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A bee starts flying from point $P_0$. She flies $1$ inch due east to point $P_1$. For $j \ge 1$, once the bee reaches point $P_j$, she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$. When the bee reaches $P_{2015},$ how far from $P_0$ is she, in inches? |
Let $\omega = e^{\pi i/6}.$ Then assuming the bee starts at the origin, $P_{2015}$ is at the point \[z = 1 + 2 \omega + 3 \omega^2 + 4 \omega^3 + \dots + 2015 \omega^{2014}.\]Then \[\omega z = \omega + 2 \omega^2 + 3 \omega^3 + 4 \omega^4 + \dots + 2015 \omega^{2015}.\]Subtracting these equations, we get \begin{align*... |
Triangle $ABC$ has a right angle at $B$, and contains a point $P$ for which $PA = 10$, $PB = 6$, and $\angle APB = \angle BPC = \angle CPA$. Find $PC$. [asy] unitsize(0.2 cm); pair A, B, C, P; A = (0,14); B = (0,0); C = (21*sqrt(3),0); P = intersectionpoint(arc(B,6,0180),arc(C,33,0180)); draw(A--B--C--cycle); draw(... |
Since $\angle APB = \angle BPC = \angle CPA,$ they are all equal to $120^\circ.$ Let $z = PC.$ By the Law of Cosines on triangles $BPC,$ $APB,$ and $APC,$ \begin{align*} BC^2 &= z^2 + 6z + 36, \\ AB^2 &= 196, \\ AC^2 &= z^2 + 10z + 100. \end{align*}By the Pythagorean Theorem, $AB^2 + BC^2 = AC^2,$ so \[196 + z^2 + 6z... |
If \[\frac{\cos^4 \alpha}{\cos^2 \beta} + \frac{\sin^4 \alpha}{\sin^2 \beta} = 1,\]then find the sum of all possible values of \[\frac{\sin^4 \beta}{\sin^2 \alpha} + \frac{\cos^4 \beta}{\cos^2 \alpha}.\] |
We can write the first equation as \[\frac{\cos^4 \alpha}{\cos^2 \beta} + \frac{\sin^4 \alpha}{\sin^2 \beta} = \cos^2 \alpha + \sin^2 \alpha.\]Then \[\cos^4 \alpha \sin^2 \beta + \sin^4 \alpha \cos^2 \beta = \cos^2 \alpha \cos^2 \beta \sin^2 \beta + \sin^2 \alpha \cos^2 \beta \sin^2 \beta,\]so \[\cos^4 \alpha \sin^2 \b... |
If $x + \frac{1}{x} = \sqrt{3}$, then find $x^{18}$. |
Solution 1: We can rewrite the given equation as $x^2 - \sqrt{3} x + 1 = 0$, so by the quadratic formula, \[x = \frac{\sqrt{3} \pm \sqrt{3 - 4}}{2} = \frac{\sqrt{3} \pm i}{2},\]which means $x = e^{\pi i/6}$ or $x = e^{11 \pi i/6}$. If $x = e^{\pi i/6}$, then \[x^{18} = e^{3 \pi i} = -1,\]and if $x = e^{11 \pi i/6}$, t... |
Cube $ABCDEFGH,$ labeled as shown below, has edge length $1$ and is cut by a plane passing through vertex $D$ and the midpoints $M$ and $N$ of $\overline{AB}$ and $\overline{CG}$ respectively. The plane divides the cube into two solids. Find the volume of the larger of the two solids. [asy] import cse5; unitsize(8mm)... |
Define a coordinate system with $D$ at the origin and $C,$ $A,$ and $H$ on the $x$-, $y$-, and $z$-axes respectively. Then $D=(0,0,0),$ $M=\left(\frac{1}{2},1,0\right),$ and $N=\left(1,0,\frac{1}{2}\right).$ The plane going through $D,$ $M,$ and $N$ has equation \[2x-y-4z=0.\]This plane intersects $\overline{BF}$ at $Q... |
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 transformation $T,$ taking vectors to vectors, has the following properties: (i) $T(a \mathbf{v} + b \mathbf{w}) = a T(\mathbf{v}) + b T(\mathbf{w})$ for all vectors $\mathbf{v}$ and $\mathbf{w},$ and for all scalars $a$ and $b.$ (ii) $T(\mathbf{v} \times \mathbf{w}) = T(\mathbf{v}) \times T(\mathbf{w})$ for all v... |
From (ii), (iii), and (iv), \[T \left( \begin{pmatrix} 6 \\ 6 \\ 3 \end{pmatrix} \times \begin{pmatrix} -6 \\ 3 \\ 6 \end{pmatrix} \right) = \begin{pmatrix} 4 \\ -1 \\ 8 \end{pmatrix} \times \begin{pmatrix} 4 \\ 8 \\ -1 \end{pmatrix}.\]This reduces to \[T \begin{pmatrix} 27 \\ -54 \\ 54 \end{pmatrix} = \begin{pmatrix} ... |
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... |
Since $AP:PB = 10:3,$ we can write \[\frac{\overrightarrow{P} - \overrightarrow{A}}{10} = \frac{\overrightarrow{P} - \overrightarrow{B}}{7}.\]Isolating $\overrightarrow{P},$ we find \[\overrightarrow{P} = -\frac{3}{7} \overrightarrow{A} + \frac{10}{7} \overrightarrow{B}.\]Thus, $(t,u) = \boxed{\left( -\frac{3}{7}, \fra... |
Find the integer $n,$ $0 \le n \le 180,$ such that $\cos n^\circ = \cos 758^\circ.$ |
Since the cosine function has period $360^\circ,$ \[\cos 758^\circ = \cos (758^\circ - 2 \cdot 360^\circ) = \cos 38^\circ,\]so $n = \boxed{38}.$ |
In equilateral triangle $ABC,$ let points $D$ and $E$ trisect $\overline{BC}$. Find $\sin \angle DAE.$ |
Without loss of generality, let the triangle sides have length 6. [asy] pair A = (1, sqrt(3)), B = (0, 0), C= (2, 0); pair M = (1, 0); pair D = (2/3, 0), E = (4/3, 0); draw(A--B--C--cycle); label("$A$", A, N); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, S); label("$E$", E, S); label("$M$", M, S); draw(A-... |
Find the matrix $\mathbf{M}$ that triples the second row of a matrix. In other words, \[\mathbf{M} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} a & b \\ 3c & 3d \end{pmatrix}.\]If no such matrix $\mathbf{M}$ exists, then enter the zero matrix. |
Let $\mathbf{M} = \begin{pmatrix} p & q \\ r & s \end{pmatrix}.$ Then \[\mathbf{M} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} p & q \\ r & s \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} pa + qc & pb + qd \\ ra + sc & rb + sd \end{pmatrix}.\]We want this to be equal t... |
Find the number of ordered pairs $(a,b)$ of complex numbers such that \[a^3 b^5 = a^7 b^2 = 1.\] |
From the equation $a^3 b^5 = 1,$ $a^6 b^{10} = 1.$ From the equation $a^7 b^2 = 1,$ $a^{35} b^{10} = 1.$ Dividing these equations, we get \[a^{29} = 1.\]Therefore, $a$ must be a 29th root of unity. From the equation $a^7 b^2 = 1,$ $a^{14} b^4 = 1.$ Hence, \[\frac{a^3 b^5}{a^{14} b^4} = 1.\]This leads to $b = a^{11}... |
Let $\mathbf{M}$ be a matrix, and let $\mathbf{v}$ and $\mathbf{w}$ be vectors, such that \[\mathbf{M} \mathbf{v} = \begin{pmatrix} 2 \\ 3 \end{pmatrix} \quad \text{and} \quad \mathbf{M} \mathbf{w} = \begin{pmatrix} -2 \\ -5 \end{pmatrix}.\]Compute $\mathbf{M} (\mathbf{v} + 3 \mathbf{w}).$ |
We can distribute, to get \begin{align*} \mathbf{M} (\mathbf{v} + 3 \mathbf{w}) &= \mathbf{M} \mathbf{v} + \mathbf{M} (3 \mathbf{w}) \\ &= \mathbf{M} \mathbf{v} + 3 \mathbf{M} \mathbf{w} \\ &= \begin{pmatrix} 2 \\ 3 \end{pmatrix} + 3 \begin{pmatrix} -2 \\ -5 \end{pmatrix} \\ &= \boxed{\begin{pmatrix} -4 \\ -12 \end{pma... |
Let $x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}$. What is the greatest integer that does not exceed $100x$? |
Note that $\frac{\sum_{n=1}^{44} \cos n}{\sum_{n=1}^{44} \sin n} = \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\cos 89 + \cos 88 + \dots + \cos 46}$ Now use the sum-product formula $\cos x + \cos y = 2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})$ We want to pair up $[1, 44]$, $[2, 43]$, $[3, 42]$, etc. from the numerator and $... |
Below is the graph of $y = a \sin (bx + c)$ for some positive constants $a,$ $b,$ and $c.$ Find the smallest possible value of $c.$ [asy]import TrigMacros; size(300); real f(real x) { return 2*sin(4*x + pi/2); } draw(graph(f,-pi,pi,n=700,join=operator ..),red); trig_axes(-pi,pi,-3,3,pi/2,1); layer(); rm_trig_labe... |
We see that the graph reaches a maximum at $x = 0.$ The graph of $y = \sin x$ first reaches a maximum at $x = \frac{\pi}{2}$ for positive values of $x,$ so $c = \boxed{\frac{\pi}{2}}.$ |
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... |
Below is the graph of $y = a \sin (bx + c) + d$ for some positive constants $a,$ $b,$ $c,$ and $d.$ Find $a.$ [asy]import TrigMacros; size(400); real f(real x) { return 2*sin(3*x + pi) + 1; } draw(graph(f,-3*pi,3*pi,n=700,join=operator ..),red); trig_axes(-3*pi,3*pi,-4,4,pi/2,1); layer(); rm_trig_labels(-5,5, 2);... |
The graph oscillates between 3 and $-1,$ so $a = \frac{3 - (-1)}{2} = \boxed{2}.$ |
Let $a$ and $b$ be angles such that $\cos a + \cos b = \frac{1}{2}$ and $\sin a + \sin b = \frac{3}{11}.$ Find \[\tan \left( \frac{a + b}{2} \right).\] |
From the equation $\cos a + \cos b = \frac{1}{2},$ by sum-to-product, \[2 \cos \left( \frac{a + b}{2} \right) \cos \left( \frac{a - b}{2} \right) = \frac{1}{2}.\]Similarly, from the equation $\sin a + \sin b = \frac{3}{11},$ \[2 \sin \left( \frac{a + b}{2} \right) \cos \left( \frac{a - b}{2} \right) = \frac{3}{11}.\]Di... |
Find the curve defined by the equation \[r = 4 \tan \theta \sec \theta.\](A) Line (B) Circle (C) Parabola (D) Ellipse (E) Hyperbola Enter the letter of the correct option. |
From $r = 4 \tan \theta \sec \theta,$ \[r = 4 \cdot \frac{\sin \theta}{\cos \theta} \cdot \frac{1}{\cos \theta}.\]Then $r \cos^2 \theta = 4 \sin \theta,$ so \[r^2 \cos^2 \theta = 4r \sin \theta.\]Hence, $x^2 = 4y.$ This is the equation of a parabola, so the answer is $\boxed{\text{(C)}}.$ [asy] unitsize(0.15 cm); pa... |
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... |
Let $\mathcal{C}$ be the hyperbola $y^2 - x^2 = 1$. Given a point $P_0$ on the $x$-axis, we construct a sequence of points $(P_n)$ on the $x$-axis in the following manner: let $\ell_n$ be the line with slope 1 passing passing through $P_n$, then $P_{n+1}$ is the orthogonal projection of the point of intersection of $\... |
Let $P_n = (x_n, 0)$. Then the $\ell_n$ meet $\mathcal{C}$ at $(x_{n+1}, x_{n+1} - x_n)$. Since this point lies on the hyperbola, we have $(x_{n+1} - x_n)^2 - x_{n+1}^2 = 1$. Rearranging this equation gives \[ x_{n+1} = \frac{x_n^2 - 1}{2x_n}. \]Choose a $\theta_0 \in (0, \pi)$ with $\cot\theta_0 = x_0$, and define $\t... |
Let \[f(x) = (\arccos x)^3 + (\arcsin x)^3.\]Find the range of $f(x).$ All functions are in radians. |
First, we claim that $\arccos x + \arcsin x = \frac{\pi}{2}$ for all $x \in [-1,1].$ Note that \[\cos \left( \frac{\pi}{2} - \arcsin x \right) = \cos (\arccos x) = x.\]Furthermore, $-\frac{\pi}{2} \le \arcsin x \le \frac{\pi}{2},$ so $0 \le \frac{\pi}{2} - \arcsin x \le \pi.$ Therefore, \[\frac{\pi}{2} - \arcsin x = ... |
Below is the graph of $y = a \sin bx$ for some constants $a < 0$ and $b > 0.$ Find $a.$ [asy]import TrigMacros; size(400); real g(real x) { return (-2*sin(x/3)); } draw(graph(g,-3*pi,3*pi,n=700,join=operator ..),red); trig_axes(-3*pi,3*pi,-3,3,pi/2,1); layer(); rm_trig_labels(-5, 5, 2); label("$1$", (0,1), E); l... |
The maximum value of $a \sin bx$ is $|a|,$ so $a = \boxed{-2}.$ |
Compute $\arctan ( \tan 65^\circ - 2 \tan 40^\circ )$. (Express your answer in degrees as an angle between $0^\circ$ and $180^\circ$.) |
From the identity $\tan (90^\circ - x) = \frac{1}{\tan x},$ we have that \[\tan 65^\circ - 2 \tan 40^\circ = \frac{1}{\tan 25^\circ} - \frac{2}{\tan 50^\circ}.\]By the double-angle formula, \[\frac{1}{\tan 25^\circ} - \frac{2}{\tan 50^\circ} = \frac{1}{\tan 25^\circ} - \frac{1 - \tan^2 25^\circ}{\tan 25^\circ} = \tan 2... |
Convert the point $(6,2 \sqrt{3})$ in rectangular coordinates to polar coordinates. Enter your answer in the form $(r,\theta),$ where $r > 0$ and $0 \le \theta < 2 \pi.$ |
We have that $r = \sqrt{6^2 + (2 \sqrt{3})^2} = 4 \sqrt{3}.$ Also, if we draw the line connecting the origin and $(6,2 \sqrt{3}),$ this line makes an angle of $\frac{\pi}{6}$ with the positive $x$-axis. [asy] unitsize(0.6 cm); draw((-1,0)--(8,0)); draw((0,-1)--(0,4)); draw(arc((0,0),4*sqrt(3),0,30),red,Arrow(6)); dr... |
Compute $\begin{pmatrix} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{pmatrix}^6.$ |
We see that \[\begin{pmatrix} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{pmatrix} = 2 \begin{pmatrix} \sqrt{3}/2 & -1/2 \\ 1/2 & \sqrt{3}/2 \end{pmatrix} = 2 \begin{pmatrix} \cos \frac{\pi}{6} & -\sin \frac{\pi}{6} \\ \sin \frac{\pi}{6} & \cos \frac{\pi}{6} \end{pmatrix}.\]Note that $\begin{pmatrix} \cos \frac{\pi}{6} & -\sin ... |
If $\sec x + \tan x = \frac{5}{2},$ then find $\sec x - \tan x.$ |
Note that \begin{align*} (\sec x + \tan x)(\sec x - \tan x) &= \sec^2 x - \tan^2 x \\ &= \frac{1}{\cos^2 x} - \frac{\sin^2 x}{\cos^2 x} \\ &= \frac{1 - \sin^2 x}{\cos^2 x} = \frac{\cos^2 x}{\cos^2 x} = 1. \end{align*}Therefore, $\sec x - \tan x = \boxed{\frac{2}{5}}.$ |
Find all values of $x$ with $0 \le x < 2 \pi$ that satisfy $\sin x + \cos x = \sqrt{2}.$ Enter all the solutions, separated by commas. |
Squaring both sides, we get \[\sin^2 x + 2 \sin x \cos x + \cos x^2 = 2.\]Then $2 \sin x \cos x = 1,$ so $\sin 2x = 1.$ Since $0 \le x < 2 \pi,$ $2x = \frac{\pi}{2}$ or $2x = \frac{5 \pi}{2},$ so $x = \frac{\pi}{4}$ or $x = \frac{5 \pi}{4}.$ We check that only $\boxed{\frac{\pi}{4}}$ works. |
What is the period of $y = \sin 5x $? |
The graph of $y = \sin 5x$ passes through one full period as $5x$ ranges from $0$ to $2\pi$, which means $x$ ranges from $0$ to $\boxed{\frac{2\pi}{5}}.$ The graph of $y = \sin 5x$ is shown below: [asy] import TrigMacros; size(400); real g(real x) { return sin(5*x); } draw(graph(g,-3*pi,3*pi,n=700,join=operator .... |
Find the range of the function \[f(x) = \left( \arccos \frac{x}{2} \right)^2 + \pi \arcsin \frac{x}{2} - \left( \arcsin \frac{x}{2} \right)^2 + \frac{\pi^2}{12} (x^2 + 6x + 8).\] |
First, we claim that $\arccos x + \arcsin x = \frac{\pi}{2}$ for all $x \in [-1,1].$ Note that \[\cos \left( \frac{\pi}{2} - \arcsin x \right) = \cos (\arccos x) = x.\]Furthermore, $-\frac{\pi}{2} \le \arcsin x \le \frac{\pi}{2},$ so $0 \le \frac{\pi}{2} - \arcsin x \le \pi.$ Therefore, \[\frac{\pi}{2} - \arcsin x = ... |
Find the ordered pair $(a,b)$ of integers such that \[\sqrt{9 - 8 \sin 50^\circ} = a + b \csc 50^\circ.\] |
We write \[9 - 8 \sin 50^\circ = \frac{9 \sin^2 50^\circ - 8 \sin^3 50^\circ}{\sin^2 50^\circ} = \frac{9 \sin^2 50^\circ - 6 \sin 50^\circ + 6 \sin 50^\circ - 8 \sin^3 50^\circ}{\sin^2 50^\circ}.\]By the triple angle identity, \begin{align*} 6 \sin 50^\circ - 8 \sin^3 50^\circ &= 2 \sin (3 \cdot 50^\circ) \\ &= 2 \sin ... |
In triangle $ABC,$ $D,$ $E,$ and $F$ are points on sides $\overline{BC},$ $\overline{AC},$ and $\overline{AB},$ respectively, so that $BD:DC = CE:EA = AF:FB = 1:2.$ [asy] unitsize(0.8 cm); pair A, B, C, D, E, F, P, Q, R; A = (2,5); B = (0,0); C = (7,0); D = interp(B,C,1/3); E = interp(C,A,1/3); F = interp(A,B,1/3); ... |
Let $\mathbf{a}$ denote $\overrightarrow{A},$ etc. Then from the given information, \begin{align*} \mathbf{d} &= \frac{2}{3} \mathbf{b} + \frac{1}{3} \mathbf{c}, \\ \mathbf{e} &= \frac{1}{3} \mathbf{a} + \frac{2}{3} \mathbf{c}, \\ \mathbf{f} &= \frac{2}{3} \mathbf{a} + \frac{1}{3} \mathbf{b}. \end{align*}From the firs... |
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... |
In triangle $ABC,$ $\angle B = 60^\circ$ and $\angle C = 45^\circ.$ The point $D$ divides $\overline{BC}$ in the ratio $1:3$. Find \[\frac{\sin \angle BAD}{\sin \angle CAD}.\] |
By the Law of Sines on triangle $ABC,$ \[\frac{BD}{\sin \angle BAD} = \frac{AD}{\sin 60^\circ} \quad \Rightarrow \quad \quad \sin \angle BAD = \frac{BD \sqrt{3}}{2 AD}.\]By the Law of Sines on triangle $ACD,$ \[\frac{CD}{\sin \angle CAD} = \frac{AD}{\sin 45^\circ} \quad \Rightarrow \quad \quad \sin \angle CAD = \frac{C... |
Find the positive integer $n$ such that \[\sin \left( \frac{\pi}{2n} \right) + \cos \left (\frac{\pi}{2n} \right) = \frac{\sqrt{n}}{2}.\] |
Squaring both sides, we get \[\sin^2 \left( \frac{\pi}{2n} \right) + 2 \sin \left( \frac{\pi}{2n} \right) \cos \left( \frac{\pi}{2n} \right) + \cos^2 \left( \frac{\pi}{2n} \right) = \frac{n}{4},\]which we can re-write as \[\sin \frac{\pi}{n} + 1 = \frac{n}{4},\]so \[\sin \frac{\pi}{n} = \frac{n}{4} - 1.\]Since $-1 \le ... |
In triangle $ABC,$ points $D$ and $E$ are on $\overline{AB}$ and $\overline{AC},$ respectively, and angle bisector $\overline{AT}$ intersects $\overline{DE}$ at $F.$ If $AD = 1,$ $DB = 3,$ $AE = 2,$ and $EC = 4,$ compute $\frac{AF}{AT}.$ [asy] unitsize(1 cm); pair A, B, C, D, E, F, T; B = (0,0); C = (5,0); A = inte... |
Let $\mathbf{a}$ denote $\overrightarrow{A},$ etc. Then from the given information, \[\mathbf{d} = \frac{3}{4} \mathbf{a} + \frac{1}{4} \mathbf{b}\]and \[\mathbf{e} = \frac{2}{3} \mathbf{a} + \frac{1}{3} \mathbf{c}.\]Hence, $\mathbf{b} = 4 \mathbf{d} - 3 \mathbf{a}$ and $\mathbf{c} = 3 \mathbf{e} - 2 \mathbf{a}.$ By ... |
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 matrix \[\begin{pmatrix} a & b \\ -\frac{4}{5} & \frac{3}{5} \end{pmatrix}\]corresponds to a reflection. Enter the ordered pair $(a,b).$ |
Let $\mathbf{R}$ be the matrix, let $\mathbf{v}$ be a vector, and let $\mathbf{r} = \mathbf{R} \mathbf{v}.$ Then $\mathbf{R} \mathbf{r} = \mathbf{v},$ which means $\mathbf{R}^2 \mathbf{v} = \mathbf{v}.$ (In geometrical terms, if we reflect a vector, and reflect it again, then we get back the same vector as the origin... |
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, \[[ADE] = \frac{1}{2} \cdot 4 \cdot 7 \cdot \frac{20 \sqrt{2}}{33} = \boxed{\frac{280 \sqrt{2}}{33}}.\] |
The curve $y = \sin x$ cuts the line whose equation is $y = \sin 70^\circ$ into segments having the successive ratios \[\dots p : q : p : q \dots\]with $p < q.$ Compute the ordered pair of relatively prime positive integers $(p,q).$ |
The graph of $y = \sin x$ intersects the line $y = \sin 70^\circ$ at points of the form $(70^\circ + 360^\circ n, \sin 70^\circ)$ and $(110^\circ + 360^\circ n, \sin 70^\circ),$ where $n$ is an integer. [asy] unitsize(1.2 cm); real func (real x) { return(sin(x)); } draw(graph(func,-2*pi,2*pi),red); draw((-2*pi,Sin... |
In coordinate space, a particle starts at the point $(2,3,4)$ and ends at the point $(-1,-3,-3),$ along the line connecting the two points. Along the way, the particle intersects the unit sphere centered at the origin at two points. Then the distance between these two points can be expressed in the form $\frac{a}{\sq... |
The line can be parameterized by \[\begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix} + t \left( \begin{pmatrix} -1 \\ -3 \\ -3 \end{pmatrix} - \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix} \right) = \begin{pmatrix} 2 - 3t \\ 3 - 6t \\ 4 - 7t \end{pmatrix}.\]Then the particle intersects the sphere when \[(2 - 3t)^2 + (3 - 6t)^2 + (... |
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... |
Find the matrix that corresponds to rotating about the origin by an angle of $120^\circ$ counter-clockwise. |
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 \[\boxed{\... |
The matrices \[\begin{pmatrix} 3 & -8 \\ a & 11 \end{pmatrix} \quad \text{and} \quad \begin{pmatrix} 11 & b \\ 4 & 3 \end{pmatrix}\]are inverses. Enter the ordered pair $(a,b).$ |
The product of the matrices is \[\begin{pmatrix} 3 & -8 \\ a & 11 \end{pmatrix} \begin{pmatrix} 11 & b \\ 4 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 3b - 24 \\ 11a + 44 & ab + 33 \end{pmatrix}.\]We want this to be the identity matrix, so $3b - 24 = 0,$ $11a + 44 = 0,$ and $ab + 33 = 1.$ Solving, we find $(a,b) = \boxed... |
Find \[\sin \left( \sin^{-1} \frac{3}{5} + \tan^{-1} 2 \right).\] |
Let $a = \sin^{-1} \frac{3}{5}$ and $b = \tan^{-1} 2.$ Then $\sin a = \frac{3}{5}$ and $\tan b = 2.$ With the usual technique of constructing right triangles, we can find that $\cos a = \frac{4}{5},$ $\cos b = \frac{1}{\sqrt{5}},$ and $\sin b = \frac{2}{\sqrt{5}}.$ Therefore, from the angle addition formula, \begin{... |
Let $x$ and $y$ be real numbers such that \[\frac{\sin x}{\cos y} + \frac{\sin y}{\cos x} = 1 \quad \text{and} \quad \frac{\cos x}{\sin y} + \frac{\cos y}{\sin x} = 6.\]Compute \[\frac{\tan x}{\tan y} + \frac{\tan y}{\tan x}.\] |
Let us refer to the two given equations as equations (1) and (2), respectively. We can write them as \[\frac{\sin x \cos x + \sin y \cos y}{\cos y \cos x} = 1\]and \[\frac{\cos x \sin x + \cos y \sin y}{\sin y \sin x} = 6.\]Dividing these equations, we get $\frac{\sin x \sin y}{\cos x \cos y} = \frac{1}{6},$ so \[\tan... |
There exists a positive real number $x$ such that $ \cos (\arctan (x)) = x $. Find the value of $x^2$. |
Construct a right triangle with legs 1 and $x.$ Let the angle opposite the side length $x$ be $\theta.$ [asy] unitsize(1 cm); pair A, B, C; A = (2,1.8); B = (0,0); C = (2,0); draw(A--B--C--cycle); draw(rightanglemark(A,C,B,8)); label("$\theta$", B + (0.7,0.3)); label("$1$", (B + C)/2, S); label("$x$", (A + C)/2, ... |
The orthocenter of triangle $ABC$ divides altitude $\overline{CF}$ into segments with lengths $HF = 6$ and $HC = 15.$ Calculate $\tan A \tan B.$ [asy] unitsize (1 cm); pair A, B, C, D, E, F, H; A = (0,0); B = (5,0); C = (4,4); D = (A + reflect(B,C)*(A))/2; E = (B + reflect(C,A)*(B))/2; F = (C + reflect(A,B)*(C))/2;... |
Draw altitudes $\overline{BE}$ and $\overline{CF}.$ [asy] unitsize (1 cm); pair A, B, C, D, E, F, H; A = (0,0); B = (5,0); C = (4,4); D = (A + reflect(B,C)*(A))/2; E = (B + reflect(C,A)*(B))/2; F = (C + reflect(A,B)*(C))/2; H = extension(A,D,B,E); draw(A--B--C--cycle); draw(A--D); draw(B--E); draw(C--F); label("$A... |
A line is expressed in the form \[\begin{pmatrix} 1 \\ 3 \end{pmatrix} \cdot \left( \begin{pmatrix} x \\ y \end{pmatrix} - \begin{pmatrix} -2 \\ 8 \end{pmatrix} \right) = 0.\]The equation of the line can be expressed in the form $y = mx + b.$ Enter the ordered pair $(m,b).$ |
Expanding, we get \[\begin{pmatrix} 1 \\ 3 \end{pmatrix} \cdot \left( \begin{pmatrix} x \\ y \end{pmatrix} - \begin{pmatrix} -2 \\ 8 \end{pmatrix} \right) = \begin{pmatrix} 1 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} x + 2 \\ y - 8 \end{pmatrix} = (x + 2) + 3(y - 8) = 0.\]Solving for $y,$ we find \[y = -\frac{1}{3} x + ... |
The point $(1,1,1)$ is rotated $180^\circ$ about the $y$-axis, then reflected through the $yz$-plane, reflected through the $xz$-plane, rotated $180^\circ$ about the $y$-axis, and reflected through the $xz$-plane. Find the coordinates of the point now. |
After $(1,1,1)$ is rotated $180^\circ$ about the $y$-axis, it goes to $(-1,1,-1).$ After $(-1,1,-1)$ is reflected through the $yz$-plane, it goes to $(1,1,-1).$ After $(1,1,-1)$ is reflected through the $xz$-plane, it goes to $(1,-1,-1).$ After $(1,-1,-1)$ is rotated $180^\circ$ about the $y$-axis, it goes to $(-1,-... |
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