Внутри квадрата со стороной а=(sqrt(12)+sqrt(6))/s...

Criado em: 6 de dezembro de 2025

Respondido usando GPT-5.1 Thinking por Chat01

Pergunta

Внутри квадрата со стороной а=(sqrt(12)+sqrt(6))/sqrt(π) расположены два круга так, что расстояние между их центрами не меньше суммы радиусов. Какое наибольшее значение может принимать сумма площадей кругов

Pensamento

Finalizing the solution for maximal area sum

Great, the area sum equals 9 – that's the expected answer. Now, we need to rigorously argue why this is indeed the maximum. Since the problem is in Russian, I’ll structure the solution accordingly, using the right math language. Instead of a web search, I’ll focus on presenting the reasoning here, starting from scaling and maximizing the sum of squared radii.

Maximizing the sum of radii

First, we're scaling the problem: when we find the dimensionless solution for the circle sum, we'll just multiply by π a². To do this, we need to first optimize the sum of squared radii of two non-intersecting circles in a unit square. We’ll prove an optimal configuration and show why other arrangements result in smaller sums. Starting with the largest circle, we label its radius R_max.

Considering circle placement and limits

If the largest circle’s radius (R_max) is smaller than half the side of the square (a/2), we can keep inflating it until it touches at least one side or the other circle. The optimal configuration would have it touching both neighbors and sides. I’m thinking of using a known inequality for disjoint circles inside a square for a more systematic approach.

Exploring circle configurations within a square

I’m thinking about two non-overlapping discs in a square and their radii. If we know the sum of their radii (r1 + r2), we could try bounding their sum of squares. The sum of squares reaches its maximum when one radius is as large as possible, and the other is the smallest. To get stronger bounds, I need individual constraints on each radius too. I can scan through corners to evaluate the worst-case scenario.

Exploring geometric inequalities and bounding areas

I’m thinking about deriving an inequality for the sum of the radii squared. Maybe we can use a formula involving the square’s side and the circle radii. There's also the classical inequality that the sum of areas of inscribed circles must be less than or equal to the domain's area. The problem specifies a value for the side, so the total area is expected to be a bit over 80% of the square's area, which is plausible.

Crafting a moderately detailed solution

For this problem, while calculus could be useful, the exam task likely doesn't require full rigor. I’ll aim for a concise, manageable explanation.

The maximal sum of areas for two non-overlapping circles inside a square is 9. The approach is to solve for an arbitrary side length of the square, and then substitute the specific value. The configuration involves one circle inscribed and the other tangent to two sides.

Exploring a geometric solution for circle arrangement

I’m considering a strategy where I inflate a circle until it’s tangent to all four sides, adjusting the other circle’s radius to keep separation. I’m trying to derive upper bounds for the circle radii. I’m also checking if using a general inequality for the sum of radii can work. Introducing the Apollonius circle might help here— it seems like a useful way to derive this maximum sum constraint.

Exploring positioning and overlap constraints for circles

I'm considering how to resize the larger circle until it touches a side, ensuring it doesn't overlap with the smaller circle. Alternatively, I could fix the radii and reposition the circles to satisfy the conditions. This seems messy, but we don't need full rigor and could show a plausible solution outline.

Considering optimization and constraints for circles

I'm thinking of using Lagrange multipliers to maximize the sum of square radii under certain constraints. The two circles must be tangent to each other for non-overlap. If their centers are far apart, we can enlarge both radii as long as we keep them within square boundaries. This isn't trivial, but it could work for the exam.

Outlining optimal solution for circle configuration

I’m considering the optimal configuration for two circles within a square. First, the two circles must be tangent to each other, and the larger one must be inscribed. If we enlarge the smaller circle, we would contradict the maximality condition. Then, I’ll show how the center of the large circle coincides with the center of the square, making it inscribed.

Outlining possible configurations for circle arrangement

To solve this optimally, the solution seems to involve showing each circle must be tangent either to another circle or to a side. Then, the larger circle will be inscribed, which means it's tangent to every side. We need to compute the areas involved, simplify the problem, and assume the optimal configuration follows a typical pattern, like a combination of a center inscribed and corner tangent arrangements.

Refining circle tangency and scaling arguments

To maximize the sum of areas, the configuration will always have at least one circle tangent to the boundary, with the larger circle usually inscribed. If the larger circle isn't tangent, scaling the circles will increase the total area but won't improve the solution. Ultimately, we focus on the configuration with the largest possible radii, verifying that enlarging the circles within the square will still maintain the tangency conditions, ensuring that no larger configuration exists.

Thought for 3m 45s

Resposta

Ответ: наибольшая возможная сумма площадей двух кругов равна 99.

Разберёмся, откуда она берётся.


1. Сведём задачу к геометрии

Пусть сторона квадрата — aa.

Ищем максимум

S=π(r12+r22),S = \pi(r_1^2 + r_2^2),

где r1,r2r_1, r_2 — радиусы двух непересекающихся кругов внутри квадрата.

В «предельной» (оптимальной) конфигурации:

  • оба круга будут упираться либо в стороны квадрата, либо друг в друга (иначе можно их немного увеличить);
  • один из кругов окажется наибольшим и «зажмётся» квадратом максимально — т.е. он будет вписанным кругом квадрата.

Поэтому разумно рассмотреть конфигурацию:

  • Первый круг — вписанный в квадрат.
  • Второй круг — в одном из углов, касается двух сторон квадрата и первого круга.

Оказывается, именно в такой конфигурации достигается максимум суммы площадей (подробное обоснование в олимпиадных решениях обычно проводится методом «перетаскивания» и «раздувания» кругов, но здесь ограничимся этим естественным экстремальным случаем).


2. Радиусы кругов при оптимальном расположении

Рассмотрим квадрат со стороной aa. Введём координаты:

  • Центр квадрата — (a2,a2)\left(\dfrac a2,\dfrac a2\right).
  • Первый круг (большой) — вписанный: R=a2,O1(a2,a2).R = \frac a2,\qquad O_1\left(\frac a2,\frac a2\right).
  • Второй круг (малый) расположен в углу (0,0)(0,0), касается осей: r=радиус малого круга,O2(r,r).r = \text{радиус малого круга},\qquad O_2(r,r).

Условие касания кругов:

O1O2=R+r.O_1O_2 = R + r.

Считаем расстояние между центрами:

O1O2=(a2r)2+(a2r)2=2(a2r).O_1O_2 = \sqrt{\left(\frac a2 - r\right)^2 + \left(\frac a2 - r\right)^2} = \sqrt{2}\left(\frac a2 - r\right).

При касании:

2(a2r)=a2+r.\sqrt{2}\left(\frac a2 - r\right) = \frac a2 + r.

Решаем это уравнение относительно rr:

2a22r=a2+r\sqrt{2}\cdot\frac a2 - \sqrt{2}r = \frac a2 + r 2rr=a22a2-\sqrt{2}r - r = \frac a2 - \sqrt{2}\cdot\frac a2 r(2+1)=a2(12)-r(\sqrt{2} + 1) = \frac a2(1 - \sqrt{2}) r=a2(21)2+1=a2(21)221=a2(322)=a(322).r = \frac{\frac a2(\sqrt{2} - 1)}{\sqrt{2} + 1} = \frac a2\cdot\frac{(\sqrt{2} - 1)^2}{2 - 1} = \frac a2(\,3 - 2\sqrt{2}\,) = a\left(\frac{3}{2} - \sqrt{2}\right).

Итак:

R=a2,r=a(322).R = \frac a2,\qquad r = a\left(\frac{3}{2} - \sqrt{2}\right).

3. Максимальная сумма площадей для квадрата со стороной aa

Тогда

Smax=π(R2+r2)=π[(a2)2+(a(322))2]=πa2(9232).S_{\max} = \pi\left( R^2 + r^2 \right) = \pi\left[\left(\frac a2\right)^2 + \left(a\left(\frac{3}{2} - \sqrt{2}\right)\right)^2\right] = \pi a^2\left(\frac{9}{2} - 3\sqrt{2}\right).

4. Подставляем данную сторону квадрата

По условию:

a=12+6π.a = \frac{\sqrt{12} + \sqrt{6}}{\sqrt{\pi}}.

Сначала найдём a2a^2:

a2=(12+6)2π=12+6+272π=18+122π.a^2 = \frac{(\sqrt{12} + \sqrt{6})^2}{\pi} = \frac{12 + 6 + 2\sqrt{72}}{\pi} = \frac{18 + 12\sqrt{2}}{\pi}.

Теперь:

Smax=πa2(9232)=(9232)(18+122).S_{\max} = \pi a^2\left(\frac{9}{2} - 3\sqrt{2}\right) = \left(\frac{9}{2} - 3\sqrt{2}\right)(18 + 12\sqrt{2}).

Посчитаем произведение:

(9232)(18+122)=9218+92122321832122.\left(\frac{9}{2} - 3\sqrt{2}\right)(18 + 12\sqrt{2}) = \frac{9}{2}\cdot 18 + \frac{9}{2}\cdot 12\sqrt{2} - 3\sqrt{2}\cdot 18 - 3\sqrt{2}\cdot 12\sqrt{2}. =81+54254272=8172=9.= 81 + 54\sqrt{2} - 54\sqrt{2} - 72 = 81 - 72 = 9.

Итак,

Smax=9.\boxed{S_{\max} = 9}.

Ответ: максимальная сумма площадей двух кругов, расположенных описанным образом внутри данного квадрата, равна 99.

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