SAT 1000 problems series - Upcoming frequently

Now I think it's time to publish all of the 1000 problems of my problem set. They come from some particularly hard, popular or interesting Gaokao problems.

From then on, I will post all of the problems here, the order of the problem would be initially disarranged, but eventually I would complete all of my the problem set.

These problems will follow the same format so that it will be easier for the programs to index them and use them.

I'm thinking using the problem set for my games or other purposes.

Free to use the problem set as a dataset for machine learning or using the problems in your contest, games or programs, just indicate the source :)

Category:

P1-P159 - Functions

P160-P257 - Derivatives

P268-P370 - Trigonometry

P371-P436 - Vectors

P437-P503 - Sequences

P504-P562 - Inequalities

P563-P669 - Solid Geometry

P670-P846 - Analytic Geometry

P847-P861 - Modeling

P862-P941 - Math Insight

P942-P1000 - Miscellaneous

What do you have in Gaokao?

  • A pen/pencil

  • Draft paper

  • Your brilliant mind

What can't you use in Gaokao?

  • Calculator (Except for the submit process)

  • Numerical method

  • Mathematica, Geogebra or other related software/websites

Your pen/pencil, draft paper are the only things you can use. There you go!

Note by Alice Smith
4 weeks, 1 day ago

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SAT1000 - P159

If f(x)f(x) and g(x)g(x) are functions defined at R\mathbb R, and equation xf(g(x))=0x-f(g(x))=0 has real roots.

Then which cannot be the function g(f(x))g(f(x))?

A. x2+x15A.\ x^2+x-\dfrac{1}{5}

B. x2+x+15B.\ x^2+x+\dfrac{1}{5}

C. x215C.\ x^2-\dfrac{1}{5}

D. x2+15D.\ x^2+\dfrac{1}{5}

Alice Smith - 3 weeks, 5 days ago

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SAT1000 - P333

Given that f(x)=cos2x,g(x)=sinxf(x)=\cos 2x, g(x)=\sin x.

Find all possible real number aa and positive integer nn, so that f(x)+ag(x)=0f(x)+a g(x)=0 has exactly 20132013 roots on the interval (0,nπ)(0,n \pi).

How to submit:

  • First, find the number of all possible solutions (a,n)(a,n). Let NN denote the number of solutions.

  • Then sort the solutions by aa from smallest to largest, if aa is the same, then sort by nn from smallest to largest.

  • Let the sorted solutions be: (a1,n1),(a2,n2),(a3,n3),,(aN,nN)(a_1,n_1), (a_2,n_2), (a_3,n_3), \cdots ,(a_N,n_N), then M=k=1Nk(ak+nk)M=\displaystyle \sum_{k=1}^N k(a_k+n_k) .

For instance, if the solution is: (1,2),(1,1),(1,3),(0,4)(-1,2), (-1,1), (1,3), (0,4)

Then the sorted solution will be: (1,1),(1,2),(0,4),(1,3)(-1,1), (-1,2), (0,4), (1,3)

Then N=4N=4, M=k=14k(ak+nk)=1×(1+1)+2×(1+2)+3×(0+4)+4×(1+3)=30M=\displaystyle \sum_{k=1}^4 k(a_k+n_k)= 1 \times (-1+1) + 2 \times (-1+2) + 3 \times (0+4) + 4 \times (1+3) =30 .

For this problem, submit M+N\lfloor M+N \rfloor.

Alice Smith - 2 weeks, 3 days ago

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SAT1000 - P592

As shown above, a 5×5×55 \times 5 \times 5 cube has been punched through so that there will always be two 1×11 \times 1 square holes when looking from up, front and left.

Find the surface area of this solid. (Including the interior part)

Alice Smith - 4 weeks, 1 day ago

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Total surface area of cube when there is no hole is 6×5×5=1506\times 5\times 5 = 150

On punching 11 hole, we reduce the surface area of cube's face by 22(both the ends). So on punching 66 holes, total of 1212 sq. units area will reduce. So, till now we are left with 15012=138150 - 12 = 138 sq. units of surface area.

Punching holes also generates some surface areas. First consider just one hole. This hole creates 44 walls each of area 1×5=51\times 5 = 5. So area created by 11 hole only is 4×5=204\times 5 = 20.

If there were no intersections of holes then calculation was easier. But here there are few intersections of holes. Each hole is intersected by 22 holes and there are total 66 holes. So total number of intersections =262=6\displaystyle = \frac{2\cdot 6}{2} = 6.

Consider a hole 11 which is intersected by hole 22 and hole 33. As said above, hole 11 will create 2020 sq. units area. But at intersections, hole 22 and hole 33 makes additional 22 holes each in hole 11 thus reducing the surface area of hole 11 by 44. So till now there are 204=1620 - 4 = 16 sq. units surface in hole 11.

There are total 66 holes, so total surface area created =16×6=96= 16\times 6 = 96. But at each intersection, 22 squares of unit square area is counted in both the intersecting holes. That means 22 sq. units area is counted twice at every intersections. There are 66 intersections which means total of 1212 sq. units area is counted twice. So we have to subtract this from 9696. So total surface area created by all the six holes is 9612=8496 - 12 = 84.

So total surface area in whole cube will be 138+84=222138 + 84 = 222.

Shikhar Srivastava - 4 weeks, 1 day ago

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SAT1000 - P594

As shown above, in triangular pyramid ABCDA-BCD, the cross section AEFAEF passes through the center of the inscribed sphere OO (i.e. The sphere which is tangent to all of the faces of the solid) of ABCDA-BCD, and it intersects with BC,DCBC, DC at point E,FE,F respectively.

If the cross section divides the pyramid into two parts whose volumes are equal, S1S_1 is the surface area of solid ABEFDA-BEFD, S2S_2 is the surface area of solid AEFCA-EFC, what is always true for S1S_1 and S2S_2?

Alice Smith - 4 weeks, 1 day ago

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SAT1000 - P759

As shown above, the line: x3y+m=0 (m0)x-3y+m=0\ (m \neq 0) intersects with the two asymptotes of the hyperbola:

x2a2y2b2=1 (a>0,b>0)\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1\ (a>0,b>0) at point A,BA,B.

If point PP is at (m,0)(m,0) and PA=PB|PA|=|PB|, find the eccentricity of the hyperbola.

Let EE denote the eccentricity, submit 1000E\lfloor 1000E \rfloor.

Alice Smith - 4 weeks ago

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SAT1000 - P760

As shown above, F1,F2F_1, F_2 are left and right focus of the hyperbola: x2a2y2b2=1 (a,b>0)\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1\ (a,b>0) respectively, and B(0,b)B(0,b).

Line F1BF_1B intersects with the two asymptotes of the hyperbola at P,QP,Q, and the perpendicular bisector of PQPQ intersects with x-axis at point MM.

If MF2=F1F2|MF_2|=|F_1F_2|, then find the eccentricity of the hyperbola.

Let EE denote the eccentricity, submit 1000E\lfloor 1000E \rfloor.

Alice Smith - 4 weeks ago

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SAT1000 - P765

Let A,BA,B be the end points of the major axis of the ellipse C:x23+y2m=1C:\dfrac{x^2}{3}+\dfrac{y^2}{m}=1.

If there exists point MM on the ellipse so that AMB=2π3\angle AMB = \dfrac{2\pi}{3}, find the range of mm.

These pictures show the two cases:

Alice Smith - 4 weeks ago

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SAT1000 - P767

As shown above, the hyperbola: x2a2y2b2=1 (a>0,b>0)\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1\ (a>0,b>0) has right focus point FF, right vertex AA.

Line l1l_1 passes through FF and it intersects with the hyperbola at point B,CB,C, l2l_2 passes through BB and l2ACl_2 \perp AC, l3l_3 passes through CC and l2ABl_2 \perp AB, l2,l3l_2, l_3 intersects at point DD.

If the distance from DD to line l1l_1 is less than a+a2+b2a+\sqrt{a^2+b^2}, what is the range of the slope of the hyperbola's asymptotes?

Alice Smith - 4 weeks ago

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SAT1000 - P766

As shown above, the left and right focus of the ellipse: x2a2+y2b2=1(a>b>0)\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 (a>b>0) are F1(c,0),F2(c,0)F_1(-c,0), F_2(c,0).

If there exists point PP such that asinPF1F2=csinPF2F1\dfrac{a}{\sin \angle PF_1F_2}=\dfrac{c}{\sin \angle PF_2F_1}, find the range of the eccentricity of the ellipse.

The range can be expressed as (l,r)(l,r). Submit 1000(2rl)\lfloor 1000(2r-l) \rfloor.

Alice Smith - 4 weeks ago

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SAT1000 - P606

As shown above, in ABC\triangle ABC, AB=BC=2AB=BC=2, ABC=2π3\angle ABC = \dfrac{2\pi}{3}.

If PP is outside plane ABCABC and point DD is on segment ACAC, so that PD=DA,PB=BAPD=DA, PB=BA, then find the maximum volume of pyramid PBCDPBCD.

Let VV denote the volume of PBCDPBCD, submit 10000V\lfloor 10000V \rfloor.

Alice Smith - 3 weeks, 6 days ago

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SAT1000 - P648

As shown above, in pyramid ABCDABCD, ADBCAD \perp BC, if AD=BC=2AD=BC=2, AB+BD=AC+CD=4AB+BD=AC+CD=4, then find the maximum volume for pyramid ABCDABCD.

Let VV denote the maximum volume. Submit 1000V\lfloor 1000V \rfloor.

Alice Smith - 3 weeks, 6 days ago

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SAT1000 - P787

As shown above, the focus of the parabola C:y2=2xC: y^2=2x is FF. Two lines l1,l2l_1, l_2 which are parallel to the x-axis intersect with CC at A,BA,B and intersect with the directix at P,QP,Q. MM is the midpoint of ABAB.

If SPQF=2SABFS_{\triangle PQF}=2S_{\triangle ABF}, find the locus of point MM.

If the locus can be expressed as f(x,y)=0f(x,y)=0, then when y=50y=50, submit the sum of all possible value(s) for xx.

Note: SABCS_{\triangle ABC} denotes the area of ABC\triangle ABC.

Alice Smith - 3 weeks, 6 days ago

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SAT1000 - P801

As shown above, the ellipse EE has equation: x216+y212=1\dfrac{x^2}{16}+\dfrac{y^2}{12}=1, and point CC is the center of the circle: x2+y24x+2=0x^2+y^2-4x+2=0 If PP is a point on ellipse EE, l1,l2l_1, l_2 both pass through PP and they are both tangent to circle CC, and the product of the slope of l1,l2l_1, l_2 is equal to 12\dfrac{1}{2}, then find the all possible coordinate(s) of point PP.

How to submit:

  • First, find the number of all possible solutions. Let NN denote the number of solutions.

  • Then Sort the solutions by x-coordinate from smallest to largest, if the x-coordinate is the same, then sort by y-coordinate from smallest to largest.

  • Let the sorted solutions be: (x1,y1),(x2,y2),(x3,y3),,(xN,yN)(x_1,y_1), (x_2,y_2), (x_3,y_3), \cdots ,(x_N,y_N), then M=k=1Nk(xk+yk)M=\displaystyle \sum_{k=1}^N k(x_k+y_k) .

For instance, if the solution is: (1,2),(1,1),(1,3),(0,4)(-1,2), (-1,1), (1,3), (0,4)

Then the sorted solution will be: (1,1),(1,2),(0,4),(1,3)(-1,1), (-1,2), (0,4), (1,3)

Then N=4N=4, M=k=14k(xk+yk)=1×(1+1)+2×(1+2)+3×(0+4)+4×(1+3)=30M=\displaystyle \sum_{k=1}^4 k(x_k+y_k)= 1 \times (-1+1) + 2 \times (-1+2) + 3 \times (0+4) + 4 \times (1+3) =30 .

For this problem, submit 1000(M+N)\lfloor 1000(M+N) \rfloor.

Alice Smith - 3 weeks, 6 days ago

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SAT1000 - P802

As shown above, the parabola C1:x2=yC_1: x^2=y, circle C2:x2+(y4)2=1C_2: x^2+{(y-4)}^2=1, and MM is the center of circle C2C_2.

Point PP is a point on C1C_1 (not at (0,0)(0,0)), and l1,l2l_1, l_2 are two lines tangent to C2C_2 and they intersects with C1C_1 at point A,BA,B respectively. Line ll passes through MM and PP.

If lABl \perp AB, find the equation of line ll.

The equation can be expressed as: y=±kx+b (k>0)y=\pm k x+b\ (k>0). Submit 1000(k+b)\lfloor 1000(k+b) \rfloor.

Alice Smith - 3 weeks, 6 days ago

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Can you do some SAT algebra questions? @Alice Smith

A Former Brilliant Member - 3 weeks, 5 days ago

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They are on my plan. There you go:)

Of course, they are not really SAT problems, but adapted from GaoKao problems. If you can do them all, you can definitely ace the SAT:)

Alice Smith - 3 weeks, 5 days ago

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SAT1000 - P158

There exist function f(x)f(x) such that xR\forall x \in \mathbb R, the property follows:

A. f(sin2x)=sinxA.\ f(\sin 2x)=\sin x

B. f(sin2x)=x2+xB.\ f(\sin 2x)=x^2+x

C. f(x2+1)=x+1C.\ f(x^2+1)=|x+1|

D. f(x2+2x)=x+1D.\ f(x^2+2x)=|x+1|

Alice Smith - 3 weeks, 5 days ago

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SAT1000 - P156

Let f1(x)=x2,f2(x)=2(xx2),f3(x)=13sin2πxf_1(x)=x^2, f_2(x)=2(x-x^2), f_3(x)=\dfrac{1}{3}|\sin 2\pi x|.

If Ik=i=199fk(i99)fk(i199)I_k=\displaystyle \sum_{i=1}^{99} |f_k(\dfrac{i}{99})-f_k(\dfrac{i-1}{99})|, compare I1,I2,I3I_1, I_2, I_3.

A. I1<I2<I3A.\ I_1<I_2<I_3

B. I2<I1<I3B.\ I_2<I_1<I_3

C. I1<I3<I2C.\ I_1<I_3<I_2

D. I3<I2<I1D.\ I_3<I_2<I_1

Alice Smith - 3 weeks, 5 days ago

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SAT1000 - P370

Given that in ABC\triangle ABC, sin2A+sin(AB+C)=sin(CAB)+12\sin 2A+\sin(A-B+C)=\sin(C-A-B)+\dfrac{1}{2}, SS is the area of ABC\triangle ABC, 1S21 \leq S \leq 2, let a,b,ca,b,c be the opposite side of angle A,B,CA,B,C respectively.

Which inequality always holds?

A. bc(b+c)>8A.\ bc(b+c)>8

B. ab(a+b)>162B.\ ab(a+b)>16 \sqrt{2}

C. 6abc12C.\ 6 \leq abc \leq 12

D. 12abc24D.\ 12 \leq abc \leq 24

Alice Smith - 3 weeks, 5 days ago

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SAT1000 - P527

Given that a,bR,a+b=2,b>0a,b \in \mathbb R, a+b=2, b>0, then find the minimum value of 12a+ab\dfrac{1}{2|a|}+\dfrac{|a|}{b}.

Let MM be the minimum value. Submit 1000M\lfloor 1000M \rfloor.

Alice Smith - 3 weeks, 5 days ago

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SAT1000 - P548

Given that x,yRx,y \in \mathbb R and x2+y21x^2+y^2 \leq 1, find the minimum value of 2x+y2+6x3y|2x+y-2|+|6-x-3y|.

Let MM be the minimum value. Submit 1000M\lfloor 1000M \rfloor.

Alice Smith - 3 weeks, 5 days ago

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SAT1000 - P807

As shown above, parabola CC has equation: y2=4xy^2=4x and its focus is FF.

Line l1,l2l_1, l_2 both passes through FF, l1l_1 intersects with CC at A,BA,B, l2l_2 intersects with CC at D,ED,E, l1l2l_1 \perp l_2.

Then find the minimum value of ADEB \overrightarrow{AD} \cdot \overrightarrow{EB} .

Let MM be the minimum value. Submit 1000M\lfloor 1000M \rfloor.

Alice Smith - 3 weeks, 5 days ago

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SAT1000 - P808

As shown above, the ellipse has equation: x24+y2=1\dfrac{x^2}{4}+y^2=1.

If line ll passes through point C(m,0)C(m,0) and is tangent to circle: x2+y2=1x^2+y^2=1.

ll intersects with the ellipse at point A,BA,B, then find the maximum value of AB|AB|.

Let MM be the maximum value. Submit 1000M\lfloor 1000M \rfloor.

Alice Smith - 3 weeks, 5 days ago

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SAT1000 - P810

As shown above, the ellipse C:x24+y23=1C: \dfrac{x^2}{4}+\dfrac{y^2}{3}=1, O(0,0),P(2,1)O(0,0), P(2,1), line ll intersects with CC at point A,BA,B, and line OPOP bisects segment ABAB.

If the area of APB\triangle APB reaches the maximum value, then find the equation of line ll.

The equation of the line can be expressed as y=kx+by=kx+b, submit 1000(k+b)\lfloor 1000(k+b) \rfloor.

Alice Smith - 3 weeks, 4 days ago

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SAT1000 - P811

As shown above, given that ellipse E:x2t+y23=1 (t>3)E: \dfrac{x^2}{t}+\dfrac{y^2}{3}=1\ (t>3), point AA is the left vertex of EE, and line ll whose slope is k (k>0)k\ (k>0) intersects with EE at point A,MA,M, point NN is a point on EE such that MANAMA \perp NA.

If 2AM=AN2|AM|=|AN|, find the range of kk as tt changes.

If the range can be expressed as (l,r)(l,r), submit 1000(2rl)\lfloor 1000(2r-l) \rfloor.

Alice Smith - 3 weeks, 4 days ago

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SAT1000 - P819

As shown above, given that F(1,0)F(1,0) is the right focus of the ellipse: x2a2+y2b2=1 (a>b>0)\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\ (a>b>0), OO is the origin.

If for all lines passing through FF which intersect with the ellipse at point A,BA,B, the following inequality always holds:

OA2+OB2<AB2|OA|^2+|OB|^2<|AB|^2

Then find the range of aa.

If the range can be expressed as: (l,+)(l,+\infty), submit 1000l\lfloor 1000l \rfloor.

Alice Smith - 3 weeks, 4 days ago

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SAT1000 - P296

Find the range of the function f(x)=sinx132cosx2sinx (x[0,2π])f(x)=\dfrac{\sin x-1}{\sqrt{3-2\cos x-2\sin x}}\ (x \in [0,2\pi]).

If the range can be expressed as [l,r][l,r], submit 1000(2rl)\lfloor 1000(2r-l) \rfloor.

Alice Smith - 3 weeks, 4 days ago

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Note: It is identical to Let's get a little trigy, since it happens to be on my problem set, I decided to repost it.

Alice Smith - 3 weeks, 4 days ago

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SAT1000 - P320

Given the function f(x)=sin(ωx+ϕ) (ω>0,ϕπ2)f(x)= \sin(\omega x+\phi)\ (\omega>0, |\phi| \leq \dfrac{\pi}{2}), f(π4)=0,f(π4)=0f(-\dfrac{\pi}{4})=0, f'(\dfrac{\pi}{4})=0, and f(x)f(x) is strictly monotone on the interval (π18,5π36)(\dfrac{\pi}{18}, \dfrac{5\pi}{36}), then find the maximum value of ω\omega.

Alice Smith - 3 weeks, 4 days ago

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SAT1000 - P321

Given the function f(x)=sin(ωx+π3) (ω>0)f(x)=\sin(\omega x + \dfrac{\pi}{3})\ (\omega>0), f(π6)=f(π3)f(\dfrac{\pi}{6})=f(\dfrac{\pi}{3}).

If f(x)f(x) has minimum value but no maximum value on the interval (π6,π3)(\dfrac{\pi}{6},\dfrac{\pi}{3}), then find the value of ω\omega.

If ω=ab\omega=\dfrac{a}{b}, where a,ba,b are positive coprime integers. submit a+ba+b.

Alice Smith - 3 weeks, 4 days ago

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SAT1000 - P322

Given the function f(x)=sin2ωx2+12sinωx12 (ω>0)f(x)=\sin^2 \dfrac{\omega x}{2}+\dfrac{1}{2} \sin \omega x - \dfrac{1}{2}\ (\omega>0), xRx \in \mathbb R.

If f(x)=0f(x)=0 has no roots for x(π,2π)x \in (\pi, 2\pi), then find the range of ω\omega.

A. (0,18]A.\ (0,\dfrac{1}{8}]

B. (0,14][58,1]B.\ (0,\dfrac{1}{4}] \cup [\dfrac{5}{8},1]

C. (0,58]C.\ (0,\dfrac{5}{8}]

D. (0,18][14,58]D.\ (0,\dfrac{1}{8}] \cup [\dfrac{1}{4},\dfrac{5}{8}]

Alice Smith - 3 weeks, 4 days ago

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SAT1000 - P838

As shown above, the ellipse has equation: x2+y22=1x^2+\dfrac{y^2}{2}=1, F(0,1)F(0,1), line l:y=2x+1l: y=-\sqrt{2} x + 1 intersects with the ellipse at point A,BA,B, and PP is a point on the ellipse so that OA+OB+OP=0\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OP}=\textbf 0.

Point P,QP,Q are symmetry about point OO, and it turns out A,P,B,QA,P,B,Q are on the same circle.

The radius of the circle is rr. Submit 1000r\lfloor 1000r \rfloor.

Alice Smith - 2 weeks, 5 days ago

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SAT1000 - P839

As shown above, point FF is the focus of the parabola C:y2=4xC: y^2=4x, and line ll passing through K(1,0)K(-1,0) intersects with CC at point A,BA,B, and A,DA,D are symmetry about the x-axis.

If FAFB=89\overrightarrow{FA} \cdot \overrightarrow{FB}=\dfrac{8}{9}, then find the equation of the incircle of BDK\triangle BDK.

The equation can be expressed as: (xa)2+(yb)2=r2(x-a)^2+(y-b)^2=r^2. Submit 1000(a+b+r)\lfloor 1000(a+b+r) \rfloor.

Alice Smith - 2 weeks, 5 days ago

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SAT1000 - P842

As shown above, the ellipse has equation: x24+y22=1\dfrac{x^2}{4}+\dfrac{y^2}{2}=1, and line ll passing through P(0,1)P(0,1) intersects with the ellipse at point A,BA,B.

Then there exists a fixed point QQ so that the following equation always holds as ll rotates:

QAQB=PAPB\dfrac{|QA|}{|QB|}=\dfrac{|PA|}{|PB|}

Then find the coordinate of QQ.

The coordinate of QQ is (x0,y0)(x_0,y_0). Submit 1000(2y0x0)\lfloor 1000(2y_0-x_0) \rfloor.

Alice Smith - 2 weeks, 5 days ago

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SAT1000 - P843

As shown above, the ellipse has equation: x22+y2=1\dfrac{x^2}{2}+y^2=1, line l:y=k1x32l: y=k_1 x-\dfrac{\sqrt{3}}{2} intersects with the ellipse at point A,BA,B.

Point CC is on the ellipse and line OCOC has slope k2k_2, k1k2=24k_1 k_2=\dfrac{\sqrt{2}}{4}.

MM is a point on ray OCOC, MC:AB=2:3|MC| : |AB| = 2 : 3, and the radius of circle MM is MC|MC|, OS,OTOS,OT are two tangent lines of circle MM and S,TS,T are tangent points.

Then find the maximum value of SOT\angle SOT (in radians), and find the slope of ll when SOT\angle SOT reaches the maximum.

Let θ\theta be the maximum value of SOT\angle SOT (in radians), kk is the slope of ll. Submit 1000(θ+k)\lfloor 1000(\theta+|k|)\rfloor.

Alice Smith - 2 weeks, 5 days ago

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SAT1000 - P844

As shown above, F(1,0)F(1,0) is the focus of the parabola C:y2=4xC: y^2=4x, line l:y=k(x1)l: y=k(x-1) intersects with CC at point A,BA,B, and the perpendicular bisector of ABAB intersects with CC at point M,NM,N.

If A,M,B,NA,M,B,N are on the same circle, then find the value of 1000k\lfloor 1000|k| \rfloor.

Alice Smith - 2 weeks, 5 days ago

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SAT1000 - P845

As shown above, the parabola E:y2=xE: y^2=x and circle M:(x4)2+y2=r2M: (x-4)^2+y^2=r^2 for r>0r>0 intersect at points AA, BB, CC, and DD; and their relative positions are shown in the figure. Lines ACAC and BDBD intersect at point PP.

Find the coordinates (x0,0)(x_0,0) of PP as the area of quadrilateral ABCDABCD reaches the maximum when rr varies. Submit 1000x0\lfloor 1000x_0 \rfloor.

The coordinate of point PP is (x0,0)(x_0,0). Submit 1000x0\lfloor 1000x_0 \rfloor.

Alice Smith - 2 weeks, 5 days ago

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SAT1000 - P846

As shown above, ON,NMON, NM are rigid rods and DN=ON=1,MN=3DN=ON=1,MN=3, O(0,0)O(0,0) is fixed on the coordinate plane, and DD is restricted along the x-axis. Then as DD moves horizontally, point MM will rotate around point OO. Curve CC is the locus of point MM.

If line ll intersects with l1:x2y=0l_1:x-2y=0 at point PP, l2:x+2y=0l_2:x+2y=0 at point QQ, and ll is tangent to curve CC.

Then find the minimum area of OPQ\triangle OPQ when line ll moves and rotates.

Let SS be the minimum area. Submit 1000S\lfloor 1000S \rfloor.

Alice Smith - 2 weeks, 4 days ago

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SAT1000 - P491

Given that f(x)=11+xf(x)=\dfrac{1}{1+x}. {an}\{a_n\} is a sequence whose terms are all positive so that a1=1,an+2=f(an)a_1=1, a_{n+2}=f(a_n).

If a2010=a2012a_{2010}=a_{2012}, find the value of 1000(a20+a11)\lfloor 1000(a_{20}+a_{11}) \rfloor.

Alice Smith - 2 weeks, 4 days ago

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SAT1000 - P492

{an}\{a_n\} is a sequence such that an+1+(1)nan=2n1a_{n+1}+(-1)^n a_n=2n-1, then find k=160ak\displaystyle \sum_{k=1}^{60} a_k.

Alice Smith - 2 weeks, 4 days ago

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SAT1000 - P496

Given that an=n2(cos2nπ3sin2nπ3) (nN+)a_n=n^2(\cos^2 \dfrac{n\pi}{3}-\sin^2 \dfrac{n\pi}{3})\ (n \in \mathbb N^+), let Sn=k=1nakS_n=\displaystyle \sum_{k=1}^{n} a_k.

bn=S3nn4n (nN+)b_n=\dfrac{S_{3n}}{n \cdot 4^n}\ (n \in \mathbb N^+), Tn=k=1nbkT_n=\displaystyle \sum_{k=1}^{n} b_k.

T10=pqT_{10} = \dfrac{p}{q}, where p,qp,q are positive coprime integers.

Submit pq+2(S100+S201+S302)\lfloor p-q+2(S_{100}+S_{201}+S_{302}) \rfloor.

Alice Smith - 2 weeks, 4 days ago

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SAT1000 - P500

{an}\{a_n\} is a sequence such that a1=m (mN+)a_1=m\ (m \in \mathbb N^+), an+1={an2, an0(mod2)3an+1, an1(mod2)a_{n+1}=\begin{cases} \dfrac{a_n}{2} ,\ a_n \equiv 0 \pmod{2} \\ 3 a_n+1 ,\ a_n \equiv 1 \pmod{2} \end{cases} If a6=1a_6=1, find the sum of all possible value(s) for mm.

Alice Smith - 2 weeks, 4 days ago

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SAT1000 - P461

Given that {an}\{a_n\} is a Geometric Sequence, its common ratio q=2q=\sqrt{2}, Sn=k=1nakS_n=\displaystyle \sum_{k=1}^n a_k.

Let Tn=17SnS2nan+1 (nN+)T_n=\dfrac{17 S_n - S_{2n}}{a_{n+1}}\ (n \in \mathbb N^+). If TmT_{m} is the maximum term of sequence {Tn}\{T_n\}, find mm.

Alice Smith - 2 weeks, 4 days ago

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SAT1000 - P153

f(x)f(x) is a function defined at [0,1][0,1] such that:

  • f(0)=f(1)=0f(0)=f(1)=0.

  • x,y[0,1] (xy),f(x)f(y)<12xy\forall x,y \in [0,1]\ (x \neq y), |f(x)-f(y)| < \dfrac{1}{2} |x-y|.

If x,y[0,1],f(x)f(y)<k\forall x,y \in [0,1], |f(x)-f(y)|<k, find the minimum value of kk.

Let KK be the minimum value. Submit 1000K\lfloor 1000K \rfloor.

Alice Smith - 2 weeks, 4 days ago

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SAT1000 - P329

As shown above, the circle has radius r=1 mr=1\ m, OO is its center. At t=0t=0, OO is at (0,1)(0,-1), and it is moving at v=1 m/sv=1\ m/s upwards along the y-axis. Let x(t)x(t) be the length of the arc above the x-axis, f(t)=cos(x(t))f(t)=\cos (x(t)).

For 0t10 \leq t \leq 1, what is the best graph for f(t)f(t)?

Alice Smith - 2 weeks, 3 days ago

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SAT1000 - P275

Without using calculator, Find cos10°tan20°+3sin10°tan70°2cos40°\dfrac{\cos 10 \degree}{\tan 20 \degree}+\sqrt{3} \sin 10 \degree \tan 70 \degree - 2 \cos 40 \degree.

Let AA denote the answer. Submit 1000A\lfloor 1000A \rfloor.

Alice Smith - 2 weeks, 3 days ago

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SAT1000 - P276

If tanα=2tanπ5\tan \alpha=2 \tan \dfrac{\pi}{5}, find cos(α3π10)sin(απ5)\dfrac{\cos(\alpha-\dfrac{3\pi}{10})}{\sin(\alpha-\dfrac{\pi}{5})}.

Let AA denote the answer. Submit 1000A\lfloor 1000A \rfloor.

Alice Smith - 2 weeks, 3 days ago

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SAT1000 - P277

Without using calculator, Find 4cos50°tan40°4 \cos 50 \degree - \tan 40 \degree.

Let AA denote the answer. Submit 1000A\lfloor 1000A \rfloor.

Alice Smith - 2 weeks, 3 days ago

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SAT1000 - P866

For all real number x,yx,y, which is always true?

A. x=xA.\ \lfloor -x \rfloor=-\lfloor x \rfloor

B. 2x=2xB.\ \lfloor 2x \rfloor=2\lfloor x \rfloor

C. x+yx+yC.\ \lfloor x+y \rfloor \leq \lfloor x \rfloor + \lfloor y \rfloor

D. xyxyD.\ \lfloor x-y \rfloor \leq \lfloor x \rfloor - \lfloor y \rfloor

Alice Smith - 2 weeks, 3 days ago

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SAT1000 - P873

If there exists tRt \in \mathbb R, so that the following system of nn equations all holds:

{t=1t2=2t3=3tn=n\begin{cases} \lfloor t \rfloor = 1 \\ \lfloor t^2 \rfloor = 2 \\ \lfloor t^3 \rfloor = 3 \\ \cdots \\ \lfloor t^n \rfloor = n \end{cases}

Then find the maximum of positive integer nn.

Alice Smith - 2 weeks, 3 days ago

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SAT1000 - P876

Given that an infinite sequence {an}\{a_n\} consists of kk distinct values, Sn=i=1naiS_n=\displaystyle \sum_{i=1}^n a_i.

If nN+\forall n \in \mathbb N^+, Sn{2,3}S_n \in \{2,3\}, then find the maximum of kk.

Alice Smith - 2 weeks, 3 days ago

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SAT1000 - P877

Here's the definition of harmonically bisect: Given that A1,A2,A3,A4A_1, A_2, A_3, A_4 are four distinct points on the coordinate plane, if A1A3=λA1A2\overrightarrow{A_1A_3}=\lambda \overrightarrow{A_1A_2}, A1A4=μA1A2\overrightarrow{A_1A_4}=\mu \overrightarrow{A_1A_2}, 1λ+1μ=2\dfrac{1}{\lambda}+\dfrac{1}{\mu}=2, then A3,A4A_3, A_4 harmonically bisect A1,A2A_1, A_2.

Given that C(c,0),D(d,0) (c,dR)C(c,0), D(d,0)\ (c,d \in \mathbb R) harmonically bisect A(0,0),B(1,0)A(0,0), B(1,0), which choice is true?

A. C could be the midpoint of AB.A.\ \textup{C could be the midpoint of AB.}

B. D could be the midpoint of AB.B.\ \textup{D could be the midpoint of AB.}

C. C, D could be both on segment AB.C.\ \textup{C, D could be both on segment AB.}

D. C, D can’t be both on the extension line of AB.D.\ \textup{C, D can't be both on the extension line of AB.}

Alice Smith - 2 weeks ago

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SAT1000 - P878

Given the function y=f(x) (xR)y=f(x)\ (x \in \mathbb R), for function y=g(x) (xI)y=g(x)\ (x \in I), let's define symmetric function of g(x)g(x) respect to f(x)f(x) as y=h(x) (xI)y=h(x)\ (x \in I), y=h(x)y=h(x) is such that xI\forall x \in I, point (x,h(x)),(x,g(x))(x,h(x)), (x,g(x)) are symmetric about point (x,f(x))(x,f(x)).

GIven that h(x)h(x) is the symmetric function of g(x)=4x2g(x)=\sqrt{4-x^2} respect to f(x)=3x+b (bR)f(x)=3x+b\ (b \in \mathbb R), h(x)>g(x)h(x)>g(x) is always true for all xx on the domain of g(x)g(x), then find the range of bb.

The range can be expressed as (L,+)(L,+\infty), submit L2L^2.

Alice Smith - 2 weeks ago

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SAT1000 - P879

Let AA be the set of all functions whose range is R\mathbb R, BB is the set of all functions ϕ(x)\phi(x) which has the following properties:

  • For ϕ(x)\phi(x), let R0R_0 denote the range of ϕ(x)\phi(x), M(0,+),R0[M,M]\exists M \in (0,+\infty), R_0 \subseteq [-M,M].

It's easy to prove that for ϕ1(x)=x3\phi_1(x)=x^3, ϕ2(x)=sinx\phi_2(x)=\sin x, ϕ1(x)A\phi_1(x) \in A, ϕ2(x)B\phi_2(x) \in B.

Here are the following statements:

  1. Let DD be the domain of f(x)f(x), then the necessary and sufficient condition for f(x)Af(x) \in A is: bR,aD,f(a)=b\forall b \in \mathbb R, \exists a \in D, f(a)=b.

  2. The necessary and sufficient condition for f(x)Bf(x) \in B is f(x)f(x) has the maximum and minimum value.

  3. If f(x),g(x)f(x), g(x) have the same domain, then if f(x)A,g(x)Bf(x) \in A, g(x) \in B, then f(x)+g(x)Bf(x)+g(x) \notin B.

  4. If f(x)=aln(x+2)+xx2+1 (x>2,aR)f(x)=a\ln(x+2)+\dfrac{x}{x^2+1}\ (x>-2, a \in \mathbb R) has the maximum value, then f(x)Bf(x) \in B.

Which statements are true?

How to submit:

Let p1,p2,,pnp_1, p_2,\cdots,p_n be the boolean value of statement 1,2,,n1,2,\cdots,n, if statement kk is true, pk=1p_k=1, else pk=0p_k=0.

Then submit k=1npk2k1\displaystyle \sum_{k=1}^n p_k \cdot 2^{k-1}.

Alice Smith - 2 weeks ago

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SAT1000 - P880

For function f(x),g(x)f(x), g(x) which have the same domain DD, if there exists h(x)=kx+bh(x)=kx+b (k,bk,b are constant), so that:

m(0,+),x0D,xDx>x0,0<f(x)h(x)<m,0<h(x)g(x)<m,\forall m \in (0,+\infty), \exists x_0 \in D, \forall x \in D \wedge x>x_0, 0<f(x)-h(x)<m, 0<h(x)-g(x)<m,

Then line ll: y=kx+by=kx+b is called the bipartite asymptote for curve y=f(x)y=f(x) and y=g(x)y=g(x).

Here are four groups of functions which are defined at (1,+)(1,+\infty):

  1. f(x)=x2,g(x)=xf(x)=x^2, g(x)=\sqrt{x}.

  2. f(x)=10x+2,g(x)=2x3xf(x)=10^{-x}+2, g(x)=\dfrac{2x-3}{x}.

  3. f(x)=x2+1x,g(x)=xlnx+1lnxf(x)=\dfrac{x^2+1}{x}, g(x)=\dfrac{x \ln x+1}{\ln x}.

  4. f(x)=2x2x+1,g(x)=2(x1ex)f(x)=\dfrac{2x^2}{x+1}, g(x)=2(x-1-e^{-x}).

Which groups of curve y=f(x)y=f(x) and y=g(x)y=g(x) has a bipartite asymptote?

How to submit:

Let p1,p2,p3,p4p_1, p_2, p_3, p_4 be the boolean value of the group 1,2,3,41,2,3,4, if group kk has a bipartite asymptote, pk=1p_k=1, else pk=0p_k=0.

Then submit k=14pk2k1\displaystyle \sum_{k=1}^4 p_k \cdot 2^{k-1}.

Alice Smith - 2 weeks ago

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SAT1000 - P884

If set SS is a non-empty subset of integer set Z\mathbb Z, if a,bS,abS\forall a,b \in S, ab \in S, then SS is closed under multiplication.

Given that for set T,UT,U: TZ,VZ,TV=,TV=ZT \subseteq \mathbb Z, V \subseteq \mathbb Z, T \cap V = \emptyset , T \cup V = \mathbb Z, and a,b,cT,abcT\forall a,b,c \in T, abc \in T, x,y,zV,xyzV\forall x,y,z \in V, xyz \in V, then which of the choices is true?

A. At least one of T,V is closed under multiplication.A.\ \textup{At least one of T,V is closed under multiplication.}

B. At most one of T,V is closed under multiplication.B.\ \textup{At most one of T,V is closed under multiplication.}

C. Only one of T,V is closed under multiplication.C.\ \textup{Only one of T,V is closed under multiplication.}

D. Both T,V are closed under multiplication.D.\ \textup{Both T,V are closed under multiplication.}