\overrightarrow{H} = \overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C}
Step 3: $ Then
\begin{align*}
PA^2 &= \|\overrightarrow{P} - \overrightarrow{A}\|^2 \\
&= (\overrightarrow{P} - \overright... |
Evaluate
\[\begin{vmatrix} \cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \end{vmatrix}.\] | Precalculus | Level 2 | Step 1: We can expand the determinant as follows:
\begin{align*}
\begin{vmatrix} \cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \end{vmatrix} &= \cos \alpha \cos \beta \begin{vmatrix} \cos \beta & 0 \\ \sin ... |
Let point $O$ be the origin of a three-dimensional coordinate system, and let points $A,$ $B,$ and $C$ be located on the positive $x,$ $y,$ and $z$ axes, respectively. If $OA = \sqrt[4]{75}$ and $\angle BAC = 30^\circ,$ then compute the area of triangle $ABC.$ | Precalculus | Level 5 | Step 1: Let $b = OB$ and $c = OC
Step 2: $
[asy]
import three;
size(250);
currentprojection = perspective(6,3,2);
triple A, B, C, O;
A = (3,0,0);
B = (0,4,0);
C = (0,0,2);
O = (0,0,0);
draw(O--(5,0,0));
draw(O--(0,5,0));
draw(O--(0,0,3));
draw(A--B--C--cycle);
label("$A$", A, S);
label("$B$", B, S);
label("$C$", ... |
If the angle between the vectors $\mathbf{a}$ and $\mathbf{b}$ is $43^\circ,$ what is the angle between the vectors $-\mathbf{a}$ and $\mathbf{b}$? | Precalculus | Level 1 | Step 1: Since $\mathbf{a}$ and $-\mathbf{a}$ point in opposite directions, the angle between them is $180^\circ
Step 2: $ Then the angle between $-\mathbf{a}$ and $\mathbf{b}$ is $180^\circ - 43^\circ = \boxed{137^\circ}
|
Right triangle $ABC$ (hypotenuse $\overline{AB}$) is inscribed in equilateral triangle $PQR,$ as shown. If $PC = 3$ and $BP = CQ = 2,$ compute $AQ.$
[asy]
unitsize(0.8 cm);
pair A, B, C, P, Q, R;
P = (0,0);
Q = (5,0);
R = 5*dir(60);
A = Q + 8/5*dir(120);
B = 2*dir(60);
C = (3,0);
draw(A--B--C--cycle);
draw(P--Q--... | Precalculus | Level 3 | Step 1: We see that the side length of equilateral triangle $PQR$ is 5
Step 2: Let $x = AQ
Step 3: $
By the Law of Cosines on triangle $BCP,$
\[BC^2 = 2^2 + 3^2 - 2 \cdot 2 \cdot 3 \cdot \cos 60^\circ = 7
Step 4: \]Then by the Law of Cosines on triangle $ACQ,$
\[AC^2 = x^2 + 2^2 - 2 \cdot x \cdot 2 \cdot \cos 60^\circ... |
If the three points $(1,a,b),$ $(a,2,b),$ $(a,b,3)$ are collinear, what is the value of $a + b$? | Precalculus | Level 2 | Step 1: Note that the $z$-coordinate of both $(1,a,b)$ and $(a,2,b)$ is $b,$ so the whole line must lie in the plane $z = b
Step 2: $ Hence, $b = 3
Step 3: $
Similarly, the $x$-coordinate of both $(a,2,b)$ and $(a,b,3)$ is $a,$ so the whole line must lie in the plane $x = a
Step 4: $ Hence, $a = 1,$ so $a + b = \box... |
The lengths of the sides of a triangle are consecutive integers, and the largest angle is twice the smallest angle. Find the cosine of the smallest angle. | Precalculus | Level 3 | Step 1: Let the side lengths be $n,$ $n + 1,$ $n + 2
Step 2: $ Then the smallest angle $x$ is opposite the side of length $n,$ and its cosine is
\[\cos x = \frac{(n + 1)^2 + (n + 2)^2 - n^2}{2(n + 1)(n + 2)} = \frac{n^2 + 6n + 5}{2(n + 1)(n + 2)} = \frac{(n + 1)(n + 5)}{2(n + 1)(n + 2)} = \frac{n + 5}{2(n + 2)}
Step 3... |
A projection takes $\begin{pmatrix} 1 \\ -2 \end{pmatrix}$ to $\begin{pmatrix} \frac{3}{2} \\ -\frac{3}{2} \end{pmatrix}.$ Which vector does the projection take $\begin{pmatrix} -4 \\ 1 \end{pmatrix}$ to? | Precalculus | Level 4 | Step 1: Since the projection of $\begin{pmatrix} 1 \\ -2 \end{pmatrix}$ is $\begin{pmatrix} \frac{3}{2} \\ -\frac{3}{2} \end{pmatrix},$ the vector being projected onto is a scalar multiple of $\begin{pmatrix} \frac{3}{2} \\ -\frac{3}{2} \end{pmatrix}
Step 2: $ Thus, we can assume that the vector being projected onto i... |
Find the number of solutions to the equation
\[\tan (5 \pi \cos \theta) = \cot (5 \pi \sin \theta)\]where $\theta \in (0, 2 \pi).$ | Precalculus | Level 5 | Step 1: From the given equation,
\[\tan (5 \pi \cos \theta) = \frac{1}{\tan (5 \pi \sin \theta)},\]so $\tan (5 \pi \cos \theta) \tan (5 \pi \sin \theta) = 1
Step 2: $
Then from the angle addition formula,
\begin{align*}
\cot (5 \pi \cos \theta + 5 \pi \sin \theta) &= \frac{1}{\tan (5 \pi \cos \theta + 5 \pi \sin \thet... |
Find all angles $\theta,$ $0 \le \theta \le 2 \pi,$ with the following property: For all real numbers $x,$ $0 \le x \le 1,$
\[x^2 \cos \theta - x(1 - x) + (1 - x)^2 \sin \theta > 0.\] | Precalculus | Level 5 | Step 1: Taking $x = 0,$ we get $\sin \theta > 0
Step 2: $ Taking $x = 1,$ we get $\cos \theta > 0
Step 3: $ Hence, $0 < \theta < \frac{\pi}{2}
Step 4: $
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 \... |
Let $\mathbf{a} = \begin{pmatrix} 7 \\ -4 \\ -4 \end{pmatrix}$ and $\mathbf{c} = \begin{pmatrix} -2 \\ -1 \\ 2 \end{pmatrix}.$ Find the vector $\mathbf{b}$ such that $\mathbf{a},$ $\mathbf{b},$ and $\mathbf{c}$ are collinear, and $\mathbf{b}$ bisects the angle between $\mathbf{a}$ and $\mathbf{c}.$
[asy]
unitsize(0.5... | Precalculus | Level 5 | Step 1: The line through $\mathbf{a}$ and $\mathbf{c}$ can be parameterized by
\[\begin{pmatrix} 7 - 9t \\ -4 + 3t \\ -4 + 6t \end{pmatrix}
Step 2: \]Then $\mathbf{b}$ is of this form
Step 3: Furthermore, the angle between $\mathbf{a}$ and $\mathbf{b}$ is equal to the angle between $\mathbf{b}$ and $\mathbf{c}
Step 4: ... |
There is an angle $\theta$ in the range $0^\circ < \theta < 45^\circ$ which satisfies
\[\tan \theta + \tan 2 \theta + \tan 3 \theta = 0.\]Calculate $\tan \theta$ for this angle. | Precalculus | Level 5 | Step 1: Let $t = \tan \theta
Step 2: $ Then $\tan 2 \theta = \frac{2t}{1 - t^2}$ and $\tan 3 \theta = \frac{3t - t^3}{1 - 3t^2},$ so
\[t + \frac{2t}{1 - t^2} + \frac{3t - t^3}{1 - 3t^2} = 0
Step 3: \]This simplifies to $4t^5 - 14t^3 + 6t = 0
Step 4: $ This factors as $2t(2t^2 - 1)(t^2 - 3) = 0
Step 5: $
Since $0^\ci... |
Compute
\[
\begin{vmatrix} \cos 1 & \cos 2 & \cos 3 \\ \cos 4 & \cos 5 & \cos 6 \\ \cos 7 & \cos 8 & \cos 9 \end{vmatrix}
.\]All the angles are in radians. | Precalculus | Level 2 | Step 1: The entries in each row are $\cos n,$ $\cos (n + 1),$ and $\cos (n + 2)$ for some integer $n
Step 2: $ From the angle addition formula,
\[\cos n + \cos (n + 2) = 2 \cos (n + 1) \cos 1
Step 3: \]Then
\[\cos (n + 2) = 2 \cos 1 \cos (n + 1) - \cos n
Step 4: \]Thus, we can obtain the third column of the matrix by ... |
In triangle $ABC,$ $\sin A = \frac{3}{5}$ and $\cos B = \frac{5}{13}.$ Find $\cos C.$ | Precalculus | Level 4 | Step 1: We have that
\[\cos^2 A = 1 - \sin^2 A = \frac{16}{25},\]so $\cos A = \pm \frac{4}{5}
Step 2: $
Also,
\[\sin^2 B = 1 - \cos^2 B = \frac{144}{169}
Step 3: \]Since $\sin B$ is positive, $\sin B = \frac{12}{13}
Step 4: $
Then
\begin{align*}
\sin C &= \sin (180^\circ - A - B) \\
&= \sin (A + B) \\
&= \sin A \cos ... |
Line segment $\overline{AB}$ is extended past $B$ to $P$ such that $AP:PB = 10:3.$ Then
\[\overrightarrow{P} = t \overrightarrow{A} + u \overrightarrow{B}\]for some constants $t$ and $u.$ Enter the ordered pair $(t,u).$
[asy]
unitsize(1 cm);
pair A, B, P;
A = (0,0);
B = (5,1);
P = interp(A,B,10/7);
draw(A--P);
d... | Precalculus | Level 4 | Step 1: Since $AP:PB = 10:3,$ we can write
\[\frac{\overrightarrow{P} - \overrightarrow{A}}{10} = \frac{\overrightarrow{P} - \overrightarrow{B}}{7}
Step 2: \]Isolating $\overrightarrow{P},$ we find
\[\overrightarrow{P} = -\frac{3}{7} \overrightarrow{A} + \frac{10}{7} \overrightarrow{B}
Step 3: \]Thus, $(t,u) = \boxed{\... |
Let $\mathbf{A}$ and $\mathbf{B}$ be matrices such that
\[\mathbf{A} + \mathbf{B} = \mathbf{A} \mathbf{B}.\]If $\mathbf{A} \mathbf{B} = \begin{pmatrix} 20/3 & 4/3 \\ -8/3 & 8/3 \end{pmatrix},$ find $\mathbf{B} \mathbf{A}.$ | Precalculus | Level 5 | Step 1: From $\mathbf{A} \mathbf{B} = \mathbf{A} + \mathbf{B},$
\[\mathbf{A} \mathbf{B} - \mathbf{A} - \mathbf{B} = \mathbf{0}
Step 2: \]Then $\mathbf{A} \mathbf{B} - \mathbf{A} - \mathbf{B} + \mathbf{I} = \mathbf{I}
Step 3: $ In the style of Simon's Favorite Factoring Trick, we can write this as
\[(\mathbf{A} - \math... |
Compute $\tan \left (\operatorname{arccot} \frac{4}{7} \right).$ | Precalculus | Level 1 | Step 1: Consider a right triangle where the adjacent side is 4 and the opposite side is 7
Step 2: [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} \fra... |
The area of the parallelogram generated by the vectors $\mathbf{a}$ and $\mathbf{b}$ is 8. Find the area of the parallelogram generated by the vectors $2 \mathbf{a} + 3 \mathbf{b}$ and $\mathbf{a} - 5 \mathbf{b}.$ | Precalculus | Level 4 | Step 1: Since the area of the parallelogram generated by the vectors $\mathbf{a}$ and $\mathbf{b}$ is 8,
\[\|\mathbf{a} \times \mathbf{b}\| = 8
Step 2: \]Then the area of the parallelogram generated by the vectors $2 \mathbf{a} + 3 \mathbf{b}$ and $\mathbf{a} - 5 \mathbf{b}$ is
\[\|(2 \mathbf{a} + 3 \mathbf{b}) \times ... |
Find the distance from the point $(1,2,3)$ to the line described by
\[\begin{pmatrix} 6 \\ 7 \\ 7 \end{pmatrix} + t \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix}.\] | Precalculus | Level 4 | Step 1: A point on the line is given by
\[\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 6 \\ 7 \\ 7 \end{pmatrix} + t \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} = \begin{pmatrix} 3t + 6 \\ 2t + 7 \\ -2t + 7 \end{pmatrix}
Step 2: \][asy]
unitsize (0.6 cm);
pair A, B, C, D, E, F, H;
A = (2,5);
B = (0,0);
... |
The matrix for projecting onto a certain line $\ell,$ which passes through the origin, is given by
\[\renewcommand{\arraystretch}{1.5} \begin{pmatrix} \frac{2}{15} & -\frac{1}{15} & -\frac{1}{3} \\ -\frac{1}{15} & \frac{1}{30} & \frac{1}{6} \\ -\frac{1}{3} & \frac{1}{6} & \frac{5}{6} \end{pmatrix} \renewcommand{\arrays... | Precalculus | Level 5 | Step 1: Let $\mathbf{P}$ denote the given matrix, so $\mathbf{P} \mathbf{v}$ is the projection of $\mathbf{v}$ onto $\ell
Step 2: $ In particular, $\mathbf{P} \mathbf{v}$ lies on $\ell$ for any vector $\mathbf{v}
Step 3: $ So, we can take $\mathbf{v} = \mathbf{i}
Step 4: $ Then
\[\mathbf{P} \mathbf{i} = \begin{pmatr... |
What is the smallest positive integer $n$ such that all the roots of $z^4 - z^2 + 1 = 0$ are $n^{\text{th}}$ roots of unity? | Precalculus | Level 4 | Step 1: Multiplying the equation $z^4 - z^2 + 1 = 0$ by $z^2 + 1$, we get $z^6 + 1 = 0$
Step 2: Multiplying this equation by $z^6 - 1 = 0$, we get $z^{12} - 1 = 0$
Step 3: Therefore, every root of $z^4 - z^2 + 1 = 0$ is a $12^{\text{th}}$ root of unity
Step 4: We can factor $z^{12} - 1 = 0$ as
\[(z^6 - 1)(z^6 + 1) = (z... |
Find the volume of the region in space defined by
\[|x + y + z| + |x + y - z| \le 8\]and $x,$ $y,$ $z \ge 0.$ | Precalculus | Level 4 | Step 1: Let $a$ and $b$ be real numbers
Step 2: If $a \ge b,$ then
\[|a + b| + |a - b| = (a + b) + (a - b) = 2a
Step 3: \]If $a \le b,$ then
\[|a + b| + |a - b| = (a + b) + (b - a) = 2b
Step 4: \]In either case, $|a + b| + |a - b| = 2 \max\{a,b\}
Step 5: $
Thus, the condition $|x + y + z| + |x + y - z| \le 8$ is equiv... |
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