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棱长为 4 的正方体 ABCD−A1B1C1D1A B C D-A_{1} B_{1} C_{1} D_{1} 中， M,NM, N 分别为 BD，C1B1B D ， C_{1} B_{1} 的...
Mar 8, 2024
棱长为 4 的正方体 ABCD−A1B1C1D1A B C D-A_{1} B_{1} C_{1} D_{1} 中， M,NM, N 分别为 BD，C1B1B D ， C_{1} B_{1} 的中点，点 PP 在正方体 ABCD−A1B1C1D1A B C D-A_{1} B_{1} C_{1} D_{1} 的表面上运动，若 MP⊥CNM P \perp C N ，则 APA P 的最大值为 ( ) A. 4 B. 6 c. 33\sqrt{33} D. 41\sqrt{41}
Solution by Steps
step 2
Let M M and N N be the midpoints of BD BD and C1B1 C_1B_1 respectively
step 3
Since MP⊥CN MP \perp CN , we can use the Pythagorean theorem in triangle MNP MNP to express MP MP in terms of NP NP and MN MN
step 4
The length of MN MN is the diagonal of the square base, which is 42 4\sqrt{2}
step 5
The maximum length of NP NP occurs when P P is at the vertex of the cube farthest from N N , which is A A
step 6
The length of AN AN is the space diagonal of the cube, which is 43 4\sqrt{3}
step 7
Using the Pythagorean theorem, MP2=AN2−MN2 MP^2 = AN^2 - MN^2
step 8
Substituting the lengths, MP2=(43)2−(42)2 MP^2 = (4\sqrt{3})^2 - (4\sqrt{2})^2
step 9
Simplifying, MP2=48−32=16 MP^2 = 48 - 32 = 16 , so MP=4 MP = 4
step 10
The maximum length of AP AP occurs when P P coincides with A A , so AP=MP AP = MP
step 11
Therefore, the maximum length of AP AP is 4
A
Key Concept
Pythagorean theorem in 3D geometry
Explanation
To find the maximum length of AP AP , we consider the right triangle MNP MNP and use the Pythagorean theorem to relate MP MP , NP NP , and MN MN . The maximum AP AP occurs when P P is at the vertex A A , which is also the point M M in this scenario.
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