※ 引述《will2004 (夏天的風)》之銘言：
: 請問矩形ABCD，AB:AD=1:2，求矩形周長?
: https://imgur.com/a/tpbofgA
: 謝謝
來列式
假設此矩形左下角位於(0,0)
且長2k, 高k
一點(x,y)
x^2+y^2 = 18 (1)
x^2+(k-y)^2 = 10 (2)
(2k-x)^2 + (k-y)^2 = 26 (3)
(3)-(2)
(2k-x)^-x^2 = 16
4k^2-4kx = 16
k^2-kx = 4
kx = k^2-4
x = k-4/k
(1)-(2)
y^2 - k^2 +2ky -y^2 = 8
2ky = 8+k^2
y = k/2 + 4/k
x,y代入(1)
(k-4/k)^2 + (k/2+4/k)^2 = k^2 - 8 + 16/k^2 + k^2/4 + 4 + 16/k^2 = 18
1.25k^2+32/k^2 = 22
let k^2 = A
5A/4 + 32/A = 22
5A^2 + 128 - 88A = 0
A = (88+sqrt(88^2-4*5*128))/10 = (88+72)/10 = 16(只能取正)
k = 4, 周長=2*(4+8) = 24#
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