Реши математическую задачу по планиметрии наиболее...

बनाया गया: 27 जनवरी 2026

का उपयोग करके उत्तर दिया गया GPT-5.2 Thinking द्वारा Chat01

प्रश्न

Реши математическую задачу по планиметрии наиболее простым способом: На окружности $\Omega$ радиуса $\sqrt{37}$ последовательно расположены точки $P$ , $Q$ , $R$ , $S$ так, что $QS\perp PR$ . Известно, что $PS:RS=6:1$ , а в четырёхугольник $PQRS$ можно вписать окружность. Найдите площадь этого четырёхугольника и длину биссектрисы $QB$ треугольника $PQR$ .

विचार

Solving geometry problem with tangential quadrilateral

I need to solve a geometry problem in Russian involving a cyclic, tangential quadrilateral (PQRS) inscribed in a circle with radius √37. The diagonals QS and PR are perpendicular. I also know the ratio PS:RS = 6:1, so PS = 6k and RS = k. The goal is to find the area of the quadrilateral and the length of the angle bisector QB from Q to PR.

Approaching area of tangential cyclic quadrilateral

The quadrilateral is tangential and cyclic with perpendicular diagonals. For tangential quadrilaterals, the area formula involves the inradius and semiperimeter, while for cyclic ones, Brahmagupta’s formula applies. Given perpendicular diagonals, the area can also be calculated as (1/2) * PR * QS, and diagonals’ relationship might help connect side lengths for further calculations.

Verifying properties of cyclic quadrilateral with perpendicular diagonals

In a cyclic quadrilateral with perpendicular diagonals (orthodiagonal), the opposite sides satisfy the relation a² + c² = b² + d². The area is also given by 1/2 * d₁ * d₂. I’m thinking of verifying this using Ptolemy’s theorem, law of cosines, and Pythagorean identities for the intersections of diagonals.

Analyzing cyclic tangential quadrilateral and conditions

I’m using the relationship between side lengths and central angles for the cyclic quadrilateral. The tangential condition gives a key equation involving angle sines, leading to Eq1. The diagonals are perpendicular, so I’m thinking the sum of opposite arcs (α+γ) should equal (β+δ?). It's interesting to express angles by arcs and explore their geometry.

Confirming condition for perpendicular diagonals in cyclic quadrilateral

To ensure the diagonals are perpendicular, the sum of the arcs between the intersections of chords PR and QS must be 180°. I’ve confirmed that if the diagonals are perpendicular, then the central angles α+γ = π and β+δ = π are true, ensuring orthogonality. This relationship holds for perpendicular diagonals in cyclic quadrilaterals.

Solving for angles and relationships in cyclic quadrilateral

I've got the relationships set up: For chords PQ and RS, they subtend angles α and γ, respectively, and are supplementary. Connecting lengths and angles with sine and cosine, I've derived equations linking the angles. The goal now is to solve for t and u, ensuring they satisfy the system of trigonometric relations involving sin and cos, with constraints on their ranges.

Solving for s and verifying values

I'm solving for s, converting the original equation into something algebraic by using trigonometric identities. After squaring both sides and simplifying, I end up with s² = 121/6205. This simplifies to s = 11/√6205, and I verify the values to ensure they fit the initial equation. Finally, I determine cos(t) is √(6084/6205).

Verifying chord length calculations and tangential condition

I’m checking through my calculations. I verified that A and B represent chord lengths and should relate to sine functions based on their respective arcs. There seems to be inconsistency with the results from my earlier algebra. I'll need to reconsider the sign or possibly my understanding of the relevant arcs. The tangential condition suggests PQ+RS = QR+SP.

Revisiting tangential condition and correcting algebra

I realized I needed to adjust my previous approach, correcting the equation for the tangential quadrilateral. The relationship should have been sin(α/2) + sin(γ/2) = sin(β/2) + sin(δ/2), leading to the corrected equation cos t - sin u = 5 sin t. Now I’m solving this with the correct values for A and B.

Revisiting quadrilateral symmetry and calculating dimensions

I’ve made the necessary corrections and found the arcs: α = δ = π - 2t, β = γ = 2t. This symmetry gives equal chord lengths: PQ = SP = 12, and QR = RS = 2. The quadrilateral seems like an isosceles trapezoid, though the opposite sides aren’t parallel. Now, I'm focusing on calculating the area.

Thought for 2m 52s

उत्तर

Обозначим через $O$ центр окружности, $R=\sqrt{37}$ . Пусть центральные углы, соответствующие хордам

PQ,\,QR,\,RS,\,SP

равны соответственно $\alpha,\beta,\gamma,\delta$ (в этом порядке), тогда $\alpha+\beta+\gamma+\delta=2\pi$ .

1) Перпендикулярность диагоналей $\;QS\perp PR$

Диагонали $PR$ и $QS$ пересекаются внутри окружности, и угол между ними равен половине суммы соответствующих дуг:

\angle(\,PR,QS\,)=\frac12(\overset{\frown}{PQ}+\overset{\frown}{RS})=\frac12(\alpha+\gamma).

Так как угол $90^\circ$ , получаем

\alpha+\gamma=\pi,\qquad \beta+\delta=\pi.

2) Хорды через центральные углы и условие $PS:RS=6:1$

Длины хорд:

PQ=2R\sin\frac\alpha2,\quad QR=2R\sin\frac\beta2,\quad RS=2R\sin\frac\gamma2,\quad PS=2R\sin\frac\delta2.

Из $\beta+\delta=\pi$ имеем $\dfrac\delta2=\dfrac\pi2-\dfrac\beta2$ , значит

\sin\frac\delta2=\cos\frac\beta2.

А из $\alpha+\gamma=\pi$ аналогично

\sin\frac\alpha2=\cos\frac\gamma2.

По условию $PS=6\,RS$ :

2R\sin\frac\delta2=6\cdot 2R\sin\frac\gamma2 \;\Rightarrow\; \cos\frac\beta2=6\sin\frac\gamma2.

Обозначим $t=\dfrac\gamma2,\;u=\dfrac\beta2$ . Тогда

\cos u=6\sin t. \tag{A}

3) Условие вписанной окружности (касательный четырёхугольник)

Для касательного четырёхугольника верно:

PQ+RS=QR+PS.

Делим на $2R$ :

\sin\frac\alpha2+\sin\frac\gamma2=\sin\frac\beta2+\sin\frac\delta2.

Подставляем $\sin\frac\alpha2=\cos t,\;\sin\frac\gamma2=\sin t,\;\sin\frac\beta2=\sin u,\;\sin\frac\delta2=\cos u$ :

\cos t+\sin t=\sin u+\cos u.

Но $\cos u=6\sin t$ из (A), значит

\cos t+\sin t=\sin u+6\sin t \;\Rightarrow\; \cos t-\sin u=5\sin t. \tag{B}

Теперь заметим: из (A) $\cos u=6\sin t\le 1$ , значит $\sin t\le \frac16$ , и $t,u\in(0,\frac\pi2)$ . Тогда $\sin u=\sqrt{1-\cos^2u}=\sqrt{1-36\sin^2t}$ , $\cos t=\sqrt{1-\sin^2t}$ . Подставляем в (B):

\sqrt{1-s^2}-\sqrt{1-36s^2}=5s,\quad s=\sin t.

Дважды возводя в квадрат, получаем $s^2=\dfrac1{37}$ , то есть

\sin t=\frac1{\sqrt{37}},\qquad \cos t=\frac6{\sqrt{37}}.

Тогда из $\cos u=6\sin t=\frac6{\sqrt{37}}=\cos t$ и $u,t\in(0,\frac\pi2)$ следует $u=t$ , то есть $\beta=\gamma$ , а значит и $\alpha=\delta$ .

4) Стороны и диагонали

Теперь легко считаем стороны:

RS=2R\sin t=2\sqrt{37}\cdot\frac1{\sqrt{37}}=2,

PS=6RS=12,

QR=2R\sin u=2,\qquad PQ=2R\sin\frac\alpha2=2R\cos t=2\sqrt{37}\cdot\frac6{\sqrt{37}}=12.

Диагональ $PR$ соответствует центральному углу $\alpha+\beta=\pi$ , значит $PR$ — диаметр:

PR=2R=2\sqrt{37}.

Диагональ $QS$ соответствует центральному углу $\beta+\gamma=4t$ , поэтому

QS=2R\sin\frac{4t}{2}=2R\sin(2t)=2\sqrt{37}\cdot 2\sin t\cos t =2\sqrt{37}\cdot 2\cdot\frac1{\sqrt{37}}\cdot\frac6{\sqrt{37}} =\frac{24}{\sqrt{37}}.

5) Площадь четырёхугольника

Так как диагонали перпендикулярны,

[ PQRS ]=\frac12\cdot PR\cdot QS =\frac12\cdot (2\sqrt{37})\cdot\frac{24}{\sqrt{37}}=24.

6) Длина биссектрисы $QB$ в $\triangle PQR$

Так как $PR$ — диаметр, $\angle PQR=90^\circ$ . В треугольнике $PQR$ :

PQ=12,\quad QR=2,\quad PR=2\sqrt{37}.

Длина биссектрисы из вершины $Q$ :

QB=\frac{\sqrt{PQ\cdot QR\left((PQ+QR)^2-PR^2\right)}}{PQ+QR} =\frac{\sqrt{12\cdot2\,(14^2-(2\sqrt{37})^2)}}{14} =\frac{\sqrt{24\,(196-148)}}{14} =\frac{\sqrt{24\cdot48}}{14} =\frac{24\sqrt2}{14} =\frac{12\sqrt2}{7}.

Ответ:

[ PQRS ]=24,\qquad QB=\frac{12\sqrt2}{7}.