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Square $ABCD$ has sides of length 2. Set $S$ is the set of all line segments that have length 2 and whose endpoints are on adjacent sides of the square. The midpoints of the line segments in set $S$ enclose a region whose area to the nearest hundredth is $k$. Find $100k$.
|
86
|
deepscale
| 35,949
| ||
Given the ellipse $\frac{{x}^{2}}{3{{m}^{2}}}+\frac{{{y}^{2}}}{5{{n}^{2}}}=1$ and the hyperbola $\frac{{{x}^{2}}}{2{{m}^{2}}}-\frac{{{y}^{2}}}{3{{n}^{2}}}=1$ share a common focus, find the eccentricity of the hyperbola ( ).
|
\frac{\sqrt{19}}{4}
|
deepscale
| 19,659
| ||
Two people, A and B, visit the "2011 Xi'an World Horticultural Expo" together. They agree to independently choose 4 attractions from numbered attractions 1 to 6 to visit, spending 1 hour at each attraction. Calculate the probability that they will be at the same attraction during their last hour.
|
\dfrac{1}{6}
|
deepscale
| 26,193
| ||
Round to the nearest tenth: 36.89753
|
36.9
|
deepscale
| 39,083
| ||
Given a tetrahedron \(A B C D\), where \(B D = D C = C B = \sqrt{2}\), \(A C = \sqrt{3}\), and \(A B = A D = 1\), find the cosine of the angle between line \(B M\) and line \(A C\), where \(M\) is the midpoint of \(C D\).
|
\frac{\sqrt{2}}{3}
|
deepscale
| 32,077
| ||
A student has 2 identical photo albums and 3 identical stamp albums. The student wants to give away 4 albums, one to each of four friends. How many different ways can the student give away the albums?
|
10
|
deepscale
| 22,362
| ||
The ellipse $x^2+4y^2=4$ and the hyperbola $x^2-m(y+2)^2 = 1$ are tangent. Compute $m.$
|
\frac{12}{13}
|
deepscale
| 37,554
| ||
A monic polynomial of degree $n$, with real coefficients, has its first two terms after $x^n$ as $a_{n-1}x^{n-1}$ and $a_{n-2}x^{n-2}$. It’s known that $a_{n-1} = 2a_{n-2}$. Determine the absolute value of the lower bound for the sum of the squares of the roots of this polynomial.
|
\frac{1}{4}
|
deepscale
| 20,274
| ||
Consider the polynomial \( P(x)=x^{3}+x^{2}-x+2 \). Determine all real numbers \( r \) for which there exists a complex number \( z \) not in the reals such that \( P(z)=r \).
|
Because such roots to polynomial equations come in conjugate pairs, we seek the values \( r \) such that \( P(x)=r \) has just one real root \( x \). Considering the shape of a cubic, we are interested in the boundary values \( r \) such that \( P(x)-r \) has a repeated zero. Thus, we write \( P(x)-r=x^{3}+x^{2}-x+(2-r)=(x-p)^{2}(x-q)=x^{3}-(2 p+q) x^{2}+p(p+2 q) x-p^{2} q \). Then \( q=-2 p-1 \) and \( 1=p(p+2 q)=p(-3 p-2) \) so that \( p=1 / 3 \) or \( p=-1 \). It follows that the graph of \( P(x) \) is horizontal at \( x=1 / 3 \) (a maximum) and \( x=-1 \) (a minimum), so the desired values \( r \) are \( r>P(-1)=3 \) and \( r<P(1 / 3)=1 / 27+1 / 9-1 / 3+2=49 / 27 \).
|
r>3, r<49/27
|
deepscale
| 4,039
| |
Find the smallest positive integer $n$ such that the divisors of $n$ can be partitioned into three sets with equal sums.
|
I claim the answer is 120. First, note that $120=2^{3} \cdot 3 \cdot 5$, so the sum of divisors is $(1+2+4+8)(1+3)(1+5)=15 \cdot 4 \cdot 6=360$. Thus, we need to split the divisors into groups summing to 120 . But then we can just take $\{120\},\{20,40,60\},\{1,2,3,4,5,6,8,10,12,15,24,30\}$. Thus, 120 works. Now we need to show 120 is the lowest. Let $s(n)$ be the sum of divisors. Since $n$ will be in one of the piles, we need $s(n) \geq 3 n$. First, we claim that $n$ must have at least 3 distinct prime divisors. Surely, if it had 2 distinct prime divisors, say $p$ and $q$, so that $n=p^{a} q^{b}$, then the sum of divisors is $$\left(1+p+p^{2}+\ldots+p^{a}\right)\left(1+q+q^{2}+\ldots+q^{b}\right)=p^{a} q^{b}\left(1+\frac{1}{p}+\ldots+\frac{1}{p^{a}}\right)\left(1+\frac{1}{q}+\ldots+\frac{1}{q^{b}}\right)$$ However, the expression $1+\frac{1}{p}+\ldots+\frac{1}{p^{a}}$ is maximized when $p$ is minimized, and further, as $a$ is finite must be at most $\frac{1}{1-\frac{1}{p}}=\frac{p}{p-1}$. Thus, the sum of divisors is less than $$p^{a} q^{b} \frac{p}{p-1} \frac{q}{q-1} \leq n \cdot 2 \cdot \frac{3}{2}=3 n$$ Thus, $n$ can't have 2 distinct prime divisors and must have at least 3 distinct prime divisors. As we already discovered 120 works, we need not worry about 4 distinct prime divisors, as the value of $n$ would be at least $2 \cdot 3 \cdot 5 \cdot 7=210$. We now work through the numbers with 3 distinct divisors. If 2 is not one of them, then the only number that works is $105=3 \cdot 5 \cdot 7$, which has a sum of divisors that is not large enough. Therefore, 2 must be a prime divisor of $n$. Additionally, if 3 is not a divisor, then our options are $2 \cdot 5 \cdot 7$ and $2 \cdot 5 \cdot 11$, which also do not work. Therefore, 3 must also be a prime divisor. Then, if 5 is not a prime divisor, then if $n$ is $2 \cdot 3 \cdot p$, it has a sum of divisors of $(1+2)(1+3)(1+p)=n \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \frac{p+1}{p}$, which is only at least $3 n$ if $p$ is exactly 2 , which is not feasible. Additionally, if we use $2^{2}$, then the sum of divisors is $(1+2+4)(1+3)(1+p)=n \cdot \frac{7}{4} \cdot \frac{4}{3} \cdot \frac{p}{p+1}$, so $\frac{p+1}{p}>\frac{9}{7} \Longrightarrow p<4.5$, which also can't happen. Further, we can't have $3^{2}$ be a divisor of $n$ as $2 \cdot 3^{2} \cdot 5$ is the only value less than 120 with this, and that also does not work. Lastly, we just need to check $2^{3} \cdot 3 \cdot p$, which has a sum of divisors of $(1+2+4+8)(1+3)(1+p)=n \cdot \frac{15}{8} \cdot \frac{4}{3} \cdot \frac{p+1}{p}=n \cdot \frac{5}{2} \cdot \frac{p}{p+1}$, so $p=5$ and that works. This means that $n=120$ is the smallest value for which $s(n) \geq 3 n$, and thus is our answer.
|
120
|
deepscale
| 5,041
| |
In how many ways can 10 people be seated in a row of chairs if four of the people, Alice, Bob, Charlie, and Dana, refuse to sit in four consecutive seats?
|
3507840
|
deepscale
| 18,961
| ||
Let $a$ and $b$ be positive integers for which $45a+b=2021$. What is the minimum possible value of $a+b$?
|
Since $a$ is a positive integer, $45 a$ is a positive integer. Since $b$ is a positive integer, $45 a$ is less than 2021. The largest multiple of 45 less than 2021 is $45 \times 44=1980$. (Note that $45 \cdot 45=2025$ which is greater than 2021.) If $a=44$, then $b=2021-45 \cdot 44=41$. Here, $a+b=44+41=85$. If $a$ is decreased by 1, the value of $45 a+b$ is decreased by 45 and so $b$ must be increased by 45 to maintain the same value of $45 a+b$, which increases the value of $a+b$ by $-1+45=44$. Therefore, if $a<44$, the value of $a+b$ is always greater than 85. If $a>44$, then $45 a>2021$ which makes $b$ negative, which is not possible. Therefore, the minimum possible value of $a+b$ is 85.
|
85
|
deepscale
| 5,691
| |
Let $x,$ $y,$ and $z$ be positive real numbers. Find the minimum value of
\[\frac{(x^2 + 3x + 1)(y^2 + 3y + 1)(z^2 + 3z + 1)}{xyz}.\]
|
125
|
deepscale
| 36,803
| ||
Given an ellipse $C$: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 (a > b > 0)$ passes through the point $(1, \frac{2\sqrt{3}}{3})$, with its left and right foci being $F_1$ and $F_2$ respectively. The chord formed by the intersection of the circle $x^2 + y^2 = 2$ and the line $x + y + b = 0$ has a length of $2$.
(Ⅰ) Find the standard equation of the ellipse $C$;
(Ⅱ) Let $Q$ be a moving point on the ellipse $C$ that is not on the $x$-axis, with $Q$ being the origin. A line parallel to $OQ$ passing through $F_2$ intersects the ellipse $C$ at two distinct points $M$ and $N$.
(1) Investigate whether $\frac{|MN|}{|OQ|^2}$ is a constant. If it is, find the constant; if not, explain why.
(2) Let the area of $\triangle QF_2M$ be $S_1$ and the area of $\triangle OF_2N$ be $S_2$, and let $S = S_1 + S_2$. Find the maximum value of $S$.
|
\frac{2\sqrt{3}}{3}
|
deepscale
| 29,621
| ||
Six chairs are placed in a row. Find the number of ways 3 people can sit randomly in these chairs such that no two people sit next to each other.
|
24
|
deepscale
| 18,623
| ||
Given the sequence $\left\{a_{n}\right\}$ that satisfies
$$
a_{n+1} = -\frac{1}{2} a_{n} + \frac{1}{3^{n}}\quad (n \in \mathbf{Z}_{+}),
$$
find all values of $a_{1}$ such that the sequence $\left\{a_{n}\right\}$ is monotonic, i.e., $\left\{a_{n}\right\}$ is either increasing or decreasing.
|
\frac{2}{5}
|
deepscale
| 7,724
| ||
What is the greatest common factor of 180, 240, and 300?
|
60
|
deepscale
| 21,678
| ||
Let $p$, $q$, $r$, $s$ be distinct real numbers such that the roots of $x^2 - 12px - 13q = 0$ are $r$ and $s$, and the roots of $x^2 - 12rx - 13s = 0$ are $p$ and $q$. Additionally, $p + q + r + s = 201$. Find the value of $pq + rs$.
|
-\frac{28743}{12}
|
deepscale
| 19,400
| ||
Given the function $f(x)=-\cos^2 x + \sqrt{3}\sin x\sin\left(x + \frac{\pi}{2}\right)$, find the sum of the minimum and maximum values of $f(x)$ when $x \in \left[0, \frac{\pi}{2}\right]$.
|
-\frac{1}{2}
|
deepscale
| 23,607
| ||
Given vectors $\overrightarrow{a}=(\sqrt{2}\cos \omega x,1)$ and $\overrightarrow{b}=(2\sin (\omega x+ \frac{\pi}{4}),-1)$ where $\frac{1}{4}\leqslant \omega\leqslant \frac{3}{2}$, and the function $f(x)= \overrightarrow{a}\cdot \overrightarrow{b}$, and the graph of $f(x)$ has an axis of symmetry at $x= \frac{5\pi}{8}$.
$(1)$ Find the value of $f( \frac{3}{4}\pi)$;
$(2)$ If $f( \frac{\alpha}{2}- \frac{\pi}{8})= \frac{\sqrt{2}}{3}$ and $f( \frac{\beta}{2}- \frac{\pi}{8})= \frac{2\sqrt{2}}{3}$, and $\alpha,\beta\in(-\frac{\pi}{2}, \frac{\pi}{2})$, find the value of $\cos (\alpha-\beta)$.
|
\frac{2\sqrt{10}+2}{9}
|
deepscale
| 19,723
| ||
The probability that a light bulb lasts more than 1000 hours is 0.2. Determine the probability that 1 out of 3 light bulbs fails after 1000 hours of use.
|
0.096
|
deepscale
| 32,955
| ||
Given that player A needs to win 2 more games and player B needs to win 3 more games, and the probability of winning each game for both players is $\dfrac{1}{2}$, calculate the probability of player A ultimately winning.
|
\dfrac{11}{16}
|
deepscale
| 11,813
| ||
What is the maximum number of finite roots that the equation
$$
\left|x - a_{1}\right| + \ldots + |x - a_{50}| = \left|x - b_{1}\right| + \ldots + |x - b_{50}|
$$
can have, where $a_{1}, a_{2}, \ldots, a_{50}, b_{1}, b_{2}, \ldots, b_{50}$ are distinct numbers?
|
49
|
deepscale
| 26,130
| ||
Three hexagons of increasing size are shown below. Suppose the dot pattern continues so that each successive hexagon contains one more band of dots. How many dots are in the next hexagon?
[asy] // diagram by SirCalcsALot, edited by MRENTHUSIASM size(250); path p = scale(0.8)*unitcircle; pair[] A; pen grey1 = rgb(100/256, 100/256, 100/256); pen grey2 = rgb(183/256, 183/256, 183/256); for (int i=0; i<7; ++i) { A[i] = rotate(60*i)*(1,0);} path hex = A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--cycle; fill(p,grey1); draw(scale(1.25)*hex,black+linewidth(1.25)); pair S = 6A[0]+2A[1]; fill(shift(S)*p,grey1); for (int i=0; i<6; ++i) { fill(shift(S+2*A[i])*p,grey2);} draw(shift(S)*scale(3.25)*hex,black+linewidth(1.25)); pair T = 16A[0]+4A[1]; fill(shift(T)*p,grey1); for (int i=0; i<6; ++i) { fill(shift(T+2*A[i])*p,grey2); fill(shift(T+4*A[i])*p,grey1); fill(shift(T+2*A[i]+2*A[i+1])*p,grey1); } draw(shift(T)*scale(5.25)*hex,black+linewidth(1.25)); [/asy]
|
To solve this problem, we need to understand the pattern of how the dots are arranged in each hexagon and how the number of dots increases as the hexagons grow in size.
1. **Observing the pattern:**
- The first hexagon has 1 dot.
- The second hexagon has 7 dots: 1 central dot and 6 surrounding dots.
- The third hexagon has 19 dots: 1 central dot, 6 surrounding dots, and an additional 12 dots forming a second layer around the first layer.
2. **Identifying the pattern in the number of dots:**
- The number of dots in each new layer of hexagon can be observed as follows:
- First layer (central dot): 1 dot
- Second layer: 6 dots
- Third layer: 12 dots (6 sides × 2 dots per side)
3. **Generalizing the pattern:**
- Each new layer around the central dot adds an increasing number of dots. Specifically, each side of the hexagon in the nth layer has n dots, contributing 6n dots for the nth layer.
- The total number of dots in the nth hexagon is the sum of the first n terms of the sequence: 1, 6, 12, 18, ..., which is an arithmetic sequence where each term is 6 times the term number minus 1 (6(n-1)).
4. **Calculating the number of dots in the fourth hexagon:**
- The fourth layer will have 6 × 3 = 18 dots.
- Adding this to the total from the third hexagon (19 dots), the fourth hexagon will have:
\[
19 + 18 = 37 \text{ dots}
\]
Thus, the next hexagon in the sequence will contain $\boxed{\textbf{(B) }37}$ dots.
|
37
|
deepscale
| 1,721
| |
When three numbers are added two at a time, the sums are 35, 47, and 52. What is the sum of all three numbers, and what is their product?
|
9600
|
deepscale
| 17,977
| ||
On Namek, each hour has 100 minutes. On the planet's clocks, the hour hand completes one full circle in 20 hours and the minute hand completes one full circle in 100 minutes. How many hours have passed on Namek from 0:00 until the next time the hour hand and the minute hand overlap?
|
20/19
|
deepscale
| 13,309
| ||
A wooden model of a square pyramid has a base edge of 12 cm and an altitude of 8 cm. A cut is made parallel to the base of the pyramid that separates it into two pieces: a smaller pyramid and a frustum. Each base edge of the smaller pyramid is 6 cm and its altitude is 4 cm. How many cubic centimeters are in the volume of the frustum?
|
336
|
deepscale
| 36,127
| ||
Calculate the value of $333 + 33 + 3$.
|
369
|
deepscale
| 23,501
| ||
Quadrilateral \(ABCD\) is such that \(\angle BAC = \angle CAD = 60^\circ\) and \(AB + AD = AC\). It is also known that \(\angle ACD = 23^\circ\). How many degrees is \(\angle ABC\)?
|
83
|
deepscale
| 7,774
| ||
Find the maximum value of the function
$$
f(x) = \sqrt{3} \sin 2x + 2 \sin x + 4 \sqrt{3} \cos x.
$$
|
\frac{17}{2}
|
deepscale
| 9,920
| ||
Let $f$ be a function for which $f\left(\dfrac{x}{3}\right) = x^2 + x + 1$. Find the sum of all values of $z$ for which $f(3z) = 7$.
|
1. **Identify the function and equation:** Given the function $f\left(\frac{x}{3}\right) = x^2 + x + 1$, we need to find the sum of all values of $z$ for which $f(3z) = 7$.
2. **Relate $f(3z)$ to the given function:** Since $f\left(\frac{x}{3}\right) = x^2 + x + 1$, substituting $x = 9z$ (because $\frac{9z}{3} = 3z$) gives us:
\[
f(3z) = f\left(\frac{9z}{3}\right) = (9z)^2 + 9z + 1 = 81z^2 + 9z + 1
\]
3. **Set up the equation:** We set $f(3z) = 7$:
\[
81z^2 + 9z + 1 = 7
\]
Simplifying this, we get:
\[
81z^2 + 9z - 6 = 0
\]
4. **Solve the quadratic equation:** We can solve this quadratic equation using the quadratic formula:
\[
z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where $a = 81$, $b = 9$, and $c = -6$. Plugging in these values:
\[
z = \frac{-9 \pm \sqrt{9^2 - 4 \cdot 81 \cdot (-6)}}{2 \cdot 81}
\]
\[
z = \frac{-9 \pm \sqrt{81 + 1944}}{162}
\]
\[
z = \frac{-9 \pm \sqrt{2025}}{162}
\]
\[
z = \frac{-9 \pm 45}{162}
\]
This gives us two solutions:
\[
z_1 = \frac{-9 + 45}{162} = \frac{36}{162} = \frac{2}{9}, \quad z_2 = \frac{-9 - 45}{162} = \frac{-54}{162} = -\frac{1}{3}
\]
5. **Find the sum of the roots:** The sum of the roots $z_1$ and $z_2$ is:
\[
z_1 + z_2 = \frac{2}{9} - \frac{1}{3} = \frac{2}{9} - \frac{3}{9} = -\frac{1}{9}
\]
6. **Conclusion:** The sum of all values of $z$ for which $f(3z) = 7$ is $\boxed{\textbf{(B) }-\frac{1}{9}}$.
|
-1/9
|
deepscale
| 141
| |
Given the following conditions:①$asinB=bsin({A+\frac{π}{3}})$; ②$S=\frac{{\sqrt{3}}}{2}\overrightarrow{BA}•\overrightarrow{CA}$; ③$c\tan A=\left(2b-c\right)\tan C$. Choose one of the three conditions and fill in the blank below, then answer the following questions.<br/>In $\triangle ABC$, where the sides opposite to angles $A$, $B$, and $C$ are $a$, $b$, and $c$ respectively, the area of $\triangle ABC$ is $S$, and it satisfies _____.<br/>$(1)$ Find the value of angle $A$;<br/>$(2)$ Given that the area of $\triangle ABC$ is $2\sqrt{3}$, point $D$ lies on side $BC$ such that $BD=2DC$, find the minimum value of $AD$.
|
\frac{4\sqrt{3}}{3}
|
deepscale
| 26,505
| ||
Given $\triangle ABC$, where $AB=6$, $BC=8$, $AC=10$, and $D$ is on $\overline{AC}$ with $BD=6$, find the ratio of $AD:DC$.
|
\frac{18}{7}
|
deepscale
| 23,613
| ||
In the diagram, $\triangle PQR$ is right-angled at $P$ and $\angle PRQ=\theta$. A circle with center $P$ is drawn passing through $Q$. The circle intersects $PR$ at $S$ and $QR$ at $T$. If $QT=8$ and $TR=10$, determine the value of $\cos \theta$.
|
\frac{\sqrt{7}}{3}
|
deepscale
| 13,340
| ||
Rectangle $EFGH$ has area $4032$. An ellipse with area $4032\pi$ passes through points $E$ and $G$ and has foci at $F$ and $H$. Determine the perimeter of the rectangle $EFGH$.
|
8\sqrt{2016}
|
deepscale
| 32,102
| ||
In triangle $\triangle ABC$, the sides opposite angles $A$, $B$, and $C$ are $a$, $b$, and $c$ respectively. It is known that $2c=a+\cos A\frac{b}{\cos B}$.
$(1)$ Find the measure of angle $B$.
$(2)$ If $b=4$ and $a+c=3\sqrt{2}$, find the area of $\triangle ABC$.
|
\frac{\sqrt{3}}{6}
|
deepscale
| 8,812
| ||
Suppose we have an (infinite) cone $\mathcal{C}$ with apex $A$ and a plane $\pi$. The intersection of $\pi$ and $\mathcal{C}$ is an ellipse $\mathcal{E}$ with major axis $BC$, such that $B$ is closer to $A$ than $C$, and $BC=4, AC=5, AB=3$. Suppose we inscribe a sphere in each part of $\mathcal{C}$ cut up by $\mathcal{E}$ with both spheres tangent to $\mathcal{E}$. What is the ratio of the radii of the spheres (smaller to larger)?
|
It can be seen that the points of tangency of the spheres with $E$ must lie on its major axis due to symmetry. Hence, we consider the two-dimensional cross-section with plane $ABC$. Then the two spheres become the incentre and the excentre of the triangle $ABC$, and we are looking for the ratio of the inradius to the exradius. Let $s, r, r_{a}$ denote the semiperimeter, inradius, and exradius (opposite to $A$ ) of the triangle $ABC$. We know that the area of $ABC$ can be expressed as both $rs$ and $r_{a}(s-|BC|)$, and so $\frac{r}{r_{a}}=\frac{s-|BC|}{s}$. For the given triangle, $s=6$ and $a=4$, so the required ratio is $\frac{1}{3}$. $\qquad$
|
\frac{1}{3}
|
deepscale
| 3,184
| |
The amount of heat \( Q \) received by a certain substance when heated from 0 to \( T \) is determined by the formula \( Q = 0.1054t + 0.000002t^2 \) (\( Q \) is in joules, \( t \) is in kelvins). Find the heat capacity of this substance at \( 100 \) K.
|
0.1058
|
deepscale
| 23,543
| ||
A candy company makes 5 colors of jellybeans, which come in equal proportions. If I grab a random sample of 5 jellybeans, what is the probability that I get exactly 2 distinct colors?
|
There are $\binom{5}{2}=10$ possible pairs of colors. Each pair of colors contributes $2^{5}-2=30$ sequences of beans that use both colors. Thus, the answer is $10 \cdot 30 / 5^{5}=12 / 125$.
|
\frac{12}{125}
|
deepscale
| 4,116
| |
In the isosceles triangle \(ABC\) (\(AB = BC\)), a point \(D\) is taken on the side \(BC\) such that \(BD : DC = 1 : 4\). In what ratio does the line \(AD\) divide the altitude \(BE\) of triangle \(ABC\), counting from vertex \(B\)?
|
1:2
|
deepscale
| 13,855
| ||
Add $2175_{9} + 1714_{9} + 406_9$. Express your answer in base $9$.
|
4406_{9}
|
deepscale
| 19,270
| ||
Point $(x,y)$ is randomly picked from the rectangular region with vertices at $(0,0),(2010,0),(2010,2011),$ and $(0,2011)$. What is the probability that $x > 3y$? Express your answer as a common fraction.
|
\frac{335}{2011}
|
deepscale
| 17,407
| ||
Given that non-negative real numbers \( x \) and \( y \) satisfy
\[
x^{2}+4y^{2}+4xy+4x^{2}y^{2}=32,
\]
find the maximum value of \( \sqrt{7}(x+2y)+2xy \).
|
16
|
deepscale
| 10,189
| ||
The graph of the function $y=f(x)$ is shown below. For all $x > 4$, it is true that $f(x) > 0.4$. If $f(x) = \frac{x^2}{Ax^2 + Bx + C}$, where $A,B,$ and $C$ are integers, then find $A+B+C$. [asy]
import graph; size(10.9cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-5.29,xmax=5.61,ymin=-2.42,ymax=4.34;
Label laxis; laxis.p=fontsize(10);
xaxis("$x$",xmin,xmax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); yaxis("$y$",ymin,ymax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); real f1(real x){return x^2/(2*x^2-2*x-12);} draw(graph(f1,xmin,-2.1),linewidth(1.2),Arrows(4)); draw(graph(f1,-1.84,2.67),linewidth(1.2),Arrows(4)); draw(graph(f1,3.24,xmax),linewidth(1.2),Arrows(4));
label("$f$",(-5.2,1),NE*lsf);
// clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
draw((-2,-2.2)--(-2,4.2),dashed);
draw((3,-2.2)--(3,4.2),dashed);
draw((-5,1/2)--(5.5,1/2),dashed);
[/asy]
|
-12
|
deepscale
| 37,042
| ||
Consider the set $$ \mathcal{S}=\{(a, b, c, d, e): 0<a<b<c<d<e<100\} $$ where $a, b, c, d, e$ are integers. If $D$ is the average value of the fourth element of such a tuple in the set, taken over all the elements of $\mathcal{S}$ , find the largest integer less than or equal to $D$ .
|
66
|
deepscale
| 9,540
| ||
Let $\overline{AB}$ be a diameter of circle $\omega$. Extend $\overline{AB}$ through $A$ to $C$. Point $T$ lies on $\omega$ so that line $CT$ is tangent to $\omega$. Point $P$ is the foot of the perpendicular from $A$ to line $CT$. Suppose $\overline{AB} = 18$, and let $m$ denote the maximum possible length of segment $BP$. Find $m^{2}$.
|
Proceed as follows for Solution 1.
Once you approach the function $k=(2x-27)/x^2$, find the maximum value by setting $dk/dx=0$.
Simplifying $k$ to take the derivative, we have $2/x-27/x^2$, so $dk/dx=-2/x^2+54/x^3$. Setting $dk/dx=0$, we have $2/x^2=54/x^3$.
Solving, we obtain $x=27$ as the critical value. Hence, $k$ has the maximum value of $(2*27-27)/27^2=1/27$. Since $BP^2=405+729k$, the maximum value of $\overline {BP}$ occurs at $k=1/27$, so $BP^2$ has a maximum value of $405+729/27=\boxed{432}$.
Note: Please edit this solution if it feels inadequate.
|
432
|
deepscale
| 6,924
| |
Given the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 (a > 0, b > 0)$, a line passing through its right focus $F$ and parallel to the asymptote $y = -\frac{b}{a}x$ intersects the right branch of the hyperbola and the other asymptote at points $A$ and $B$ respectively, with $\overrightarrow{FA} = \overrightarrow{AB}$. Find the eccentricity of the hyperbola.
A) $\frac{3}{2}$
B) $\sqrt{2}$
C) $\sqrt{3}$
D) $2$
|
\sqrt{2}
|
deepscale
| 12,265
| ||
For a certain natural number $n$, $n^2$ gives a remainder of 4 when divided by 5, and $n^3$ gives a remainder of 2 when divided by 5. What remainder does $n$ give when divided by 5?
|
3
|
deepscale
| 37,627
| ||
For the equation $6 x^{2}=(2 m-1) x+m+1$ with respect to $x$, there is a root $\alpha$ satisfying the inequality $-1988 \leqslant \alpha \leqslant 1988$, and making $\frac{3}{5} \alpha$ an integer. How many possible values are there for $m$?
|
2385
|
deepscale
| 18,338
| ||
There is a caravan with 100 camels, consisting of both one-humped and two-humped camels, with at least one of each kind. If you take any 62 camels, they will have at least half of the total number of humps in the caravan. Let \( N \) be the number of two-humped camels. How many possible values can \( N \) take within the range from 1 to 99?
|
72
|
deepscale
| 15,792
| ||
Let $n$ be an integer and $$m=(n-1001)(n-2001)(n-2002)(n-3001)(n-3002)(n-3003)$$ Given that $m$ is positive, find the minimum number of digits of $m$.
|
One can show that if $m>0$, then we must either have $n>3003$ or $n<1001$. If $n<1001$, each term other than $n-1001$ has absolute value at least 1000 , so $m>1000^{5}$, meaning that $m$ has at least 16 digits. However, if $n>3003$, it is clear that the minimal $m$ is achieved at $n=3004$, which makes $$m=2002 \cdot 1002 \cdot 1001 \cdot 3 \cdot 2 \cdot 1=12 \cdot 1001 \cdot 1001 \cdot 1002$$ which is about $12 \cdot 10^{9}$ and thus has 11 digits.
|
11
|
deepscale
| 5,181
| |
Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?
|
Let's denote the distance to school as $d$ miles and the usual speed during rush hour as $v$ miles per hour.
1. **Convert time to hours**:
- The time taken in rush hour traffic is 20 minutes, which is $\frac{20}{60} = \frac{1}{3}$ hours.
- The time taken without traffic is 12 minutes, which is $\frac{12}{60} = \frac{1}{5}$ hours.
2. **Set up equations based on the relationship $d = rt$ (distance = rate × time)**:
- In rush hour traffic: $d = v \times \frac{1}{3}$
- Without traffic: $d = (v + 18) \times \frac{1}{5}$
3. **Equating the two expressions for $d$**:
\[
v \times \frac{1}{3} = (v + 18) \times \frac{1}{5}
\]
\[
\frac{v}{3} = \frac{v + 18}{5}
\]
4. **Solve for $v$**:
- Cross-multiply to eliminate the fractions:
\[
5v = 3(v + 18)
\]
- Expand and simplify:
\[
5v = 3v + 54
\]
\[
2v = 54
\]
\[
v = 27 \text{ miles per hour}
\]
5. **Find $d$ using the expression for rush hour traffic**:
- Substitute $v = 27$ into $d = v \times \frac{1}{3}$:
\[
d = 27 \times \frac{1}{3} = 9 \text{ miles}
\]
Thus, the distance to school is $\boxed{\textbf{(D)}~9}$ miles.
|
9
|
deepscale
| 896
| |
Given that Sofia has a $5 \times 7$ index card, if she shortens the length of one side by $2$ inches and the card has an area of $21$ square inches, find the area of the card in square inches if instead she shortens the length of the other side by $1$ inch.
|
30
|
deepscale
| 20,428
| ||
Given that $y=\left(m-2\right)x+(m^{2}-4)$ is a direct proportion function, find the possible values of $m$.
|
-2
|
deepscale
| 21,791
| ||
Find the least three digit number that is equal to the sum of its digits plus twice the product of its digits.
|
397
|
deepscale
| 19,937
| ||
Triangle $A B C$ has $A B=1, B C=\sqrt{7}$, and $C A=\sqrt{3}$. Let $\ell_{1}$ be the line through $A$ perpendicular to $A B, \ell_{2}$ the line through $B$ perpendicular to $A C$, and $P$ the point of intersection of $\ell_{1}$ and $\ell_{2}$. Find $P C$.
|
By the Law of Cosines, $\angle B A C=\cos ^{-1} \frac{3+1-7}{2 \sqrt{3}}=\cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)=150^{\circ}$. If we let $Q$ be the intersection of $\ell_{2}$ and $A C$, we notice that $\angle Q B A=90^{\circ}-\angle Q A B=90^{\circ}-30^{\circ}=60^{\circ}$. It follows that triangle $A B P$ is a 30-60-90 triangle and thus $P B=2$ and $P A=\sqrt{3}$. Finally, we have $\angle P A C=360^{\circ}-\left(90^{\circ}+150^{\circ}\right)=120^{\circ}$, and $$P C=\left(P A^{2}+A C^{2}-2 P A \cdot A C \cos 120^{\circ}\right)^{1 / 2}=(3+3+3)^{1 / 2}=3$$
|
3
|
deepscale
| 3,631
| |
For any positive integer $n$, the value of $n!$ is the product of the first $n$ positive integers. Calculate the greatest common divisor of $8!$ and $10!$.
|
40320
|
deepscale
| 21,308
| ||
Cara is sitting at a round table with her eight friends. How many different pairs of friends could Cara be potentially sitting between?
|
28
|
deepscale
| 8,381
| ||
Given the sequence $\{a_n\}$ such that the sum of the first $n$ terms is $S_n$, $S_1=6$, $S_2=4$, $S_n>0$ and $S_{2n}$, $S_{2n-1}$, $S_{2n+2}$ form a geometric progression, while $S_{2n-1}$, $S_{2n+2}$, $S_{2n+1}$ form an arithmetic progression. Determine the value of $a_{2016}$.
Options:
A) $-1009$
B) $-1008$
C) $-1007$
D) $-1006$
|
-1009
|
deepscale
| 26,822
| ||
Given the graph of the function $y=\cos (x+\frac{4\pi }{3})$ is translated $\theta (\theta > 0)$ units to the right, and the resulting graph is symmetrical about the $y$-axis, determine the smallest possible value of $\theta$.
|
\frac{\pi }{3}
|
deepscale
| 29,716
| ||
Find the terminating decimal expansion of $\frac{11}{125}$.
|
0.088
|
deepscale
| 38,253
| ||
There are 30 students in Mrs. Taylor's kindergarten class. If there are twice as many students with blond hair as with blue eyes, 6 students with blond hair and blue eyes, and 3 students with neither blond hair nor blue eyes, how many students have blue eyes?
|
11
|
deepscale
| 35,427
| ||
For what values of the parameter \( a \in \mathbb{R} \) does the maximum distance between the roots of the equation
\[ a \tan^{3} x + \left(2-a-a^{2}\right) \tan^{2} x + \left(a^{2}-2a-2\right) \tan x + 2a = 0 \]
belonging to the interval \(\left(-\frac{\pi}{2} , \frac{\pi}{2}\right)\), take the smallest value? Find this smallest value.
|
\frac{\pi}{4}
|
deepscale
| 31,308
| ||
All positive integers whose digits add up to 12 are listed in increasing order: $39, 48, 57, ...$. What is the twelfth number in that list?
|
165
|
deepscale
| 18,775
| ||
In triangle \(ABC\), angle bisectors \(AA_{1}\), \(BB_{1}\), and \(CC_{1}\) are drawn. \(L\) is the intersection point of segments \(B_{1}C_{1}\) and \(AA_{1}\), \(K\) is the intersection point of segments \(B_{1}A_{1}\) and \(CC_{1}\). Find the ratio \(LM: MK\) if \(M\) is the intersection point of angle bisector \(BB_{1}\) with segment \(LK\), and \(AB: BC: AC = 2: 3: 4\). (16 points)
|
11/12
|
deepscale
| 28,265
| ||
Let $x_1,$ $x_2,$ $x_3,$ $x_4,$ $x_5$ be the roots of the polynomial $f(x) = x^5 + x^2 + 1,$ and let $g(x) = x^2 - 2.$ Find
\[g(x_1) g(x_2) g(x_3) g(x_4) g(x_5).\]
|
-23
|
deepscale
| 37,000
| ||
The line $y = \frac{5}{3} x - \frac{17}{3}$ is to be parameterized using vectors. Which of the following options are valid parameterizations?
(A) $\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \end{pmatrix} + t \begin{pmatrix} -3 \\ -5 \end{pmatrix}$
(B) $\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 17 \\ 5 \end{pmatrix} + t \begin{pmatrix} 6 \\ 10 \end{pmatrix}$
(C) $\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2 \\ -7/3 \end{pmatrix} + t \begin{pmatrix} 3/5 \\ 1 \end{pmatrix}$
(D) $\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 14/5 \\ -1 \end{pmatrix} + t \begin{pmatrix} 1 \\ 3/5 \end{pmatrix}$
(E) $\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ -17/3 \end{pmatrix} + t \begin{pmatrix} 15 \\ -25 \end{pmatrix}$
Enter the letters of the correct options, separated by commas.
|
\text{A,C}
|
deepscale
| 39,920
| ||
Jack, Jill, and John play a game in which each randomly picks and then replaces a card from a standard 52 card deck, until a spades card is drawn. What is the probability that Jill draws the spade? (Jack, Jill, and John draw in that order, and the game repeats if no spade is drawn.)
|
The desired probability is the relative probability that Jill draws the spade. In the first round, Jack, Jill, and John draw a spade with probability $1 / 4,3 / 4 \cdot 1 / 4$, and $(3 / 4)^{2} \cdot 1 / 4$ respectively. Thus, the probability that Jill draws the spade is $$\frac{3 / 4 \cdot 1 / 4}{1 / 4+3 / 4 \cdot 1 / 4+(3 / 4)^{2} \cdot 1 / 4}=\frac{12}{37}$$
|
\frac{12}{37}
|
deepscale
| 3,962
| |
Twelve chess players played a round-robin tournament. Each player then wrote 12 lists. In the first list, only the player himself was included, and in the $(k+1)$-th list, the players included those who were in the $k$-th list as well as those whom they defeated. It turned out that each player's 12th list differed from the 11th list. How many draws were there?
|
54
|
deepscale
| 13,216
| ||
Let $(a_1,a_2,a_3,\ldots,a_{14})$ be a permutation of $(1,2,3,\ldots,14)$ where $a_1 > a_2 > a_3 > a_4 > a_5 > a_6 > a_7$ and $a_7 < a_8 < a_9 < a_{10} < a_{11} < a_{12} < a_{13} < a_{14}$. An example of such a permutation is $(7,6,5,4,3,2,1,8,9,10,11,12,13,14)$. Determine the number of such permutations.
|
1716
|
deepscale
| 28,981
| ||
Let \(ABCD\) be a square of side length 2. Let points \(X, Y\), and \(Z\) be constructed inside \(ABCD\) such that \(ABX, BCY\), and \(CDZ\) are equilateral triangles. Let point \(W\) be outside \(ABCD\) such that triangle \(DAW\) is equilateral. Let the area of quadrilateral \(WXYZ\) be \(a+\sqrt{b}\), where \(a\) and \(b\) are integers. Find \(a+b\).
|
\(WXYZ\) is a kite with diagonals \(XZ\) and \(WY\), which have lengths \(2 \sqrt{3}-2\) and 2, so the area is \(2 \sqrt{3}-2=\sqrt{12}-2\).
|
10
|
deepscale
| 4,904
| |
In a circle, parallel chords of lengths 5, 12, and 13 determine central angles of $\theta$, $\phi$, and $\theta + \phi$ radians, respectively, where $\theta + \phi < \pi$. If $\sin \theta$, which is a positive rational number, is expressed as a fraction in lowest terms, what is the sum of its numerator and denominator?
|
18
|
deepscale
| 21,859
| ||
Between 1000 and 9999, the number of four-digit integers with distinct digits where the absolute difference between the first and last digit is 2.
|
840
|
deepscale
| 26,609
| ||
Given the parabola $y^{2}=2x$ with focus $F$, a line passing through $F$ intersects the parabola at points $A$ and $B$. If $|AB|= \frac{25}{12}$ and $|AF| < |BF|$, determine the value of $|AF|$.
|
\frac{5}{6}
|
deepscale
| 30,540
| ||
Arrange the letters a, a, b, b, c, c into three rows and two columns, with the requirement that each row has different letters and each column also has different letters, and find the total number of different arrangements.
|
12
|
deepscale
| 10,365
| ||
A mathematician is working on a geospatial software and comes across a representation of a plot's boundary described by the equation $x^2 + y^2 + 8x - 14y + 15 = 0$. To correctly render it on the map, he needs to determine the diameter of this plot.
|
10\sqrt{2}
|
deepscale
| 24,326
| ||
Some bugs are sitting on squares of $10\times 10$ board. Each bug has a direction associated with it **(up, down, left, right)**. After 1 second, the bugs jump one square in **their associated**direction. When the bug reaches the edge of the board, the associated direction reverses (up becomes down, left becomes right, down becomes up, and right becomes left) and the bug moves in that direction. It is observed that it is **never** the case that two bugs are on same square. What is the maximum number of bugs possible on the board?
|
40
|
deepscale
| 30,617
| ||
For each positive integer $n$, find the number of $n$-digit positive integers that satisfy both of the following conditions:
[list]
[*] no two consecutive digits are equal, and
[*] the last digit is a prime.
[/list]
|
To solve this problem, we need to determine the number of \( n \)-digit positive integers that meet two criteria:
1. No two consecutive digits are equal.
2. The last digit is a prime number.
### Step 1: Count All \( n \)-Digit Numbers
The total number of \( n \)-digit numbers is \( 9 \times 10^{n-1} \). The first digit can be any non-zero digit (1 to 9), giving us 9 choices. Each subsequent digit can be any digit from 0 to 9. Thus, the total number of \( n \)-digit numbers is given by:
\[
9 \times 10^{n-1}
\]
### Step 2: Account for the Last Digit Being Prime
The last digit must be a prime number. The single-digit prime numbers are 2, 3, 5, and 7. There are 4 choices for the last digit to be prime.
### Step 3: Ensure No Two Consecutive Digits Are Equal
To ensure no two consecutive digits are equal, the first digit is chosen from 9 options (1 to 9, as it cannot be 0). Each subsequent digit is selected from 9 possibilities as well, since it cannot equal the previous digit. Therefore, for an arbitrary \( n \)-digit number, this gives us:
\[
9 \times 9^{n-1}
\]
### Step 4: Calculate the Efficient Case
Considering all \( n \)-digit numbers with no two consecutive digits equal, and ensuring the last digit is a prime, we use our last two observations to find the solution.
Since the condition of ending in a prime digit affects only the last digit, we independently multiply the choices for valid numbers by the fraction of these digits:
\[
\frac{4}{10} \cdot 9 \times 9^{n-1}
\]
The term \(\frac{4}{10}\) (or \(\frac{2}{5}\)) represents the probability that the last digit chosen is prime, within the context of all digits.
Thus, the count becomes:
\[
\frac{4}{10} \cdot 9^n = \frac{2}{5} \cdot 9^n
\]
### Step 5: Consider Alternating Sign Factor
Account for alternating sign factors based on parity of \( n \):
Upon careful analysis of different conditions on \( n \) (whether \( n \) is even or odd), we find that the adjustment \(\frac{2}{5} \cdot (-1)^n\) effectively corrects the overcounting.
The final answer is:
\[
\boxed{\frac{2}{5} \cdot 9^n - \frac{2}{5} \cdot (-1)^n}
\]
|
\frac{2}{5} \cdot 9^n - \frac{2}{5} \cdot (-1)^n
|
deepscale
| 6,122
| |
Define the function $f: \mathbb{R} \rightarrow \mathbb{R}$ by $$f(x)= \begin{cases}\frac{1}{x^{2}+\sqrt{x^{4}+2 x}} & \text { if } x \notin(-\sqrt[3]{2}, 0] \\ 0 & \text { otherwise }\end{cases}$$ The sum of all real numbers $x$ for which $f^{10}(x)=1$ can be written as $\frac{a+b \sqrt{c}}{d}$, where $a, b, c, d$ are integers, $d$ is positive, $c$ is square-free, and $\operatorname{gcd}(a, b, d)=1$. Find $1000 a+100 b+10 c+d$.
|
If $x \in(-\sqrt[3]{2}, 0]$, it is evidently not a solution, so let us assume otherwise. Then, we find $$f(x)=\frac{\sqrt{x^{4}+2 x}-x^{2}}{2 x}$$ which implies that $x f(x)^{2}+x^{2} f(x)-1 / 2=0$, by reverse engineering the quadratic formula. Therefore, if $x>0, f(x)$ is the unique positive real $t$ so that $x t^{2}+x^{2} t=1 / 2$. However, then $x$ is the unique positive real so that $x t^{2}+x^{2} t=1 / 2$, so $f(t)=x$. This implies that if $x>0$, then $f(f(x))=x$. Suppose that $f^{10}(x)=1$. Then, since $f(x)>0$, we find that $f(x)=f^{10}(f(x))=f^{11}(x)=f(1)$. Conversely, if $f(x)=f(1)$, then $f^{10}(x)=f^{9}(f(x))=f^{9}(f(1))=1$, so we only need to solve $f(x)=f(1)$. This is equivalent to $x^{2}+\sqrt{x^{4}+2 x}=1+\sqrt{3} \Longleftrightarrow \sqrt{x^{4}+2 x}=1+\sqrt{3}-x^{2} \Longrightarrow x^{4}+2 x=x^{4}-2(1+\sqrt{3}) x^{2}+(1+\sqrt{3})^{2}$, which is equivalent to $$2(1+\sqrt{3}) x^{2}+2 x-(1+\sqrt{3})^{2}=0$$ Obviously, if $x=1$ then $f(x)=f(1)$, so we already know 1 is a root. This allows us to easily factor the quadratic and find that the other root is $-\frac{1+\sqrt{3}}{2}$. This ends up not being extraneous-perhaps the shortest way to see this is to observe that if $x=-\frac{1+\sqrt{3}}{2}$, $$1+\sqrt{3}-x^{2}=(1+\sqrt{3})\left(1-\frac{1+\sqrt{3}}{4}\right)>0$$ so since we already know $$x^{4}+2 x=\left(1+\sqrt{3}-x^{2}\right)^{2}$$ we have $$\sqrt{x^{4}+2 x}=1+\sqrt{3}-x^{2}$$ Therefore, the sum of solutions is $\frac{1-\sqrt{3}}{2}$.
|
932
|
deepscale
| 5,179
| |
Circle $T$ has its center at point $T(-2,6)$. Circle $T$ is reflected across the $y$-axis and then translated 8 units down. What are the coordinates of the image of the center of circle $T$?
|
(2, -2)
|
deepscale
| 35,741
| ||
In a checkered square with a side length of 2018, some cells are painted white and the rest are black. It is known that from this square, one can cut out a 10x10 square where all the cells are white, and a 10x10 square where all the cells are black. What is the smallest value for which it is guaranteed that one can cut out a 10x10 square in which the number of black and white cells differ by no more than?
|
10
|
deepscale
| 14,473
| ||
Find all the functions $f: \mathbb{R} \to\mathbb{R}$ such that
\[f(x-f(y))=f(f(y))+xf(y)+f(x)-1\]
for all $x,y \in \mathbb{R} $.
|
We are given a functional equation for functions \( f: \mathbb{R} \to \mathbb{R} \):
\[
f(x - f(y)) = f(f(y)) + x f(y) + f(x) - 1
\]
for all \( x, y \in \mathbb{R} \). We seek to find all possible functions \( f \) that satisfy this equation.
### Step 1: Notice Special Cases
First, we test with \( x = 0 \):
\[
f(-f(y)) = f(f(y)) + f(y) + f(0) - 1
\]
This helps us to express \( f(f(y)) \) in terms of other values.
Next, try \( y = 0 \):
\[
f(x - f(0)) = f(f(0)) + x f(0) + f(x) - 1
\]
This equation depends on \( f(0) \) and helps provide information about the structure of \( f \).
### Step 2: Consider Possible Simplifications
Assume a linear form for \( f \). Consider \( f(x) = ax^2 + bx + c \) and solve it to match the equation.
### Step 3: Test Specific Guesses
Let's try a special form, like \( f(x) = 1 - \frac{x^2}{2} \), the solution given in the reference answer.
Substitute back into the original equation
Then substitute \( f \) and check if it satisfies the equation:
For \( f(x) = 1 - \frac{x^2}{2} \), we compute:
- \( f(y) = 1 - \frac{y^2}{2} \)
- \( f(f(y)) = 1 - \frac{{\left(1 - \frac{y^2}{2}\right)}^2}{2} = 1 - \frac{1}{2} \left(1 - y^2 + \frac{y^4}{4}\right) = 1 - \frac{1}{2} \left(\frac{y^4}{4} - y^2 + 1\right)\)
Place these in the left and right sides:
1. **Left-hand side**:
\[
f(x - f(y)) = f\left(x - \left(1 - \frac{y^2}{2}\right)\right) = f\left(x - 1 + \frac{y^2}{2}\right) = 1 - \frac{\left(x - 1 + \frac{y^2}{2}\right)^2}{2}
\]
2. **Right-hand side**:
\[
f(f(y)) + x f(y) + f(x) - 1 = \left(1 - \frac{{(f(y))}^2}{2}\right) + x \left(1 - \frac{y^2}{2}\right) + 1 - \frac{x^2}{2} - 1
\]
Simplify and verify that each side is equal.
### Conclusion
The function \( f(x) = 1 - \frac{x^2}{2} \) satisfies the functional equation and therefore is the solution:
\[
\boxed{1 - \frac{x^2}{2}}
\]
|
f(x)=1-\dfrac{x^2}{2}
|
deepscale
| 6,337
| |
Let $x$, $y$ and $z$ all exceed $1$ and let $w$ be a positive number such that $\log_x w = 24$, $\log_y w = 40$ and $\log_{xyz} w = 12$. Find $\log_z w$.
|
Converting all of the logarithms to exponentials gives $x^{24} = w, y^{40} =w,$ and $x^{12}y^{12}z^{12}=w.$ Thus, we have $y^{40} = x^{24} \Rightarrow z^3=y^2.$ We are looking for $\log_z w,$ which by substitution, is $\log_{y^{\frac{2}{3}}} y^{40} = 40 \div \frac{2}{3} =\boxed{60}.$
~coolmath2017
~Lucas
|
60
|
deepscale
| 6,416
| |
Given that the boat is leaking water at the rate of 15 gallons per minute, the maximum time to reach the shore is 50/15 minutes. If Amy rows towards the shore at a speed of 2 mph, then every 30 minutes she increases her speed by 1 mph, find the maximum rate at which Boris can bail water in gallons per minute so that they reach the shore safely without exceeding a maximum capacity of 50 gallons.
|
14
|
deepscale
| 26,085
| ||
If the line $ax+by-1=0$ ($a>0$, $b>0$) passes through the center of symmetry of the curve $y=1+\sin(\pi x)$ ($0<x<2$), find the smallest positive period for $y=\tan\left(\frac{(a+b)x}{2}\right)$.
|
2\pi
|
deepscale
| 18,331
| ||
Given Daphne's four friends visit her every 4, 6, 8, and 10 days respectively, all four friends visited her yesterday, calculate the number of days in the next 365-day period when exactly two friends will visit her.
|
129
|
deepscale
| 25,844
| ||
Given a square pyramid \(M-ABCD\) with a square base such that \(MA = MD\), \(MA \perp AB\), and the area of \(\triangle AMD\) is 1, find the radius of the largest sphere that can fit into this square pyramid.
|
\sqrt{2} - 1
|
deepscale
| 25,434
| ||
The Seattle weather forecast suggests a 60 percent chance of rain each day of a five-day holiday. If it does not rain, then the weather will be sunny. Stella wants exactly two days to be sunny during the holidays for a gardening project. What is the probability that Stella gets the weather she desires? Give your answer as a fraction.
|
\frac{4320}{15625}
|
deepscale
| 12,815
| ||
How many four-digit numbers are divisible by 17?
|
530
|
deepscale
| 20,507
| ||
Given an arithmetic-geometric sequence $\{ a_{n} \}$ that satisfies $a\_1 + a\_3 = 10$, $a\_2 + a\_4 = 5$, find the maximum value of the product $a\_1 a\_2 \ldots a\_n$.
|
64
|
deepscale
| 26,612
| ||
Suppose the function $f(x) = ax + \frac{x}{x-1}$ where $x > 1$.
(1) If $a > 0$, find the minimum value of the function $f(x)$.
(2) If $a$ is chosen from the set \{1, 2, 3\} and $b$ is chosen from the set \{2, 3, 4, 5\}, find the probability that $f(x) > b$ always holds true.
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\frac{5}{6}
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| 21,606
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Denote by $\mathbb{Z}^2$ the set of all points $(x,y)$ in the plane with integer coordinates. For each integer $n \geq 0$, let $P_n$ be the subset of $\mathbb{Z}^2$ consisting of the point $(0,0)$ together with all points $(x,y)$ such that $x^2 + y^2 = 2^k$ for some integer $k \leq n$. Determine, as a function of $n$, the number of four-point subsets of $P_n$ whose elements are the vertices of a square.
|
The answer is $5n+1$.
We first determine the set $P_n$. Let $Q_n$ be the set of points in $\mathbb{Z}^2$ of the form $(0, \pm 2^k)$ or $(\pm 2^k, 0)$ for some $k \leq n$. Let $R_n$ be the set of points in $\mathbb{Z}^2$ of the form $(\pm 2^k, \pm 2^k)$ for some $k \leq n$ (the two signs being chosen independently). We prove by induction on $n$ that \[ P_n = \{(0,0)\} \cup Q_{\lfloor n/2 \rfloor} \cup R_{\lfloor (n-1)/2 \rfloor}. \] We take as base cases the straightforward computations \begin{align*} P_0 &= \{(0,0), (\pm 1, 0), (0, \pm 1)\} \\ P_1 &= P_0 \cup \{(\pm 1, \pm 1)\}. \end{align*} For $n \geq 2$, it is clear that $\{(0,0)\} \cup Q_{\lfloor n/2 \rfloor} \cup R_{\lfloor (n-1)/2 \rfloor} \subseteq P_n$, so it remains to prove the reverse inclusion. For $(x,y) \in P_n$, note that $x^2 + y^2 \equiv 0 \pmod{4}$; since every perfect square is congruent to either 0 or 1 modulo 4, $x$ and $y$ must both be even. Consequently, $(x/2, y/2) \in P_{n-2}$, so we may appeal to the induction hypothesis to conclude.
We next identify all of the squares with vertices in $P_n$. In the following discussion, let $(a,b)$ and $(c,d)$ be two opposite vertices of a square, so that the other two vertices are \[ \left( \frac{a-b+c+d}{2}, \frac{a+b-c+d}{2} \right) \] and \[ \left( \frac{a+b+c-d}{2}, \frac{-a+b+c+d}{2} \right). \]
\begin{itemize}
\item Suppose that $(a,b) = (0,0)$. Then $(c,d)$ may be any element of $P_n$ not contained in $P_0$. The number of such squares is $4n$.
\item Suppose that $(a,b), (c,d) \in Q_k$ for some $k$. There is one such square with vertices \[ \{(0, 2^k), (0, 2^{-k}), (2^k, 0), (2^{-k}, 0)\} \] for $k = 0,\dots,\lfloor \frac{n}{2} \rfloor$, for a total of $\lfloor \frac{n}{2} \rfloor + 1$. To show that there are no others, by symmetry it suffices to rule out the existence of a square with opposite vertices $(a,0)$ and $(c,0)$ where $a > \left| c \right|$. The other two vertices of this square would be $((a+c)/2, (a-c)/2)$ and $((a+c)/2, (-a+c)/2)$. These cannot belong to any $Q_k$, or be equal to $(0,0)$, because $|a+c|, |a-c| \geq a - |c| > 0$ by the triangle inequality. These also cannot belong to any $R_k$ because $(a + |c|)/2 > (a - |c|)/2$. (One can also phrase this argument in geometric terms.)
\item Suppose that $(a,b), (c,d) \in R_k$ for some $k$. There is one such square with vertices \[ \{(2^k, 2^k), (2^k, -2^k), (-2^k, 2^k), (-2^k, -2^k)\} \] for $k=0,\dots, \lfloor \frac{n-1}{2} \rfloor$, for a total of $\lfloor \frac{n+1}{2} \rfloor$. To show that there are no others, we may reduce to the previous case: rotating by an angle of $\frac{\pi}{4}$ and then rescaling by a factor of $\sqrt{2}$ would yield a square with two opposite vertices in some $Q_k$ not centered at $(0,0)$, which we have already ruled out.
\item It remains to show that we cannot have $(a,b) \in Q_k$ and $(c,d) \in R_k$ for some $k$. By symmetry, we may reduce to the case where $(a,b) = (0, 2^k)$ and $(c,d) = (2^\ell, \pm 2^\ell)$. If $d>0$, then the third vertex $(2^{k-1}, 2^{k-1} + 2^\ell)$ is impossible. If $d<0$, then the third vertex $(-2^{k-1}, 2^{k-1} - 2^\ell)$ is impossible.
\end{itemize}
Summing up, we obtain \[ 4n + \left\lfloor \frac{n}{2} \right\rfloor + 1 + \left\lfloor \frac{n+1}{2} \right\rfloor = 5n+1 \] squares, proving the claim.
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5n+1
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| 5,721
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The Fahrenheit temperature ( $F$ ) is related to the Celsius temperature ( $C$ ) by $F = \tfrac{9}{5} \cdot C + 32$ . What is the temperature in Fahrenheit degrees that is one-fifth as large if measured in Celsius degrees?
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-4
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| 23,651
| ||
On March 12, 2016, the fourth Beijing Agriculture Carnival opened in Changping. The event was divided into seven sections: "Three Pavilions, Two Gardens, One Belt, and One Valley." The "Three Pavilions" refer to the Boutique Agriculture Pavilion, the Creative Agriculture Pavilion, and the Smart Agriculture Pavilion; the "Two Gardens" refer to the Theme Carnival Park and the Agricultural Experience Park; the "One Belt" refers to the Strawberry Leisure Experience Belt; and the "One Valley" refers to the Yanshou Ecological Sightseeing Valley. Due to limited time, a group of students plans to visit "One Pavilion, One Garden, One Belt, and One Valley." How many different routes can they take for their visit? (Answer with a number)
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144
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deepscale
| 21,076
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Ten distinct positive real numbers are given and the sum of each pair is written (So 45 sums). Between these sums there are 5 equal numbers. If we calculate product of each pair, find the biggest number $k$ such that there may be $k$ equal numbers between them.
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Given ten distinct positive real numbers, consider all distinct pairs \((a_i, a_j)\) where \(1 \leq i < j \leq 10\). For each pair, we calculate the sum \(S_{ij} = a_i + a_j\). We are informed that among these 45 sums, 5 of them are equal.
Next, we need to analyze the products \(P_{ij} = a_i \cdot a_j\) of these pairs and determine the largest possible value of \(k\) such that there may be \(k\) equal products.
### Analysis
1. **Given:** There are 5 equal sums among the sums \(S_{ij} = a_i + a_j\). Let's denote these equal sums by \(c\). Thus, there exist 5 distinct pairs \((a_i, a_j)\) such that:
\[
a_i + a_j = c.
\]
2. **Number of Pairs:** With 10 distinct numbers, there are \(\binom{10}{2} = 45\) unique pairs. The problem specifies that some pairs share the same sum.
3. **Finding Equal Products:** We now consider the product set \(\{P_{ij} = a_i \cdot a_j\}\) for these 45 pairs. We need to find the largest possible \(k\) such that \(k\) products can be equal.
4. **Investigate Matching Products:** Consider the 5 pairs \((a_{i_1}, a_{j_1}), (a_{i_2}, a_{j_2}), \ldots, (a_{i_5}, a_{j_5})\) with equal sum \(c\). If any two pairs are identical (i.e., \(a_{i} = a_{j} = \frac{c}{2}\)), the product \(a_i \cdot a_j\) will also be the same. These conditions suggest potentially having multiple identical products.
5. **Combination Analysis:** Each number can appear in at most 9 pairs. Given the constraint of sums, one must analyze the overlap in pairs and potential pairwise symmetry to maximize repeated products.
6. **Solving for Maximum Equal Products:** The optimal scenario for product maximal repetition due to symmetry is when the setup allows for such pairwise balance. Given symmetry or duplication through alternative pairings:
\[
k = 4.
\]
### Conclusion
The maximum number \(k\) of equal \(P_{ij}\) is determined through strategically pairing symmetrically balanced numbers such that their products can repeat up to a degree of \(k = 4\).
Thus, the maximum value of \(k\) is:
\[
\boxed{4}
\]
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4
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| 6,099
| |
Given Tom paid $180, Dorothy paid $200, Sammy paid $240, and Alice paid $280, and they agreed to split the costs evenly, calculate the amount that Tom gave to Sammy minus the amount that Dorothy gave to Alice.
|
20
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| 28,378
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A certain function $f$ has the properties that $f(3x) = 3f(x)$ for all positive real values of $x$, and that $f(x) = 1 - |x - 2|$ for $1\leq x \leq 3$. Find the smallest $x$ for which $f(x) = f(2001)$.
|
429
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| 36,849
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The solutions to the equations $z^2 = 9 + 9\sqrt{7}i$ and $z^2 = 5 + 5\sqrt{2}i$, where $i = \sqrt{-1}$, form the vertices of a parallelogram in the complex plane. Determine the area of this parallelogram, which can be written in the form $p\sqrt{q} - r\sqrt{s}$, where $p, q, r, s$ are positive integers and neither $q$ nor $s$ is divisible by the square of any prime number.
A) $\frac{2}{3}\sqrt{35} - \frac{2}{3}\sqrt{7}$
B) $\frac{3}{4}\sqrt{35} - \frac{3}{4}\sqrt{15}$
C) $\frac{1}{2}\sqrt{21} - 3\sqrt{3}$
D) $\frac{3}{8}\sqrt{15} - \frac{3}{8}\sqrt{7}$
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\frac{3}{4}\sqrt{35} - \frac{3}{4}\sqrt{15}
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| 30,403
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A plan is to transport 1240 tons of goods A and 880 tons of goods B using a fleet of trucks to a certain location. The fleet consists of two different types of truck carriages, A and B, with a total of 40 carriages. The cost of using each type A carriage is 6000 yuan, and the cost of using each type B carriage is 8000 yuan.
(1) Write a function that represents the relationship between the total transportation cost (y, in ten thousand yuan) and the number of type A carriages used (x);
(2) If each type A carriage can carry a maximum of 35 tons of goods A and 15 tons of goods B, and each type B carriage can carry a maximum of 25 tons of goods A and 35 tons of goods B, find all possible arrangements of the number of type A and type B carriages to be used according to this requirement;
(3) Among these arrangements, which one has the minimum transportation cost, and what is the minimum cost?
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26.8
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| 27,070
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What is the least common multiple of 6, 8, and 10?
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120
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| 39,102
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