In Cartesian coordinate system x O y xOyx O y , pointPPP toxxThe distance on the x- axis is equal to the pointPPP to point( 0 , 1 2 ) (0,\dfrac 12)(0,21) distance, record pointPPThe trajectory of P is WWW
(1) RequestedWWEquation
(2) for W known rectangle ABCD ABCDA BC D has three vertices atWWAbove W , prove that the perimeter of the rectangle is greater than3 3 3\sqrt 333
Solution:
\quad(1) According to the question, x 2 + ( y − 1 2 ) 2 = y 2 x^2+(y-\dfrac 12)^2=y^2x2+(y−21)2=y2
\qquad Arranging y = x 2 + 1 4 y=x^2+\dfrac 14y=x2+41
\quad (2) May wish to set the rectangle at WWThe three points on W are A , B , CA,B,CA,B,C，andA , BA, BA,B atxxOn the same side as the x- axis,AB ⊥ AC AB\bot ACAB⊥AC
y = x 2 + 1 4 \qquad y=x^2+\dfrac 14 y=x2+41The axis of symmetry is xxx- axis
\qquad May wish to set A , BA, BA,B are all inxxOn the right side of the x- axis, if it is on the left side, setWWW aboutxxThe x- axis is symmetrical so thatA , BA,BA,B flips toxxx- axis right
\qquadThen ( p , p 2 + 1 4 ) A(p, p^2+\dfrac 14)A(p,p2+41)， B ( q , q 2 + 1 4 ) B(q,q^2+\dfrac14) B(q,q2+41)， C ( t , t 2 + 1 4 ) C(t,t^2+\dfrac14) C(t,t2+41)
\qquad 则AB : y = ( p + q ) x − pq + 1 4 AB:y=(p+q)x-pq+\dfracAB:y=(p+q)x−pq+41
∵ A C ⊥ A B \qquad \because AC\bot AB ∵AC⊥AB
∴ AC : y = − 1 p + qx + pp + q + p 2 + 1 4 \qquad \therefore AC:y=-\dfrac{1}{p+q}x+\dfrac{p}{p+q} +p^2+\dfrac∴AC:y=−p+q1x+p+qp+p2+41
t 2 + 1 4 = − 1 p + qt + pp + q + p 2 + 1 4 \qquad t^2+\dfrac 14=-\dfrac{1}{p+q}t+\dfrac{p}{p +q}+p^2+\dfrac 14t2+41=−p+q1t+p+qp+p2+41, the solution is t = pt=pt=p (round up) ort = − p − 1 p + qt=-p-\dfrac{1}{p+q}t=−p−p+q1
∴ C ( − p − 1 p + q , ( p + 1 p + q ) 2 + 1 4 ) \qquad \therefore C(-p-\dfrac{1}{p+q},(p+\dfrac{1 }{p+q})^2+\dfrac 14)∴C(−p−p+q1,(p+p+q1)2+41)
AB = ( p − q ) 2 + ( p 2 − q 2 ) 2 = ∣ p 2 − q 2 ∣ 1 + 1 ( p + q ) 2 \qquad AB=\sqrt{(pq)^2+(p^ 2-q^2)^2}=|p^2-q^2|\sqrt{1+\dfrac{1}{(p+q)^2}}AB=(p−q)2+(p2−q2)2=∣p2−q2∣1+(p+q)21
AC = ( 2 p + 1 p + q ) 2 + [ 2 pp + q + 1 ( p + q ) 2 ] 2 = ( 2 p + 1 p + q ) 1 + 1 ( p + q ) 2 \quad AC =\sqrt{(2p+\dfraction{1}{p+q})^2+[\dfraction{2p}{p+q}+\dfraction{1}{(p+q)^2}]^2} =(2p+\dfraction{1}{p+q})\sqrt{1+\dfraction{1}{(p+q)^2}}AC=(2p+p+q1)2+[p+q2p+(p+q)21]2=(2p+p+q1)1+(p+q)21
\qquad 令W = AB + AC = ( ∣ p 2 − q 2 ∣ + 2 p + 1 p + q ) 1 + 1 ( p + q ) 2 W=AB+AC=(|p^2-q^2|+ 2p+\dfraction{1}{p+q})\sqrt{1+\dfraction{1}{(p+q)^2}}W=AB+AC=(∣p2−q2∣+2p+p+q1)1+(p+q)21
\qquadThe circumference of the card is greater than 3 3 3\sqrt 333, which proves that W > 3 3 2 W>\dfrac{3\sqrt 3}{2}W>233
\qquad令k = p + qk=p+qk=p+q ,则W = ( k ∣ p − q ∣ + 2 p + 1 k ) 1 + 1 k 2 W=(k|pq|+2p+\dfrac 1k)\sqrt{1+\dfrac{1}{k^ 2}}W=(k∣p−q∣+2p+k1)1+k21
∵ A , B \qquad\because A,B ∵A,B are all inxxx- axis right
∴ k > 0 \qquad\therefore k>0 ∴k>0
\qquad ① 0 < k ≤ 1 0<k\leq 1 0<k≤1 o'clock
∵ k ∣ p − q ∣ + 2 p ≥ k ( q − p ) + 2 p = k ( p + q ) + ( 2 − 2 k ) p ≥ k 2 \qquad\because k|p-q|+2p\geq k(q-p)+2p=k(p+q)+(2-2k)p\geq k^2 ∵k∣p−q∣+2p≥k(q−p)+2p=k(p+q)+(2−2k ) p _≥k2
∴ W ≥ ( k 2 + 1 k ) 1 + 1 k 2 \qquad\therefore W\geq(k^2+\dfrac 1k)\sqrt{1+\dfrac{1}{k^2}}∴W≥(k2+k1)1+k21
W 2 ≥ k 4 + k 2 + 2 k + 2 k + 1 k 2 + 1 k 4 ≥ 2 k 4 ⋅ 1 k 4 + 2 k 2 ⋅ 1 k 2 + 2 × 2 k ⋅ 1 k = 8 \qquad W^2\geq k^4+k^2+2k+\dfrac 2k+\dfrac{1}{k^2}+\dfrac{1}{k^4}\geq 2\sqrt{k^4\cdot\ dfrac{1}{k^4}}+2\sqrt{k^2\cdot\dfrac{1}{k^2}}+2\times 2\sqrt{k\cdot\dfrac 1k}=8W2≥k4+k2+2 k+k2+k21+k41≥2k4⋅k41+2k2⋅k21+2×2k⋅k1=8
W 2 ≥ 8 > 27 4 \qquad W^2\geq 8>\dfrac{27}{4}W2≥8>427, ie W > 3 3 2 W>\dfrac{3\sqrt 3}{2}W>233
\qquad ② k > 1 k>1 k>1 o'clock
∵ k ∣ p − q ∣ + 2 p + 1 p + q > ∣ p − q ∣ + 2 p + 1 p + q ≥ ( q − p ) + 2 p + 1 p + q = k + 1 k \quad \because k|pq|+2p+\dfrac{1}{p+q}>|pq|+2p+\dfrac{1}{p+q}\geq(qp)+2p+\dfrac{1}{p+q }=k+\dfrac 1k∵k∣p−q∣+2p+p+q1>∣p−q∣+2p+p+q1≥(q−p)+2p+p+q1=k+k1
∴ W > ( k + 1 k ) 1 + 1 k 2 \qquad\therefore W>(k+\dfrac 1k)\sqrt{1+\dfrac{1}{k^2}} ∴W>(k+k1)1+k21， W 2 > k 2 ( 1 + 1 k 2 ) 3 W^2>k^2(1+\dfrac{1}{k^2})^3 W2>k2(1+k21)3
\qquad Prove that k 2 ( 1 + 1 k 2 ) 3 ≥ 27 4 k^2(1+\dfrac{1}{k^2})^3\geq\dfrac{27}{4}k2(1+k21)3≥427So ( 1 + 1 k 2 ) 3 ≥ 27 4 k 2 (1+\dfrac{1}{k^2})^3\geq\dfrac{27}{4k^2}(1+k21)3≥4 k227
\qquad 即1 + k − 2 ≥ 3 ⋅ ( 4 k 2 ) − 1 3 1+k^{-2}\geq3\cdot(4k^2)^{-\frac 13}1+k−2≥3⋅( 4k _2)−31, 1 + k − 2 − 3 ⋅ 4 − 1 3 ⋅ k − 2 3 ≥ 0 1+k^{-2}-3\cdot 4^{-\frac 13}\cdot k^{-\frac 23} \geq1+k−2−3⋅4−31⋅k−32≥0
\qquad令 g ( x ) = 1 + x − 2 − 3 ⋅ 4 − 1 3 ⋅ x − 2 3 g(x)=1+x^{-2}-3\cdot 4^{-\frac 13}\cdot x^{-\frac 23} g(x)=1+x−2−3⋅4−31⋅x−32
\qquad则 g ′ ( x ) = − 2 x − 3 + 4 1 6 x − 5 3 g'(x)=-2x^{-3}+4^{\frac 16}x^{-\frac 53} g′(x)=−2x _ _−3+461x−35
x = 2 \qquad x=2 x=At 2 g ′ ( x ) = 0 g'(x)=0g′(x)=0
∴ g ( x ) \qquad\therefore g(x) ∴g ( x ) in[ 2 , + ∞ ) [\sqrt 2,+\infty)[2,+ ∞ ) , monotonically increasing at( − ∞ , 2 ) (-\infty,\sqrt 2)(−∞,2) on monotonically decreasing
∵ g ( 2 ) = 1 + 1 2 − 3 ⋅ 4 − 1 3 ⋅ 4 − 1 6 = 0 \qquad\because g(\sqrt 2)=1+\dfrac 12-3\cdot4^{-\frac 13}\cdot 4^{-\frac 16}=0 ∵g(2)=1+21−3⋅4−31⋅4−61=0
g ( x ) ≥ g ( 2 ) = 0 \qquad g(x)\geq g(\sqrt 2)=0g(x)≥g(2)=0， x ∈ ( 1 , + ∞ ) x\in(1,+\infty) x∈(1,+∞)
∴ k 2 ( 1 + 1 k 2 ) 3 ≥ 27 4 \qquad\therefore k^2(1+\dfrac{1}{k^2})^3\geq\dfrac{27}{4}∴k2(1+k21)3≥427
∵ W 2 > k 2 ( 1 + 1 k 2 ) 3 ≥ 27 4 \qquad\because W^2>k^2(1+\dfrac{1}{k^2})^3\geq\dfrac{27}{4} ∵W2>k2(1+k21)3≥427
∴ W > 3 3 2 \qquad\therefore W>\dfrac{3\sqrt 3}{2} ∴W>233
\qquad In summary, W > 3 3 2 W>\dfrac{3\sqrt 3}{2}W>233
\qquadProve that the perimeter of a rectangle is greater than 3 3 3\sqrt 333