Cho tứ diện OABC có ba mặt OAB,OAC,OBC là những tam giác vuông tại đỉnh O.Gọi α,β,γ lần lượt là góc giữa mặt phẳng (ABC) với các mặt phẳng (OAB),(OAC),(OBC) Chứng minh rằng : cos2α+cos2β+cos2γ=1

Gọi H là hình chiếu vuông góc của O lên (ABC)
{OAOBOAOCOA(OBC)OABC
Mặt khác vì :
OH(ABC)OHBC(OHA)BCHABC
Chứng minh tương tự ta có HBAC do đó H là trực tâm ΔABC
Đặt  OA=a,OB=b,OC=c và giả sử
AHBC=A1,BHAC=B1,CHAB=C1
suy ra OA1A=α,OBB1=β,OCC1=γ
Trong ΔAOA1 vuông tại O ta có :
cosα=cos^OA1A=cos(900^OA1A)=sin^OAA1=OHOA=OHa
Tương tự ta cũng có : cosβ=OHbcosγ=OHc
Trong các ΔAOA1ΔOBC vuông tại O ta có :
1OH2=1OA2+1OA21=1OA2+1OB2+1OC2=1a2+1b2+1c2
OH2a2+OH2b2+OH2c2=1cos2α+cos2β+cos2γ=1  (đpcm)

Thẻ

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