一個正n邊形邊長為x,今截去n個角使之成為正2n邊形,求此正2n邊形邊長?
2006-08-11 12:24:12 · 5 個解答 · 發問者 ? 7 in 科學 ➔ 數學
1. 正 n 邊形的內角和 = π ( n - 2 ) 所以,單一內角之角度為 π ( n - 2 ) / n = π - 2π / n2. 假設截去 n 個角的每個角為以 y 為腰邊、 以 π - 2π / n 為頂角的等腰三角形。由餘弦定理得知, 此等腰三角形之底邊 = √( y² + y² - 2.y.y.cos(π - 2π / n) ) = √[ 2y² ( 1 + cos( 2π / n ) ) ]3. 按題意 x - 2y = √[ 2y² ( 1 + cos( 2π / n ) ) ] ⇒ x = 2y + y.√( 2 + 2cos( 2π / n ) ) ⇒ y = x / ( 2 + √( 2 + 2cos( 2π / n ) ) ) = x / ( 2 + √( 2 + 4 cos²( π / n ) - 2 ) ) = x / ( 2 + 2 cos( π / n ) )4. 正三邊形 → 正六邊形 ⇒ y = x / ( 2 + 2.1/2 ) = x / 3 正四邊形 → 正八邊形 ⇒ y = x / ( 2 + 2.1/√2 ) = x / ( 2 + √2 ) n → ∞ 時 , 正 n 邊形 → 正 2n 邊形, y → x / 4
2006-08-11 18:03:34 補充:
真是遭糕,預覽後,忘了打上最後的結果。5. 新正 n 邊形的邊長為 x - 2y = x - 2( x / ( 2 + 2 cos(π/n) ) ) = ( 1 - 1 / ( 1 + cos(π/n) ) ) x = x [ cos(π/n) / ( 1 + cos(π/n) ) ]
2006-08-11 22:53:24 補充:
5. 新正 "2n" 邊形的邊長為 x - 2y 才對
2006-08-11 13:47:25 · answer #1 · answered by 我的日子只有混 5 · 0⤊ 0⤋
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2014-08-07 17:00:03 · answer #2 · answered by Anonymous · 0⤊ 0⤋
bdref43的答案似乎不對ㄝ
2006-08-14 18:24:24 · answer #3 · answered by ? 7 · 0⤊ 0⤋
一個正n邊形和截去n個角使之成為正2n邊形有一相同的內切圓, 半徑為r.
正n邊形邊長為x,x=n*a,
正2n邊形邊長為y,y=2n*b
截角Area:
(a-b)*r/2-b*r/2=(a-2b)*r/2=r(1/cos(Pi/n)-1)*b/2
a-2b=b/cos(pi/n)-b
x/n=(y/2n)(1/cos(pi/n)+1)
y=2*x/(1/cos(Pi/n)+1), Pi=3.1415926 or 180 deg.
Answer: 2x/(1/cos(Pi/n)+1), Pi=3.1415926 or 180 deg.
2006-08-14 21:58:29 補充:
In calculations, x and y were used as circumference. But the answer was not affected.
2006-08-16 18:43:08 補充:
正n邊形邊長為x,內切圓半徑r=x/(2*tan(2Pi/n))正2n邊形邊長y=2r*tan(2Pi/2n)=x*tan(Pi/n)/tan(2Pi/n)Let Pi/n=s, y=x*tan(s)/tan(2s) [1]tan(2s)=2tan(s)/(1-tan(s)^2) [2][2] to [1]: y=x*(1-tan(s)^2)/2 for n>4or y=x*cos(2s)/cos^2(s)/2
2006-08-16 18:46:03 補充:
e.g.n=5, s=36, cos(2s)=0.309, cos(s)=0.654, y=0.361xn=6, s=30, cos(2s)=0.5, cos(s)=0.751, y=0.444xn=8, s=22.5, cos(2s)=0.707, cos(s)=0.812, y=0.485xCorrect answer: x*cos(2Pi/n)/cos^2(Pi/n)/2
2006-08-17 16:14:16 補充:
Try one more time.正n邊形邊長為x,內切圓半徑r=x/(2*tan(2Pi/2n))正2n邊形邊長y=2r*tan(2Pi/4n)=x*tan(Pi/2n)/tan(Pi/n)Let Pi/2n=s, y=x*tan(s)/tan(2s) [1]tan(2s)=2tan(s)/(1-tan(s)^2) [2][2] to [1]: y=x*(1-tan(s)^2)/2 for n>4or y=x*cos(2s)/cos^2(s)/2
2006-08-17 16:41:32 補充:
correct answer: x*cos(Pi/n)/(cos(Pi/2n))^0.5/2Examples:n=3, cos(Pi/n)=0.5, (cos(Pi/2n)=0.866, y=0.333xn=5, cos(Pi/n)=0.809, (cos(Pi/2n)=0.951, y=0.447xn=8, cos(Pi/n)=0.924, (cos(Pi/2n)=0.981, y=0.48xn=20, cos(Pi/n)=0.988, (cos(Pi/2n)=0.997, y=0.497x
2006-08-17 16:46:40 補充:
y varies between 0.3333x and 0.5x, consistent with theory.In the equation, Pi=3.1415926
2006-08-20 00:01:41 補充:
1. 公式應適用於任何n值.2. y應介於最小值(正三角形)0.3333和最大值0.5(n無限大,接近於圓,邊長與弧角成正比)之間.所以答案是y=x*cos(Pi/n)/(cos(Pi/2n))^0.5/2,請忽略先前的兩個不實的答案.
2006-08-14 17:40:48 · answer #4 · answered by ? 7 · 0⤊ 0⤋
答案應該是(x^2+4r^2)^0.5-x
過程太囉嗦+沒啥把握,只好用意見XD
2006-08-11 13:15:51 · answer #5 · answered by 李 3 · 0⤊ 0⤋