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1.
Sketch, very simple.
Cut the triangular pyramid along the side AB and flatten these three sides.
The link BB'(B' is point b),
The length of the line segment BB' is the minimum value of the perimeter of the section triangle.
Let ∠BAC=α, then COS α = (4a2+4a2-A2)/(2 * 2a * 2a) = 7/8.
Because ∠BAB'=3∠BAC=3α.
So cos3alpha = 4 (cosα) 3-3cosα = 4 (7/8) 3-3 * (7/8) = 7/128.
So bb' = √ (4a2+4a2-2 * 2a * (7/128)) = (1/4) a.
That is to say, the minimum perimeter of a cross-sectional triangle is (11/4) a.
Calculate the area of △ABB' with Helen formula;
x=2a，y=2a，z=( 1 1/4)a
p=(x+y+z)/2
s(△abb')=((p*(p-x)*(p-y)*(p-z)))^( 1/2)
=[(33√ 15)/64]a^2
So the height of BB' side is 2 * [(33 √15)/64] a 2/[(11/4) a].
h =( 1/ 1 1)*[(33√ 15)/8]a
△ ACD = (1/2) * (√15) A Height h' on the bottom CD.
And because △AEF∽△ACD
So EF/CD=h/h'
So EF=CD*(h/h')=a*(3/4)
So be = FB' = [(11/4) a-(3/4) a]/2 = a.
△BEF (after folding back) According to three sides, BE=a, FB'=a, EF = (3/4) A.
According to Helen's formula, calculate:
S(△BEF)=[(3/64)*√55]a^2
2.
In the sketch, O is the center of the bottom surface, the surface A1b1c1d1is located between the apex and the bottom surface of the cone, and O 1 is the plane A1d/kloc-0.
Let a regular quadrangular prism ABCD-a1b1c1d1be inscribed in a cone SO.
The cross section SEF of the cone passes through the diagonal surface of a regular quadrangular prism.
Then △SEF is the axial section of the cone S.
The quadrilateral AA 1C 1C is the inscribed rectangle of △SEF.
Let the SO intersection of cone A 1C 1 be O 1.
Because OE=R, it =√3R.
So ∠ ESO = 30.
So1= √ 3o1a1.
Let the base length of a regular quadrangular prism be a and the height h.
Then a 1c 1 = √ 2a, oo1= h.
Because o1a1=1/2a1c1= a √ 2/2.
So 1 = a √ 6/2.
So h = √ 3r-a √ 6/2.
Therefore, the surface area of a regular prism s = 2a2+4ah = 2a2+4a (√ 3r-a √ 6/2) = (2-2 √ 6) a2+4 √ 3ar.
Because 2-2 √ 6
So s has a maximum.
When a =-4 √ 3r/2 (2-2 √ 6) = (3 √ 2+√ 3)/5r ∈ (0, √ 2r)
S gets the maximum value = 6 (√ 6+ 1) r 2/5.
3.
Let the center of gravity be: (x0, y0)
Because x0=(xA+xB+xC)/3, y0=(yA+yB+yC)/3.
So xc = 3x0-2-(-1) = 3x0-1,yC=3y0-0-2=3y0-2.
Because (xC, yC) is 2x+y-3=0.
So 2*(3x0- 1)+(3y0-2)-3=0.
6x0-2+3y0-2-3=0
That is, 6x0+3y0-7=0.
So the trajectory equation of the center of gravity G is: 6x+3y-7=0.
4.
Let the coordinates of point n be (a, b).
The coordinates of point m are (c, d).
Because n is a point on the light.
So vector om = k vector on(k >；; 0)
Vector OM=(c, d) vector ON=(a, b)
(c，d)=k(a，b)
c=ka
d=kb
So the coordinate of point m is (ka, kb).
|OM|=√[(ka)？ +(kb)？ ]
=√[k？ (a？ +b？ )]
=k√(a？ +b？ )
|ON|=√(a？ +b？ )
Then | om || open |
=k(a？ +b？ )
= 120
Because the point m is on the circle.
The coordinates of point m satisfy the equation of circle: x? +y？ -6x-8y=0
(ka)？ +(kb)？ -6ka-8kb=0
k(ka？ +kb？ -6a-8b)=0
120-6a-8b=0
3a+4b-60=0
So the trajectory equation of point n is 3x+4y-60=0.
5.
You forgot to write a function in your topic.
6.
f(x)=(√a)sin[( 1-a)x]+cos[( 1-a)x]
= √ (a+1) sin {[(1-a) x]+φ}, where tanφ= 1/(√a).
≤√(a+ 1)
Because the title says that the maximum value of f(x) is 2.
That is, √(a+ 1)=2, that is, a=3.
So f(x)=2sin(-2x+φ), where tanφ= 1/(√3).
So its minimum positive period is: 2π/|-2|=π.
7.
This question is quoted from my brother upstairs.
Sina +sinb=-sinc
cosa+cosb=-cosc
Straighten both sides at the same time
(sina+sinb)^2=(-sinc)^2
(cosa+cosb)^2=(-cosc)^2
Fully expand
(sina)^2+(sinb)^2+2sina*sinb=(sinc)^2
(cosa)^2+(cosb)^2+2cosa*cosb=(cosc)^2
Add it all up.
2(cosa * cosb+Sina * sinb)=- 1
cos(a-b)=- 1/2
Because it is an even function.
cos(b-a)=- 1/2
From 0 < a < b < c < 2π, we get 0 < b-a < 2π, where b-a is the second quadrant or the third quadrant.
So b-a=2π/3 or 4π/3.