좌표평면에서 A ( 0 , 0 ) \displaystyle \mathrm A(0,0) A ( 0 , 0 ) B ( c , 0 ) \displaystyle \mathrm B(c,0) B ( c , 0 ) C ( α , β ) \displaystyle \mathrm C(\alpha,\beta) C ( α , β ) P ( x , y ) \displaystyle \mathrm P(x,y) P ( x , y ) P A → = ( − x , − y ) \displaystyle \overrightarrow{\mathrm{PA}}=(-x,-y) PA = ( − x , − y ) P B → = ( c − x , − y ) \displaystyle \overrightarrow{\mathrm{PB}}=(c-x,-y) PB = ( c − x , − y ) P C → = ( α − x , β − y ) \displaystyle \overrightarrow{\mathrm{PC}}=(\alpha-x,\beta-y) PC = ( α − x , β − y )
Q = P A → ⋅ P B → + P B → ⋅ P C → + P C → ⋅ P A → \displaystyle Q=\overrightarrow{\mathrm{PA}}\cdot\overrightarrow{\mathrm{PB}}+\overrightarrow{\mathrm{PB}}\cdot\overrightarrow{\mathrm{PC}}+\overrightarrow{\mathrm{PC}}\cdot\overrightarrow{\mathrm{PA}} Q = PA ⋅ PB + PB ⋅ PC + PC ⋅ PA Q = ( − x ) ( c − x ) + ( − y ) ( − y ) + ( c − x ) ( α − x ) + ( − y ) ( β − y ) + ( α − x ) ( − x ) + ( β − y ) ( − y ) = 3 x 2 − 2 ( c + α ) x + 3 y 2 − 2 β y + c α = 3 ( x − c + α 3 ) 2 + 3 ( y − β 3 ) 2 + c α − ( c + α ) 2 + β 2 3 \displaystyle \begin{aligned}Q&=(-x)(c-x)+(-y)(-y)+(c-x)(\alpha-x)+(-y)(\beta-y)+(\alpha-x)(-x)+(\beta-y)(-y)\\&=3x^2-2(c+\alpha)x+3y^2-2\beta y+c\alpha\\&=3\left(x-\frac{c+\alpha}3\right)^2+3\left(y-\frac\beta3\right)^2+c\alpha-\frac{(c+\alpha)^2+\beta^2}3\end{aligned} Q = ( − x ) ( c − x ) + ( − y ) ( − y ) + ( c − x ) ( α − x ) + ( − y ) ( β − y ) + ( α − x ) ( − x ) + ( β − y ) ( − y ) = 3 x 2 − 2 ( c + α ) x + 3 y 2 − 2 β y + c α = 3 ( x − 3 c + α ) 2 + 3 ( y − 3 β ) 2 + c α − 3 ( c + α ) 2 + β 2 이다. 점 G ( c + α 3 , β 3 ) \displaystyle \mathrm G\left(\frac{c+\alpha}3,\frac\beta3\right) G ( 3 c + α , 3 β ) A B C \displaystyle \mathrm{ABC} ABC
따라서 Q \displaystyle Q Q P = G \displaystyle \mathrm P=\mathrm G P = G c α − ( c + α ) 2 + β 2 3 \displaystyle c\alpha-\frac{(c+\alpha)^2+\beta^2}3 c α − 3 ( c + α ) 2 + β 2
또한 b 2 = α 2 + β 2 \displaystyle b^2=\alpha^2+\beta^2 b 2 = α 2 + β 2 a 2 = ( α − c ) 2 + β 2 = α 2 + β 2 − 2 c α + c 2 \displaystyle a^2=(\alpha-c)^2+\beta^2=\alpha^2+\beta^2-2c\alpha+c^2 a 2 = ( α − c ) 2 + β 2 = α 2 + β 2 − 2 c α + c 2 α 2 + β 2 = b 2 \displaystyle \alpha^2+\beta^2=b^2 α 2 + β 2 = b 2 2 c α = b 2 + c 2 − a 2 \displaystyle 2c\alpha=b^2+c^2-a^2 2 c α = b 2 + c 2 − a 2
따라서 Q \displaystyle Q Q c α − ( c + α ) 2 + β 2 3 = c α − c 2 − ( α 2 + β 2 ) 3 = b 2 + c 2 − a 2 2 − c 2 − b 2 3 = − a 2 + b 2 + c 2 6 \displaystyle c \alpha - \frac{{( c + \alpha )^{2} + \beta^{2}}}{3} = \frac{{c \alpha - c^{2} - ( \alpha^{2} + \beta^{2} )}}{3} =\frac{{\frac{{b^{2} + c^{2} - a^{2}}}{2} - c^{2} - b^{2}}}{3} = - \frac{{a^{2} + b^{2} + c^{2}}}{6} c α − 3 ( c + α ) 2 + β 2 = 3 c α − c 2 − ( α 2 + β 2 ) = 3 2 b 2 + c 2 − a 2 − c 2 − b 2 = − 6 a 2 + b 2 + c 2
G P ‾ 2 = ( x − c + α 3 ) 2 + ( y − β 3 ) 2 \displaystyle \overline{\mathrm{GP}}^2=\left(x-\frac{c+\alpha}3\right)^2+\left(y-\frac\beta3\right)^2 GP 2 = ( x − 3 c + α ) 2 + ( y − 3 β ) 2 P \displaystyle \mathrm P P Q \displaystyle Q Q
P = A \displaystyle \mathrm P=\mathrm A P = A Q = b c cos A = b 2 + c 2 − a 2 2 \displaystyle Q=bc\cos A=\frac{b^2+c^2-a^2}2 Q = b c cos A = 2 b 2 + c 2 − a 2
P = B \displaystyle \mathrm P=\mathrm B P = B Q = c a cos B = c 2 + a 2 − b 2 2 \displaystyle Q=ca\cos B=\frac{c^2+a^2-b^2}2 Q = c a cos B = 2 c 2 + a 2 − b 2
P = C \displaystyle \mathrm P=\mathrm C P = C Q = a b cos C = a 2 + b 2 − c 2 2 \displaystyle Q=ab\cos C=\frac{a^2+b^2-c^2}2 Q = ab cos C = 2 a 2 + b 2 − c 2
그런데 a ≤ b ≤ c \displaystyle a\le b\le c a ≤ b ≤ c Q \displaystyle Q Q P = A \displaystyle \mathrm P=\mathrm A P = A b 2 + c 2 − a 2 2 \displaystyle \frac{b^2+c^2-a^2}2 2 b 2 + c 2 − a 2
수리논술, 배움에서 논증의 완성까지.