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Câu hỏi

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Final area confirmation and explanation in Russian

After rounding to two decimal places, we get an area of 0.68. But being more precise, the exact result is 2√26 / 15. Now, I should explain this clearly in Russian, including the math derivation, but keeping it concise without unnecessary intricacies. Also, confirming that the cross-section remains a quadrilateral and does not intersect other edges.

Final area presentation and calculations

The cross-section is confirmed as a quadrilateral, formed by intersections at A, Q, T, and R. To compute the area, we can break it down by showing each step: place the base square, determine the apex, and compute the necessary coordinates. Using the vector method, we apply the formula to get the final area.

Analytic explanation and coordinate setup

We’ll keep the explanation brief but insightful. First, I set up the coordinates: A(-½,-½,0), B(½,-½,0), C(½,½,0), D(-½,½,0), with edge AB equal to 1. For the vertex P, we calculate height hh, finding it to be 12\frac{1}{\sqrt{2}}. From there, I compute the position of point T and the plane equation, ultimately simplifying it to z=210(x+y+1)z = \frac{\sqrt{2}}{10}(x+y+1).

Quick calculation of the normal vector and plane equation

To get the normal vector, we start by computing the vector AT from A(-½,-½,0) to T(1/3,1/3,h/3), which gives (5/6,5/6,h/3) = (5/6,5/6,√2/6). After multiplying by 6, we get (5,5,√2). The cross product with vector BD gives the normal vector n = (-√2,-√2,10). We then use this to solve for the plane equation: 10z = √2(x + y + 1), which simplifies to z = (√2/10)(x + y + 1).

Finding the area of triangle AQT

To find the area of triangle AQT, I calculate triangle vectors AQ and AT. After determining the coordinates of Q and T, I compute the cross product of vectors AQ and AT, which simplifies to a value of √26/15. We could showcase the reasoning behind this, but we’ll focus on the final computed area formula, keeping the math elegant without extensive steps.

Thought for 4m 12s

Trả lời

Возьмём декартову систему координат.

1. Модель пирамиды

Пусть основание – квадрат со стороной 1 в плоскости z=0z=0:

A(−12,−12,0),  B(12,−12,0),  C(12,12,0),  D(−12,12,0).A\left(-\frac12,-\frac12,0\right),\; B\left(\frac12,-\frac12,0\right),\; C\left(\frac12,\frac12,0\right),\; D\left(-\frac12,\frac12,0\right).

Правильная пирамида: вершина PP над центром квадрата O(0,0,0)O(0,0,0), причём PA=1PA = 1.

AO=22,PO=h,PA2=AO2+h2⇒1=12+h2⇒h=12.AO = \frac{\sqrt2}{2},\quad PO = h,\quad PA^2 = AO^2 + h^2 \Rightarrow 1 = \frac12 + h^2 \Rightarrow h=\frac1{\sqrt2}.

Значит

P(0,0,12).P(0,0,\tfrac1{\sqrt2}).

2. Точка TT на ребре PCPC

P(0,0,h),  C(12,12,0),  h=12.P(0,0,h),\; C\left(\tfrac12,\tfrac12,0\right),\; h=\frac1{\sqrt2}.

Условие PT:TC=2:1PT:TC = 2:1 даёт

T=P+23(C−P)=(13,13,h3)=(13,13,26).T = P + \frac{2}{3}(C-P) = \left(\frac13,\frac13,\frac{h}{3}\right) = \left(\frac13,\frac13,\frac{\sqrt2}{6}\right).

3. Уравнение секущей плоскости

Плоскость проходит через точку AA и TT и параллельна прямой BDBD.

Вектор

AT⃗=T−A=(56,56,26),BD⃗=D−B=(−1,1,0).\vec{AT} = T-A = \left(\frac56,\frac56,\frac{\sqrt2}{6}\right), \quad \vec{BD} = D-B = (-1,1,0).

Нормаль плоскости:

n⃗=AT⃗×BD⃗=(−2,−2,10).\vec{n} = \vec{AT}\times\vec{BD}=(-\sqrt2,-\sqrt2,10).

Уравнение плоскости через A(−12,−12,0)A(-\tfrac12,-\tfrac12,0):

n⃗⋅((x,y,z)−A)=0⇒2x+2y−10z+2=0\vec{n}\cdot\bigl((x,y,z)-A\bigr)=0 \Rightarrow \sqrt2 x + \sqrt2 y - 10z + \sqrt2 = 0

или

z=210(x+y+1).(*)z = \frac{\sqrt2}{10}(x+y+1). \tag{*}

4. Точки пересечения плоскости с рёбрами

  • С ребром PBPB:
PB: (x,y,z)=(0,0,h)+s(12,−12,−h).PB:\ (x,y,z) = (0,0,h) + s\left(\tfrac12,-\tfrac12,-h\right).

Подставляем в (∗)(*) и решаем по ss, получаем s=45s=\frac45.
Тогда

Q=P+45(B−P)=(25,−25,210).Q = P + \frac45(B-P) = \left(\frac25,-\frac25,\frac{\sqrt2}{10}\right).
  • С ребром PDPD аналогично:
R=P+45(D−P)=(−25,25,210).R = P + \frac45(D-P) = \left(-\frac25,\frac25,\frac{\sqrt2}{10}\right).

С основанием (плоскость z=0z=0) секущая плоскость пересекается по прямой x+y=−1x+y=-1, которая проходит только через вершину AA. Значит сечение – четырёхугольник AQTRA Q T R.

5. Площадь сечения

Разобьём четырёхугольник на два треугольника: AQTAQT и ATRATR. Они равны по симметрии, поэтому достаточно найти площадь одного.

Векторы:

AQ⃗=Q−A=(910,110,210),AT⃗=(56,56,26).\vec{AQ} = Q-A = \left(\frac{9}{10},\frac{1}{10},\frac{\sqrt2}{10}\right), \quad \vec{AT} = \left(\frac56,\frac56,\frac{\sqrt2}{6}\right).

Площадь треугольника

S△AQT=12∥AQ⃗×AT⃗∥=2615.S_{\triangle AQT} = \frac12\left\lVert \vec{AQ}\times\vec{AT}\right\rVert = \frac{\sqrt{26}}{15}.

Тогда площадь сечения:

S=2S△AQT=22615≈0.679869…S = 2 S_{\triangle AQT} = \frac{2\sqrt{26}}{15} \approx 0.679869\ldots

Округляя до двух знаков после запятой,

S≈0,68.\boxed{S \approx 0{,}68}.

Câu hỏi

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Evaluating probability of equal distribution of worms

The problem involves distributing 7 worms between 2 chicks. Since there are equal chances for each chick each time, it’s a problem of binomial distribution. I’ll calculate the probability for cases where both chicks get at least 2 worms, using the binomial coefficient C(7, k). Let’s compute this with careful attention to fractions and arithmetic.

Verifying the final result and considering complements

I got the probability of both chicks getting at least 2 worms as 0.875. Now, I should check the boundary conditions, making sure I'm excluding k=0 and k=7. Everything checks out. To double-check, I calculated the complement probability: the chance that a chick gets fewer than 2 worms, which is 1/8. This confirms my original answer.

Thought for 25s

Trả lời

Обозначим через XX количество червяков, полученных первым птенцом за день. Тогда второй получит 7−X7-X червяков.

Каждый рейс мама выбирает птенца случайно, независимо, с вероятностью 1/21/2. Значит, X∼Bin(n=7,p=1/2)X \sim \text{Bin}(n=7, p=1/2).

Нужно, чтобы каждый птенец получил по крайней мере 2 червяка:

X≥2и7−X≥2⇒2≤X≤5.X \ge 2 \quad \text{и} \quad 7-X \ge 2 \Rightarrow 2 \le X \le 5.

Тогда

P(2≤X≤5)=∑k=25(7k)(12)7=127[(72)+(73)+(74)+(75)].P(2 \le X \le 5) = \sum_{k=2}^{5} \binom{7}{k} \left(\frac12\right)^7 = \frac{1}{2^7}\left[\binom{7}{2}+\binom{7}{3}+\binom{7}{4}+\binom{7}{5}\right]. (72)=21, (73)=35, (74)=35, (75)=21,\binom{7}{2}=21,\ \binom{7}{3}=35,\ \binom{7}{4}=35,\ \binom{7}{5}=21, ⇒P=21+35+35+21128=112128=78=0,875.\Rightarrow P = \frac{21+35+35+21}{128} = \frac{112}{128} = \frac{7}{8} = 0{,}875.

Ответ, округлённый до трёх знаков после запятой:

0,875\boxed{0{,}875}

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