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Problem 28 Consider the tetrahedronABCDwith AB= CD= 4 and BC= BD= 2 √ 5. LetHbe the foot of the perpendicular fromAto lineCD. Suppose that the planesABHandBCD are perpendicular to each other and that AH= 4. LetGbe the centroid of triangleABH, and letSbe the sphere with centerGtangent to the planeACD. LetTbe the locus of all pointsPonSsuch that∠AP G= π 2 . Find the area of the orthogonal projection ofT onto the planeABC. [4 points] ① π 7 ② π 6 ③ π 5 ④ π 4 ⑤ π 3 18 Numerical answer 더보기

Let me solve this step by step.

1. Coordinate setup

Place 𝐻 H at the origin and let 𝐶 𝐷 CD lie on the 𝑥 x-axis. Since 𝐴 𝐻 ⊥ 𝐶 𝐷 AH⊥CD and 𝐴 𝐻

4 AH=4, we can put 𝐴

( 0 , 4 , 0 ) A=(0,4,0). Let 𝐶

( − 2 , 0 , 0 ) C=(−2,0,0), 𝐷

( 2 , 0 , 0 ) D=(2,0,0) (so 𝐶 𝐷

4 CD=4).

Because 𝐵 𝐶

𝐵 𝐷 BC=BD, point 𝐵 B lies on the perpendicular bisector plane of 𝐶 𝐷 CD, i.e. 𝑥

0 x=0. Write 𝐵

( 0 , 𝑦 𝐵 , 𝑧 𝐵 ) B=(0,y B ​

,z B ​

).

From 𝐵 𝐶 2

20 BC 2 =20:

( 0 + 2 ) 2 + 𝑦 𝐵 2 + 𝑧 𝐵 2

20    ⟹    𝑦 𝐵 2 + 𝑧 𝐵 2

(0+2) 2 +y B 2 ​

+z B 2 ​

=20⟹y B 2 ​

+z B 2 ​

=16.

From 𝐴 𝐵 2

16 AB 2 =16:

( 0 − 0 ) 2 + ( 𝑦 𝐵 − 4 ) 2 + 𝑧 𝐵 2

16    ⟹    𝑦 𝐵 2 − 8 𝑦 𝐵 + 16 + 𝑧 𝐵 2

(0−0) 2 +(y B ​

−4) 2 +z B 2 ​

=16⟹y B 2 ​

−8y B ​

+16+z B 2 ​

=16.

Using 𝑦 𝐵 2 + 𝑧 𝐵 2

16 y B 2 ​

+z B 2 ​

=16 gives − 8 𝑦 𝐵 + 16

0 −8y B ​

+16=0, so 𝑦 𝐵

2 y B ​

=2, 𝑧 𝐵

± 2 3 z B ​

=±2 3 ​

. Choose 𝐵

( 0 , 2 , 2 3 ) B=(0,2,2 3 ​

).

2. Centroid 𝐺 G of △ 𝐴 𝐵 𝐻 △ABH

𝐺

( 0 + 0 + 0 3 , 4 + 2 + 0 3 , 0 + 2 3 + 0 3 )

( 0 , 2 , 2 3 3 ) . G=( 3 0+0+0 ​

, 3 4+2+0 ​

, 3 0+2 3 ​

+0 ​

)=(0,2, 3 2 3 ​

​

).

3. Sphere 𝑆 S (center 𝐺 G, tangent to plane 𝐴 𝐶 𝐷 ACD)

Plane 𝐴 𝐶 𝐷 ACD contains 𝐴 ( 0 , 4 , 0 ) , 𝐶 ( − 2 , 0 , 0 ) , 𝐷 ( 2 , 0 , 0 ) A(0,4,0),C(−2,0,0),D(2,0,0); all have 𝑧

0 z=0, so 𝐴 𝐶 𝐷 ACD is the plane 𝑧

0 z=0. Distance from 𝐺 G to 𝑧

0 z=0 is 2 3 3 3 2 3 ​

​

, hence radius 𝑟

2 3 3 r= 3 2 3 ​

​

.

4. Locus 𝑇 T

𝑃 P lies on 𝑆 S and ∠ 𝐴 𝑃 𝐺

𝜋 2 ∠APG= 2 π ​

. Since 𝑃 𝐺

𝑟 PG=r (radius), the right‑angle condition gives

𝐴 𝑃 2 + 𝑃 𝐺 2

𝐴 𝐺 2    ⟹    𝐴 𝑃 2

𝐴 𝐺 2 − 𝑟 2 . AP 2 +PG 2 =AG 2 ⟹AP 2 =AG 2 −r 2 .

Thus 𝑃 P also lies on the sphere centered at 𝐴 A with radius 𝐴 𝐺 2 − 𝑟 2 AG 2 −r 2 ​

. The intersection of two spheres is a circle.

Compute 𝐴 𝐺 2 AG 2 :

𝐴 𝐺 2

( 0 − 0 ) 2 + ( 4 − 2 ) 2 + ( 0 − 2 3 3 ) 2

4 + 12 9

4 + 4 3

16 3 . AG 2 =(0−0) 2 +(4−2) 2 +(0− 3 2 3 ​

​

) 2 =4+ 9 12 ​

=4+ 3 4 ​

= 3 16 ​

.

Then

𝐴 𝐺 2 − 𝑟 2

16 3 − 12 9

16 3 − 4 3

12 3

4 , AG 2 −r 2

3 16 ​

− 9 12 ​

= 3 16 ​

− 3 4 ​

= 3 12 ​

=4,

so the second sphere has radius 2 2.

The circle 𝑇 T lies in the plane perpendicular to 𝐴 𝐺 AG (the radical plane of the two spheres). Vector 𝐴 𝐺 →

( 0 , − 2 , 2 3 3 ) AG =(0,−2, 3 2 3 ​

​

). The plane of 𝑇 T has normal 𝐴 𝐺 → AG .

5. Projection onto plane 𝐴 𝐵 𝐶 ABC

Plane 𝐴 𝐵 𝐶 ABC contains 𝐴 ( 0 , 4 , 0 ) , 𝐵 ( 0 , 2 , 2 3 ) , 𝐶 ( − 2 , 0 , 0 ) A(0,4,0),B(0,2,2 3 ​

),C(−2,0,0). Vectors 𝐴 𝐵 →

( 0 , − 2 , 2 3 ) AB =(0,−2,2 3 ​

), 𝐴 𝐶 →

( − 2 , − 4 , 0 ) AC =(−2,−4,0). Normal to 𝐴 𝐵 𝐶 ABC:

𝑛 → 𝐴 𝐵 𝐶

𝐴 𝐵 → × 𝐴 𝐶 →

( 8 3 ,   − 4 3 ,   − 4 ) . n ABC ​

= AB × AC =(8 3 ​

,−4 3 ​

,−4).

The angle 𝜃 θ between the plane of 𝑇 T (normal 𝐴 𝐺 → AG ) and plane 𝐴 𝐵 𝐶 ABC (normal 𝑛 → 𝐴 𝐵 𝐶 n ABC ​

) satisfies

cos ⁡ 𝜃

∣ 𝐴 𝐺 → ⋅ 𝑛 → 𝐴 𝐵 𝐶 ∣ ∥ 𝐴 𝐺 → ∥   ∥ 𝑛 → 𝐴 𝐵 𝐶 ∥ . cosθ= ∥ AG ∥∥ n ABC ​

∥ ∣ AG ⋅ n ABC ​

∣ ​

. 𝐴 𝐺 → ⋅ 𝑛 → 𝐴 𝐵 𝐶

( 0 ) ( 8 3 ) + ( − 2 ) ( − 4 3 ) + ( 2 3 3 ) ( − 4 )

8 3 − 8 3 3

16 3 3 . AG ⋅ n ABC ​

=(0)(8 3 ​

)+(−2)(−4 3 ​

)+( 3 2 3 ​

​

)(−4)=8 3 ​

− 3 8 3 ​

​

= 3 16 3 ​

​

. ∥ 𝐴 𝐺 → ∥

16 3

4 3 , ∥ 𝑛 → 𝐴 𝐵 𝐶 ∥

( 8 3 ) 2 + ( − 4 3 ) 2 + ( − 4 ) 2

192 + 48 + 16

256

∥ AG ∥= 3 16 ​

​

= 3 ​

4 ​

,∥ n ABC ​

∥= (8 3 ​

) 2 +(−4 3 ​

) 2 +(−4) 2 ​

= 192+48+16 ​

= 256 ​

=16. cos ⁡ 𝜃

16 3 3 4 3 ⋅ 16

16 3 3 64 3

16 3 3 ⋅ 3 64

48 192

1 4 . cosθ= 3 ​

4 ​

⋅16 3 16 3 ​

​

​

= 3 ​

64 ​

3 16 3 ​

​

​

= 3 16 3 ​

​

⋅ 64 3 ​

​

= 192 48 ​

= 4 1 ​

.

6. Area of projection

The circle 𝑇 T has radius 𝑅

2 R=2 (from 𝐴 𝑃

2 AP=2). Area of 𝑇 T is 𝜋 𝑅 2

4 𝜋 πR 2 =4π. Orthogonal projection area

4 𝜋 ⋅ cos ⁡ 𝜃

4 𝜋 ⋅ 1 4

𝜋 =4π⋅cosθ=4π⋅ 4 1 ​

=π.

𝜋 π ​

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