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human
|
The polynomial
$$g(x) = x^3 - x^2 - (m^2 + m) x + 2m^2 + 4m + 2$$is divisible by $x-4$ and all of its zeroes are integers. Find all possible values of $m$.
|
gpt
|
Since $g(x)$ is divisible by $x-4$, we have $g(4)=0$. We also have
\begin{align*}
g(4) &= 4^3 - 4^2 - (m^2+m)(4) + 2m^2+4m+2 \\
&= 50 - 2m^2,
\end{align*}so $0=50-2m^2$. Thus $m$ can only be $5$ or $-5$. We check both possibilities.
If $m=5$, then $g(x)=x^3-x^2-30x+72=(x-4)(x^2+3x-18)=(x-4)(x+6)(x-3)$, so all zeroes are integers.
If $m=-5$, then $g(x)=x^3-x^2-20x+32=(x-4)(x^2+3x-8)$, but $x^2+3x-8$ does not have integer zeroes.
Therefore, the only solution is $m=\boxed{5}$.
|
human
|
The function $f$ takes nonnegative integers to real numbers, such that $f(1) = 1,$ and
\[f(m + n) + f(m - n) = \frac{f(2m) + f(2n)}{2}\]for all nonnnegative integers $m \ge n.$ Find the sum of all possible values of $f(10).$
|
gpt
|
Setting $m = n = 0,$ we get
\[2f(0) = f(0),\]so $f(0) = 0.$
Setting $n = 0,$ we get
\[2f(m) = \frac{f(2m)}{2}.\]Thus, we can write the given functional equation as
\[f(m + n) + f(m - n) = 2f(m) + 2f(n).\]In particular, setting $n = 1,$ we get
\[f(m + 1) + f(m - 1) = 2 + 2f(m),\]so
\[f(m + 1) = 2f(m) - f(m - 1) + 2\]for all $m \ge 1.$
Then
\begin{align*}
f(2) &= 2f(1) - f(0) + 2 = 4, \\
f(3) &= 2f(2) - f(1) + 2 = 9, \\
f(4) &= 2f(3) - f(2) + 2 = 16,
\end{align*}and so on.
By a straight-forward induction argument,
\[f(m) = m^2\]for all nonnegative integers $m.$ Note that this function satisfies the given functional equation, so the sum of all possible values of $f(10)$ is $\boxed{100}.$
|
human
|
Real numbers $x,$ $y,$ and $z$ satisfy the following equality:
\[4(x + y + z) = x^2 + y^2 + z^2.\]Let $M$ be the maximum value of $xy + xz + yz,$ and let $m$ be the minimum value of $xy + xz + yz.$ Find $M + 10m.$
|
gpt
|
Let $A = x + y + z,$ $B = x^2 + y^2 + z^2,$ and $C = xy + xz + yz.$ We are told that
\[4A = B.\]Then
\[A^2 = (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + xz + yz) = B + 2C = 4A + 2C.\]Hence,
\[C = \frac{1}{2} (A - 2)^2 - 2.\]Also,
\[B - C = x^2 + y^2 + z^2 - (xy + xz + yz) = \frac{(x - y)^2 + (x - z)^2 + (y - z)^2}{2} \ge 0,\]so $C \le B.$ Then $A^2 = B + 2C \le 3B = 12A.$ Hence, $0 \le A \le 12,$ so $-2 \le C \le 48.$
We see that $C = -2$ when $(x,y,z) = (2,-\sqrt{2},\sqrt{2}),$ and $C = 48$ when $(x,y,z) = (4,4,4),$ so $M = 48$ and $m = -2,$ and $M + 10m = \boxed{28}.$
|
human
|
Find the range of the function \[f(x) = \frac{x}{x^2-x+1},\]where $x$ can be any real number. (Give your answer in interval notation.)
|
gpt
|
Let $y$ be a number in the range of $f.$ This means that there is a real number $x$ such that \[y = \frac{x}{x^2-x+1}.\]Multiplying both sides by $x^2-x+1$ and rearranging, we get the equation \[yx^2-(y+1)x+y=0.\]Since $x^2-x+1 = (x-\tfrac12)^2 + \tfrac34 > 0$ for all $x,$ our steps are reversible, so $y$ is in the range of $f$ if and only if this equation has a real solution for $x.$ In turn, this equation has a real solution for $x$ if and only if the discriminant of this quadratic is nonnegative. Therefore, the range of $f$ consists exactly of the values of $y$ which satisfy \[(y+1)^2 - 4y^2 \ge 0,\]or \[0 \ge 3y^2 - 2y - 1.\]This quadratic factors as \[0 \ge (3y+1)(y-1),\]which means that the solutions to the inequality are given by $-\tfrac13 \le y \le 1.$ Therefore, the range of $g$ is the closed interval $\boxed{[-\tfrac13, 1]}.$
|
human
|
Let $a$, $b$, $c$, $d$, and $e$ be positive integers with $a+b+c+d+e=2010$ and let $M$ be the largest of the sum $a+b$, $b+c$, $c+d$ and $d+e$. What is the smallest possible value of $M$?
|
gpt
|
We have that
\[M = \max \{a + b, b + c, c + d, d + e\}.\]In particular, $a + b \le M,$ $b + c \le M,$ and $d + e \le M.$ Since $b$ is a positive integer, $c < M.$ Hence,
\[(a + b) + c + (d + e) < 3M.\]Then $2010 < 3M,$ so $M > 670.$ Since $M$ is an integer, $M \ge 671.$
Equality occurs if $a = 669,$ $b = 1,$ $c = 670,$ $d = 1,$ and $e = 669,$ so the smallest possible value of $M$ is $\boxed{671}.$
|
human
|
Find the ordered pair $(a,b)$ of positive integers, with $a < b,$ for which
\[\sqrt{1 + \sqrt{21 + 12 \sqrt{3}}} = \sqrt{a} + \sqrt{b}.\]
|
gpt
|
First, we simplify $\sqrt{21 + 12 \sqrt{3}}.$ Let
\[\sqrt{21 + 12 \sqrt{3}} = x + y.\]Squaring both sides, we get
\[21 + 12 \sqrt{3} = x^2 + 2xy + y^2.\]To make the right-hand side look like the left-hand side, we set $x^2 + y^2 = 21$ and $2xy = 12 \sqrt{3},$ so $xy = 6 \sqrt{3}.$ Then $x^2 y^2 = 108,$ so by Vieta's formulas, $x^2$ and $y^2$ are the roots of the quadratic
\[t^2 - 21t + 108 = 0.\]This factors as $(t - 9)(t - 12) = 0,$ whose solutions are 9 and 12. Therefore,
\[\sqrt{21 + 12 \sqrt{3}} = \sqrt{9} + \sqrt{12} = 3 + 2 \sqrt{3}.\]Now we must simplify
\[\sqrt{1 + 3 + 2 \sqrt{3}} = \sqrt{4 + 2 \sqrt{3}}.\]Performing the same technique gives us
\[\sqrt{4 + 2 \sqrt{3}} = 1 + \sqrt{3},\]so $(a,b) = \boxed{(1,3)}.$
|
human
|
If $x$ is real, compute the maximum integer value of
\[\frac{3x^2 + 9x + 17}{3x^2 + 9x + 7}.\]
|
gpt
|
First, we can write
\[\frac{3x^2 + 9x + 17}{3x^2 + 9x + 7} = \frac{(3x^2 + 9x + 7) + 10}{3x^2 + 9x + 7} = 1 + \frac{10}{3x^2 + 9x + 7}.\]Thus, we want to minimize $3x^2 + 9x + 7.$
Completing the square, we get
\[3x^2 + 9x + 7 = 3 \left( x + \frac{3}{2} \right)^2 + \frac{1}{4},\]so the minimum value of $3x^2 + 9x + 7$ is $\frac{1}{4}.$
Therefore, the maximum integer value of
\[1 + \frac{10}{3x^2 + 9x + 7}\]is $1 + \frac{10}{1/4} = \boxed{41}.$
|
human
|
There are two values of $a$ for which the equation $4x^2 + ax + 8x + 9 = 0$ has only one solution for $x$. What is the sum of these values of $a$?
|
gpt
|
We can write the quadratic as
\[4x^2 + (a + 8)x + 9 = 0.\]If the quadratic has one solution, then its discriminant must be zero:
\[(a + 8)^2 - 4 \cdot 4 \cdot 9 = 0.\]Expanding, we get $a^2 + 16a - 80 = 0.$ By Vieta's formulas, the sum of the roots is $\boxed{-16}.$
|
human
|
Calculate $\frac{1}{4} \cdot \frac{2}{5} \cdot \frac{3}{6} \cdot \frac{4}{7} \cdots \frac{49}{52} \cdot \frac{50}{53}$. Express your answer as a common fraction.
|
gpt
|
Note that from $\frac{4}{7}$ to $\frac{50}{53},$ the numerator of each fraction cancels with the denominator of the fraction three terms before it. Thus, the product simplifies to \[\frac{1 \cdot 2 \cdot 3}{51\cdot 52\cdot 53 }= \boxed{\frac{1}{23426}}.\]
|
human
|
Determine the largest positive integer $n$ such that there exist positive integers $x, y, z$ so that \[
n^2 = x^2+y^2+z^2+2xy+2yz+2zx+3x+3y+3z-6
\]
|
gpt
|
The given equation rewrites as $n^2 = (x+y+z+1)^2+(x+y+z+1)-8$. Writing $r = x+y+z+1$, we have $n^2 = r^2+r-8$. Clearly, one possibility is $n=r=\boxed{8}$, which is realized by $x=y=1, z=6$. On the other hand, for $r > 8$, we have $r^2 < r^2+r-8 < (r+1)^2.$
|
human
|
The function $f$ satisfies \[
f(x) + f(2x+y) + 5xy = f(3x - y) + 2x^2 + 1
\]for all real numbers $x,y$. Determine the value of $f(10)$.
|
gpt
|
Setting $x = 10$ and $y=5$ gives $f(10) + f(25) + 250 = f(25) + 200 + 1$, from which we get $f(10) = \boxed{-49}$.
$\text{Remark:}$ By setting $y = \frac x 2$, we see that the function is $f(x) = -\frac 1 2 x^2 + 1$, and it can be checked that this function indeed satisfies the given equation.
|
human
|
If $a$, $b$, $c$, $d$, $e$, and $f$ are integers for which $1000x^3+27= (ax^2 + bx +c )(d x^2 +ex + f)$ for all $x$, then what is $a^2+b^2+c^2+d^2+e^2+f^2$?
|
gpt
|
Apply the sum of cubes factorization to the expression $1000x^3+27 = (10x)^3+3^3$ to obtain \[
1000x^3+27 = (10x+3)(100x^2-30x+9).
\]Thus $a^2+b^2+c^2+d^2+e^2+f^2=0^2+10^2+3^2+100^2+(-30)^2+9^2=\boxed{11,\!090}$. Note that the fundamental theorem of algebra implies that the factorization we have given is unique, since the discriminant $(-30)^2-4(100)(9)$ of the quadratic $100x^2-30x+9$ is negative.
|
human
|
For positive integers $n$, define $S_n$ to be the minimum value of the sum
\[\sum_{k=1}^n \sqrt{(2k-1)^2+a_k^2},\]where $a_1,a_2,\ldots,a_n$ are positive real numbers whose sum is $17$. Find the unique positive integer $n$ for which $S_n$ is also an integer.
|
gpt
|
For $k = 0, 1, 2, \ldots, n,$ let $P_k = (k^2,a_1 + a_2 + \dots + a_k).$ Note that $P_0 = (0,0)$ and $P_n = (n^2,a_1 + a_2 + \dots + a_n) = (n^2,17).$
[asy]
unitsize(0.4 cm);
pair[] A, P;
P[0] = (0,0);
A[0] = (5,0);
P[1] = (5,1);
A[1] = (9,1);
P[2] = (9,3);
P[3] = (12,6);
A[3] = (15,6);
P[4] = (15,10);
draw(P[0]--A[0]--P[1]--cycle);
draw(P[1]--A[1]--P[2]--cycle);
draw(P[3]--A[3]--P[4]--cycle);
draw(P[0]--P[4],dashed);
label("$P_0$", P[0], W);
label("$P_1$", P[1], N);
label("$P_2$", P[2], N);
label("$P_{n - 1}$", P[3], W);
label("$P_n$", P[4], NE);
label("$a_1$", (A[0] + P[1])/2, E);
label("$a_2$", (A[1] + P[2])/2, E);
label("$a_n$", (A[3] + P[4])/2, E);
dot((21/2 - 0.5,9/2 - 0.5));
dot((21/2,9/2));
dot((21/2 + 0.5,9/2 + 0.5));
[/asy]
Then for each $k = 1, 2, \ldots, n,$ we have \[\begin{aligned} P_{k-1}P_k &= \sqrt{(k^2-(k-1)^2)+((a_1+a_2+\dots+a_{k-1}+a_{k})-(a_1+a_2+\dots+a_{k-1}))^2} \\ &= \sqrt{(2k-1)^2+a_k^2}, \end{aligned}\]so that $S_n$ is the minimum value of the sum $P_0P_1 + P_1P_2 + \dots + P_{n-1}P_n.$ By the triangle inequality, \[P_0P_1 + P_1P_2 + \dots + P_{n-1}P_n \ge P_0P_n = \sqrt{n^4 + 289}.\]Furthemore, equality occurs when all the $P_i$ are collinear, so $S_n = \sqrt{n^4+289}$ for each $n.$
It remains to find the $n$ for which $S_n$ is an integer, or equivalently, $n^4+289$ is a perfect square. Let $n^4+289=m^2$ for some positive integer $m.$ Then $m^2-n^4=289,$ which factors as \[(m-n^2)(m+n^2) = 289.\]Since $n^2$ is positive and $289 = 17^2,$ the only possibility is $m-n^2=1$ and $m+n^2=289,$ giving $m = 145$ and $n^2 = 144.$ Thus $n = \sqrt{144} = \boxed{12}.$
|
human
|
Let $x,$ $y,$ $z$ be real numbers, all greater than 3, so that
\[\frac{(x + 2)^2}{y + z - 2} + \frac{(y + 4)^2}{z + x - 4} + \frac{(z + 6)^2}{x + y - 6} = 36.\]Enter the ordered triple $(x,y,z).$
|
gpt
|
By Cauchy-Schwarz,
\[(y + z - 2) + (z + x - 4) + (x + y - 6)] \left[ \frac{(x + 2)^2}{y + z - 2} + \frac{(y + 4)^2}{z + x - 4} + \frac{(z + 6)^2}{x + y - 6} \right] \ge [(x + 2) + (y + 4) + (z + 6)]^2.\]This simplifies to
\[36(2x + 2y + 2z - 12) \ge (x + y + z + 12)^2.\]Let $s = x + y + z.$ Then $36(2s - 12) \ge (s + 12)^2.$ This simplifies to $s^2 - 48s + 576 \le 0,$ which then factors as $(s - 24)^2 \le 0.$ Hence, $s = 24.$
Thus, the inequality above turns into an equality, which means
\[\frac{x + 2}{y + z - 2} = \frac{y + 4}{z + x - 4} = \frac{z + 6}{x + y - 6}.\]Since $x + y + z = 24,$
\[\frac{x + 2}{22 - x} = \frac{y + 4}{20 - y} = \frac{z + 6}{18 - z}.\]Each fraction must then be equal to
\[\frac{(x + 2) + (y + 4) + (z + 6)}{(22 - x) + (20 - y) + (18 - z)} = \frac{x + y + z + 12}{60 - (x + y + z)} = 1.\]From here, it is easy to solve for $x,$ $y,$ and $z,$ to find $x = 10,$ $y = 8,$ and $z = 6.$
Hence, $(x,y,z) = \boxed{(10,8,6)}.$
|
human
|
The set of points $(x,y)$ such that $|x - 3| \le y \le 4 - |x - 1|$ defines a region in the $xy$-plane. Compute the area of this region.
|
gpt
|
Plotting $y = |x - 3|$ and $y = 4 - |x - 1|,$ we find that the two graphs intersect at $(0,3)$ and $(4,1).$
[asy]
unitsize(1 cm);
real funcone (real x) {
return(abs(x - 3));
}
real functwo (real x) {
return(4 - abs(x - 1));
}
fill((3,0)--(4,1)--(1,4)--(0,3)--cycle,gray(0.7));
draw(graph(funcone,-0.5,4.5));
draw(graph(functwo,-0.5,4.5));
draw((-0.5,0)--(4.5,0));
draw((0,-0.5)--(0,4.5));
label("$y = |x - 3|$", (3.5,3));
label("$y = 4 - |x - 1|$", (0,1), UnFill);
dot("$(0,3)$", (0,3), W);
dot("$(4,1)$", (4,1), E);
dot("$(3,0)$", (3,0), S);
dot("$(1,4)$", (1,4), N);
[/asy]
The region then is a rectangle with side lengths $\sqrt{2}$ and $3 \sqrt{2},$ so its area is $(\sqrt{2})(3 \sqrt{2}) = \boxed{6}.$
|
human
|
The planet Xavier follows an elliptical orbit with its sun at one focus. At its nearest point (perigee), it is 2 astronomical units (AU) from the sun, while at its furthest point (apogee) it is 12 AU away. When Xavier is midway along its orbit, as shown, how far is it from the sun, in AU?
[asy]
unitsize(1 cm);
path ell = xscale(2)*arc((0,0),1,-85,265);
filldraw(Circle((0,-1),0.1));
filldraw(Circle((-1.4,0),0.2),yellow);
draw(ell,Arrow(6));
[/asy]
|
gpt
|
Let $A$ be the perigee, let $B$ be the apogee, let $F$ be the focus where the sun is, let $O$ be the center of the ellipse, and let $M$ be the current position of Xavier.
[asy]
unitsize(1 cm);
pair A, B, F, M, O;
path ell = xscale(2)*Circle((0,0),1);
A = (-2,0);
B = (2,0);
F = (-sqrt(3),0);
O = (0,0);
M = (0,-1);
draw(ell);
draw(A--M);
draw(O--M);
draw(F--M);
draw(A--B);
dot("$A$", A, W);
dot("$B$", B, E);
dot("$F$", F, N);
dot("$M$", M, S);
dot("$O$", O, N);
[/asy]
Then $AB$ is a major axis of the ellipse, and $AB = 2 + 12 = 14.$ Since $M$ is the midway point, $MF = AO = \frac{14}{2} = \boxed{7}.$
|
human
|
For some integer $m$, the polynomial $x^3 - 2011x + m$ has the three integer roots $a$, $b$, and $c$. Find $|a| + |b| + |c|.$
|
gpt
|
By Vieta's formulas, \[\left\{ \begin{aligned} a + b + c &= 0 \\ ab+bc+ac&=-2011. \end{aligned} \right.\]Since $a+b=-c,$ the second equation becomes $ab+(-c)c = -2011$, or \[c^2 - ab= 2011.\]At least two of $a, b, c$ must have the same sign; without loss of generality, let $a$ and $b$ have the same sign. Furthermore, since we can negate all of $a, b, c$ and still satisfy the two above equations, assume that $c \ge 0.$ (Note that we only want the sum $|a| + |b| + |c|$, which does not change if we swap or negate the variables.)
Now, we have $ab \ge 0,$ so $c^2 \ge 2011$, giving $c \ge 44.$ We also have \[\frac{c^2}{4} = \left(\frac{a+b}{2}\right)^2 \ge ab\]by AM-GM, so $2011 = c^2 - ab \ge 3c^2/4,$ giving $c \le 51.$
Finally, we have $(a-b)^2 = (a+b)^2 - 4ab = (-c)^2 - 4(c^2-2011) = 8044 - 3c^2$, which must be a perfect square.
Testing $c = 44, 45, \ldots, 51$, we find that $8044 - 3c^2$ is a perfect square only when $c = 49$. Therefore, $c = 49$, and so \[\left\{ \begin{aligned} a+b&= -c = -49, \\ ab &= c^2 - 2011 = 390. \end{aligned} \right.\]Thus, $a$ and $b$ are the roots of $t^2 + 49t + 390 = 0$, which factors as $(t+10)(t+39) = 0$. Thus, $\{a, b\} = \{-10, -39\}$.
The answer is \[|a| + |b| + |c| = 39 + 10 + 49 = \boxed{98}.\]
|
human
|
Let $x,$ $y,$ $z$ be real numbers such that
\begin{align*}
x + y + z &= 4, \\
x^2 + y^2 + z^2 &= 6.
\end{align*}Let $m$ and $M$ be the smallest and largest possible values of $x,$ respectively. Find $m + M.$
|
gpt
|
From the given equations, $y + z = 4 - x$ and $y^2 + z^2 = 6 - x^2.$ By Cauchy-Schwarz,
\[(1 + 1)(y^2 + z^2) \ge (y + z)^2.\]Hence, $2(6 - x^2) \ge (4 - x)^2.$ This simplifies to $3x^2 - 8x + 4 \le 0,$ which factors as $(x - 2)(3x - 2) \le 0.$ Hence, $\frac{2}{3} \le x \le 2.$
For $x = \frac{3}{2},$ we can take $y = z = \frac{5}{3}.$ For $x = 2,$ we can take $y = z = 1.$ Thus, $m = \frac{2}{3}$ and $M = 2,$ so $m + M = \boxed{\frac{8}{3}}.$
|
human
|
Find the distance between the foci of the ellipse
\[\frac{x^2}{20} + \frac{y^2}{4} = 7.\]
|
gpt
|
First, we divide both sides by 7, to get
\[\frac{x^2}{140} + \frac{y^2}{28} = 1.\]Thus, $a^2 = 140$ and $b^2 = 28,$ so $c^2 = a^2 - b^2 = 140 - 28 = 112.$ Thus, $c = \sqrt{112} = 4 \sqrt{7},$ so the distance between the foci is $2c = \boxed{8 \sqrt{7}}.$
|
human
|
One focus of the ellipse $\frac{x^2}{2} + y^2 = 1$ is at $F = (1,0).$ There exists a point $P = (p,0),$ where $p > 0,$ such that for any chord $\overline{AB}$ that passes through $F,$ angles $\angle APF$ and $\angle BPF$ are equal. Find $p.$
[asy]
unitsize(2 cm);
pair A, B, F, P;
path ell = xscale(sqrt(2))*Circle((0,0),1);
F = (1,0);
A = (sqrt(2)*Cos(80),Sin(80));
B = intersectionpoint(interp(A,F,0.1)--interp(A,F,5),ell);
P = (2,0);
draw(ell);
draw(A--B);
draw(A--P--B);
draw(F--P);
dot("$A$", A, N);
dot("$B$", B, SE);
dot("$F$", F, SW);
dot("$P$", P, E);
[/asy]
|
gpt
|
First, we consider a particular line, $y = x - 1,$ which passes through $F.$ Substituting, we get
\[\frac{x^2}{2} + (x - 1)^2 = 1.\]This simplifies to $3x^2 - 4x = x(3x - 4) = 0,$ so $x = 0$ or $x = \frac{4}{3}.$ Thus, we can let $A = \left( \frac{4}{3}, \frac{1}{3} \right)$ and $B = (0,-1).$
The slope of line $AP$ is then $\frac{1/3}{4/3 - p} = \frac{1}{4 - 3p},$ and the slope of line $BP$ is $\frac{-1}{-p} = \frac{1}{p}.$ Since $\angle APF = \angle BPF,$ these slopes are negatives of each other, so
\[\frac{1}{3p - 4} = \frac{1}{p}.\]Then $p = 3p - 4,$ so $p = \boxed{2}.$
For a complete solution, we prove that this works for all chords $\overline{AB}$ that pass through $F.$ Let $A = (x_a,y_a)$ and $B = (x_b,y_b).$ Then the condition $\angle APF = \angle BPF$ is equivalent to
\[\frac{y_a}{x_a - 2} + \frac{y_b}{x_b - 2} = 0,\]or $y_a (x_b - 2) + y_b (x_a - 2) = 0.$ Then $y_a x_b - 2y_a + y_b x_a - 2y_b = 0.$
Let $y = m(x - 1)$ be the equation of line $AB.$ Substituting, we get
\[\frac{x^2}{2} + m^2 (x - 1)^2 = 1.\]This simplifies to $(2m^2 + 1) x^2 - 4m^2 x + 2m^2 - 2 = 0.$ By Vieta's formulas,
\[x_a + x_b = \frac{4m^2}{2m^2 + 1} \quad \text{and} \quad x_a x_b = \frac{2m^2 - 2}{2m^2 + 1}.\]Then
\begin{align*}
y_a x_b - 2y_a + y_b x_a - 2y_b &= m(x_a - 1) x_b - 2m(x_a - 1) + m(x_b - 1) x_a - 2m(x_b - 1) \\
&= 2mx_a x_b - 3m (x_a + x_b) + 4m \\
&= 2m \cdot \frac{2m^2 - 2}{2m^2 + 1} - 3m \cdot \frac{4m^2}{2m^2 + 1} + 4m \\
&= 0.
\end{align*}Thus, $\angle APF = \angle BPF$ for all chords $\overline{AB}$ that pass through $F.$
|
human
|
The sequence $(a_n)$ is defined recursively by $a_0=1$, $a_1=\sqrt[19]{2}$, and $a_n=a_{n-1}a_{n-2}^2$ for $n\geq 2$. What is the smallest positive integer $k$ such that the product $a_1a_2\cdots a_k$ is an integer?
|
gpt
|
Let $b_n = 19 \log_2 a_n.$ Then $a_n = 2^{\frac{b_n}{19}},$ so
\[2^{\frac{b_n}{19}} = 2^{\frac{b_{n - 1}}{19}} \cdot 2^{\frac{2b_{n - 2}}{19}} = 2^{\frac{b_{n - 1} + 2b_{n - 2}}{19}},\]which implies
\[b_n = b_{n - 1} + 2b_{n - 2}.\]Also, $b_0 = 0$ and $b_1 = 1.$
We want
\[a_1 a_2 \dotsm a_k = 2^{\frac{b_1 + b_2 + \dots + b_k}{19}}\]to be an integer. In other words, we want $b_1 + b_2 + \dots + b_k$ to be a multiple of 19.
So, let $s_k = b_1 + b_2 + \dots + b_k.$ Using the recurrence $b_n = b_{n - 1} + 2b_{n - 2},$ we can compute the first few terms of $(b_n)$ and $(s_n)$ modulo 19:
\[
\begin{array}{c|c|c}
n & b_n & s_n \\ \hline
1 & 1 & 1 \\
2 & 1 & 2 \\
3 & 3 & 5 \\
4 & 5 & 10 \\
5 & 11 & 2 \\
6 & 2 & 4 \\
7 & 5 & 9 \\
8 & 9 & 18 \\
9 & 0 & 18 \\
10 & 18 & 17 \\
11 & 18 & 16 \\
12 & 16 & 13 \\
13 & 14 & 8 \\
14 & 8 & 16 \\
15 & 17 & 14 \\
16 & 14 & 9 \\
17 & 10 & 0
\end{array}
\]Thus, the smallest such $k$ is $\boxed{17}.$
Alternatively, we can solve the recursion $b_0 = 0,$ $b_1 = 1,$ $b_n = b_{n - 1} + 2b_{n - 2}$ to get
\[b_n = \frac{2^n - (-1)^n}{3}.\]
|
human
|
Let $x,$ $y,$ $z$ be real numbers such that $4x^2 + y^2 + 16z^2 = 1.$ Find the maximum value of
\[7x + 2y + 8z.\]
|
gpt
|
By Cauchy-Schwarz
\[\left( \frac{49}{4} + 4 + 4 \right) (4x^2 + y^2 + 16z^2) \ge (7x + 2y + 8z)^2.\]Since $4x^2 + y^2 + 16z^2 = 1,$
\[(7x + 2y + 8z)^2 \le \frac{81}{4}.\]Hence, $7x + 2y + 8z \le \frac{9}{2}.$
For equality to occur, we must have $\frac{2x}{7/2} = \frac{y}{2} = \frac{4z}{2}$ and $4x^2 + y^2 + 16z^2 = 1.$ We can solve, to find $x = \frac{7}{18},$ $y = \frac{4}{9},$ and $z = \frac{1}{9},$ so the maximum value of $7x + 2y + 8z$ is $\boxed{\frac{9}{2}}.$
|
human
|
For what values of $x$ is $\frac{\log{(3-x)}}{\sqrt{x-1}}$ defined?
|
gpt
|
The expression inside the square root must be greater than 0 because the denominator cannot be equal to 0. Therefore, $x-1>0$, so $x>1$. The expression inside the logarithm must be greater than 0, so $3-x>0$, which gives $x<3$. Therefore, the interval of $x$ for which the expression $\frac{\log{(3-x)}}{\sqrt{x-1}}$ is defined is $1<x<3$, which is $\boxed{(1,3)}$.
|
human
|
Let $a,$ $b,$ $c$ be positive real numbers. Find the smallest possible value of
\[6a^3 + 9b^3 + 32c^3 + \frac{1}{4abc}.\]
|
gpt
|
By AM-GM,
\[6a^3 + 9b^3 + 32c^3 \ge 3 \sqrt[3]{6a^3 \cdot 9b^3 \cdot 32c^3} = 36abc.\]Again by AM-GM,
\[36abc + \frac{1}{4abc} \ge 2 \sqrt{36abc \cdot \frac{1}{4abc}} = 6.\]Equality occurs when $6a^3 = 9b^3 = 32c^3$ and $36abc = 3.$ We can solve, to get $a = \frac{1}{\sqrt[3]{6}},$ $b = \frac{1}{\sqrt[3]{9}},$ and $c = \frac{1}{\sqrt[3]{32}}.$ Therefore, the minimum value is $\boxed{6}.$
|
human
|
Let $a > 0$, and let $P(x)$ be a polynomial with integer coefficients such that
\[P(1) = P(3) = P(5) = P(7) = a\]and
\[P(2) = P(4) = P(6) = P(8) = -a.\]What is the smallest possible value of $a$?
|
gpt
|
There must be some polynomial $Q(x)$ such that $$P(x)-a=(x-1)(x-3)(x-5)(x-7)Q(x).$$Then, plugging in values of $2,4,6,8,$ we get
$$P(2)-a=(2-1)(2-3)(2-5)(2-7)Q(2) = -15Q(2) = -2a,$$$$P(4)-a=(4-1)(4-3)(4-5)(4-7)Q(4) = 9Q(4) = -2a,$$$$P(6)-a=(6-1)(6-3)(6-5)(6-7)Q(6) = -15Q(6) = -2a,$$$$P(8)-a=(8-1)(8-3)(8-5)(8-7)Q(8) = 105Q(8) = -2a.$$That is,
$$-2a=-15Q(2)=9Q(4)=-15Q(6)=105Q(8).$$Thus, $a$ must be a multiple of $\text{lcm}(15,9,15,105)=315$.
Now we show that there exists $Q(x)$ such that $a=315.$ Inputting this value into the above equation gives us
$$Q(2)=42, \quad Q(4)=-70, \quad Q(6)=42, \quad Q(8)=-6.$$From $Q(2) = Q(6) = 42,$ $Q(x)=R(x)(x-2)(x-6)+42$ for some $R(x).$ We can take $R(x) = -8x + 60,$ so that $Q(x)$ satisfies both $Q(4) = -70$ and $Q(8) = -6.$
Therefore, our answer is $ \boxed{ 315}. $
|
human
|
Let $a,$ $b,$ $c$ be complex numbers such that
\begin{align*}
ab + 4b &= -16, \\
bc + 4c &= -16, \\
ca + 4a &= -16.
\end{align*}Enter all possible values of $abc,$ separated by commas.
|
gpt
|
Adding the equations, we get
\[ab + ac + bc + 4(a + b + c) = -48.\]Multiplying the equations by $c,$ $a,$ $b,$ respectively, we get
\begin{align*}
abc + 4bc &= -16c, \\
abc + 4ac &= -16a, \\
abc + 4ab &= -16b.
\end{align*}Adding all these equations, we get
\[3abc + 4(ab + ac + bc) = -16(a + b + c).\]Then
\begin{align*}
3abc &= -4(ab + ac + bc) - 16(a + b +c) \\
&= -4(ab + ac + bc + 4(a + b + c)) \\
&= (-4)(-48) = 192,
\end{align*}so $abc = \boxed{64}.$
|
human
|
The polynomial equation \[x^3 + bx + c = 0,\]where $b$ and $c$ are rational numbers, has $5-\sqrt{2}$ as a root. It also has an integer root. What is it?
|
gpt
|
Because the coefficients of the polynomial are rational, the radical conjugate of $5-\sqrt{2},$ which is $5+\sqrt{2},$ must also be a root of the polynomial. By Vieta's formulas, the sum of the roots of this polynomial is $0$; since $(5-\sqrt2) + (5+\sqrt2) = 10,$ the third, integer root must be $0 - 10 = \boxed{-10}.$
|
human
|
The function $y=\frac{x^3+8x^2+21x+18}{x+2}$ can be simplified into the function $y=Ax^2+Bx+C$, defined everywhere except at $x=D$. What is the sum of the values of $A$, $B$, $C$, and $D$?
|
gpt
|
The fact that the function can be simplified to a quadratic means we can probably divide $(x+2)$ out of the numerator after factoring the numerator into $(x+2)$ and the quadratic $Ax^2+Bx+C$. Using long division or synthetic division, we find that the numerator breaks down into $(x+2)$ and $(x^2+6x+9)$.
Now, we have
\[y=\frac{(x+2)(x^2+6x+9)}{x+2}.\]After we divide out the $x+2$, we're left with $x^2+6x+9$, so $A=1$, $B=6$, and $C=9$.
The domain of the quadratic function is all real numbers, but our original function was undefined when the denominator $x+2$ equaled 0. After dividing out the $x+2$ we still have to take into account that the function is undefined at $x+2=0$. So, the function is not defined at $x=-2$, giving us our value for $D$.
Therefore, $A+B+C+D=1+6+9+(-2)=\boxed{14}$.
|
human
|
The function $f(x)$ satisfies
\[f(x - y) = f(x) f(y)\]for all real numbers $x$ and $y,$ and $f(x) \neq 0$ for all real numbers $x.$ Find $f(3).$
|
gpt
|
Setting $x = 3$ and $y = \frac{3}{2},$ we get
\[f \left( \frac{3}{2} \right) = f(3) f \left( \frac{3}{2} \right).\]Since $f \left( \frac{3}{2} \right) \neq 0,$ we can divide both sides by $f \left( \frac{3}{2} \right),$ to get $f(3) = \boxed{1}.$
|
human
|
Find all solutions to the equation\[ \sqrt[4]{x} = \frac{12}{7 - \sqrt[4]{x}}.\]
|
gpt
|
Let $y = \sqrt[4]{x}.$ Then we have $y = \frac{12}{7-y},$ or $y(7-y) = 12.$ Rearranging and factoring, we get \[(y-3)(y-4) = 0.\]Therefore, $y = 3$ or $y = 4.$ Since $x = y^4,$ we have $x = 3^4 = 81$ or $x = 4^4 = 256,$ so the values for $x$ are $x = \boxed{81, 256}.$
|
human
|
Find the number of ordered quadruples $(a,b,c,d)$ of nonnegative real numbers such that
\begin{align*}
a^2 + b^2 + c^2 + d^2 &= 4, \\
(a + b + c + d)(a^3 + b^3 + c^3 + d^3) &= 16.
\end{align*}
|
gpt
|
Note that
\[(a^2 + b^2 + c^2 + d^2)^2 = 16 = (a + b + c + d)(a^3 + b^3 + c^3 + d^3),\]which gives us the equality case in the Cauchy-Schwarz Inequality. Hence,
\[(a + b + c + d)(a^3 + b^3 + c^3 + d^3) - (a^2 + b^2 + c^2 + d^2)^2 = 0.\]This expands as
\begin{align*}
&a^3 b - 2a^2 b^2 + ab^3 + a^3 c - 2a^2 c^2 + ac^3 + a^3 d - 2a^2 d^2 + ad^2 \\
&\quad + b^3 c - 2b^2 c^2 + bc^3 + b^3 d - 2b^2 d^2 + bd^3 + c^3 d - 2c^2 d^2 + cd^3 = 0.
\end{align*}We can write this as
\[ab(a - b)^2 + ac(a - c)^2 + ad(a - d)^2 + bc(b - c)^2 + bd(b - d)^2 + cd(c - d)^2 = 0.\]Since $a,$ $b,$ $c,$ $d$ are all nonnegative, each term must be equal to 0. This means for any two variables among $a,$ $b,$ $c,$ $d,$ either one of them is 0, or they are equal. (For example, either $b = 0,$ $d = 0,$ or $b = d.$) In turn, this means that among $a,$ $b,$ $c,$ $d,$ all the positive values must be equal.
Each variable $a,$ $b,$ $c,$ $d$ can be 0 or positive, leading to $2^4 = 16$ possible combinations. However, since $a^2 + b^2 + c^2 + d^2 = 4,$ not all of them can be equal to 0, leaving $16 - 1 = 15$ possible combinations.
For any of the 15 combinations, the quadruple $(a,b,c,d)$ is uniquely determined. For example, suppose we set $a = 0,$ and $b,$ $c,$ $d$ to be positive. Then $b = c = d,$ and $b^2 + c^2 + d^2 = 4,$ so $b = c = d = \frac{2}{\sqrt{3}}.$
Hence, there are $\boxed{15}$ possible quadruples $(a,b,c,d).$
|
human
|
Let $x,$ $y,$ $z$ be positive real numbers such that $x + y + z = 1.$ Find the minimum value of
\[\frac{1}{x + y} + \frac{1}{x + z} + \frac{1}{y + z}.\]
|
gpt
|
By Cauchy-Schwarz,
\[[(x + y) + (x + z) + (y + z)] \left( \frac{1}{x + y} + \frac{1}{x + z} + \frac{1}{y + z} \right) \ge (1 + 1 + 1)^2 = 9,\]so
\[\frac{1}{x + y} + \frac{1}{x + z} + \frac{1}{y + z} \ge \frac{9}{2(x + y + z)} = \frac{9}{2}.\]Equality occurs when $x = y = z = \frac{1}{3},$ so the minimum value is $\boxed{\frac{9}{2}}.$
|
human
|
An ellipse has foci at $F_1 = (0,2)$ and $F_2 = (3,0).$ The ellipse intersects the $x$-axis at the origin, and one other point. What is the other point of intersection?
|
gpt
|
The distance between the origin and $F_1$ is 2, and the distance between the origin and $F_2$ is 3, so every point $P$ on the ellipse satisfies
\[PF_1 + PF_2 = 5.\]So, if $(x,0)$ is an intercept of the ellipse, then
\[\sqrt{x^2 + 4} + \sqrt{(x - 3)^2} = 5.\]We can write this as
\[\sqrt{x^2 + 4} + |x - 3| = 5.\]If $x \le 3,$ then
\[\sqrt{x^2 + 4} + (3 - x) = 5,\]so $\sqrt{x^2 + 4} = x + 2.$ Squaring both sides, we get
\[x^2 + 4 = x^2 + 4x + 4,\]which leads to $x = 0.$ This solution corresponds to the origin.
If $x \ge 3,$ then
\[\sqrt{x^2 + 4} + (x - 3) = 5,\]so $\sqrt{x^2 + 4} = 8 - x.$ Squaring both sides, we get
\[x^2 + 4 = 64 - 16x + x^2,\]which leads to $x = \frac{15}{4}.$ Thus, the other $x$-intercept is $\boxed{\left( \frac{15}{4}, 0 \right)}.$
|
human
|
If
\[\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 3 \quad \text{and} \quad \frac{a}{x} + \frac{b}{y} + \frac{c}{z} = 0,\]find $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}.$
|
gpt
|
Let $p = \frac{x}{a},$ $q = \frac{y}{b},$ $r = \frac{z}{c}.$ Then $p + q + r = 3$ and $\frac{1}{p} + \frac{1}{q} + \frac{1}{r} = 0,$ so $pq + pr + qr = 0.$
We want $p^2 + q^2 + r^2.$ Squaring the equation $p + q + r = 3,$ we get
\[p^2 + q^2 + r^2 + 2(pq + pr + qr) = 9,\]so $p^2 + q^2 + r^2 = \boxed{9}.$
|
human
|
The partial fraction decomposition of
\[\frac{x^2 - 19}{x^3 - 2x^2 - 5x + 6}\]is
\[\frac{A}{x - 1} + \frac{B}{x + 2} + \frac{C}{x - 3}.\]Find the product $ABC.$
|
gpt
|
We have that
\[\frac{x^2 - 19}{x^3 - 2x^2 - 5x + 6} = \frac{A}{x - 1} + \frac{B}{x + 2} + \frac{C}{x - 3}.\]Multiplying both sides by $x^3 - 2x^2 - 5x + 6 = (x - 1)(x + 2)(x - 3),$ we get
\[x^2 - 19 = A(x + 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x + 2).\]Setting $x = 1,$ we get $-6A = -18$, so $A = 3.$
Setting $x = -2,$ we get $15B = -15,$ so $B = -1.$
Setting $x = 3,$ we get $10C = -10,$ so $C = -1.$ Hence, $ABC = \boxed{3}.$
|
human
|
Suppose that all four of the numbers \[2 - \sqrt{5}, \;4+\sqrt{10}, \;14 - 2\sqrt{7}, \;-\sqrt{2}\]are roots of the same nonzero polynomial with rational coefficients. What is the smallest possible degree of the polynomial?
|
gpt
|
Because the polynomial has rational coefficients, the radical conjugate of each of the four roots must also be roots of the polynomial. Therefore, the polynomial has at least $4 \times 2 = 8$ roots, so its degree is at least 8.
Note that for each of the four numbers, the monic quadratic with that number and its conjugate has rational coefficients. For example, the quadratic with roots $2 - \sqrt{5}$ and $2 + \sqrt{5}$ is
\[(x - 2 + \sqrt{5})(x - 2 - \sqrt{5}) = (x - 2)^2 - 5 = x^2 - 4x - 1.\]Thus, there exists such a polynomial of degree $\boxed{8},$ so this is the minimum.
|
human
|
Let $x$ and $y$ be real numbers such that $x + y = 3.$ Find the maximum value of
\[x^4 y + x^3 y + x^2 y + xy + xy^2 + xy^3 + xy^4.\]
|
gpt
|
First, we can factor out $xy,$ to get
\[xy (x^3 + x^2 + x + 1 + y + y^2 + y^3) = xy(x^3 + y^3 + x^2 + y^2 + x + y + 1).\]We know $x + y = 3.$ Let $p = xy.$ Then
\[9 = (x + y)^2 = x^2 + 2xy + y^2 = x^2 + 2xy + y^2,\]so $x^2 + y^2 = 9 - 2p.$
Also,
\[27 = (x + y)^3 = x^3 + 3x^2 y + 3xy^2 + y^3,\]so $x^3 + y^3 = 27 - 3xy(x + y) = 27 - 9p.$
Thus,
\begin{align*}
xy (x^3 + y^3 + x^2 + y^2 + x + y + 1) &= p (27 - 9p + 9 - 2p + 3 + 1) \\
&= p(40 - 11p) \\
&= -11p^2 + 40p \\
&= -11 \left( p - \frac{20}{11} \right)^2 + \frac{400}{11} \\
&\le \frac{400}{11}.
\end{align*}Equality occurs when $xy = p = \frac{20}{11}.$ By Vieta's formulas, $x$ and $y$ are the roots of
\[t^2 - 3t + \frac{20}{11} = 0.\]The discriminant of this quadratic is positive, so equality is possible. Thus, the maximum value is $\boxed{\frac{400}{11}}.$
|
human
|
Determine the value of the expression
\[\log_2 (27 + \log_2 (27 + \log_2 (27 + \cdots))),\]assuming it is positive.
|
gpt
|
Let
\[x = \log_2 (27 + \log_2 (27 + \log_2 (27 + \dotsb))).\]Then
\[x = \log_2 (27 + x),\]so $2^x = x + 27.$
To solve this equation, we plot $y = 2^x$ and $y = x + 27.$
[asy]
unitsize(0.15 cm);
real func (real x) {
return(2^x);
}
draw(graph(func,-30,log(40)/log(2)),red);
draw((-30,-3)--(13,40),blue);
draw((-30,0)--(13,0));
draw((0,-5)--(0,40));
dot("$(5,32)$", (5,32), SE);
label("$y = 2^x$", (10,16));
label("$y = x + 27$", (-18,18));
[/asy]
By inspection, the graphs intersect at $(5,32).$ Beyond this point, the graph of $y = 2^x$ increases much faster than the graph of $y = x + 27,$ so the only positive solution is $x = \boxed{5}.$
|
human
|
Find the largest positive integer $n$ such that
\[\sin^n x + \cos^n x \ge \frac{1}{n}\]for all real numbers $x.$
|
gpt
|
Setting $x = \pi,$ we get
\[(-1)^n \ge \frac{1}{n},\]so $n$ must be even. Let $n = 2m.$
Setting $x = \frac{\pi}{4},$ we get
\[\left( \frac{1}{\sqrt{2}} \right)^{2m} + \left( \frac{1}{\sqrt{2}} \right)^{2m} \ge \frac{1}{2m}.\]This simplifies to
\[\frac{1}{2^{m - 1}} \ge \frac{1}{2m},\]so $2^{m - 2} \le m.$ We see that $m = 4$ is a solution, and the function $2^{m - 2}$ grows faster than $m,$ so $m = 4$ is the largest possible value of $m.$
We must then prove that
\[\sin^8 x + \cos^8 x \ge \frac{1}{8}\]for all real numbers $x.$
By QM-AM,
\[\sqrt{\frac{\sin^8 x + \cos^8 x}{2}} \ge \frac{\sin^4 x + \cos^4 x}{2},\]so
\[\sin^8 x + \cos^8 x \ge \frac{(\sin^4 x + \cos^4 x)^2}{2}.\]Again by QM-AM,
\[\sqrt{\frac{\sin^4 x + \cos^4 x}{2}} \ge \frac{\sin^2 x + \cos^2 x}{2} = \frac{1}{2},\]so
\[\sin^4 x + \cos^4 x \ge \frac{1}{2}.\]Therefore,
\[\sin^8 x + \cos^8 x \ge \frac{(1/2)^2}{2} = \frac{1}{8}.\]We conclude that the largest such positive integer $n$ is $\boxed{8}.$
|
human
|
Let $a$ and $b$ be real numbers. Consider the following five statements:
$\frac{1}{a} < \frac{1}{b}$
$a^2 > b^2$
$a < b$
$a < 0$
$b < 0$
What is the maximum number of these statements that can be true for any values of $a$ and $b$?
|
gpt
|
Suppose $a < 0,$ $b < 0,$ and $a < b.$ Then
\[\frac{1}{a} - \frac{1}{b} = \frac{b - a}{ab} > 0,\]so $\frac{1}{a} > \frac{1}{b}.$ Thus, not all five statements can be true.
If we take $a = -2$ and $b = -1,$ then all the statements are true except the first statement. Hence, the maximum number of statements that can be true is $\boxed{4}.$
|
human
|
Integers $x$ and $y$ with $x>y>0$ satisfy $x+y+xy=80$. What is $x$?
|
gpt
|
Applying Simon's Favorite Factoring Trick, we add 1 to both sides to get $xy + x + y + 1 = 81,$ so
\[(x + 1)(y + 1) = 81.\]The only possibility is then $x + 1 = 27$ and $y + 1 = 3,$ so $x = \boxed{26}.$
|
human
|
Find the roots of $z^2 - z = 5 - 5i.$
Enter the roots, separated by commas.
|
gpt
|
We can write $z^2 - z - (5 - 5i) = 0.$ By the quadratic formula,
\[z = \frac{1 \pm \sqrt{1 + 4(5 - 5i)}}{2} = \frac{1 \pm \sqrt{21 - 20i}}{2}.\]Let $21 - 20i = (a + bi)^2,$ where $a$ and $b$ are real numbers. This expands as
\[a^2 + 2abi - b^2 = 21 - 20i.\]Equating the real and imaginary parts, we get $a^2 - b^2 = 21$ and $ab = -10,$ so $b = -\frac{10}{a}.$ Substituting, we get
\[a^2 - \frac{100}{a^2} = 21.\]Then $a^4 - 21a^2 - 100 = 0,$ which factors as $(a^2 - 25)(a^2 + 4) = 0.$ Since $a$ is real, $a^2 = 25,$ which means $a = 5$ or $a = -5.$
If $a = 5,$ then $b = -2,$ so
\[z = \frac{1 + 5 - 2i}{2} = 3 - i.\]If $a = -5,$ then $b = 2,$ so
\[z = \frac{1 - 5 + 2i}{2} = -2 + i.\]Therefore, the solutions are $\boxed{3 - i, -2 + i}.$
|
human
|
Suppose $f(x) = 6x - 9$ and $g(x) = \frac{x}{3} + 2$. Find $f(g(x)) - g(f(x))$.
|
gpt
|
We have that
$$\begin{aligned} f(g(x)) &= f\left(\frac{x}{3} + 2\right) = 6\left(\frac{x}{3} + 2\right) - 9 \\
&= 2x + 12 - 9\\
&= 2x + 3
\end{aligned}$$and
$$\begin{aligned} g(f(x)) &= g(6x-9) = \frac{6x-9}{3} + 2 \\
&= 2x -3 +2\\
&= 2x -1.
\end{aligned}$$So
$$f(g(x)) - g(f(x)) = 2x+3 - (2x-1) = 2x + 3 - 2x +1 = \boxed{4}.$$
|
human
|
Which type of conic section is described by the equation \[\sqrt{x^2 + (y-1)^2} + \sqrt{(x-5)^2 + (y+3)^2} = 10?\]Enter "C" for circle, "P" for parabola, "E" for ellipse, "H" for hyperbola, and "N" for none of the above.
|
gpt
|
This doesn't look like any of the standard forms of any of the conic sections. Instead, we appeal to the definitions of the conic sections. Note that the two terms on the left-hand side represent the distances in the $xy-$plane from $(x, y)$ to $(0, 1)$ and $(5, -3),$ respectively. So the given equation really says that the sum of the distances from $(x, y)$ to $(0, 1)$ and $(5, -3)$ is a constant (namely, $10$). So the graph of this equation should be an ellipse.
To check that the ellipse is non-degenerate, we compute the distance between $(0,1)$ and $(5,-3)$ to be \[\sqrt{(5-0)^2 + (-3-1)^2} = \sqrt{41},\]which is less than $10.$ Therefore, the given equation satisfies the triangle inequality, so the ellipse is non-degenerate. The answer is $\boxed{\text{(E)}}.$
|
human
|
Find all $t$ such that $x-t$ is a factor of $6x^2+13x-5.$
Enter your answer as a list separated by commas.
|
gpt
|
From the Factor theorem, if $x-t$ is a factor of $ 6x^2+13x-5$ we know that
$$6t^2+13t - 5 = 0$$Factoring gives us
$$(2t+5)(3t-1) = 0$$Hence $t = \boxed{\frac{1}{3}}$ or $t = \boxed{-\frac{5}{2}}$.
|
human
|
There are integers $b,c$ for which both roots of the polynomial $x^2-x-1$ are also roots of the polynomial $x^5-bx-c$. Determine the product $bc$.
|
gpt
|
Let $r$ be a root of $x^2-x-1$. Then, rearranging, we have
$$r^2 = r+1.$$Multiplying both sides by $r$ and substituting gives
\begin{align*}
r^3 &= r^2+r \\
&= (r+1)+r \\
&= 2r+1.
\end{align*}Repeating this process twice more, we have
\begin{align*}
r^4 &= r(2r+1) \\
&= 2r^2+r \\
&= 2(r+1)+r \\
&= 3r+2
\end{align*}and
\begin{align*}
r^5 &= r(3r+2) \\
&= 3r^2+2r \\
&= 3(r+1)+2r \\
&= 5r+3.
\end{align*}Thus, each root of $x^2-x-1$ is also a root of $x^5-5x-3$, which gives $bc = 5\cdot 3 = \boxed{15}$.
(It is left to the reader to investigate why this answer is unique.)
|
human
|
When the graph of $y = 2x^2 - x + 7$ is shifted four units to the right, we obtain the graph of $y = ax^2 + bx + c$. Find $a + b + c$.
|
gpt
|
When we shift the graph of $y = 2x^2 - x + 7$ four units to the right, we obtain the graph of $y = 2(x - 4)^2 - (x - 4) + 7$, which simplifies to $y = 2x^2 - 17x + 43$. Therefore, $a + b + c = 2 - 17 + 43 = \boxed{28}$.
Another way to solve the problem is as follows: The graph of $y = ax^2 + bx + c$ always passes through the point $(1, a + b + c)$. In other words, $a + b + c$ is the $y$-coordinate of the point on the parabola whose $x$-coordinate is 1. But this parabola is obtained by shifting the graph of $y = 2x^2 - x + 7$ four units to the right, so $a + b + c$ is also the $y$-coordinate of the point on the original parabola whose $x$-coordinate is $1 - 4 = -3$. This $y$-coordinate is equal to $2 \cdot (-3)^2 - (-3) + 7 = 28$.
|
human
|
Let $a$ and $b$ be the roots of $k(x^2 - x) + x + 5 = 0.$ Let $k_1$ and $k_2$ be the values of $k$ for which $a$ and $b$ satisfy
\[\frac{a}{b} + \frac{b}{a} = \frac{4}{5}.\]Find
\[\frac{k_1}{k_2} + \frac{k_2}{k_1}.\]
|
gpt
|
The quadratic equation in $x$ is $kx^2 - (k - 1) x + 5 = 0,$ so by Vieta's formulas, $a + b = \frac{k - 1}{k}$ and $ab = \frac{5}{k}.$ Then
\begin{align*}
\frac{a}{b} + \frac{b}{a} &= \frac{a^2 + b^2}{ab} \\
&= \frac{(a + b)^2 - 2ab}{ab} \\
&= \frac{(a + b)^2}{ab} - 2 \\
&= \frac{(\frac{k - 1}{k})^2}{\frac{5}{k}} - 2 \\
&= \frac{(k - 1)^2}{5k} - 2.
\end{align*}So
\[\frac{(k - 1)^2}{5k} - 2 = \frac{4}{5}.\]This equation simplifies to $k^2 - 16k + 1 = 0.$ Again by Vieta's formulas, $k_1 + k_2 = 16$ and $k_1 k_2 = 1,$ so
\begin{align*}
\frac{k_1}{k_2} + \frac{k_2}{k_1} &= \frac{k_1^2 + k_2^2}{k_1 k_2} \\
&= \frac{(k_1 + k_2)^2 - 2k_1 k_2}{k_1 k_2} \\
&= \frac{(k_1 + k_2)^2}{k_1 k_2} - 2 \\
&= 16^2 - 2 = \boxed{254}.
\end{align*}
|
human
|
The function $f$ is linear and satisfies $f(d+1)-f(d) = 3$ for all real numbers $d$. What is $f(3)-f(5)$?
|
gpt
|
Taking $d = 3,$ we get
\[f(4) - f(3) = 3.\]Taking $d = 4,$ we get
\[f(5) - f(4) = 3.\]Adding these equations, we get $f(5) - f(3) = 6,$ so $f(3) - f(5) = \boxed{-6}.$
|
human
|
A function $f$ is defined by $f(z) = (4 + i) z^2 + \alpha z + \gamma$ for all complex numbers $z$, where $\alpha$ and $\gamma$ are complex numbers and $i^2 = - 1$. Suppose that $f(1)$ and $f(i)$ are both real. What is the smallest possible value of $| \alpha | + |\gamma |$?
|
gpt
|
Let $\alpha = a + bi$ and $\gamma = c + di,$ where $a,$ $b,$ $c,$ and $d$ are real numbers. Then
\begin{align*}
f(1) &= (4 + i) + \alpha + \gamma = (a + c + 4) + (b + d + 1)i, \\
f(i) &= (4 + i)(-1) + \alpha i + \gamma = (-b + c - 4) + (a + d - 1)i.
\end{align*}Since $f(1)$ and $f(i)$ are both real, $b + d + 1 = 0$ and $a + d - 1 = 0,$ so $a = -d + 1$ and $b = -d - 1.$ Then
\begin{align*}
|\alpha| + |\gamma| &= \sqrt{a^2 + b^2} + \sqrt{c^2 + d^2} \\
&= \sqrt{(-d + 1)^2 + (-d - 1)^2} + \sqrt{c^2 + d^2} \\
&= \sqrt{2d^2 + 2} + \sqrt{c^2 + d^2} \\
&\ge \sqrt{2}.
\end{align*}Equality occurs when $a = 1,$ $b = -1,$ $c = 0,$ and $d = 0.$ Therefore, the minimum value is $\boxed{\sqrt{2}}.$
|
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