算額(その527)
和算図形問題あれこれ - 令和4年8月の問題-No.2
https://gunmawasan.web.fc2.com/kongetu-no-mondai.html
(7) 滋賀県 吉田博氏宅 p.116
近畿数学史学会:近畿の算額「数学の絵馬を訪ねて」,平成4年5月16日 初版第一刷,大阪教育図書株式会社,大阪市.
キーワード:円10個,累円
外円内に甲円,乙円,丙円,丁円,戊円が入っている。外円が 40 寸のとき,丙円を最も大きくしたときの戊円の直径を求めよ。
外円の半径と中心座標を r0, (0, 0)
甲円の半径と中心座標を r1, (0, r0 - r1)
乙円の半径と中心座標を r2, (r2, y2)
丙円の半径と中心座標を r3, (x3, y3)
丁円の半径と中心座標を r4, (x4, y4)
戊円の半径と中心座標を r5, (x5, y5)
とおき,以下の連立方程式を解く r1 について解く。その後 r3についての式をr1で微分し導関数が0になるときの r1 を求める。そのときの r1 で表される r5 が求める値である。
しかし,連立方程式の記号解が求まらないので,数値微分を用いて同じことをする。
include("julia-source.txt");
using SymPy
@syms r0::positive, r1::positive, y1::positive,
r2::positive, y2,
r3::positive, x3::positive, y3,
r4::positive, x4::positive, y4,
r5::positive, x5::positive, y5;
r0 = 40//2
y1 = r0 - r1
x2 = r2
eq1 = r2^2 + y2^2 - (r0 - r2)^2
eq3 = x3^2 + y3^2 - (r0 - r3)^2
eq6 = x4^2 + y4^2 - (r0 - r4)^2
eq9 = x5^2 + y5^2 - (r0 - r5)^2
eq2 = r2^2 + (y1 - y2)^2 - (r1 + r2)^2
eq4 = x3^2 + (y1 - y3)^2 - (r1 + r3)^2
eq7 = x4^2 + (y1 - y4)^2 - (r1 + r4)^2
eq10 = x5^2 + (y1 - y5)^2 - (r1 + r5)^2
eq5 = (x3 - x2)^2 + (y3 - y2)^2 - (r3 + r2)^2
eq8 = (x4 - x3)^2 + (y4 - y3)^2 - (r4 + r3)^2
eq11 = (x5 - x4)^2 + (y5 - y4)^2 - (r5 + r4)^2;
# res = solve([eq1, eq2, eq3, eq4, eq5, eq6, eq7, eq8, eq9, eq10, eq11], (r2, y2, r3, x3, y3, r4, x4, y4, r5, x5, y5))
using NLsolve
function nls(r1, func, params...; ini = [0.0])
if typeof(ini) <: Number
r = nlsolve((vout, vin) -> vout[1] = func(r1, vin[1], params..., [ini]), ftol=big"1e-40")
v = r.zero[1]
else
r = nlsolve((vout, vin)->vout .= func(r1, vin, params...), ini, ftol=big"1e-40")
v = r.zero
end
return v, r.f_converged
end;
function H(r1, u)
(r2, y2, r3, x3, y3, r4, x4, y4, r5, x5, y5) = u
return [
r2^2 + y2^2 - (20 - r2)^2, # eq1
r2^2 - (r1 + r2)^2 + (-r1 - y2 + 20)^2, # eq2
x3^2 + y3^2 - (20 - r3)^2, # eq3
x3^2 - (r1 + r3)^2 + (-r1 - y3 + 20)^2, # eq4
(-r2 + x3)^2 - (r2 + r3)^2 + (-y2 + y3)^2, # eq5
x4^2 + y4^2 - (20 - r4)^2, # eq6
x4^2 - (r1 + r4)^2 + (-r1 - y4 + 20)^2, # eq7
-(r3 + r4)^2 + (-x3 + x4)^2 + (-y3 + y4)^2, # eq8
x5^2 + y5^2 - (20 - r5)^2, # eq9
x5^2 - (r1 + r5)^2 + (-r1 - y5 + 20)^2, # eq10
-(r4 + r5)^2 + (-x4 + x5)^2 + (-y4 + y5)^2, # eq11
]
end;
numerical_derivative(f, x, h=1e-10) = (f(x + h) - f(x - h)) / (2 * h)
numerical_derivative2(f, x, h=1e-10) = (f(x + h) - 2 * f(x) + f(x - h)) / h^2
r1 = big"10"
while true
iniv = BigFloat[6.1, -13.3, 4.8, 15.3, -5.7, 3.4, 16.0, 4.6, 2.3, 15.2, 9.5]
h = 1e-10
r3 = nls(r1, H, ini=iniv)[1][3]
r3l = nls(r1 - h, H, ini=iniv)[1][3]
r3r = nls(r1 + h, H, ini=iniv)[1][3]
numerical_dervative = (r3r - r3l)/2h
numerical_dervative2 = (nls(r1 + h, H, ini=iniv)[1][3] -2nls(r1, H, ini=iniv)[1][3] + nls(r1 - h, H, ini=iniv)[1][3])/h^2
abs(numerical_dervative/numerical_dervative2) < big"1e-20" && break
r1 -= numerical_dervative/numerical_dervative2
println("r1 = $r1")
end
println("r1 = $r1")
r1 = 12.48538011695906438403271202135121844125836007027810600838640181720678979984704
r1 = 12.01201496172095097990312251445735848809684416949930658448600519799368331897538
r1 = 12.00000898032969429019237242846760671180689989079357424987938861517199168254016
r1 = 12.00000000000504037748540527008744540545918004544761142950056638766185544102556
r1 = 11.99999999999999999999979325438981227049694304973714727017681791708893764281108
r1 = 11.99999999999999999999979325438981227049694304973714727017681791708893764281108
甲円の半径が 12 のとき,乙円が最大で,戊円の半径を求める。
r1 = 12
iniv = BigFloat[6.1, -13.3, 4.8, 15.3, -5.7, 3.4, 16.0, 4.6, 2.3, 15.2, 9.5]
res = nls(r1, H, ini=iniv)
(BigFloat[7.500000000000000000000000000000000000000000000000000000000000000420264272338678, -10.00000000000000000000000000000000000000000000000000000000000000168105708935471, 4.999999999999999999999999999999999999999999999995121239468409802261948427817372, 15.00000000000000000000000000000000000000000000004390884457514065793291839983194, 1.951504212636079095220628871952364045461575089304458517289222454862992166675505e-47, 2.999999999999999999999999999999999999999999998473295509659893374550847421982661, 15.00000000000000000000000000000000000000000000329535771623506103989092567879276, 8.000000000000000000000000000000000000000000006106817961360426501796610311927722, 1.874999999999999999999999999999999999999999998197629693673087468497217798802652, 13.12499999999999999999999999999999999999999999841365663452270670651293440891112, 12.5000000000000000000000000000000000000000000072094812253076501260111288044371], true)
戊円の半径は 1.875(直径は 3.75)である。
その他のパラメータは以下の通り。
r2 = 7.5; y2 = -10; r3 = 5; x3 = 15; y3 = 1.9515e-47; r4 = 3; x4 = 15; y4 = 8; r5 = 1.875; x5 = 13.125; y5 = 12.5
甲円の直径 = 24; 丙円の直径 = 10; 戊円の直径 = 3.75
function draw(more=false)
pyplot(size=(500, 500), grid=false, aspectratio=1, label="", fontfamily="IPAMincho")
(r2, y2, r3, x3, y3, r4, x4, y4, r5, x5, y5) = res[1]
r0 = 40//2
x1 = 0
y1 = r0 - r1
x2 = r2
@printf("r2 = %g; y2 = %g; r3 = %g; x3 = %g; y3 = %g; r4 = %g; x4 = %g; y4 = %g; r5 = %g; x5 = %g; y5 = %g\n", r2, y2, r3, x3, y3, r4, x4, y4, r5, x5, y5)
@printf("甲円の直径 = %g; 丙円の直径 = %g; 戊円の直径 = %g\n", 2r1, 2r3, 2r5)
plot()
circle(-r2, y2, r2, :blue)
circle(x3, y3, r3, :magenta)
circle(-x3, y3, r3, :magenta)
circle(x4, y4, r4, :brown)
circle(-x4, y4, r4, :brown)
circle(x5, y5, r5, :orange)
circle(-x5, y5, r5, :orange)
if more
delta = (fontheight = (ylims()[2]- ylims()[1]) / 500 * 10 * 2) /3 # size[2] * fontsize * 2
hline!([0], color=:black, lw=0.5)
vline!([0], color=:black, lw=0.5)
point(x1, y1, " 甲円", mark=false)
point(x2, y2, "乙円", mark=false)
point(x3, y3, "丙円", mark=false)
point(x4, y4, "丁円", mark=false)
point(x5, y5, "戊円", mark=false)
end
end;