(1) 2辺の長さを 1 , a 1,a 1 , a a > 0 a>0 a > 0 θ \theta θ S = a 2 sin θ ≦ a 2 \displaystyle
S=\frac{a}{2}\sin\theta\leq\frac{a}{2} S = 2 a sin θ ≦ 2 a θ = π / 2 \theta=\pi/2 θ = π /2 c c c c 2 = 1 + a 2 − 2 a cos θ = 1 + a 2 , c = a 2 + 1 . c^2=1+a^2-2a\cos\theta=1+a^2,
\qquad c=\sqrt{a^2+1}. c 2 = 1 + a 2 − 2 a cos θ = 1 + a 2 , c = a 2 + 1 .
(2) 残りの辺を x > 0 x>0 x > 0 p , q , r , s p,q,r,s p , q , r , s A C = z AC=z A C = z ϕ = ∠ A B C \phi=\angle ABC ϕ = ∠ A B C ψ = ∠ A D C \psi=\angle ADC ψ = ∠ A D C K ( z ) K(z) K ( z ) cos ϕ = p 2 + q 2 − z 2 2 p q , cos ψ = r 2 + s 2 − z 2 2 r s . \displaystyle
\cos\phi=\frac{p^2+q^2-z^2}{2pq},
\qquad
\cos\psi=\frac{r^2+s^2-z^2}{2rs}. cos ϕ = 2 pq p 2 + q 2 − z 2 , cos ψ = 2 r s r 2 + s 2 − z 2 . z z z ϕ , ψ \phi,\psi ϕ , ψ π \pi π π \pi π z z z K ′ ( z ) = z 2 ( cos ϕ sin ϕ + cos ψ sin ψ ) = z 2 sin ( ϕ + ψ ) sin ϕ sin ψ . \displaystyle
K'(z)=\frac z2\left(\frac{\cos\phi}{\sin\phi}+\frac{\cos\psi}{\sin\psi}\right)
=\frac z2\,\frac{\sin(\phi+\psi)}{\sin\phi\sin\psi}. K ′ ( z ) = 2 z ( sin ϕ cos ϕ + sin ψ cos ψ ) = 2 z sin ϕ sin ψ sin ( ϕ + ψ ) . K K K ϕ + ψ < π \phi+\psi<\pi ϕ + ψ < π ϕ + ψ > π \phi+\psi>\pi ϕ + ψ > π ϕ + ψ = π \phi+\psi=\pi ϕ + ψ = π
このとき余弦定理を2つの三角形に用いると z 2 = p 2 + q 2 − 2 p q cos ϕ = s 2 + r 2 + 2 r s cos ϕ , cos ϕ = p 2 + q 2 − r 2 − s 2 2 ( p q + r s ) . \displaystyle
z^2=p^2+q^2-2pq\cos\phi
=s^2+r^2+2rs\cos\phi,
\quad
\cos\phi=\frac{p^2+q^2-r^2-s^2}{2(pq+rs)}. z 2 = p 2 + q 2 − 2 pq cos ϕ = s 2 + r 2 + 2 r s cos ϕ , cos ϕ = 2 ( pq + r s ) p 2 + q 2 − r 2 − s 2 . K = 1 2 ( p q + r s ) sin ϕ , \displaystyle
K=\frac12(pq+rs)\sin\phi, K = 2 1 ( pq + r s ) sin ϕ , m = ( p + q + r + s ) / 2 m=(p+q+r+s)/2 m = ( p + q + r + s ) /2 K 2 = 4 ( p q + r s ) 2 − ( p 2 + q 2 − r 2 − s 2 ) 2 16 = ( ( r + s ) 2 − ( p − q ) 2 ) ( ( p + q ) 2 − ( r − s ) 2 ) 16 = ( m − p ) ( m − q ) ( m − r ) ( m − s ) . \displaystyle
K^2=\frac{4(pq+rs)^2-(p^2+q^2-r^2-s^2)^2}{16}
=\frac{\bigl((r+s)^2-(p-q)^2\bigr)\bigl((p+q)^2-(r-s)^2\bigr)}{16}
=(m-p)(m-q)(m-r)(m-s). K 2 = 16 4 ( pq + r s ) 2 − ( p 2 + q 2 − r 2 − s 2 ) 2 = 16 ( ( r + s ) 2 − ( p − q ) 2 ) ( ( p + q ) 2 − ( r − s ) 2 ) = ( m − p ) ( m − q ) ( m − r ) ( m − s ) . 1 1 1 a a a x x x K 2 = ( a + x 2 ) 2 ( 2 + x − a 2 ) ( 2 + a − x 2 ) = ( a + x ) 2 ( 4 − ( x − a ) 2 ) 16 . \displaystyle
K^2=\left(\frac{a+x}{2}\right)^2
\left(\frac{2+x-a}{2}\right)
\left(\frac{2+a-x}{2}\right)
=\frac{(a+x)^2\bigl(4-(x-a)^2\bigr)}{16}. K 2 = ( 2 a + x ) 2 ( 2 2 + x − a ) ( 2 2 + a − x ) = 16 ( a + x ) 2 ( 4 − ( x − a ) 2 ) . x > 0 x>0 x > 0 ∣ x − a ∣ < 2 |x-a|<2 ∣ x − a ∣ < 2 t = x − a t=x-a t = x − a K 2 = ( 2 a + t ) 2 ( 4 − t 2 ) 16 , d ( K 2 ) d t = 2 a + t 4 ( 2 − a t − t 2 ) . \displaystyle
K^2=\frac{(2a+t)^2(4-t^2)}{16},
\qquad
\frac{d(K^2)}{dt}=\frac{2a+t}{4}(2-at-t^2). K 2 = 16 ( 2 a + t ) 2 ( 4 − t 2 ) , d t d ( K 2 ) = 4 2 a + t ( 2 − a t − t 2 ) . 2 a + t = a + x > 0 2a+t=a+x>0 2 a + t = a + x > 0 2 − a t − t 2 = 0 2-at-t^2=0 2 − a t − t 2 = 0 t ± = − a ± a 2 + 8 2 . \displaystyle
t_\pm=\frac{-a\pm\sqrt{a^2+8}}{2}. t ± = 2 − a ± a 2 + 8 . t − < − a t_-<-a t − < − a t + > 0 t_+>0 t + > 0 t + 2 + a t + = 2 t_+^2+at_+=2 t + 2 + a t + = 2 t + < 2 < 2 t_+<\sqrt2<2 t + < 2 < 2 t + t_+ t + t + t_+ t + x = a + t + = a + a 2 + 8 2 . \displaystyle
x=a+t_+=\boxed{\frac{a+\sqrt{a^2+8}}{2}}. x = a + t + = 2 a + a 2 + 8 .
