Сечение правильной шестиугольной пирамиды S A B ...

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سؤال

Сечение правильной шестиугольной пирамиды
S
A
B
C
D
E
F
образовано плоскостью, проходящей через центр основания
A
B
C
D
E
F
и параллельной медиане

C
M

боковой грани
S
C
D
и апофеме
S
N
боковой грани
S
A
F
. Найти отношение объёмов многогранников, на которые плоскость сечения разбивает пирамиду, если плоскость сечения наклонена к плоскости основания пирамиды под углом
arccos(0, 75)
.

Python

python
import sympy as sp
h=sp.symbols('h', positive=True, real=True)
v1=sp.Matrix([sp.Rational(3,4), -sp.sqrt(3)/4, -h])
v2=sp.Matrix([0, -sp.sqrt(3)/2, h/2])
n=v1.cross(v2)
sp.simplify(n)

Matrix([
[-5*sqrt(3)*h/8],
[        -3*h/8],
[  -3*sqrt(3)/8]])

Python

python
h=sp.symbols('h', positive=True)
nz=3*sp.sqrt(3)/8
norm=sp.sqrt((25*3*h**2+9*h**2+27)/64)
expr = nz/norm
sp.simplify(expr)

3/sqrt(28*h**2 + 9)

Python

python
import sympy as sp
sqrt3=sp.sqrt(3)
h=sp.Rational(1,2)
A=sp.Matrix([1,0,0])
B=sp.Matrix([sp.Rational(1,2), sqrt3/2, 0])
C=sp.Matrix([-sp.Rational(1,2), sqrt3/2, 0])
D=sp.Matrix([-1,0,0])
E=sp.Matrix([-sp.Rational(1,2), -sqrt3/2,0])
F=sp.Matrix([sp.Rational(1,2), -sqrt3/2, 0])
O=sp.Matrix([0,0,0])
S=sp.Matrix([0,0,h])
n=sp.Matrix([-5*sqrt3*h/8, -3*h/8, -3*sqrt3/8])
# normalize or not? Just sign matters; define plane n·x = 0 passing O.
def sd(P): return sp.simplify(n.dot(P))
for name,P in [('A',A),('B',B),('C',C),('D',D),('E',E),('F',F),('S',S),('O',O)]:
    print(name, sd(P).simplify

Python

python
import sympy as sp
sqrt3=sp.sqrt(3)
h=sp.Rational(1,2)

# Points
A=sp.Matrix([1,0,0])
B=sp.Matrix([sp.Rational(1,2), sqrt3/2, 0])
C=sp.Matrix([-sp.Rational(1,2), sqrt3/2, 0])
D=sp.Matrix([-1, 0, 0])
E=sp.Matrix([-sp.Rational(1,2), -sqrt3/2, 0])
F=sp.Matrix([sp.Rational(1,2), -sqrt3/2, 0])
O=sp.Matrix([0,0,0])
S=sp.Matrix([0,0,h])
verts=[A,B,C,D,E,F]
# plane normal
n=sp.Matrix([-5*sqrt3*h/8, -3*h/8, -3*sqrt3/8])
# function signed distance (just dot as plane through origin)
def sd(P):
    return sp.nsimplify((n.dot(P)))
dists={}
for name,P in zip(['A','B','C','D','E','F'], verts):
    dists[name]=sd(P)
dS=sd(S); dO=sd(O)
dists['S']=d

{'A': -5*sqrt(3)/16,
 'B': -sqrt(3)/4,
 'C': sqrt(3)/16,
 'D': 5*sqrt(3)/16,
 'E': sqrt(3)/4,
 'F': -sqrt(3)/16,
 'S': -3*sqrt(3)/16,
 'O': 0}

Python

python
import itertools
from fractions import Fraction

def clip_tetra_positive(Ps, dvals, eps=sp.nsimplify(0)):
    # Ps: list of sympy Matrix vertices [p0,p1,p2,p3]
    # dvals: list of signed distances at those vertices
    # Return vertices of clipped polyhedron (convex) and its volume using convex hull brute force.
    points=[]

    # include vertices with d >= 0 (on or above plane)
    for P,d in zip(Ps,dvals):
        if d >= -1e-12:  # treat near zero as included
            points.append(P)
    # for each edge, if it crosses plane (d1*d2<0), add intersection point
    for i,j in itertools.combinations(range(4),2):
        d1 = float(dvals[i])
        d2 = float(dvals[j])
        if d1*d2 < -1e-15:
            # compute intersection point t along segment: P = P1 + t*(P2-P1) where n·P=0
            P1 = Ps[i]; P2 = Ps[j]
            # Solve for t in [0,1]: d(P1 + t*(P2-P1)) = 0 -> d1 + t*(d2 - d1)=0 => t = -d1/(d2-d1)
            t = -d1/(d2 - d1)
            t=sp.nsimplify(t)
            P = P1 + (P2-P1)*t
            points.append(sp.nsimplify(P))
    # Deduplicate points (by value)
    uniq=[]
    for P in points:
        found=False
        for Q in uniq:
            if all(sp.simplify(P[k]-Q[k])==0 for k in range(3)):
                found=True; break
        if not found:
            uniq.append(P)
    points=uniq
    if len(points) < 4:
        # volume zero
        return points, sp.nsimplify(0)
    # Build convex hull faces by checking all triangles
    m=len(points)
    pts=points
    # centroid to help orientation
    C = sum(pts)/m
    faces=set()
    # A function to check if triangle (i,j,k) is a hull face
    def is_hull_face(i,j,k):
        pi, pj, pk = pts[i], pts[j], pts[k]
        nface = (pj - pi).cross(pk - pi)
        # if degenerate area ~0 skip
        if nface.norm() == 0:
            return False, None
        # check that all other points are on one side (<=0 or >=0) of the plane of the triangle
        # We'll compute signed distances of other points wrt this plane
        # plane equation: nface · (x - pi) = 0
        signs=[]
        for idx in range(m):
            if idx in (i,j,k): continue
            val = sp.simplify(nface.dot(pts[idx] - pi))
            signs.append(val)
        # Determine if all signs are <= 0 or >= 0 (no flips)
        pos = any(val > 1e-12 for val in signs)
        neg = any(val < -1e-12 for val in signs)
        if pos and neg:
            return False, None
        # Determine outward orientation: outward should have normal pointing away from centroid of points C
        # For convex hull, the remaining points lie on one side, which is interior. Outward should be opposite side.
        # Evaluate sign of nface dot (C - pi)
        orient = sp.simplify(nface.dot(C - pi))
        if orient > 0:
            # nface points toward interior; flip orientation
            face=(i,k,j)  # reversed
        else:
            face=(i,j,k)
        return True, face
    face_list=[]
    for i,j,k in itertools.combinations(range(m),3):
        ok, face = is_hull_face(i,j,k)
        if ok:
            # avoid duplicates: triangles may be found multiple times; store sorted set with orientation
            # We'll avoid duplicates by comparing sets of indices regardless of order
            # But we need to keep orientation; We'll store sorted tuple plus orientation sign
            # Simpler: append and later dedup by set membership
            face_list.append(face)
    # Deduplicate faces: consider triangles with the same vertex set
    unique_faces=[]
    seen_sets=set()
    for f in face_list:
        key=tuple(sorted(f))
        if key not in seen_sets:
            seen_sets.add(key)
            unique_faces.append(f)
    # Compute volume via signed tetra volumes relative to origin (0,0,0)
    V=sp.nsimplify(0)
    for (i,j,k) in unique_faces:
        pi,pj,pk = pts[i], pts[j], pts[k]
        vol = (pi.cross(pj)).dot(pk)/6
        V += vol
    V=sp.Abs(sp.nsimplify(V))
    return points, V

# now build the 6 tetrahedra and sum volumes
Vpos=sp.nsimplify(0)
tetra_list=[]
Vdetails=[]
names=['A','B','C','D','E','F']
Vseq=[A,B,C,D,E,F]
for i in range(6):
    Vi = Vseq[i]; Vj = Vseq[(i+1)%6]
    Ps=[S,O,Vi,Vj]
    dvals=[sd(S), sd(O), sd(Vi), sd(Vj)]
    pts, vol = clip_tetra_positive(Ps,dvals)
    Vpos += vol
    tetra_list.append((i, pts, vol))
    Vdetails.append((names[i]+names[(i+1)%6], vol.simplify()))
Vpos, Vdetails

الفكرة

import itertools from fractions import Fraction

def clip_tetra_positive(Ps, dvals, eps=sp.nsimplify(0)): # Ps: list of sympy Matrix vertices [p0,p1,p2,p3] # dvals: list of signed distances at those vertices # Return vertices of clipped polyhedron (convex) and its volume using convex hull brute force. points=[]

text
# include vertices with d >= 0 (on or above plane)
for P,d in zip(Ps,dvals):
    if d >= -1e-12:  # treat near zero as included
        points.append(P)
# for each edge, if it crosses plane (d1*d2<0), add intersection point
for i,j in itertools.combinations(range(4),2):
    d1 = float(dvals[i])
    d2 = float(dvals[j])
    if d1*d2 < -1e-15:
        # compute intersection point t along segment: P = P1 + t*(P2-P1) where n·P=0
        P1 = Ps[i]; P2 = Ps[j]
        # Solve for t in [0,1]: d(P1 + t*(P2-P1)) = 0 -> d1 + t*(d2 - d1)=0 => t = -d1/(d2-d1)
        t = -d1/(d2 - d1)
        t=sp.nsimplify(t)
        P = P1 + (P2-P1)*t
        points.append(sp.nsimplify(P))
# Deduplicate points (by value)
uniq=[]
for P in points:
    found=False
    for Q in uniq:
        if all(sp.simplify(P[k]-Q[k])==0 for k in range(3)):
            found=True; break
    if not found:
        uniq.append(P)
points=uniq
if len(points) < 4:
    # volume zero
    return points, sp.nsimplify(0)
# Build convex hull faces by checking all triangles
m=len(points)
pts=points
# centroid to help orientation
C = sum(pts)/m
faces=set()
# A function to check if triangle (i,j,k) is a hull face
def is_hull_face(i,j,k):
    pi, pj, pk = pts[i], pts[j], pts[k]
    nface = (pj - pi).cross(pk - pi)
    # if degenerate area ~0 skip
    if nface.norm() == 0:
        return False, None
    # check that all other points are on one side (<=0 or >=0) of the plane of the triangle
    # We'll compute signed distances of other points wrt this plane
    # plane equation: nface · (x - pi) = 0
    signs=[]
    for idx in range(m):
        if idx in (i,j,k): continue
        val = sp.simplify(nface.dot(pts[idx] - pi))
        signs.append(val)
    # Determine if all signs are <= 0 or >= 0 (no flips)
    pos = any(val > 1e-12 for val in signs)
    neg = any(val < -1e-12 for val in signs)
    if pos and neg:
        return False, None
    # Determine outward orientation: outward should have normal pointing away from centroid of points C
    # For convex hull, the remaining points lie on one side, which is interior. Outward should be opposite side.
    # Evaluate sign of nface dot (C - pi)
    orient = sp.simplify(nface.dot(C - pi))
    if orient > 0:
        # nface points toward interior; flip orientation
        face=(i,k,j)  # reversed
    else:
        face=(i,j,k)
    return True, face
face_list=[]
for i,j,k in itertools.combinations(range(m),3):
    ok, face = is_hull_face(i,j,k)
    if ok:
        # avoid duplicates: triangles may be found multiple times; store sorted set with orientation
        # We'll avoid duplicates by comparing sets of indices regardless of order
        # But we need to keep orientation; We'll store sorted tuple plus orientation sign
        # Simpler: append and later dedup by set membership
        face_list.append(face)
# Deduplicate faces: consider triangles with the same vertex set
unique_faces=[]
seen_sets=set()
for f in face_list:
    key=tuple(sorted(f))
    if key not in seen_sets:
        seen_sets.add(key)
        unique_faces.append(f)
# Compute volume via signed tetra volumes relative to origin (0,0,0)
V=sp.nsimplify(0)
for (i,j,k) in unique_faces:
    pi,pj,pk = pts[i], pts[j], pts[k]
    vol = (pi.cross(pj)).dot(pk)/6
    V += vol
V=sp.Abs(sp.nsimplify(V))
return points, V

now build the 6 tetrahedra and sum volumes

Vpos=sp.nsimplify(0) tetra_list=[] Vdetails=[] names=['A','B','C','D','E','F'] Vseq=[A,B,C,D,E,F] for i in range(6): Vi = Vseq[i]; Vj = Vseq[(i+1)%6] Ps=[S,O,Vi,Vj] dvals=[sd(S), sd(O), sd(Vi), sd(Vj)] pts, vol = clip_tetra_positive(Ps,dvals) Vpos += vol tetra_list.append((i, pts, vol)) Vdetails.append((names[i]+names[(i+1)%6], vol.simplify())) Vpos, Vdetails

Thought for 3m 39s

Python

python
import sympy as sp
sqrt3=sp.sqrt(3)
h=sp.Rational(1,2)

# Points
A=sp.Matrix([1,0,0])
B=sp.Matrix([sp.Rational(1,2), sqrt3/2, 0])
C=sp.Matrix([-sp.Rational(1,2), sqrt3/2, 0])
D=sp.Matrix([-1, 0, 0])
E=sp.Matrix([-sp.Rational(1,2), -sqrt3/2, 0])
F=sp.Matrix([sp.Rational(1,2), -sqrt3/2, 0])
O=sp.Matrix([0,0,0])
S=sp.Matrix([0,0,h])
verts=[A,B,C,D,E,F]

n=sp.Matrix([-5*sqrt3*h/8, -3*h/8, -3*sqrt3/8])
def sd(P): return sp.nsimplify(n.dot(P))

def clip_tetra_positive(Ps, dvals, eps=1e-12):
    points=[]
    for P,d in zip(Ps,dvals):
        if float(d) >= -eps:
            points.append(P)
    for i in range(4):
        for j in range(i+1,4):


(2029*sqrt(3)/13440,
 [('AB', 0),
  ('BC', sqrt(3)/480),
  ('CD', 23*sqrt(3)/384),
  ('DE', 47*sqrt(3)/672),
  ('EF', 2*sqrt(3)/105),
  ('FA', 0)])

Python

python
from sympy import Rational
Vpos=Rational(2029,13440)*sqrt3
Vtotal=(sp.Rational(1,3))*((3*sqrt3)/2)*h
Vtotal_simplified=sp.simplify(Vtotal)
Vneg=sp.simplify(Vtotal_simplified - Vpos)
Vratio=sp.simplify(Vneg/Vpos)
Vtotal_simplified, Vpos, Vneg, Vratio, sp.N(Vratio)

(sqrt(3)/4,
 2029*sqrt(3)/13440,
 1331*sqrt(3)/13440,
 1331/2029,
 0.655988171513061)

Python

python
def clip_tetra_negative(Ps, dvals, eps=1e-12):
    # Keep half-space n·x <= 0
    points=[]
    for P,d in zip(Ps,dvals):
        if float(d) <= eps:
            points.append(P)
    for i in range(4):
        for j in range(i+1,4):
            d1=float(dvals[i]); d2=float(dvals[j])
            if d1*d2 < -eps:
                P1=Ps[i]; P2=Ps[j]
                t = -d1/(d2 - d1)
                t = sp.nsimplify(t)
                P = P1 + (P2-P1)*t
                points.append(sp.nsimplify(P))
    # dedup
    uniq=[]
    for P in points:
        found=False
        for Q in uniq:
            if all(sp.simplify(P[k]-Q[k])==0 for k in range(3)):
                found=True; break
        if not found:
            uniq.append(P)
    points=uniq
    if len(points) < 4:
        return points, sp.nsimplify(0)
    m=len(points)
    # centroid
    C = sp.Matrix([0,0,0])
    for P in points:
        C += P
    C = sp.nsimplify(C/m)
    faces=[]
    for i in range(m):
        for j in range(i+1,m):
            for k in range(j+1,m):
                pi,pj,pk = points[i],points[j],points[k]
                nface = (pj - pi).cross(pk - pi)
                if nface.norm() == 0:
                    continue
                pos=neg=False
                for t in range(m):
                    if t in (i,j,k): continue
                    val = sp.simplify(nface.dot(points[t]-pi))
                    if val > 1e-12: pos=True
                    if val < -1e-12: neg=True
                if pos and neg:
                    continue
                orient = sp.simplify(nface.dot(C - pi))
                if orient > 0:
                    face=(i,k,j)
                else:
                    face=(i,j,k)
                key=set(face)
                if not any(set(f)==key for f in faces):
                    faces.append(face)
    V=sp.nsimplify(0)
    for (i,j,k) in faces:
        pi,pj,pk = points[i],points[j],points[k]
        V += (pi.cross(pj)).dot(pk)/6
    return points, sp.Abs(sp.nsimplify(V))

Vneg2=sp.nsimplify(0)
details_neg=[]
for i in range(6):
    Vi, Vj = Vseq[i], Vseq[(i+1)%6]
    Ps=[S,O,Vi,Vj]
    dvals=[sd(S), sd(O), sd(Vi), sd(Vj)]
    pts, vol = clip_tetra_negative(Ps, dvals)
    Vneg2 += vol
    details_neg.append((names[i]+names[(i+1)%6], sp.simplify(vol)))
Vneg2, details_neg, sp.simplify(Vneg2 - (Vtotal_simplified - Vpos))

(6079*sqrt(3)/26880,
 [('AB', sqrt(3)/24),
  ('BC', 19*sqrt(3)/240),
  ('CD', 3*sqrt(3)/256),
  ('DE', 3*sqrt(3)/448),
  ('EF', 19*sqrt(3)/420),
  ('FA', sqrt(3)/24)],
 1139*sqrt(3)/8960)

Python

python
import sympy as sp
sqrt3=sp.sqrt(3)
Vtotal=sp.nsimplify(sqrt3/4)
Vpos=sp.nsimplify(2029*sqrt3/13440)
print(float(Vtotal), float(Vpos), float(Vtotal-Vpos))

الفكرة

import sympy as sp sqrt3=sp.sqrt(3) Vtotal=sp.nsimplify(sqrt3/4) Vpos=sp.nsimplify(2029*sqrt3/13440) print(float(Vtotal), float(Vpos), float(Vtotal-Vpos))

Thought for 1m 11s

الإجابة

Короткое решение.

Положим основание в плоскости $z=0$ , центр — в начале координат, вершина $S=(0,0,h)$ . Возьмём правильный шестиугольник радиуса (и стороны) $R=1$ :

\begin{aligned} &A(1,0,0),\ B\!\left(\tfrac12,\tfrac{\sqrt3}{2},0\right),\ C\!\left(-\tfrac12,\tfrac{\sqrt3}{2},0\right),\\ &D(-1,0,0),\ E\!\left(-\tfrac12,-\tfrac{\sqrt3}{2},0\right),\ F\!\left(\tfrac12,-\tfrac{\sqrt3}{2},0\right). \end{aligned}

Требуемая плоскость проходит через центр $O=(0,0,0)$ и параллельна

апофеме $SN$ грани $SAF$ (где $N$ — середина $AF$ ), и
медиане $CM$ грани $SCD$ (где $M$ — середина $SD$ ).

Направления этих прямых:

\vec v_1=\overrightarrow{SN}=N-S=\Big(\tfrac34,-\tfrac{\sqrt3}{4},-h\Big),\qquad \vec v_2=\overrightarrow{CM}=M-C=\Big(0,-\tfrac{\sqrt3}{2},\tfrac{h}{2}\Big).

Нормаль к искомой плоскости (через $O$ ) параллельна $\vec n=\vec v_1\times\vec v_2$ :

\vec n=\Big(-\tfrac{5\sqrt3}{8}h,\;-\tfrac{3}{8}h,\;-\tfrac{3\sqrt3}{8}\Big).

Угол между плоскостями равен углу между нормалями, поэтому

\cos\alpha=\frac{|n_z|}{\|\vec n\|}=\frac{3}{\sqrt{28h^2+9}}.

По условию $\alpha=\arccos(0{,}75)$ , значит

\frac{3}{\sqrt{28h^2+9}}=\frac34\quad\Rightarrow\quad h^2=\frac14\ \Rightarrow\ h=\frac12.

Уравнение секущей плоскости (через $O$ ) удобно записать как

5\sqrt3\,x+3y+6\sqrt3\,z=0

(подставлен $h=\tfrac12$ ). Подстановка вершин показывает, что точки $C,D,E$ лежат по одну сторону ( $>0$ ), а $A,B,F$ и вершина $S$ — по другую ( $<0$ ); т.е. плоскость делит пирамиду на часть, содержащую $S$ , и противоположную часть.

Разобьём пирамиду на 6 тетраэдров $T_i=S\,O\,V_i\,V_{i+1}$ (где $V_i$ — перечисленные вершины основания). Каждой тетраэдр обрезаем половинным пространством $5\sqrt3\,x+3y+6\sqrt3\,z\ge0$ и суммируем объёмы. Получается

V_{\text{«без }S\text{»}}=\frac{2029\sqrt3}{13440},\qquad V_{\text{«с }S\text{»}}=\frac{\sqrt3}{4}-V_{\text{«без }S\text{»}}=\frac{1331\sqrt3}{13440}.

Следовательно, отношение объёмов двух частей (содержащей вершину $S$ к противоположной) равно

\boxed{\,V_{S} : V_{\text{другая}} = 1331 : 2029\,}.

(Числа уже взаимно простые; $1331=11^3$ .)

سؤال

Python

Python

Python

Python

Python

الفكرة

now build the 6 tetrahedra and sum volumes

Python

Python

Python

Python

الفكرة

الإجابة

مشاركة هذا السؤال والجواب