题面
22．（12分）（2023•新高考Ⅰ）在直角坐标系x O y x O y x O y 1 2 ) \frac 1 2 ) 2 1 )
（1）求W的方程；
（2）已知矩形ABCD有三个顶点在W上，证明：矩形ABCD的周长大于3 3 3 \sqrt{3} 3 3
2023年高考数学试卷（新高考Ⅰ）
（1）求W的方程；
（2）已知矩形ABCD有三个顶点在W上，证明：矩形ABCD的周长大于3 3 3 \sqrt{3} 3 3
答案
见试题解答内容
解析
分析
（1）设点p坐标，结合几何条件即可得出W的方程
（2）首先利用平移性，化简 W 的方程可简化计算，核心是把两邻边的和用其他方式表示出来
解答
解：（1）设点P点坐标为( x , y ) ( x , \ y ) ( x , y ) ∣ y ∣ = x 2 + ( y − 1 2 ) 2 \lvert y \rvert = \sqrt{x^{2} + ( y - \frac{1}{2} )^{2}} ∣ y ∣ = x 2 + ( y − 2 1 ) 2
两边平方可得：y 2 = x 2 + y 2 − y + 1 4 y^{2} = x^{2} + y^{2} - y + \frac{1}{4} y 2 = x 2 + y 2 − y + 4 1
化简得：y = x 2 + 1 4 y = x^{2} + \frac{1}{4} y = x 2 + 4 1
故W的方程为y = x 2 + 1 4 . {y = x^{2} + {\frac{1}{4}} .} y = x 2 + 4 1 .
（2）解法一：不妨设A，B，C三点在W上，且A B ⊥ B C A B \bot B C A B ⊥ B C
设A ( a , a 2 + 1 4 ) , B ( b , b 2 + 1 4 ) , C ( c , c 2 + 1 4 ) A ( a , a^{2} + \frac{1}{4} ) , B ( b , b^{2} + \frac{1}{4} ) , C ( c , c^{2} + \frac{1}{4} ) A ( a , a 2 + 4 1 ) , B ( b , b 2 + 4 1 ) , C ( c , c 2 + 4 1 )
则∣ A B = ⃗ ( b − a , b 2 − a 2 ) , B C ⃗ = ( c − b , c 2 − b 2 ) \vec{| A B =} ( b - a , \ b^{2} - a^{2} ) , \ \vec{B C} = ( c - b , \ c^{2} - b^{2} ) ∣ A B = ( b − a , b 2 − a 2 ) , B C = ( c − b , c 2 − b 2 )
由题意，A B ⋅ B C = 0 \stackrel{}{A B} \cdot \stackrel{}{B C} = 0 A B ⋅ B C = 0 ( b − a ) ( c − b ) + ( b 2 − a 2 ) ( c 2 − b 2 ) = 0. \left( b - a \right) \left( c - b \right) + \left( b^{2} - a^{2} \right) \left( c^{2} - b^{2} \right) = 0 . ( b − a ) ( c − b ) + ( b 2 − a 2 ) ( c 2 − b 2 ) = 0.
显然( b − a ) ( c − b ) ≠ 0 ( b - a ) \ ( c - b ) \neq 0 ( b − a ) ( c − b ) = 0 1 + ( b + a ) ( c + b ) = 0 1 + \ ( b + a ) \ ( c + b ) \ = 0 1 + ( b + a ) ( c + b ) = 0
此时，∣ b + a ∣ . ∣ c + b ∣ = 1 | b + a | . | c + b | = 1 ∣ b + a ∣.∣ c + b ∣ = 1 m i n { ∣ b + a ∣ , ∣ c + b ∣ } ⩽ 1 m i n \{| b + a | , | c + b | \} \leqslant 1 min { ∣ b + a ∣ , ∣ c + b ∣ } ⩽ 1
不妨设⋅ ∣ c + b ∣ ⩽ 1 \cdot | c + b | \leqslant 1 ⋅ ∣ c + b ∣ ⩽ 1 a = − b − 1 b + c a {=} - b {-}{\frac{1}{b {+} c}} a = − b − b + c 1
则|AB|+|BC|＝|b﹣a|f1 + (� + �)# +|c﹣b|f1 + (� + �)#
= ∣ b − a ∣ 1 + 1 ( c + b ) 2 + ∣ c − b ∣ 1 + ( c + b ) 2 = | b - a | \sqrt{1 + \frac{1}{(c + b)^{2}}} + | c - b | \sqrt{1 + (c + b)^{2}} = ∣ b − a ∣ 1 + ( c + b ) 2 1 + ∣ c − b ∣ 1 + ( c + b ) 2
⩾ ∣ b − a ∣ 1 + ( c + b ) 2 + ∣ c − b ∣ 1 + ( c + b ) 2 \geqslant | b - a | \sqrt{1 + (c + b)^{2}} + | c - b | \sqrt{1 + (c + b)^{2}} ⩾ ∣ b − a ∣ 1 + ( c + b ) 2 + ∣ c − b ∣ 1 + ( c + b ) 2
⩾ ∣ c − a ∣ 1 + ( c + b ) 2 \geqslant | c - a | \sqrt{1 + (c + b)^{2}} ⩾ ∣ c − a ∣ 1 + ( c + b ) 2
= ∣ b + c + 1 b + c ∣ 1 + ( c + b ) 2 . = | b + c + \frac{1}{b + c} | \sqrt{1 + (c + b)^{2}}. = ∣ b + c + b + c 1 ∣ 1 + ( c + b ) 2 .
设x = ∣ b + c ∣ x^{=} | b {+} c | x = ∣ b + c ∣ f ( x ) = ( x + 1 x ) 1 + x 2 f ( x ) = ( x + {\frac{1}{x}} ) \ {\sqrt{1 + x^{2}}} f ( x ) = ( x + x 1 ) 1 + x 2 f ( x ) = ( 1 + x 2 ) 3 2 x , f \left( x \right) = \frac{{\left( 1 + x^{2} \right)}^{\frac{3}{2}}}{x} , f ( x ) = x ( 1 + x 2 ) 2 3 ,
f ′ ( x ) = ( 1 + x 2 ) 1 2 . ( 3 x 2 − 1 − x 2 ) x 2 = ( 1 + x 2 ) 1 2 . ( 2 x 2 − 1 ) x 2 . f^{\prime} \quad (x) = \frac{(1 + x^{2})^{\frac{1}{2}} . (3 x^{2} - 1 - x^{2})}{x^{2}} = \frac{(1 + x^{2})^{\frac{1}{2}} . (2 x^{2} - 1)}{x^{2}}. f ′ ( x ) = x 2 ( 1 + x 2 ) 2 1 . ( 3 x 2 − 1 − x 2 ) = x 2 ( 1 + x 2 ) 2 1 . ( 2 x 2 − 1 ) .
显然，x = 2 2 x {=} \frac{\sqrt{2}}{2} x = 2 2 f ( x ) ⩾ f ( 2 2 ) = 3 3 2 ; f \left( x \right) \geqslant f ( \frac{\sqrt{2}}{2} ) = \frac{3 \sqrt{3}}{2} ; f ( x ) ⩾ f ( 2 2 ) = 2 3 3 ;
故矩形A B C D A B C D A B C D 2 ( ∣ A B ∣ + ∣ B C ∣ ) ⩾ 2 f ( x ) ⩾ 3 3 2 \ ( | A B | {+} | B C | ) \ {\geqslant} 2 f \ ( x ) \ {\geqslant} 3 {\sqrt{3}} 2 ( ∣ A B ∣ + ∣ B C ∣ ) ⩾ 2 f ( x ) ⩾ 3 3
注意这里有两个取等条件，一个是∣ b + c ∣ = 1 | b + c | = 1 ∣ b + c ∣ = 1 ∣ b + c ∣ = 2 2 , | b + c | = {\frac{\sqrt{2}}{2}} , ∣ b + c ∣ = 2 2 ,
这显然是无法同时取到的，所以等号不成立，命题得证
解法二：不妨设A，B，D在抛物线W上，C不在抛物线W上，欲证命题为∣ A B ∣ + ∣ A D ∣ > 3 3 2 . | A B | {+} | A D | > \frac{3 \sqrt{3}}{2} . ∣ A B ∣ + ∣ A D ∣ > 2 3 3 .
由图象的平移可知，将抛物线W看作y = x 2 y = x^{2} y = x 2
设A ( a , a 2 ) ( a ⩾ 0 ) A ( a , a^{2} ) ( a {\geqslant} 0 ) A ( a , a 2 ) ( a ⩾ 0 )
则新抛物线方程为y^{\prime} = x^{'}^{2} + 2 a x^{\prime} ，写为极坐标方程，
即p s i n Θ = ρ 2 cos 2 Θ + 2 a ρ c o s Θ \mathrm{p s i n} \Theta = \rho^{2} \cos^{2} \Theta + 2 a \rho \mathrm{c o s} \Theta psin Θ = ρ 2 cos 2 Θ + 2 a ρ cos Θ ρ = s i n θ − 2 a c o s θ c o s 2 θ \rho{=} \frac{s i n \theta{-} 2 a c o s \theta}{c o s^{2} \theta} ρ = co s 2 θ s in θ − 2 a cos θ
欲证明的结论为∣ s i n θ − 2 a c o s θ c o s 2 θ ∣ + ∣ s i n ( θ + π 2 ) − 2 a c o s ( θ + π 2 ) c o s 2 ( θ + π 2 ) ∣ > 3 3 2 , | \frac{s i n \theta - 2 a c o s \theta}{c o s^{2} \theta} | + | \frac{s i n ( \theta + \frac{\pi}{2} ) - 2 a c o s ( \theta + \frac{\pi}{2} )}{c o s^{2} ( \theta + \frac{\pi}{2} )} | > \frac{3 \sqrt{3}}{2} , ∣ co s 2 θ s in θ − 2 a cos θ ∣ + ∣ co s 2 ( θ + 2 π ) s in ( θ + 2 π ) − 2 a cos ( θ + 2 π ) ∣ > 2 3 3 ,
也即∣ 2 a c o s θ − s i n θ c o s 2 θ ∣ + ∣ 2 a s i n θ + c o s θ s i n 2 θ ∣ > 3 3 2 . | \frac{2 a}{c o s \theta} - \frac{s i n \theta}{c o s^{2} \theta} | + | \frac{2 a}{s i n \theta} + \frac{c o s \theta}{s i n^{2} \theta} | > \frac{3 \sqrt{3}}{2} . ∣ cos θ 2 a − co s 2 θ s in θ ∣ + ∣ s in θ 2 a + s i n 2 θ cos θ ∣ > 2 3 3 .
不妨设∣ 2 c o s θ ∣ ≥ ∣ 2 s i n θ ∣ | \frac{2}{c o s {\theta}} | \mathcal{\geq} | \frac{2}{s i n {\theta}} | ∣ cos θ 2 ∣ ≥ ∣ s in θ 2 ∣
其最小值当{\frac{2}{c o s \theta}} \cdot a - {\frac{s i n \theta}{c o s^{2} \theta}} = 0 \mathbb{H}{\mathbb{\}} a = {\frac{s i n \theta}{2 c o s \theta}} 时取得，
因此欲证不等式为∣ 1 c o s θ + c o s θ s i n 2 θ ∣ > 3 3 2 | \frac{1}{c o s {\theta}} + \frac{c o s {\theta}}{s i n^{2}{\theta}} | > \frac{3 \sqrt{3}}{2} ∣ cos θ 1 + s i n 2 θ cos θ ∣ > 2 3 3 ∣ 1 c o s θ s i n 2 θ ∣ > 3 3 2 , | \frac{1}{c o s \theta s i n^{2} \theta} | > \frac{3 \sqrt{3}}{2} , ∣ cos θ s i n 2 θ 1 ∣ > 2 3 3 ,
根据均值不等式，有∣ cos θ sin 2 θ ∣ | \cos \theta \sin^{2} \theta | ∣ cos θ sin 2 θ ∣
= 1 2 ⋅ 2 cos 2 θ ( 1 − cos 2 θ ) ( 1 − cos 2 θ ) = \frac{1}{\sqrt{2}} \cdot \sqrt{2 \cos^{2} \theta (1 - \cos^{2} \theta) (1 - \cos^{2} \theta)} = 2 1 ⋅ 2 cos 2 θ ( 1 − cos 2 θ ) ( 1 − cos 2 θ )
≤ 1 2 ⋅ ( 2 3 ) 3 = 2 3 3 , \leq \frac{1}{\sqrt{2}} \cdot \sqrt{(\frac{2}{3})^{3}} = \frac{2}{3 \sqrt{3}}, ≤ 2 1 ⋅ ( 3 2 ) 3 = 3 3 2 ,
由题意，等号不成立，故原命题得证
