设直角边分别为a和b,斜边c=√（a^2 + b^2),周长C=a+b+c,则三角形面积S
=ab/2 = ch/2 =(a+b+c)·r/2
则 c·h=(a+b+c)·r;
则 r/h=c/(a+b+c)
= 1/[1+(a+b)/c]
= 1/[1 + (a+b)/√(a^2 + b^2)]
= 1/[1 + √(a^2 + b^2 +2ab)/(a^2 + b^2)]
= 1/{1 + √[1 +2ab/(a^2 + b^2)]}
其中2ab/(a^2 + b^2)的分子和分母同时除以a·b,得
= 1/{1 + √[1 + 2/ ( a/b + b/a )]}
由于 a/b + b/a ≥2√[(a/b)·(b/a )]=2,故
r/h= 1/{1 + √[1 + 2/ ( a/b + b/a )]}
≥ 1/{1 + √[1 + 2/ 2]}
=1/(1+√2)
=√2-1;
而当a/b(或b/a)趋于无穷大时,b/a(或a/b)趋于无穷小,而 a/b + b/a 趋于无穷大,则 2/ ( a/b + b/a )趋于0; 因此r/h= 1/{1 + √[1 + 2/ ( a/b + b/a )]}趋于 1/2,但达不到(否则斜边就会与一条直角边平行)
故r/h的取值范围为
√2-1≤r/h＜ 1/2
或者:
r/h=c/(a+b+c) = 1/[1+(a/c)+(b/c)]
= 1/(1+ sinB + cosB)
= 1/[1+ √2·(√2/2·sinB + √2/2·cosB)]
= 1/[1+ √2·(cos(π/4)·sinB + sin(π/4)·cosB)]
= 1/[1+ √2·sin(B + π/4)]
由于B是锐角,则 B + π/4 ∈(π/4,3π/4)
则sin(B + π/4)∈(√2/2,1]
则 r/h∈[√2-1,1/2)