[Math] Weekend 1

1-18

Title: D D1 respectively is the midpoint of the line segment BC B1C1 AD midpoint p as the parallel line of BC

Parsing: (1) ab=ac D is BC-to-BC vertical AD

M N is the midpoint mn parallel BC

MN Vertical AD

AA1 vertical Plane ABC MN belongs to plane ABC--AA1 Vertical MN

MN Vertical ADD1A1

(2) Set aa1=1 A1 as a1e parallel to the B1C1 with the A1 bit coordinate origin respectively with E D1 A as XYZ A1 (0,0,0) A (0,0,1) because P is ad--and M N is the midpoint of AB AC

M (gh3/2,1/2,1) N (-gh3/2,1/2,1) a1m= (gh3/2,1/2,1) a1a= (0,0,1) nm= (gh3,0,0)

aa1m normal vector n1 N1 a1m=0 N1 a1a=0--X=1 Y=-GH3

Normal vector (1,-gh3,0)

a1mn normal vector n2 n2 a1m=0 n2 nm=0--y=2 Z=-1

Normal vector (0,2,-1)

Cos=gh15/5

1-20

(1) E2=C2/A2=3/4-a2=4b2-X2+4Y2=4B2

M (x, y)-| Mq|=gh ( -3 (y+1) 2+4b2+12)-Y=-1 | Mq|max=4--b2=1 a2=4-x2/4+y2=1

(2) c:y=x2 N (T,T2) y/=2x--ybc-t2=2t (X-T)--Y=2tx-t2

--(1+16T2) x2-16t3x+4t4-4=0--delta=16 (-t4+16t2+1), X1+x2=16t3/1+16t2 x1x2=4t4-4/(1+16T2)

--|bc|=gh (1+4T2) |x1-x2|=4gh (1+4t2) GH (-t4+16t2+1)/(1+16T2)

BC = = d--d=1+16t2/(16gh (1+4T2))

sabc=1/2| Bc|d=1/8gh (-(t2-8) 2+65) <=GH65/8

T=+-2GH2-SABC=GH65/8

Sabcmax=gh65/8

1-21

(3) x belongs to (0,+WQ) constant has 1/3x3<ex

Make h (x) =1/3x3-ex--H ' (x) =x2-ex

X>0 ex>x2--H ' (x) <0--H (x) in (0,+WQ) monotonically decreasing

-H (x)

Take x0=3/c x>x0-1/cx2<1/cx3<ex

For any c always present x0 when x belongs to (X0,+WQ) constant has X2<cex

1-22

(1) Acb=90 CD Vertical AB--and Cd2=adxdb CD in ACB is tangent------and CD2=CEXCB-cexcb=adxdb

(2) on vertical NF---Nf=gh (of2-on2) of length is fixed, the segment on length min must be solved

The distance from the midpoint of the chord to the center is the shortest n is the be midpoint--nfmax=1/2be=2

[Math] Weekend 1