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μ•„ λ¬Έμžλ§Žλ„€ v0=1, d=1 μ˜€μΌ€μ΄? 쑰였치

첨속 (0, 1)

O~pμ—μ„œ yλ³€μœ„/yμ†μœΌλ‘œ κ΅¬ν•œ μ΄λ™μ‹œκ°„ 2μ΄λ―€λ‘œ p속은 (2a1, 1)이고 xλ³€μœ„κ°€ 1μ΄λ―€λ‘œ 2a1=1, a1=1/2μž„

q속(1, -1) 인데 p~q yμ†λ³€ν™”λŸ‰μ„ κ°€μ†λ„λ‘œ λ‚˜λˆ μ„œ κ΅¬ν•œ μ΄λ™μ‹œκ°„μ€ 2/a2이고 xλ³€μœ„λŠ” 2/a2μž„

q~(8,0)κΉŒμ§€ yλ³€μœ„/yμ†μœΌλ‘œ κ΅¬ν•œ μ΄λ™μ‹œκ°„μ΄ 2μ΄λ―€λ‘œ (0.8d)μ—μ„œ 속(2, -1)이고 xλ³€μœ„λŠ” 3μž„

1+2/a2+3=8, a2=1/2 γ„±λ§žκ³  γ„·λ§žκ³  v=||(2, -1)||=√5 γ„·λ§žλ„€




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A, B, C의 μ „λ₯˜λ₯Ό x y z라고 ν•˜μ…ˆ

Ap = (0,-2√3) => (2√3, 0) ==> (x2√3/12, 0)

Bp = (1, -√3) => (√3, 1) ==> (y√3/4, y/4)

Cp = (-3, -√3) => (√3, -3) ==> (z√3/12, -3z/12)


** μ„€λͺ…: (a, b) => (-b, a)이고

Β ===>λŠ” μ „λ₯˜λ₯Ό κ³±ν•œλ‹€μŒμ— 크기 제곱으둜 λ‚˜λˆˆ 것을 λœ»ν•¨~

λ‹€λ”ν•˜λ©΄ 0이래yo

y/4-z/4=0μ—μ„œ y=z

x2√3/12+y√3/3=0, x=-2y

끝났넀 x=-2, y=1, z=1

Oμ—μ„œ

A = (x/√3, 0)=(-2/√3, 0)

B = (0, y)=(0, 1), y=B0=1

C = (0, -z/3)=(0, -1/3)

|A+B+C|=|(-2/√3, 2/3)|=2|(-1/√3, 1/3)|=4/3



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νž˜ν‘œμ‹œν•΄λ³΄μ…ˆ xμ„±λΆ„, yμ„±λΆ„ λ‚˜λˆŒ λ•Œ λ§‰λŒ€ 쀑심좕을 μ–΄λ””λ‹€ 두든 돌림힘 계산할 λ•Œ λ³΅μž‘ν•΄μ§€κ² λ„€

그럼 좕을 돌렀

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μ„Έ 문제의 곡톡점은? μ’Œν‘œ