すべての角
α , \alpha, α , β \beta β に対して
cos ( α ± β ) = cos α cos β ∓ sin α sin β , sin ( α ± β ) = sin α cos β ± cos α sin β \begin{aligned}
\cos (\alpha\pm\beta ) &= \cos\alpha\cos\beta\mp\sin\alpha\sin\beta, \\
\sin (\alpha\pm\beta ) &= \sin\alpha\cos\beta\pm\cos\alpha\sin\beta
\end{aligned} cos ( α ± β ) sin ( α ± β ) = cos α cos β ∓ sin α sin β , = sin α cos β ± cos α sin β
が成り立つ.
また,
α , \alpha, α , β , \beta, β , α ± β \alpha\pm\beta α ± β が
π 2 \dfrac{\pi}{2} 2 π の倍数でないとき,
tan ( α ± β ) = tan α ± tan β 1 ∓ tan α tan β \tan (\alpha\pm\beta ) = \frac{\tan\alpha\pm\tan\beta}{1\mp\tan\alpha\tan\beta} tan ( α ± β ) = 1 ∓ tan α tan β tan α ± tan β
が成り立つ.
証明
(A) 座標平面において原点を O \mathrm O O A , \mathrm A, A , B \mathrm B B O A , \mathrm{OA}, O A , O B \mathrm{OB} O B α , \alpha, α , β \beta β △ O A B \triangle\mathrm{OAB} △ O A B cos ( α − β ) = cos α cos β + sin α sin β \cos (\alpha -\beta ) = \cos\alpha\cos\beta +\sin\alpha\sin\beta cos ( α − β ) = cos α cos β + sin α sin β (B) 外接円の直径が 1 1 1 △ A B C \triangle\mathrm{ABC} △ A B C α = ∠ A , \alpha = \angle\mathrm A, α = ∠ A , β = ∠ B \beta = \angle\mathrm B β = ∠ B 1 1 1 A B = B C cos β + C A cos α ⋯ [ B ] \mathrm{AB} = \mathrm{BC}\cos\beta +\mathrm{CA}\cos\alpha \quad \cdots [\text B] A B = B C cos β + C A cos α ⋯ [ B ] sin ( α + β ) = sin α cos β + cos α sin β \sin (\alpha +\beta ) = \sin\alpha\cos\beta +\cos\alpha\sin\beta sin ( α + β ) = sin α cos β + cos α sin β (C) α , \alpha, α , β \beta β A B C D \mathrm{ABCD} A B C D A C \mathrm{AC} A C ∠ A C B = α , \angle\mathrm{ACB} = \alpha, ∠ A C B = α , ∠ A C D = β , \angle\mathrm{ACD} = \beta, ∠ A C D = β , A C = 1 \mathrm{AC} = 1 A C = 1 A C ⋅ B D = A B ⋅ C D + B C ⋅ D A ⋯ [ C ] \mathrm{AC}\cdot\mathrm{BD} = \mathrm{AB}\cdot\mathrm{CD}+\mathrm{BC}\cdot\mathrm{DA} \quad \cdots [\text C] A C ⋅ B D = A B ⋅ C D + B C ⋅ D A ⋯ [ C ] sin ( α + β ) = sin α cos β + cos α sin β \sin (\alpha +\beta ) = \sin\alpha\cos\beta +\cos\alpha\sin\beta sin ( α + β ) = sin α cos β + cos α sin β
2018 / 09 / 04 2018/09/04 2 0 1 8 / 0 9 / 0 4 2022 / 05 / 27 2022/05/27 2 0 2 2 / 0 5 / 2 7 解答例
(A) O A , \mathrm{OA}, O A , O B \mathrm{OB} O B θ \theta θ ( 0 ≦ θ ≦ π ) (0 \leqq \theta \leqq \pi ) ( 0 ≦ θ ≦ π ) n n n α − β = θ + 2 n π \alpha -\beta = \theta +2n\pi α − β = θ + 2 n π A ( cos α , sin α ) , \mathrm A(\cos\alpha,\sin\alpha ), A ( cos α , sin α ) , B ( cos β , sin β ) \mathrm B(\cos\beta,\sin\beta ) B ( cos β , sin β ) cos ( α − β ) = cos θ = O A 2 + O B 2 − A B 2 2 O A ⋅ O B = 1 2 + 1 2 − { ( cos β − cos α ) 2 + ( sin β − sin α ) 2 } 2 ⋅ 1 ⋅ 1 = 2 − ( 1 + 1 − 2 cos α cos β − 2 sin α sin β ) 2 = cos α cos β + sin α sin β \begin{aligned}
&\cos (\alpha -\beta ) = \cos\theta = \frac{\mathrm{OA}^2+\mathrm{OB}^2-\mathrm{AB}^2}{2\mathrm{OA}\cdot\mathrm{OB}} \\
&= \frac{1^2+1^2-\{ (\cos\beta -\cos\alpha )^2+(\sin\beta -\sin\alpha )^2\}}{2\cdot 1\cdot 1} \\
&= \frac{2-(1+1-2\cos\alpha\cos\beta -2\sin\alpha\sin\beta )}{2} \\
&= \cos\alpha\cos\beta +\sin\alpha\sin\beta
\end{aligned} cos ( α − β ) = cos θ = 2 O A ⋅ O B O A 2 + O B 2 − A B 2 = 2 ⋅ 1 ⋅ 1 1 2 + 1 2 − { ( cos β − cos α ) 2 + ( sin β − sin α ) 2 } = 2 2 − ( 1 + 1 − 2 cos α cos β − 2 sin α sin β ) = cos α cos β + sin α sin β (B) sin ∠ C = sin ( π − α − β ) = sin ( α + β ) \sin\angle\mathrm C = \sin (\pi -\alpha -\beta ) = \sin (\alpha +\beta ) sin ∠ C = sin ( π − α − β ) = sin ( α + β ) B C = sin α , C A = sin β , A B = sin ( α + β ) \mathrm{BC} = \sin\alpha, \quad \mathrm{CA} = \sin\beta, \quad \mathrm{AB} = \sin (\alpha\!+\!\beta ) B C = sin α , C A = sin β , A B = sin ( α + β ) [ B ] [\text B] [ B ] sin ( α + β ) = sin α cos β + cos α sin β \sin (\alpha +\beta ) = \sin\alpha\cos\beta +\cos\alpha\sin\beta sin ( α + β ) = sin α cos β + cos α sin β (C) 正弦定理により
B D = sin ( α + β ) \mathrm{BD} = \sin (\alpha +\beta ) B D = sin ( α + β ) △ A B C , \triangle\mathrm{ABC}, △ A B C , △ A D C \triangle\mathrm{ADC} △ A D C A B = sin α , B C = cos α , C D = cos β , D A = sin β \mathrm{AB} = \sin\alpha,\ \mathrm{BC} = \cos\alpha,\ \mathrm{CD} = \cos\beta,\ \mathrm{DA} = \sin\beta A B = sin α , B C = cos α , C D = cos β , D A = sin β [ C ] [\text C] [ C ] sin ( α + β ) = sin α cos β + cos α sin β \sin (\alpha +\beta ) = \sin\alpha\cos\beta +\cos\alpha\sin\beta sin ( α + β ) = sin α cos β + cos α sin β
参考
「第一余弦定理」 と区別するとき, 余弦定理を「第二余弦定理」 と呼ぶ.
古代ローマの天文学者トレミー (プトレマイオス) は「トレミーの定理」を使って「弦」 c r d θ = 2 sin θ 2 \mathrm{crd}\,\theta = 2\sin\dfrac{\theta}{2} c r d θ = 2 sin 2 θ
△ A B C \triangle\mathrm{ABC} △ A B C において,
a = B C , a = \mathrm{BC}, a = B C , b = C A , b = \mathrm{CA}, b = C A , c = A B , c = \mathrm{AB}, c = A B , A = ∠ A , A = \angle\mathrm A, A = ∠ A , B = ∠ B , B = \angle\mathrm B, B = ∠ B , C = ∠ C C = \angle\mathrm C C = ∠ C とおく.
「加比の理」
p q = r s = k ⟹ p x + r y q x + s y = k \frac{p}{q} = \frac{r}{s} = k \Longrightarrow \frac{px+ry}{qx+sy} = k q p = s r = k ⟹ q x + s y p x + r y = k
(証明は
こちら ) と正弦の加法定理を使って, 正弦定理から「第一余弦定理」
a = b cos C + c cos B , b = c cos A + a cos C , c = a cos B + b cos A \begin{aligned}
a &= b\cos C+c\cos B, \\
b &= c\cos A+a\cos C, \\
c &= a\cos B+b\cos A
\end{aligned} a b c = b cos C + c cos B , = c cos A + a cos C , = a cos B + b cos A
を導け.
2019 / 05 / 10 2019/05/10 2 0 1 9 / 0 5 / 1 0 2022 / 05 / 27 2022/05/27 2 0 2 2 / 0 5 / 2 7 解答例
正弦定理
a sin A = b sin B = c sin C , \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}, sin A a = sin B b = sin C c ,
により,
a sin A = b cos C + c cos B sin B cos C + cos B sin C ( ∵ 「加比の理」 ) = b cos C + c cos B sin ( B + C ) ( ∵ 正弦の加法定理 ) = b cos C + c cos B sin A ( ∵ A + B + C = π ) \begin{aligned}
\frac{a}{\sin A} &= \frac{b\cos C+c\cos B}{\sin B\cos C+\cos B\sin C} \quad (\because\text{「加比の理」}) \\
&= \frac{b\cos C+c\cos B}{\sin (B+C)} \quad (\because\text{正弦の加法定理})\\
&= \frac{b\cos C+c\cos B}{\sin A} \quad (\because A+B+C = \pi )
\end{aligned} sin A a = sin B cos C + cos B sin C b cos C + c cos B ( ∵ 「加比の理」 ) = sin ( B + C ) b cos C + c cos B ( ∵ 正弦の加法定理 ) = sin A b cos C + c cos B ( ∵ A + B + C = π )
が得られる.
よって,
a = b cos C + c cos B a = b\cos C+c\cos B a = b cos C + c cos B が成り立つ.
同様に,
b = c cos A + a cos C , b = c\cos A+a\cos C, b = c cos A + a cos C , c = a cos B + b cos A c = a\cos B+b\cos A c = a cos B + b cos A が成り立つ.
△ A B C \triangle\mathrm{ABC} △ A B C において,
a = B C , a = \mathrm{BC}, a = B C , b = C A , b = \mathrm{CA}, b = C A , c = A B , c = \mathrm{AB}, c = A B , A = ∠ A , A = \angle\mathrm A, A = ∠ A , B = ∠ B , B = \angle\mathrm B, B = ∠ B , C = ∠ C C = \angle\mathrm C C = ∠ C とおく.
余弦の加法定理を用いて, 正弦定理から「第二余弦定理」
a 2 = b 2 + c 2 − 2 b c cos A , b 2 = c 2 + a 2 − 2 c a cos B , c 2 = a 2 + b 2 − 2 a b cos C \begin{aligned}
a^2 &= b^2+c^2-2bc\cos A, \\
b^2 &= c^2+a^2-2ca\cos B, \\
c^2 &= a^2+b^2-2ab\cos C
\end{aligned} a 2 b 2 c 2 = b 2 + c 2 − 2 b c cos A , = c 2 + a 2 − 2 c a cos B , = a 2 + b 2 − 2 a b cos C
を導け.
2021 / 10 / 11 2021/10/11 2 0 2 1 / 1 0 / 1 1 2022 / 05 / 27 2022/05/27 2 0 2 2 / 0 5 / 2 7 解答例
正弦定理により
a sin A = b sin B = c sin C = k \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k sin A a = sin B b = sin C c = k
とおくと,
a = k sin A , a = k\sin A, a = k sin A , b = k sin B , b = k\sin B, b = k sin B , c = k sin C c = k\sin C c = k sin C となるから,
a 2 − b 2 − c 2 = k 2 sin 2 A − k 2 sin 2 B − k 2 sin 2 C = k 2 { sin 2 ( B + C ) − sin 2 B − sin 2 C } = k 2 { ( sin B cos C + cos B sin C ) 2 − sin 2 B − sin 2 C } = k 2 { ( sin 2 B cos 2 C + cos 2 B sin 2 C + 2 sin B cos B sin C cos C ) − sin 2 B − sin 2 C } = k 2 { 2 sin B sin C cos B cos C − sin 2 B ( 1 − cos 2 C ) − ( 1 − cos 2 B ) sin 2 C } = k 2 ( 2 sin B sin C cos B cos C − 2 sin 2 B sin 2 C ) = 2 ( k sin B ) ( k sin C ) ( cos B cos C − sin B sin C ) = 2 b c cos ( B + C ) = − 2 b c cos A \begin{aligned}
&a^2-b^2-c^2 \\
&= k^2\sin ^2A-k^2\sin ^2B-k^2\sin ^2C \\
&= k^2\{\sin ^2(B+C)-\sin ^2B-\sin ^2C\} \\
&= k^2\{ (\sin B\cos C+\cos B\sin C)^2-\sin ^2B-\sin ^2C\} \\
&= k^2\{ (\sin ^2B\cos^2 C+\cos ^2B\sin ^2C \\
&\qquad +2\sin B\cos B\sin C\cos C)-\sin ^2B-\sin ^2C\} \\
&= k^2\{ 2\sin B\sin C\cos B\cos C \\
&\qquad -\sin ^2B(1-\cos ^2C)-(1-\cos ^2B)\sin ^2C\} \\
&= k^2(2\sin B\sin C\cos B\cos C-2\sin ^2B\sin ^2C) \\
&= 2(k\sin B)(k\sin C)(\cos B\cos C-\sin B\sin C) \\
&= 2bc\cos (B+C) \\
&= -2bc\cos A
\end{aligned} a 2 − b 2 − c 2 = k 2 sin 2 A − k 2 sin 2 B − k 2 sin 2 C = k 2 { sin 2 ( B + C ) − sin 2 B − sin 2 C } = k 2 { ( sin B cos C + cos B sin C ) 2 − sin 2 B − sin 2 C } = k 2 { ( sin 2 B cos 2 C + cos 2 B sin 2 C + 2 sin B cos B sin C cos C ) − sin 2 B − sin 2 C } = k 2 { 2 sin B sin C cos B cos C − sin 2 B ( 1 − cos 2 C ) − ( 1 − cos 2 B ) sin 2 C } = k 2 ( 2 sin B sin C cos B cos C − 2 sin 2 B sin 2 C ) = 2 ( k sin B ) ( k sin C ) ( cos B cos C − sin B sin C ) = 2 b c cos ( B + C ) = − 2 b c cos A
となり,
a 2 = b 2 + c 2 − 2 b c cos A a^2 = b^2+c^2-2bc\cos A a 2 = b 2 + c 2 − 2 b c cos A が得られる. 残りの式も, 同様に証明できる.
参考
三角関数の加法定理のもとで, 正弦定理, 余弦定理 (「第一余弦定理」,「第二余弦定理」) は同値である (
こちら と
こちら を参照).
ここで, 加法定理は三角関数を特徴付ける内在的性質の
1 1 1 つであるから, 正弦定理と余弦定理は本質的に同値な定理であると言える.
なお, ここでは, 正弦定理を「内角の正弦に対する対辺の長さの比が外接円の直径に等しい」という定理ではなく, 単に「内角の正弦に対する対辺の長さの比が互いに等しい」という定理として同値性を証明している.
a , a, a , b b b を
b ≠ ± a b \neq \pm a b = ± a なる実数とする.
2 2 2 直線
y = a x , y = ax, y = a x , y = b x y = bx y = b x のなす角の二等分線の傾き
c c c を
a , a, a , b b b で表す公式を, 正接の加法定理から導け.
2022 / 09 / 13 2022/09/13 2 0 2 2 / 0 9 / 1 3 2024 / 12 / 10 2024/12/10 2 0 2 4 / 1 2 / 1 0 解答例
y = a x , y = ax, y = a x , y = b x , y = bx, y = b x , 角の二等分線
y = c x y = cx y = c x と
x x x 軸の正方向のなす角をそれぞれ
α , \alpha, α , β \beta β γ \gamma γ とおく.
このとき,
β − γ , \beta -\gamma, β − γ , γ − α \gamma -\alpha γ − α は
π \pi π の倍数の差を除いて等しい.
つまり, ある整数
n n n について,
2 γ = α + β + n π 2\gamma = \alpha +\beta +n\pi 2 γ = α + β + n π
となる.
よって,
tan 2 γ = tan ( α + β ) \tan 2\gamma = \tan (\alpha +\beta ) tan 2 γ = tan ( α + β ) であるから,
2 2 2 倍角の公式, 加法定理により
2 c 1 − c 2 = a + b 1 − a b \frac{2c}{1-c^2} = \frac{a+b}{1-ab} 1 − c 2 2 c = 1 − a b a + b
が成り立つ.
分母を払い,
c c c について整理すると
( a + b ) c 2 − 2 ( a b − 1 ) c − ( a + b ) = 0 ⋯ [ ∗ ] ′ (a+b)c^2-2(ab-1)c-(a+b) = 0 \quad \cdots [\ast ]' ( a + b ) c 2 − 2 ( a b − 1 ) c − ( a + b ) = 0 ⋯ [ ∗ ] ′
となるから,
2 2 2 次方程式の解の公式により
c = a b − 1 ± ( a b − 1 ) 2 + ( a + b ) 2 a + b = a b − 1 ± ( a 2 + 1 ) ( b 2 + 1 ) a + b \begin{aligned}
c &= \frac{ab-1\pm\sqrt{(ab-1)^2+(a+b)^2}}{a+b} \\
&= \frac{ab-1\pm\sqrt{(a^2+1)(b^2+1)}}{a+b}
\end{aligned} c = a + b a b − 1 ± ( a b − 1 ) 2 + ( a + b ) 2 = a + b a b − 1 ± ( a 2 + 1 ) ( b 2 + 1 )
が得られる.
参考
上記の公式は, 点と直線の公式, ベクトルの内積の公式からも求められる (こちら とこちら を参照).
2 2 2 「有理二等分問題」 (rational bisection problem) と呼ばれる.
この問題は近年, 筆者により, 方程式 [ ∗ ] ′ [\ast ]' [ ∗ ] ′ 2 2 2 こちら とこちら を参照).
△ A B C \triangle\mathrm{ABC} △ A B C の外側に正三角形
A ′ B C , \mathrm A'\mathrm{BC}, A ′ B C , B ′ C A , \mathrm B'\mathrm{CA}, B ′ C A , C ′ A B \mathrm C'\mathrm{AB} C ′ A B をかき,
それらの重心をそれぞれ
A ′ ′ , \mathrm A'', A ′ ′ , B ′ ′ , \mathrm B'', B ′ ′ , C ′ ′ \mathrm C'' C ′ ′ とおく.
(1) △ A B C \triangle\mathrm{ABC} △ A B C a = B C , a = \mathrm{BC}, a = B C , b = C A , b = \mathrm{CA}, b = C A , c = A B c = \mathrm{AB} c = A B R R R A ′ ′ B ′ ′ 2 \mathrm A''\mathrm B''^2 A ′ ′ B ′ ′ 2 (2) △ A ′ ′ B ′ ′ C ′ ′ \triangle\mathrm A''\mathrm B''\mathrm C'' △ A ′ ′ B ′ ′ C ′ ′
2020 / 10 / 12 2020/10/12 2 0 2 0 / 1 0 / 1 2 2020 / 10 / 19 2020/10/19 2 0 2 0 / 1 0 / 1 9 解答例
(1) C A ′ ′ , \mathrm{CA}'', C A ′ ′ , B ′ ′ C \mathrm B''\mathrm C B ′ ′ C A ′ B C , \mathrm A'\mathrm{BC}, A ′ B C , B ′ C A \mathrm B'\mathrm{CA} B ′ C A C A ′ ′ = a 2 sin 6 0 ∘ = a 3 , B ′ ′ C = b 2 sin 6 0 ∘ = b 3 \mathrm{CA}'' = \frac{a}{2\sin 60^\circ} = \frac{a}{\sqrt 3}, \quad \mathrm B''\mathrm C = \frac{b}{2\sin 60^\circ} = \frac{b}{\sqrt 3} C A ′ ′ = 2 sin 6 0 ∘ a = 3 a , B ′ ′ C = 2 sin 6 0 ∘ b = 3 b C = ∠ A C B C = \angle\mathrm{ACB} C = ∠ A C B ∠ A ′ ′ C B ′ ′ = C + 6 0 ∘ \angle\mathrm A''\mathrm{CB}'' = C+60^\circ ∠ A ′ ′ C B ′ ′ = C + 6 0 ∘ △ A ′ ′ B ′ ′ C \triangle\mathrm A''\mathrm B''\mathrm C △ A ′ ′ B ′ ′ C △ A B C \triangle\mathrm{ABC} △ A B C A ′ ′ B ′ ′ 2 = ( a 3 ) 2 + ( b 3 ) 2 − 2 ⋅ a 3 ⋅ b 3 cos ( C + 6 0 ∘ ) = a 2 3 + b 2 3 − 2 a b 3 ( cos C cos 6 0 ∘ − sin C sin 6 0 ∘ ) = a 2 3 + b 2 3 − 2 a b 3 ( a 2 + b 2 − c 2 2 a b ⋅ 1 2 − c 2 R ⋅ 3 2 ) = R ( a 2 + b 2 + c 2 ) + 3 a b c 6 R ⋯ [ 1 ] \begin{aligned}
&\mathrm A''\mathrm B''^2 \\
&= \left(\frac{a}{\sqrt 3}\right) ^2+\left(\frac{b}{\sqrt 3}\right) ^2-2\cdot\frac{a}{\sqrt 3}\cdot\frac{b}{\sqrt 3}\cos (C+60^\circ ) \\
&= \frac{a^2}{3}+\frac{b^2}{3}-\frac{2ab}{3}(\cos C\cos 60^\circ -\sin C\sin 60^\circ ) \\
&= \frac{a^2}{3}+\frac{b^2}{3}-\frac{2ab}{3}\left(\frac{a^2+b^2-c^2}{2ab}\cdot\frac{1}{2}-\frac{c}{2R}\cdot\frac{\sqrt 3}{2}\right) \\
&= \frac{R(a^2+b^2+c^2)+\sqrt 3abc}{6R} \quad \cdots [1]
\end{aligned} A ′ ′ B ′ ′ 2 = ( 3 a ) 2 + ( 3 b ) 2 − 2 ⋅ 3 a ⋅ 3 b cos ( C + 6 0 ∘ ) = 3 a 2 + 3 b 2 − 3 2 a b ( cos C cos 6 0 ∘ − sin C sin 6 0 ∘ ) = 3 a 2 + 3 b 2 − 3 2 a b ( 2 a b a 2 + b 2 − c 2 ⋅ 2 1 − 2 R c ⋅ 2 3 ) = 6 R R ( a 2 + b 2 + c 2 ) + 3 a b c ⋯ [ 1 ] (2) [ 1 ] [1] [ 1 ] a , a, a , b , b, b , c c c A , \mathrm A, A , B , \mathrm B, B , C \mathrm C C A ′ ′ B ′ ′ 2 = B ′ ′ C ′ ′ 2 = C ′ ′ A ′ ′ 2 = R ( a 2 + b 2 + c 2 ) + 3 a b c 6 R \mathrm A''\mathrm B''^2 \!=\! \mathrm B''\mathrm C''^2 \!=\! \mathrm C''\mathrm A''^2 \!=\! \frac{R(a^2\!+\!b^2\!+\!c^2)\!+\!\sqrt 3abc}{6R} A ′ ′ B ′ ′ 2 = B ′ ′ C ′ ′ 2 = C ′ ′ A ′ ′ 2 = 6 R R ( a 2 + b 2 + c 2 ) + 3 a b c △ A ′ ′ B ′ ′ C ′ ′ \triangle\mathrm A''\mathrm B''\mathrm C'' △ A ′ ′ B ′ ′ C ′ ′
参考
本問において △ A ′ ′ B ′ ′ C ′ ′ \triangle\mathrm A''\mathrm B''\mathrm C'' △ A ′ ′ B ′ ′ C ′ ′ 「ナポレオンの定理」 (Napoleon's theorem) として知られている.
複素数平面を利用した証明については, こちら を参照されたい.
△ A B C \triangle\mathrm{ABC} △ A B C A ′ B C , \mathrm A'\mathrm{BC}, A ′ B C , B ′ C A , \mathrm B'\mathrm{CA}, B ′ C A , C ′ A B \mathrm C'\mathrm{AB} C ′ A B A ′ ′ B ′ ′ C ′ ′ \mathrm A''\mathrm B''\mathrm C'' A ′ ′ B ′ ′ C ′ ′ C + 6 0 ∘ , C+60^\circ, C + 6 0 ∘ , 3 a b c \sqrt 3abc 3 a b c C − 6 0 ∘ , C-60^\circ, C − 6 0 ∘ , − 3 a b c -\sqrt 3abc − 3 a b c 「ナポレオンの定理」 として知られている.
また, 前者の正三角形と後者の正三角形の面積の差は,
1 2 ⋅ R ( a 2 + b 2 + c 2 ) + 3 a b c 6 R sin 6 0 ∘ − 1 2 ⋅ R ( a 2 + b 2 + c 2 ) − 3 a b c 6 R sin 6 0 ∘ = 2 ⋅ 1 2 ⋅ 3 a b c 6 R ⋅ 3 2 = a b c 4 R \begin{aligned}
&\frac{1}{2}\cdot\frac{R(a^2+b^2+c^2)+\sqrt 3abc}{6R}\,\sin 60^\circ \\
&-\frac{1}{2}\cdot\frac{R(a^2+b^2+c^2)-\sqrt 3abc}{6R}\,\sin 60^\circ \\
&= 2\cdot\frac{1}{2}\cdot\frac{\sqrt 3abc}{6R}\cdot\frac{\sqrt 3}{2} \\
&= \frac{abc}{4R}
\end{aligned} 2 1 ⋅ 6 R R ( a 2 + b 2 + c 2 ) + 3 a b c sin 6 0 ∘ − 2 1 ⋅ 6 R R ( a 2 + b 2 + c 2 ) − 3 a b c sin 6 0 ∘ = 2 ⋅ 2 1 ⋅ 6 R 3 a b c ⋅ 2 3 = 4 R a b c △ A B C \triangle\mathrm{ABC} △ A B C
次のことを示せ.
(1) 0 0 0 π \pi π π 2 \dfrac{\pi}{2} 2 π α , \alpha, α , β , \beta, β , γ \gamma γ α + β + γ = π \alpha +\beta +\gamma = \pi α + β + γ = π tan α tan β tan γ = tan α + tan β + tan γ \tan\alpha\tan\beta\tan\gamma = \tan\alpha +\tan\beta +\tan\gamma tan α tan β tan γ = tan α + tan β + tan γ (2) △ A B C \triangle\mathrm{ABC} △ A B C a = B C , a = \mathrm{BC}, a = B C , b = C A , b = \mathrm{CA}, b = C A , c = A B , c = \mathrm{AB}, c = A B , s = a + b + c 2 , s = \dfrac{a+b+c}{2}, s = 2 a + b + c , A = ∠ A , A = \angle\mathrm A, A = ∠ A , B = ∠ B , B = \angle\mathrm B, B = ∠ B , C = ∠ C , C = \angle\mathrm C, C = ∠ C , S = △ A B C S = \triangle\mathrm{ABC} S = △ A B C I , \mathrm I, I , r r r B C , \mathrm{BC}, B C , C A , \mathrm{CA}, C A , A B \mathrm{AB} A B A ′ , \mathrm A', A ′ , B ′ , \mathrm B', B ′ , C ′ \mathrm C' C ′
① A B ′ = s − a \mathrm{AB}' = s-a A B ′ = s − a ② tan ( π 2 − A 2 ) = s − a r \displaystyle\tan\left(\frac{\pi}{2}-\frac{A}{2}\right) = \frac{s-a}{r} tan ( 2 π − 2 A ) = r s − a ③ S = s ( s − a ) ( s − b ) ( s − c ) S = \sqrt{s(s-a)(s-b)(s-c)} S = s ( s − a ) ( s − b ) ( s − c )
が成り立つ.
(3) 円に内接する四角形 A B C D \mathrm{ABCD} A B C D a = A B , a = \mathrm{AB}, a = A B , b = B C , b = \mathrm{BC}, b = B C , c = C D , c = \mathrm{CD}, c = C D , d = D A , d = \mathrm{DA}, d = D A , s = a + b + c + d 2 s = \dfrac{a+b+c+d}{2} s = 2 a + b + c + d A B C D \mathrm{ABCD} A B C D S S S A B \mathrm{AB} A B A \mathrm A A C D \mathrm{CD} C D D \mathrm D D E \mathrm E E p = E A , p = \mathrm{EA}, p = E A , q = E D q = \mathrm{ED} q = E D △ E D A \triangle\mathrm{EDA} △ E D A T T T
① S : T = ( b 2 − d 2 ) : d 2 S:T = (b^2-d^2):d^2 S : T = ( b 2 − d 2 ) : d 2 ② ( b − d ) ( p + q ) = d ( c + a ) , (b-d)(p+q) = d(c+a), ( b − d ) ( p + q ) = d ( c + a ) , ( b + d ) ( p − q ) = d ( c − a ) (b+d)(p-q) = d(c-a) ( b + d ) ( p − q ) = d ( c − a ) ③ S = ( s − a ) ( s − b ) ( s − c ) ( s − d ) S = \sqrt{(s-a)(s-b)(s-c)(s-d)} S = ( s − a ) ( s − b ) ( s − c ) ( s − d )
が成り立つ.
2023 / 03 / 02 2023/03/02 2 0 2 3 / 0 3 / 0 2 2023 / 12 / 21 2023/12/21 2 0 2 3 / 1 2 / 2 1 解答例
(1) α + β + γ = π \alpha +\beta +\gamma = \pi α + β + γ = π tan γ = tan ( π − α − β ) = − tan ( α + β ) = − tan α + tan β 1 − tan α tan β \begin{aligned}
\tan\gamma &= \tan (\pi -\alpha -\beta ) = -\tan (\alpha +\beta ) \\
&= -\frac{\tan\alpha +\tan\beta}{1-\tan\alpha\tan\beta}
\end{aligned} tan γ = tan ( π − α − β ) = − tan ( α + β ) = − 1 − tan α tan β tan α + tan β ( tan α tan β − 1 ) tan γ = tan α + tan β tan α tan β tan γ = tan α + tan β + tan γ \begin{aligned}
(\tan\alpha\tan\beta -1)\tan\gamma &= \tan\alpha +\tan\beta \\
\tan\alpha\tan\beta\tan\gamma &= \tan\alpha +\tan\beta +\tan\gamma
\end{aligned} ( tan α tan β − 1 ) tan γ tan α tan β tan γ = tan α + tan β = tan α + tan β + tan γ (2)
① A B ′ = A C ′ , \mathrm{AB}' = \mathrm{AC}', A B ′ = A C ′ , B C ′ = B A ′ , \mathrm{BC}' = \mathrm{BA}', B C ′ = B A ′ , C A ′ = C B ′ \mathrm{CA}' = \mathrm{CB}' C A ′ = C B ′ 2 s − a = b + c = C B ′ + A B ′ + A C ′ + B C ′ = 2 A B ′ + B A ′ + C A ′ = 2 A B ′ + a \begin{aligned}
2s-a &= b+c = \mathrm{CB}'+\mathrm{AB}'+\mathrm{AC}'+\mathrm{BC}' \\
&= 2\mathrm{AB}'+\mathrm{BA}'+\mathrm{CA}' = 2\mathrm{AB}'+a
\end{aligned} 2 s − a = b + c = C B ′ + A B ′ + A C ′ + B C ′ = 2 A B ′ + B A ′ + C A ′ = 2 A B ′ + a A B ′ = s − a \mathrm{AB}' = s-a A B ′ = s − a ② △ A I B ′ \triangle\mathrm{AIB}' △ A I B ′ tan ( π 2 − A 2 ) = tan ∠ A I B ′ = s − a r \tan\left(\frac{\pi}{2}-\frac{A}{2}\right) = \tan\angle\mathrm{AIB}' = \frac{s-a}{r} tan ( 2 π − 2 A ) = tan ∠ A I B ′ = r s − a ③ 同様に,
tan ( π 2 − B 2 ) = s − b r , tan ( π 2 − C 2 ) = s − c r \tan\left(\frac{\pi}{2}-\frac{B}{2}\right) = \frac{s-b}{r}, \quad \tan\left(\frac{\pi}{2}-\frac{C}{2}\right) = \frac{s-c}{r} tan ( 2 π − 2 B ) = r s − b , tan ( 2 π − 2 C ) = r s − c ( π 2 − A 2 ) + ( π 2 − B 2 ) + ( π 2 − C 2 ) = π \displaystyle\left(\frac{\pi}{2}-\frac{A}{2}\right) +\left(\frac{\pi}{2}-\frac{B}{2}\right) +\left(\frac{\pi}{2}-\frac{C}{2}\right) = \pi ( 2 π − 2 A ) + ( 2 π − 2 B ) + ( 2 π − 2 C ) = π tan ( π 2 − A 2 ) tan ( π 2 − B 2 ) tan ( π 2 − C 2 ) = tan ( π 2 − A 2 ) + tan ( π 2 − B 2 ) + tan ( π 2 − C 2 ) \begin{aligned}
&\tan\left(\frac{\pi}{2}-\frac{A}{2}\right)\tan\left(\frac{\pi}{2}-\frac{B}{2}\right)\tan\left(\frac{\pi}{2}-\frac{C}{2}\right) \\
&= \tan\left(\frac{\pi}{2}-\frac{A}{2}\right) +\tan\left(\frac{\pi}{2}-\frac{B}{2}\right) +\tan\left(\frac{\pi}{2}-\frac{C}{2}\right)
\end{aligned} tan ( 2 π − 2 A ) tan ( 2 π − 2 B ) tan ( 2 π − 2 C ) = tan ( 2 π − 2 A ) + tan ( 2 π − 2 B ) + tan ( 2 π − 2 C ) s − a r ⋅ s − b r ⋅ s − c r = s − a r + s − b r + s − c r s ( s − a ) ( s − b ) ( s − c ) = r 2 s 2 ( ∵ 3 s − a − b − c = s ) \begin{aligned}
\frac{s-a}{r}\cdot\frac{s-b}{r}\cdot\frac{s-c}{r} &= \frac{s-a}{r}+\frac{s-b}{r}+\frac{s-c}{r} \\
s(s-a)(s-b)(s-c) &= r^2s^2 \quad (\because 3s-a-b-c = s)
\end{aligned} r s − a ⋅ r s − b ⋅ r s − c s ( s − a ) ( s − b ) ( s − c ) = r s − a + r s − b + r s − c = r 2 s 2 ( ∵ 3 s − a − b − c = s ) S = a r 2 + b r 2 + c r 2 = r ⋅ a + b + c 2 = r s = s ( s − a ) ( s − b ) ( s − c ) \begin{aligned}
S &= \frac{ar}{2}+\frac{br}{2}+\frac{cr}{2} = r\cdot\frac{a+b+c}{2} = rs \\
&= \sqrt{s(s-a)(s-b)(s-c)}
\end{aligned} S = 2 a r + 2 b r + 2 c r = r ⋅ 2 a + b + c = r s = s ( s − a ) ( s − b ) ( s − c )
(3)
① ∠ E D A = ∠ E B C \angle\mathrm{EDA} = \angle\mathrm{EBC} ∠ E D A = ∠ E B C △ E D A , \triangle\mathrm{EDA}, △ E D A , △ E B C \triangle\mathrm{EBC} △ E B C d : b d:b d : b T : ( T + S ) = d 2 : b 2 T:(T+S) = d^2:b^2 T : ( T + S ) = d 2 : b 2 S : T = ( b 2 − d 2 ) : d 2 S:T = (b^2-d^2):d^2 S : T = ( b 2 − d 2 ) : d 2 ② p : ( q + c ) = d : b , q : ( p + a ) = d : b p:(q+c) = d:b, \quad q:(p+a) = d:b p : ( q + c ) = d : b , q : ( p + a ) = d : b b p = d ( q + c ) , b q = d ( p + a ) bp = d(q+c), \quad bq = d(p+a) b p = d ( q + c ) , b q = d ( p + a ) ( b − d ) ( p + q ) = d ( c + a ) , ( b + d ) ( p − q ) = d ( c − a ) (b-d)(p+q) = d(c+a), \quad (b+d)(p-q) = d(c-a) ( b − d ) ( p + q ) = d ( c + a ) , ( b + d ) ( p − q ) = d ( c − a ) ③ △ E D A \triangle\mathrm{EDA} △ E D A S 2 = ( b 2 − d 2 ) 2 d 4 T 2 = ( b + d ) 2 ( b − d ) 2 d 4 ⋅ d + p + q 2 ⋅ p + q − d 2 ⋅ q + d − p 2 ⋅ d + p − q 2 = ( b − d ) { ( p + q ) + d } 2 d ⋅ ( b + d ) { ( p + q ) − d } 2 d ⋅ ( b + d ) { d − ( p − q ) } 2 d ⋅ ( b − d ) { d + ( p − q ) } 2 d = d ( c + a ) + d ( b − d ) 2 d ⋅ d ( c + a ) − d ( b − d ) 2 d ⋅ d ( b + d ) − d ( c − a ) 2 d ⋅ d ( b + d ) + d ( c − a ) 2 d = c + a + b − d 2 ⋅ c + a − b + d 2 ⋅ b + d − c + a 2 ⋅ b + d + c − a 2 = ( s − a ) ( s − b ) ( s − c ) ( s − d ) \begin{aligned}
S^2 &= \frac{(b^2-d^2)^2}{d^4}T^2 \\
&= \frac{(b+d)^2(b-d)^2}{d^4}\cdot\frac{d+p+q}{2}\cdot\frac{p+q-d}{2} \\
&\qquad \cdot\frac{q+d-p}{2}\cdot\frac{d+p-q}{2} \\
&= \frac{(b-d)\{ (p+q)+d\}}{2d}\cdot\frac{(b+d)\{ (p+q)-d\}}{2d} \\
&\qquad \cdot\frac{(b+d)\{ d-(p-q)\}}{2d}\cdot\frac{(b-d)\{ d+(p-q)\}}{2d} \\
&= \frac{d(c+a)+d(b-d)}{2d}\cdot\frac{d(c+a)-d(b-d)}{2d} \\
&\qquad \cdot\frac{d(b+d)-d(c-a)}{2d}\cdot\frac{d(b+d)+d(c-a)}{2d} \\
&= \frac{c+a+b-d}{2}\cdot\frac{c+a-b+d}{2} \\
&\qquad \cdot\frac{b+d-c+a}{2}\cdot\frac{b+d+c-a}{2} \\
&= (s-a)(s-b)(s-c)(s-d)
\end{aligned} S 2 = d 4 ( b 2 − d 2 ) 2 T 2 = d 4 ( b + d ) 2 ( b − d ) 2 ⋅ 2 d + p + q ⋅ 2 p + q − d ⋅ 2 q + d − p ⋅ 2 d + p − q = 2 d ( b − d ) { ( p + q ) + d } ⋅ 2 d ( b + d ) { ( p + q ) − d } ⋅ 2 d ( b + d ) { d − ( p − q ) } ⋅ 2 d ( b − d ) { d + ( p − q ) } = 2 d d ( c + a ) + d ( b − d ) ⋅ 2 d d ( c + a ) − d ( b − d ) ⋅ 2 d d ( b + d ) − d ( c − a ) ⋅ 2 d d ( b + d ) + d ( c − a ) = 2 c + a + b − d ⋅ 2 c + a − b + d ⋅ 2 b + d − c + a ⋅ 2 b + d + c − a = ( s − a ) ( s − b ) ( s − c ) ( s − d ) S = ( s − a ) ( s − b ) ( s − c ) ( s − d ) S = \sqrt{(s-a)(s-b)(s-c)(s-d)} S = ( s − a ) ( s − b ) ( s − c ) ( s − d )
参考
(2) ③ の等式は「ヘロンの公式」 (Heron's formula) と呼ばれる (別証明はこちら を参照).
(3) ③ の等式は「ブラーマグプタの公式」 (Brahmagupta's formula) と呼ばれる (別証明はこちら を参照).
直角三角形でない
△ A B C \triangle\mathrm{ABC} △ A B C において,
a = B C , a = \mathrm{BC}, a = B C , b = C A , b = \mathrm{CA}, b = C A , c = A B , c = \mathrm{AB}, c = A B , A = ∠ A , A = \angle\mathrm A, A = ∠ A , B = ∠ B , B = \angle\mathrm B, B = ∠ B , C = ∠ C , C = \angle\mathrm C, C = ∠ C , S = △ A B C S = \triangle\mathrm{ABC} S = △ A B C とおく.
次のことを示せ.
ただし,
x , x, x , y , y, y , z ≧ 0 z \geqq 0 z ≧ 0 のとき
x + y + z ≧ 3 x y z 3 x+y+z \geqq 3\sqrt[3]{xyz} x + y + z ≧ 3 3 x y z が成り立つこと (
こちら と
こちら を参照) は, 証明なしに使ってよい.
(1) tan A tan B tan C = tan A + tan B + tan C \tan A\tan B\tan C = \tan A+\tan B+\tan C tan A tan B tan C = tan A + tan B + tan C (2) △ A B C \triangle\mathrm{ABC} △ A B C ( 4 S ) 6 ≧ 27 ( b 2 + c 2 − a 2 ) 2 ( c 2 + a 2 − b 2 ) 2 ( a 2 + b 2 − c 2 ) 2 (4S)^6 \geqq 27(b^2+c^2-a^2)^2(c^2+a^2-b^2)^2(a^2+b^2-c^2)^2 ( 4 S ) 6 ≧ 2 7 ( b 2 + c 2 − a 2 ) 2 ( c 2 + a 2 − b 2 ) 2 ( a 2 + b 2 − c 2 ) 2 2019 / 04 / 25 2019/04/25 2 0 1 9 / 0 4 / 2 5 2023 / 03 / 07 2023/03/07 2 0 2 3 / 0 3 / 0 7 解答例
(1) A + B + C = π A+B+C = \pi A + B + C = π tan C = tan ( π − A − B ) = − tan ( A + B ) = − tan A + tan B 1 − tan A tan B \begin{aligned}
\tan C &= \tan (\pi -A-B) = -\tan (A+B) \\
&= -\frac{\tan A+\tan B}{1-\tan A\tan B}
\end{aligned} tan C = tan ( π − A − B ) = − tan ( A + B ) = − 1 − tan A tan B tan A + tan B ( tan A tan B − 1 ) tan C = tan A + tan B tan A tan B tan C = tan A + tan B + tan C \begin{aligned}
(\tan A\tan B-1)\tan C &= \tan A+\tan B \\
\tan A\tan B\tan C &= \tan A+\tan B+\tan C
\end{aligned} ( tan A tan B − 1 ) tan C tan A tan B tan C = tan A + tan B = tan A + tan B + tan C (2) △ A B C \triangle\mathrm{ABC} △ A B C A , A, A , B , B, B , C ≠ π 2 C \neq \dfrac{\pi}{2} C = 2 π ( 4 S ) 6 ≧ 27 ( b 2 + c 2 − a 2 ) 2 ( c 2 + a 2 − b 2 ) 2 ( a 2 + b 2 − c 2 ) 2 ⟺ 4 S 2 b 2 c 2 4 S 2 c 2 a 2 4 S 2 a 2 b 2 ≧ 27 ( b 2 + c 2 − a 2 ) 2 4 b 2 c 2 ( c 2 + a 2 − b 2 ) 2 4 c 2 a 2 ( a 2 + b 2 − c 2 ) 2 4 a 2 b 2 ⟺ sin 2 A sin 2 B sin 2 C ≧ 27 cos 2 A cos 2 B cos 2 C ⟺ tan 2 A tan 2 B tan 2 C ≧ 27 ⋯ [ 1 ] \begin{aligned}
&(4S)^6 \geqq 27(b^2+c^2-a^2)^2(c^2+a^2-b^2)^2(a^2+b^2-c^2)^2 \\
&\iff \frac{4S^2}{b^2c^2}\frac{4S^2}{c^2a^2}\frac{4S^2}{a^2b^2} \\
&\quad \geqq 27\frac{(b^2+c^2-a^2)^2}{4b^2c^2}\frac{(c^2+a^2-b^2)^2}{4c^2a^2}\frac{(a^2+b^2-c^2)^2}{4a^2b^2} \\
&\iff \sin ^2A\sin ^2B\sin ^2C \geqq 27\cos ^2A\cos ^2B\cos ^2C \\
&\iff \tan ^2A\tan ^2B\tan ^2C \geqq 27 \quad \cdots [1]
\end{aligned} ( 4 S ) 6 ≧ 2 7 ( b 2 + c 2 − a 2 ) 2 ( c 2 + a 2 − b 2 ) 2 ( a 2 + b 2 − c 2 ) 2 ⟺ b 2 c 2 4 S 2 c 2 a 2 4 S 2 a 2 b 2 4 S 2 ≧ 2 7 4 b 2 c 2 ( b 2 + c 2 − a 2 ) 2 4 c 2 a 2 ( c 2 + a 2 − b 2 ) 2 4 a 2 b 2 ( a 2 + b 2 − c 2 ) 2 ⟺ sin 2 A sin 2 B sin 2 C ≧ 2 7 cos 2 A cos 2 B cos 2 C ⟺ tan 2 A tan 2 B tan 2 C ≧ 2 7 ⋯ [ 1 ] tan A , \tan A, tan A , tan B , \tan B, tan B , tan C > 0 \tan C > 0 tan C > 0 tan A + tan B + tan C ≧ 3 tan A tan B tan C 3 ⋯ [ 2 ] \tan A+\tan B+\tan C \geqq 3\sqrt[3]{\tan A\tan B\tan C} \quad \cdots [2] tan A + tan B + tan C ≧ 3 3 tan A tan B tan C ⋯ [ 2 ] [ 2 ] ⟺ tan A tan B tan C ≧ 3 tan A tan B tan C 3 ⟺ tan 3 A tan 3 B tan 3 C ≧ 27 tan A tan B tan C ⟺ [ 1 ] \begin{aligned}
[2] &\iff \tan A\tan B\tan C \geqq 3\sqrt[3]{\tan A\tan B\tan C} \\
&\iff \tan ^3A\tan ^3B\tan ^3C \geqq 27\tan A\tan B\tan C \\
&\iff [1]
\end{aligned} [ 2 ] ⟺ tan A tan B tan C ≧ 3 3 tan A tan B tan C ⟺ tan 3 A tan 3 B tan 3 C ≧ 2 7 tan A tan B tan C ⟺ [ 1 ]
参考
(2) の不等式は
「オノの不等式」 (Ono's inequality) として知られている.
鈍角三角形においては成り立つとは限らない.
例えば,
a = 2 , a = 2, a = 2 , b = 3 , b = 3, b = 3 , c = 4 c = 4 c = 4 なる
△ A B C \triangle\mathrm{ABC} △ A B C において,
cos C = 2 2 + 3 2 − 4 2 2 ⋅ 2 ⋅ 3 = − 1 4 , sin C = 1 − cos 2 C = 15 4 \begin{aligned}
\cos C &= \frac{2^2+3^2-4^2}{2\cdot 2\cdot 3} = -\frac{1}{4}, \\
\sin C &= \sqrt{1-\cos ^2C} = \frac{\sqrt{15}}{4}
\end{aligned} cos C sin C = 2 ⋅ 2 ⋅ 3 2 2 + 3 2 − 4 2 = − 4 1 , = 1 − cos 2 C = 4 1 5
から
S = 1 2 a b sin C = 1 2 ⋅ 2 ⋅ 3 ⋅ 15 4 = 3 15 4 S = \frac{1}{2}ab\sin C = \frac{1}{2}\cdot 2\cdot 3\cdot\frac{\sqrt{15}}{4} = \frac{3\sqrt{15}}{4} S = 2 1 a b sin C = 2 1 ⋅ 2 ⋅ 3 ⋅ 4 1 5 = 4 3 1 5
であり,
b 2 + c 2 − a 2 = 3 2 + 4 2 − 2 2 = 21 , c 2 + a 2 − b 2 = 4 2 + 2 2 − 3 2 = 11 , a 2 + b 2 − c 2 = − 3 \begin{aligned}
b^2+c^2-a^2 &= 3^2+4^2-2^2 = 21, \\
c^2+a^2-b^2 &= 4^2+2^2-3^2 = 11, \\
a^2+b^2-c^2 &= -3
\end{aligned} b 2 + c 2 − a 2 c 2 + a 2 − b 2 a 2 + b 2 − c 2 = 3 2 + 4 2 − 2 2 = 2 1 , = 4 2 + 2 2 − 3 2 = 1 1 , = − 3
であるので, 不等式の左辺
( 3 15 ) 6 = 3 9 ⋅ 5 3 = 3 7 ⋅ 1125 (3\sqrt{15})^6 = 3^9\cdot 5^3 = 3^7\cdot 1125 ( 3 1 5 ) 6 = 3 9 ⋅ 5 3 = 3 7 ⋅ 1 1 2 5
は不等式の右辺
27 ⋅ 2 1 2 ⋅ 1 1 2 ⋅ ( − 3 ) 2 = 3 7 ⋅ 7 2 ⋅ 1 1 2 = 3 7 ⋅ 5929 27\cdot 21^2\cdot 11^2\cdot (-3)^2 = 3^7\cdot 7^2\cdot 11^2 = 3^7\cdot 5929 2 7 ⋅ 2 1 2 ⋅ 1 1 2 ⋅ ( − 3 ) 2 = 3 7 ⋅ 7 2 ⋅ 1 1 2 = 3 7 ⋅ 5 9 2 9
より小さい.
△ A B C \triangle\mathrm{ABC} △ A B C において,
A = ∠ A , A = \angle\mathrm A, A = ∠ A , B = ∠ B , B = \angle\mathrm B, B = ∠ B , C = ∠ C C = \angle\mathrm C C = ∠ C とおく.
tan A , \tan A, tan A , tan B , \tan B, tan B , tan C \tan C tan C がすべて整数であるとき, それらの値を求めよ.
2019 / 03 / 27 2019/03/27 2 0 1 9 / 0 3 / 2 7 2019 / 03 / 27 2019/03/27 2 0 1 9 / 0 3 / 2 7 解答例
△ A B C \triangle\mathrm{ABC} △ A B C において,
tan A , \tan A, tan A , tan B , \tan B, tan B , tan C \tan C tan C が整数であるとする.
また,
A ≦ B ≦ C A \leqq B \leqq C A ≦ B ≦ C であるとする.
このとき,
3 A ≦ A + B + C = π 3A \leqq A+B+C = \pi 3 A ≦ A + B + C = π
から,
0 < A ≦ π 3 0 < A \leqq \dfrac{\pi}{3} 0 < A ≦ 3 π である.
よって,
0 < tan A ≦ 3 0 < \tan A \leqq \sqrt 3 0 < tan A ≦ 3
であり,
tan A \tan A tan A は整数だから,
tan A = 1 \tan A = 1 tan A = 1
である.
また,
A = π − ( B + C ) A = \pi -(B+C) A = π − ( B + C ) から,
1 = − tan ( B + C ) = − tan B + tan C 1 − tan B tan C 1 = -\tan (B+C) = -\frac{\tan B+\tan C}{1-\tan B\tan C} 1 = − tan ( B + C ) = − 1 − tan B tan C tan B + tan C
が成り立つ.
よって,
tan B tan C − tan B − tan C − 1 = 0 ( tan B − 1 ) ( tan C − 1 ) = 2 \begin{aligned}
\tan B\tan C-\tan B-\tan C-1 &= 0 \\
(\tan B-1)(\tan C-1) &= 2
\end{aligned} tan B tan C − tan B − tan C − 1 ( tan B − 1 ) ( tan C − 1 ) = 0 = 2
となる.
tan B , \tan B, tan B , tan C \tan C tan C が整数であること,
π 4 = A ≦ B ≦ C \dfrac{\pi}{4} = A \leqq B \leqq C 4 π = A ≦ B ≦ C から
B , B, B , C C C は鋭角であって
tan B ≦ tan C \tan B \leqq \tan C tan B ≦ tan C であることに注意すると,
tan B = 2 , tan C = 3 \tan B = 2, \quad \tan C = 3 tan B = 2 , tan C = 3
である.
A ≦ B ≦ C A \leqq B \leqq C A ≦ B ≦ C という条件を忘れると, 求める値は
{ tan A , tan B , tan C } = { 1 , 2 , 3 } \{\tan A,\ \tan B,\ \tan C\} = \{ 1,\ 2,\ 3\} { tan A , tan B , tan C } = { 1 , 2 , 3 }
である.
水平面上の点
O \mathrm O O の真上に相異なる
2 2 2 点
A , \mathrm A, A , B \mathrm B B をとる (
A \mathrm A A は
B \mathrm B B より上方にあるとする).
A , \mathrm A, A , B \mathrm B B を見た仰角の差が最大になるような水平面上の点
P \mathrm P P について,
a = O A , a = \mathrm{OA}, a = O A , b = O B b = \mathrm{OB} b = O B を用いて距離
O P \mathrm{OP} O P を表せ.
2015 / 04 / 23 2015/04/23 2 0 1 5 / 0 4 / 2 3 2023 / 05 / 12 2023/05/12 2 0 2 3 / 0 5 / 1 2 解答例
θ = ∠ A P B , \theta = \angle\mathrm{APB}, θ = ∠ A P B , x = O P , x = \mathrm{OP}, x = O P , α = ∠ O P A , \alpha = \angle\mathrm{OPA}, α = ∠ O P A , β = ∠ O P B \beta = \angle\mathrm{OPB} β = ∠ O P B とおく.
このとき,
0 < θ < π 2 0 < \theta < \dfrac{\pi}{2} 0 < θ < 2 π であるから
θ \theta θ が最大
⟺ \iff ⟺ tan θ \tan\theta tan θ が最大
が成り立ち,
θ = α − β , tan α = a x , tan β = b x \theta = \alpha -\beta, \quad \tan\alpha = \frac{a}{x}, \quad \tan\beta = \frac{b}{x} θ = α − β , tan α = x a , tan β = x b
であるから, 加法定理により
tan θ = tan ( α − β ) = tan α − tan β 1 + tan α tan β = a x − b x 1 + a x ⋅ b x = a − b x + a b x \begin{aligned}
\tan\theta &= \tan (\alpha -\beta ) = \frac{\tan\alpha -\tan\beta }{1+\tan\alpha\tan\beta} \\
&= \frac{\dfrac{a}{x}-\dfrac{b}{x}}{1+\dfrac{a}{x}\cdot\dfrac{b}{x}} = \frac{a-b}{x+\dfrac{ab}{x}}
\end{aligned} tan θ = tan ( α − β ) = 1 + tan α tan β tan α − tan β = 1 + x a ⋅ x b x a − x b = x + x a b a − b
が成り立つ.
よって,
θ \theta θ が最大になるのは,
x + a b x x+\dfrac{ab}{x} x + x a b が最小になるときである.
相加・相乗平均の不等式により
x + a b x x+\dfrac{ab}{x} x + x a b は
x = a b x , x = \dfrac{ab}{x}, x = x a b , x 2 = a b x^2 = ab x 2 = a b つまり
x = a b x = \sqrt{ab} x = a b の場合に限り最小値をとるから, 求める距離は
O P = a b \mathrm{OP} = \sqrt{ab} O P = a b である.
別解: 微分法 (数学 III) を利用
θ = ∠ A P B , \theta = \angle\mathrm{APB}, θ = ∠ A P B , x = O P , x = \mathrm{OP}, x = O P , α = ∠ O P A , \alpha = \angle\mathrm{OPA}, α = ∠ O P A , β = ∠ O P B \beta = \angle\mathrm{OPB} β = ∠ O P B とおく.
このとき,
0 < θ < π 2 0 < \theta < \dfrac{\pi}{2} 0 < θ < 2 π であるから
θ \theta θ が最大
⟺ \iff ⟺ tan θ \tan\theta tan θ が最大
が成り立ち,
θ = α − β , tan α = a x , tan β = b x \theta = \alpha -\beta, \quad \tan\alpha = \frac{a}{x}, \quad \tan\beta = \frac{b}{x} θ = α − β , tan α = x a , tan β = x b
であるから, 加法定理により
tan θ = tan ( α − β ) = tan α − tan β 1 + tan α tan β = a x − b x 1 + a x ⋅ b x = ( a − b ) x x 2 + a b \begin{aligned}
\tan\theta &= \tan (\alpha -\beta ) = \frac{\tan\alpha -\tan\beta }{1+\tan\alpha\tan\beta} \\
&= \frac{\dfrac{a}{x}-\dfrac{b}{x}}{1+\dfrac{a}{x}\cdot\dfrac{b}{x}} = (a-b)\frac{x}{x^2+ab}
\end{aligned} tan θ = tan ( α − β ) = 1 + tan α tan β tan α − tan β = 1 + x a ⋅ x b x a − x b = ( a − b ) x 2 + a b x
が成り立つ.
d d x tan θ = ( a − b ) a b − x 2 ( x 2 + a b ) 2 \frac{d}{dx}\tan\theta = (a-b)\frac{ab-x^2}{(x^2+ab)^2} d x d tan θ = ( a − b ) ( x 2 + a b ) 2 a b − x 2
であるから,
d d x tan θ ≧ 0 ⟺ 0 < x ≦ a b , d d x tan θ ≦ 0 ⟺ x ≧ a b \begin{aligned}
\frac{d}{dx}\tan\theta \geqq 0 &\iff 0 < x \leqq \sqrt{ab}, \\
\frac{d}{dx}\tan\theta \leqq 0 &\iff x \geqq \sqrt{ab}
\end{aligned} d x d tan θ ≧ 0 d x d tan θ ≦ 0 ⟺ 0 < x ≦ a b , ⟺ x ≧ a b
が成り立つ.
x x x 0 0 0 ⋯ \cdots ⋯ a b \sqrt{ab} a b ⋯ \cdots ⋯ d d x tan θ \dfrac{d}{dx}\tan\theta d x d tan θ + + + 0 0 0 − - − tan θ \tan\theta tan θ ↗ \nearrow ↗ 極大 ↘ \searrow ↘
よって,
tan θ \tan\theta tan θ は
x = a b x = \sqrt{ab} x = a b の場合に限って, 極大かつ最大の値をとる.
ゆえに, 求める距離は
O P = a b \mathrm{OP} = \sqrt{ab} O P = a b である.
参考
ここで解いた「レギオモンタヌスの問題」 は, 中世ドイツの天文学者・数学者のレギオモンタヌス (Regiomontanus) によって考え出された (1471 1471 1 4 7 1 方べきの定理を使った別解については, こちら を参照されたい.
θ \theta θ が
π \pi π の倍数でないとき,
cot θ = 1 tan θ \cot\theta = \dfrac{1}{\tan\theta} cot θ = tan θ 1 と定める.
(1) cot ( π − θ ) = − cot θ , cot ( α + β ) = cot α cot β − 1 cot α + cot β \cot (\pi -\theta ) = -\cot\theta,\ \cot (\alpha +\beta ) = \frac{\cot\alpha\cot\beta -1}{\cot\alpha +\cot\beta} cot ( π − θ ) = − cot θ , cot ( α + β ) = cot α + cot β cot α cot β − 1 (2) △ A B C \triangle\mathrm{ABC} △ A B C cot A + cot B + cot C ≧ 3 \cot A+\cot B+\cot C \geqq \sqrt 3 cot A + cot B + cot C ≧ 3
2020 / 11 / 19 2020/11/19 2 0 2 0 / 1 1 / 1 9 2024 / 05 / 20 2024/05/20 2 0 2 4 / 0 5 / 2 0 解答例
(1) tan ( π − θ ) = − tan θ \tan (\pi -\theta ) = -\tan\theta tan ( π − θ ) = − tan θ cot ( π − θ ) = − cot θ \cot (\pi -\theta ) = -\cot\theta cot ( π − θ ) = − cot θ tan ( α + β ) = tan α + tan β 1 − tan α tan β \tan (\alpha +\beta ) = \frac{\tan\alpha +\tan\beta}{1-\tan\alpha\tan\beta} tan ( α + β ) = 1 − tan α tan β tan α + tan β cot ( α + β ) = 1 − tan α tan β tan α + tan β = 1 tan α tan β − 1 1 tan β + 1 tan α = 1 tan α ⋅ 1 tan β − 1 1 tan α + 1 tan β = cot α cot β − 1 cot α + cot β \begin{aligned}
&\cot (\alpha +\beta ) = \frac{1-\tan\alpha\tan\beta}{\tan\alpha +\tan\beta} \\
&= \frac{\dfrac{1}{\tan\alpha\tan\beta}-1}{\dfrac{1}{\tan\beta}+\dfrac{1}{\tan\alpha}} = \frac{\dfrac{1}{\tan\alpha}\cdot\dfrac{1}{\tan\beta}-1}{\dfrac{1}{\tan\alpha}+\dfrac{1}{\tan\beta}} \\
&= \frac{\cot\alpha\cot\beta -1}{\cot\alpha +\cot\beta}
\end{aligned} cot ( α + β ) = tan α + tan β 1 − tan α tan β = tan β 1 + tan α 1 tan α tan β 1 − 1 = tan α 1 + tan β 1 tan α 1 ⋅ tan β 1 − 1 = cot α + cot β cot α cot β − 1 (2) 内角のうち C C C C \mathrm C C A B \mathrm{AB} A B H \mathrm H H A B \mathrm{AB} A B x = A H , x = \mathrm{AH}, x = A H , h = C H h = \mathrm{CH} h = C H cot A + cot B = x h + c − x h = c h \cot A+\cot B = \frac{x}{h}+\frac{c-x}{h} = \frac{c}{h} cot A + cot B = h x + h c − x = h c cot A + cot B + cot C = cot A + cot B + cot ( π − A − B ) = cot A + cot B − cot ( A + B ) = cot A + cot B − cot A cot B − 1 cot A + cot B = c h − x h ⋅ c − x h − 1 c h = x 2 − c x + c 2 + h 2 c h = ( x − c 2 ) 2 + ( 3 2 c − h ) 2 + 3 c h c h = 1 c h ( x − c 2 ) 2 + 1 c h ( 3 2 c − h ) 2 + 3 ≧ 3 \begin{aligned}
&\cot A+\cot B+\cot C \\
&= \cot A+\cot B+\cot (\pi -A-B) \\
&= \cot A+\cot B-\cot (A+B) \\
&= \cot A+\cot B-\frac{\cot A\cot B-1}{\cot A+\cot B} \\
&= \frac{c}{h}-\frac{\dfrac{x}{h}\cdot\dfrac{c-x}{h}-1}{\dfrac{c}{h}} = \frac{x^2-cx+c^2+h^2}{ch} \\
&= \frac{\left( x-\dfrac{c}{2}\right) ^2+\left(\dfrac{\sqrt 3}{2}c-h\right) ^2+\sqrt 3ch}{ch} \\
&= \frac{1}{ch}\left( x-\dfrac{c}{2}\right) ^2+\frac{1}{ch}\left(\dfrac{\sqrt 3}{2}c-h\right) ^2+\sqrt 3 \geqq \sqrt 3
\end{aligned} cot A + cot B + cot C = cot A + cot B + cot ( π − A − B ) = cot A + cot B − cot ( A + B ) = cot A + cot B − cot A + cot B cot A cot B − 1 = h c − h c h x ⋅ h c − x − 1 = c h x 2 − c x + c 2 + h 2 = c h ( x − 2 c ) 2 + ( 2 3 c − h ) 2 + 3 c h = c h 1 ( x − 2 c ) 2 + c h 1 ( 2 3 c − h ) 2 + 3 ≧ 3 x = c 2 , x = \dfrac{c}{2}, x = 2 c , h = 3 2 c , h = \dfrac{\sqrt 3}{2}c, h = 2 3 c , △ A B C \triangle\mathrm{ABC} △ A B C
参考
cot θ \cot\theta cot θ θ \theta θ 「余接」 (cotangent) と呼ぶ.
△ A B C \triangle\mathrm{ABC} △ A B C ∠ P A B = ∠ P B C = ∠ P C A = θ , \angle\mathrm{PAB} = \angle\mathrm{PBC} = \angle\mathrm{PCA} = \theta, ∠ P A B = ∠ P B C = ∠ P C A = θ , ∠ P ′ A C = ∠ P ′ C B = ∠ P ′ B A = θ \angle\mathrm P'\mathrm{AC} = \angle\mathrm P'\mathrm{CB} = \angle\mathrm P'\mathrm{BA} = \theta ∠ P ′ A C = ∠ P ′ C B = ∠ P ′ B A = θ P , \mathrm P, P , P ′ \mathrm P' P ′ 「第 1 1 1 (first Brocard point),「第 2 2 2 (second Brocard point),
θ \theta θ 「ブロカール角」 (Brocard angle) と呼ぶ.
cot θ = a 2 + b 2 + c 2 4 S \cot\theta = \frac{a^2+b^2+c^2}{4S} cot θ = 4 S a 2 + b 2 + c 2 こちら を参照).
頂点 C \mathrm C C A B \mathrm{AB} A B h h h h = 2 S c h = \dfrac{2S}{c} h = c 2 S cot A + cot B = c h = c 2 2 S \cot A+\cot B = \frac{c}{h} = \frac{c^2}{2S} cot A + cot B = h c = 2 S c 2 cot B + cot C = a 2 2 S , cot C + cot A = b 2 2 S \cot B+\cot C = \frac{a^2}{2S}, \quad \cot C+\cot A = \frac{b^2}{2S} cot B + cot C = 2 S a 2 , cot C + cot A = 2 S b 2 cot θ = a 2 + b 2 + c 2 4 S = cot A + cot B + cot C \cot\theta = \frac{a^2+b^2+c^2}{4S} = \cot A+\cot B+\cot C cot θ = 4 S a 2 + b 2 + c 2 = cot A + cot B + cot C θ ≦ 3 0 ∘ \theta \leqq 30^\circ θ ≦ 3 0 ∘ (2) の不等式は,
x 2 y 2 + y 2 z 2 + z 2 x 2 ≧ x y z ( x + y + z ) ⋯ [ ∗ ] x^2y^2+y^2z^2+z^2x^2 \geqq xyz(x+y+z) \quad \cdots [\ast ] x 2 y 2 + y 2 z 2 + z 2 x 2 ≧ x y z ( x + y + z ) ⋯ [ ∗ ] ( 1 x + 1 y + 1 z ) 2 = 1 x 2 + 1 y 2 + 1 z 2 + 2 x y + 2 y z + 2 z x = x 2 y 2 + y 2 z 2 + z 2 x 2 x 2 y 2 z 2 + 2 ⋅ x + y + z x y z ≧ x y z ( x + y + z ) x 2 y 2 z 2 + 2 ⋅ x + y + z x y z ( ∵ [ ∗ ] ) = 3 ⋅ x + y + z x y z \begin{aligned}
\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2 &= \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{zx} \\
&= \frac{x^2y^2+y^2z^2+z^2x^2}{x^2y^2z^2}+2\cdot\frac{x+y+z}{xyz} \\
&\geqq \frac{xyz(x+y+z)}{x^2y^2z^2}+2\cdot\frac{x+y+z}{xyz} \quad (\because [\ast ]) \\
&= 3\cdot\frac{x+y+z}{xyz} \\
\end{aligned} ( x 1 + y 1 + z 1 ) 2 = x 2 1 + y 2 1 + z 2 1 + x y 2 + y z 2 + z x 2 = x 2 y 2 z 2 x 2 y 2 + y 2 z 2 + z 2 x 2 + 2 ⋅ x y z x + y + z ≧ x 2 y 2 z 2 x y z ( x + y + z ) + 2 ⋅ x y z x + y + z ( ∵ [ ∗ ] ) = 3 ⋅ x y z x + y + z tan A + tan B + tan C = tan A tan B tan C \tan A+\tan B+\tan C = \tan A\tan B\tan C tan A + tan B + tan C = tan A tan B tan C 2024 2024 2 0 2 4 [ ∗ ] [\ast ] [ ∗ ] x 2 y 2 + y 2 z 2 + z 2 x 2 − x y z ( x + y + z ) = x 2 y 2 z 2 ( 1 x 2 + 1 y 2 + 1 z 2 − 1 x y − 1 y z − 1 z x ) = x 2 y 2 z 2 2 { ( 1 x − 1 y ) 2 + ( 1 y − 1 z ) 2 + ( 1 z − 1 x ) 2 } ≧ 0 \begin{aligned}
&x^2y^2+y^2z^2+z^2x^2-xyz(x+y+z) \\
&= x^2y^2z^2\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}-\frac{1}{xy}-\frac{1}{yz}-\frac{1}{zx}\right) \\
&= \frac{x^2y^2z^2}{2}\left\{\left(\frac{1}{x}-\frac{1}{y}\right) ^2+\left(\frac{1}{y}-\frac{1}{z}\right) ^2+\left(\frac{1}{z}-\frac{1}{x}\right) ^2\right\} \geqq 0
\end{aligned} x 2 y 2 + y 2 z 2 + z 2 x 2 − x y z ( x + y + z ) = x 2 y 2 z 2 ( x 2 1 + y 2 1 + z 2 1 − x y 1 − y z 1 − z x 1 ) = 2 x 2 y 2 z 2 { ( x 1 − y 1 ) 2 + ( y 1 − z 1 ) 2 + ( z 1 − x 1 ) 2 } ≧ 0
このとき, であるから