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В равнобедренном треугольнике ABC (AB=AC), в котор...

Created on: January 11, 2026

Answered using GPT-5.2 Thinking by Chat01

Question

В равнобедренном треугольнике ABC (AB=AC), в котором ∠ABC=72, отмечена середина D стороны BC. На отрезке AD выбрана точка P, а на стороне AB – точка Q так, что PQ=PC. Найдите градусную меру угла PQC.

Python

python
import sympy as sp, math
t = math.tan(math.radians(72))
t


3.077683537175253

Python

python
import numpy as np, math
t=math.tan(math.radians(72))
def q_solutions(p):
    A=1+t**2
    B=2*t*(t-p)
    C=t**2 -2*t*p -1
    roots=np.roots([A,B,C])
    return roots
for p in [0.5,1.0,2.0,3.0]:
    print(p, q_solutions(p))

Python

python
import math, numpy as np
t=math.tan(math.radians(72))
def angle_PQC(p):
    # solve for q in (-1,0) not equal -1
    roots=q_solutions(p)
    q=None
    for r in roots:
        if abs(r+1)<1e-6: 
            continue
        if -1-1e-6 <= r <= 0+1e-6:
            q=float(r)
    if q is None:
        return None
    Q=(q, t*(q+1))
    P=(0,p)
    C=(1,0)
    # vectors from Q
    v1=(P[0]-Q[0], P[1]-Q[1]) # QP
    v2=(C[0]-Q[0], C[1]-Q[1]) # QC
    # angle between v1 and v2
    dot=v1[0]*v2[0]+v1[1]*v2[1]
    ang=math.degrees(math.acos(dot/(math.hypot(*v1)*math.hypot(*v2))))
    return ang,q
for p in [0.2,0.5,0.8,1.0,1.2,1.4]:
    res=angle_PQC(p)
    print(p,res)

Thought

Proving the constant angle of 18 degrees

The angle PQC is consistently 18 degrees, which seems to be half of angle A (36 degrees). To prove this, we need to show that for a nontrivial Q, the triangle QPC is isosceles with PQ=PC. We'll also demonstrate that ∠QCP = 18° by proving P lies on the circumcenter of segment BC, making P equidistant from B and C.

Exploring the angle at Q

Since B, C, and Q lie on the circle centered at P, with P as the circumcenter of triangle BCQ, we can use central angles. The central angle BPC corresponds to the angle between the chords BQ and CQ. At Q, angle BQC is half of that central angle, but this depends on the position of P along AD, as PB and PC radii vary.

Reevaluating angle subtended by PQC

As point P moves along the perpendicular bisector, the angle subtended by chord BC changes, so it's not constant. However, this can help find the angle ∠PQC. In a circle, the angle between the radius and the chord at an endpoint relates to the angle in the opposite segment, specifically linking it to ∠QBC.

Clarifying angle relationships involving ∠PQC and ∠QBC

Okay, using the tangent-chord theorem, the angle between the tangent at Q and chord QC equals angle QBC. Since PQ is perpendicular to the tangent, the angle between PQ and QC (∠PQC) is 90° minus ∠QBC. If we confirm ∠QBC is constant, the proof will be finished. Since Q is on AB, ∠QBC might equal 72°.

Justifying ∠PQC with the tangent-chord theorem

I’ve confirmed that ∠QBC = 72° is constant when Q lies on BA. Then, using the tangent-chord theorem, the angle between the radius PQ and the chord QC equals 90° minus ∠QBC, giving us ∠PQC = 18°. This works because B, C, and Q lie on a circle centered at P, with PB = PQ = PC.

Thought for 3m 22s

Answer

Так как AB=ACAB=AC, то медиана ADAD к основанию BCBC является перпендикулярным серединным к BCBC. Значит, для любой точки P∈ADP\in AD

PB=PC.PB=PC.

По условию также PQ=PCPQ=PC. Тогда

PB=PC=PQ,PB=PC=PQ,

то есть точки B,C,QB,C,Q лежат на окружности с центром в PP (радиус PCPC).

Проведём касательную tt к этой окружности в точке QQ. Тогда t⊥PQt\perp PQ (радиус к точке касания перпендикулярен касательной).

По теореме о касательной и хорде угол между касательной tt и хордой QCQC равен углу, опирающемуся на ту же хорду QCQC в противоположной дуге, то есть

∠(t,QC)=∠QBC.\angle (t,QC)=\angle QBC.

Но Q∈ABQ\in AB, значит луч BQBQ совпадает с BABA, и потому

∠QBC=∠ABC=72∘.\angle QBC=\angle ABC=72^\circ.

Тогда искомый угол

∠PQC=90∘−∠(t,QC)=90∘−72∘=18∘.\angle PQC = 90^\circ - \angle (t,QC)=90^\circ-72^\circ=18^\circ.

Ответ: 18∘18^\circ.

Question

Отрицательное число x и действительное число y таковы, что x+xy=216 и ny+xy=144. Найдите x.

Thought

Solving system of equations

The problem seems to involve variables x and y, with potential typo in "ny" which might mean "y." The system is x + xy = 216 and y + xy = 144. I can express y from the first equation and substitute into the second equation to solve for both x and y. After simplifying, I solve the quadratic equation: x^2 - 71x - 216 = 0 and calculate the discriminant.

Reevaluating equation interpretation

It looks like I'm considering possible typos or misreadings in the equations. I've puzzled over whether "ny" in the second equation might refer to something else, like "y + xy." This leads to an irrational x, which could still be acceptable. However, I wonder if solving the system using ratios between x and y might work instead.

Verifying possible solutions for x and y

I’m rethinking the system of equations to check if "y + xy = 144" works. By solving for y in the first equation, I find x ≈ -2.92, leading to y ≈ -75, and plugging that back into the second equation gets about 144 — a decent check. Changing the second equation like "y + xy = 126" doesn’t yield a nice discrimin

Thought for 18s

Answer

Понимаем второе условие как

y+xy=144(то есть y(1+x)=144).y+xy=144 \quad\text{(то есть }y(1+x)=144\text{)}.

Тогда система:

x+xy=216  ⇒  x(1+y)=216,x+xy=216 \;\Rightarrow\; x(1+y)=216, y+xy=144  ⇒  y(1+x)=144.y+xy=144 \;\Rightarrow\; y(1+x)=144.

Из первого:

1+y=216x  ⇒  y=216x−1.1+y=\frac{216}{x}\;\Rightarrow\; y=\frac{216}{x}-1.

Подставим во второе:

(216x−1)(1+x)=144.\left(\frac{216}{x}-1\right)(1+x)=144.

Умножим на xx:

(216−x)(1+x)=144x.(216-x)(1+x)=144x.

Раскроем скобки:

216+215x−x2=144x216+215x-x^2=144x −x2+71x+216=0-x^2+71x+216=0 x2−71x−216=0.x^2-71x-216=0.

По формуле:

x=71±712+4⋅2162=71±59052.x=\frac{71\pm\sqrt{71^2+4\cdot216}}{2}=\frac{71\pm\sqrt{5905}}{2}.

Так как x<0x<0, берём минус:

 x=71−59052 .\boxed{\,x=\frac{71-\sqrt{5905}}{2}\,}.

Question

image.pngimage.png

реши

Python

python
from PIL import Image, ImageEnhance
img = Image.open('/mnt/data/image.png')
img.size


(991, 557)

Python

python
import matplotlib.pyplot as plt
plt.figure(figsize=(10,5))
plt.imshow(img)
plt.axis('off')
plt.show()

Python

python
import numpy as np, matplotlib.pyplot as plt
arr = np.array(img)
# crop around plot
crop = img.crop((250,150,800,540))
plt.figure(figsize=(6,4))
plt.imshow(crop)
plt.axis('off')
plt.show()

Thought

Breaking down and confirming the thermodynamic cycle

The diagram now shows point 1 slightly to the right of point 3, with line 3-1 curving left—this indicates adiabatic compression as volume decreases (V1 < V3). The cycle resembles Diesel/Atkinson, with isothermal, isobaric, and adiabatic processes. The heat input happens during 1-2, so Q_in = 5 kJ. Net work is the sum of W12, W23, and W31.

Figuring out work and heat in the cycle

We know that W31 is unknown, but we can use the first law (ΔU = Q - W) to find it. Energy balance across the cycle gives us ΔU23 + ΔU31 = 0. Since Q31 = 0 for an adiabatic process, W31 = -ΔU31, and since compression increases internal energy, W31 will be negative.

Computing work and efficiency

For process 3-1, the work by the gas is negative, as it's compression. Using the formula for isobaric processes, I can compute ΔU23 as ΔU23 = (3/2) * W23 = -2.55 kJ. Using that ΔU31 = +2.55 kJ, W31 = -2.55 kJ. The net work W_net = 0.75 kJ, so efficiency is 15%.

Rechecking heat and work in process 2-3

For the isobaric compression in process 2-3, it makes sense that heat is released. Using the relation ΔU23 = (3/2)W_by, I found ΔU23 = -2.55 kJ. Then, calculating heat released, Q23 = (5/2)W_by = -4.25 kJ. The net work is 0.75 kJ, confirming the efficiency: 15%.

Thought for 1m 29s

Answer

Обозначим работу газа как AA (положительна, если газ совершает работу).

1–2 (изотермическое расширение)

Для идеального газа при изотерме ΔU12=0\Delta U_{12}=0, значит

Q12=A12.Q_{12}=A_{12}.

Дано Q12=Qh=5 кДжQ_{12}=Q_h=5\text{ кДж} ⇒

A12=5 кДж.A_{12}=5\text{ кДж}.

2–3 (по графику — изобарическое сжатие)

Дано: работа, совершённая над газом, A23(над)=1,7 кДжA^{(над)}_{23}=1{,}7\text{ кДж}.
Тогда работа газа

A23=−1,7 кДж.A_{23}=-1{,}7\text{ кДж}.

При p=constp=\text{const}:

A23=nRΔT⇒ΔU23=nCVΔT=CVRA23.A_{23}=nR\Delta T \quad\Rightarrow\quad \Delta U_{23}=nC_V\Delta T=\frac{C_V}{R}A_{23}.

Для одноатомного газа CV=32RC_V=\tfrac{3}{2}R, поэтому

ΔU23=32A23=32⋅(−1,7)=−2,55 кДж.\Delta U_{23}=\frac{3}{2}A_{23}=\frac{3}{2}\cdot(-1{,}7)=-2{,}55\text{ кДж}.

3–1 (адиабатическое сжатие)

В адиабате Q31=0Q_{31}=0, значит

ΔU31=−A31.\Delta U_{31} = -A_{31}.

За цикл ΔU12+ΔU23+ΔU31=0\Delta U_{12}+\Delta U_{23}+\Delta U_{31}=0, а ΔU12=0\Delta U_{12}=0, поэтому

ΔU31=−ΔU23=2,55 кДж,\Delta U_{31}=-\Delta U_{23}=2{,}55\text{ кДж}, A31=−ΔU31=−2,55 кДж.A_{31}=-\Delta U_{31}=-2{,}55\text{ кДж}.

Работа за цикл и КПД

Aц=A12+A23+A31=5−1,7−2,55=0,75 кДж.A_{\text{ц}}=A_{12}+A_{23}+A_{31}=5-1{,}7-2{,}55=0{,}75\text{ кДж}.

Теплота, полученная от нагревателя, только на участке 1–2:

Qin=Qh=5 кДж.Q_{\text{in}}=Q_h=5\text{ кДж}.

Тогда КПД:

η=AцQin=0,755=0,15=15%.\eta=\frac{A_{\text{ц}}}{Q_{\text{in}}}=\frac{0{,}75}{5}=0{,}15=15\%.

Ответ: 15%\boxed{15\%}.

Question

image.pngimage.png

реши

Python

python
import sympy as sp
a = 2**(-sp.Rational(2,3))
PV1 = sp.Rational(1000,3) / (1 - a)  # 333.333...
PV1_s = sp.N(PV1)
a_n = sp.N(a)
PV3 = a*PV1
PV2 = 2*(3000 - PV1)  # in Joules
Qx = sp.Rational(3,2)*(PV2 - PV3)
PV1_s, a_n, sp.N(PV3), sp.N(PV2), sp.N(Qx)


(900.804794639772,
 0.629960524947437,
 567.471461306438,
 4198.39041072046,
 5446.37842412103)

Thought

Checking heat calculation consistency

The heat rejected in process 2-3, QxQ_x, is approximately 5.45 kJ, which matches the expected magnitude of heat released. This fits with the cycle's work calculations, where the net work is 1000 J. The heat input was calculated as Q12=6446 JQ_{12} = 6446 \, \text{J}, consistent with the expected results. The final value for QxQ_x is 5.4 kJ, as expected.

Formulating solution in Russian with steps

The process involves three stages: 1-2 (straight line in PV), 2-3 (isochoric, vertical line), and 3-1 (adiabatic). Using the first law and adiabatic relation, I aim to express p0V0p_0V_0 in terms of the work provided for segment 3-1. The relationship p3=p0/2γp_3 = p_0 / 2^\gamma will guide the calculation for QxQ_x.

Calculating Qx with exact terms

I’ve derived p3V3=p0V0/22/3p_3V_3 = p_0V_0 / 2^{2/3}, so by substituting into the relationship for QxQ_x, I get ( p_0V_0 = 1000 / [3(

Thought for 2m 42s

Answer

Обозначим работу газа A>0A>0, а работу внешних сил над газом Aвнеш>0A_{\text{внеш}}>0. Газ одноатомный, поэтому

U=32nRT=32 pV,γ=53.U=\frac32 nRT=\frac32\,pV,\qquad \gamma=\frac53.

По графику: 1:(V0,p0)1:(V_0,p_0), 2:(2V0,p2)2:(2V_0,p_2), 3:(2V0,p3)3:(2V_0,p_3); участок 2→32\to3 — изохора (V=constV=\text{const}), участок 3→13\to1 — адиабата.

1) Найдём p0V0p_0V_0 из адиабаты 3→13\to1

Дано: на 3→13\to1 работа внешних сил над газом ∣A31∣=500 Дж|A_{31}|=500\text{ Дж}. Это сжатие, значит работа газа

A31=−500 Дж.A_{31}=-500\text{ Дж}.

На адиабате Q31=0Q_{31}=0, поэтому по первому закону:

ΔU31=Q31−A31=0−(−500)=500 Дж.\Delta U_{31}=Q_{31}-A_{31}=0-(-500)=500\text{ Дж}.

Но

ΔU31=U1−U3=32(p0V0−p3⋅2V0)=500,\Delta U_{31}=U_1-U_3=\frac32(p_0V_0-p_3\cdot 2V_0)=500,

то есть

p0V0−2p3V0=10003.(1)p_0V_0-2p_3V_0=\frac{1000}{3}. \tag{1}

Из адиабатного соотношения pVγ=constpV^\gamma=\text{const}:

p0V0γ=p3(2V0)γ  ⇒  p3=p02γ=p025/3.p_0V_0^\gamma=p_3(2V_0)^\gamma\;\Rightarrow\; p_3=\frac{p_0}{2^\gamma}=\frac{p_0}{2^{5/3}}.

Тогда

2p3V0=2V0⋅p025/3=p0V0⋅21−5/3=p0V0⋅2−2/3.2p_3V_0=2V_0\cdot \frac{p_0}{2^{5/3}}=p_0V_0\cdot 2^{1-5/3}=p_0V_0\cdot 2^{-2/3}.

Подставляем в (1):

p0V0(1−2−2/3)=10003  ⇒  p0V0=100031−2−2/3≈900.805 Дж.p_0V_0\left(1-2^{-2/3}\right)=\frac{1000}{3} \;\Rightarrow\; p_0V_0=\frac{\frac{1000}{3}}{1-2^{-2/3}} \approx 900.805\text{ Дж}.

Также

p3V3=p3⋅2V0=2p3V0=p0V0⋅2−2/3≈567.471 Дж.p_3V_3=p_3\cdot 2V_0=2p_3V_0=p_0V_0\cdot 2^{-2/3}\approx 567.471\text{ Дж}.

2) Найдём p2V2p_2V_2 из работы на участке 1→21\to2

Участок 1→21\to2 — прямая, поэтому работа равна площади трапеции:

A12=p0+p22 (2V0−V0)=p0+p22V0=1500.A_{12}=\frac{p_0+p_2}{2}\,(2V_0-V_0)=\frac{p_0+p_2}{2}V_0=1500.

Отсюда

(p0+p2)V0=3000  ⇒  p2V0=3000−p0V0.(p_0+p_2)V_0=3000 \;\Rightarrow\; p_2V_0=3000-p_0V_0.

Тогда

p2V2=p2⋅2V0=2p2V0=2(3000−p0V0)≈4198.390 Дж.p_2V_2=p_2\cdot 2V_0=2p_2V_0=2(3000-p_0V_0)\approx 4198.390\text{ Дж}.

3) Теплота, отданная холодильнику за цикл

Тепло газ отдаёт на участке 2→32\to3 (изохорное охлаждение), причём A23=0A_{23}=0, значит

Q23=ΔU23=U3−U2=32(p3V3−p2V2).Q_{23}=\Delta U_{23}=U_3-U_2=\frac32(p_3V_3-p_2V_2).

Это отрицательно, а искомое QxQ_x — положительное количество, отданное холодильнику:

Qx=−Q23=32(p2V2−p3V3).Q_x=-Q_{23}=\frac32(p_2V_2-p_3V_3).

Подставляем числа:

Qx=32(4198.390−567.471)≈32⋅3630.919≈5446.38 Дж.Q_x=\frac32(4198.390-567.471)\approx \frac32\cdot 3630.919\approx 5446.38\text{ Дж}.

Ответ: Qx≈5.45 кДж\boxed{Q_x \approx 5.45\ \text{кДж}}.

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