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В треугольнике ABC сторона AC равна 2√21, AB=BC. В...

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سؤال

В треугольнике ABC сторона AC равна 2√21, AB=BC. Вписанная окружность ω касается стороны AB в точке K. Отрезок CK пересекает ω в точке M. Известно, что длина отрезка MK равна 4. Найдите длину стороны BC. В ответ запишите квадрат полученной величины.

الفكرة

Sorting out geometry problem with triangle and circle

I need to solve a geometry problem about triangle ABC with isosceles sides and an incircle. The segment CK passes through the tangent point K on AB, intersecting the circle at M. MK is given as 4. The problem asks for the length of BC, specifically BC^2. Let’s work through it!

Working with side lengths and chord in circle

I need to figure out the lengths of the sides of a triangle where the incircle intersects the line from vertex C. The chord length between the tangency point (K) on AB and where CK intersects the circle again (M) is given as 4. I'll use the power of a point and the distances involved to calculate the side BC. Let's dive into that!

Clarifying triangle's tangency and power of point

So, K is on AB, and line CK intersects the circle at M first, then at K. The points go C - M - K. Power of point C is related to CM and CK. The tangent lengths from C to the incircle would match. Now, I need to relate this to the sides, figuring out the lengths for BC and AB.

Analyzing tangency and semiperimeter relations

I’m working through the tangency lengths related to the incircle. The semiperimeter is s=a+21s = a + \sqrt{21}, and the tangency lengths from A and B to the circle make sense: AK=21AK = \sqrt{21} and BK=a−21BK = a - \sqrt{21}. The relationships between these lengths give insights into the sides of the triangle and the structure of the incircle.

Reviewing tangents and power of a point

I’ve confirmed that the tangents from C to the incircle along BC and AC are both √21 in length. This means the power of point C with respect to the incircle is 2121 since power equals the square of the tangent length. Now I can use this information for further calculations.

Using coordinates to find side length

I’m setting up the triangle with fixed coordinates for points A and C, assuming symmetry. With AC as the base, I can place A at (-√21, 0) and C at (√21, 0), while B’s coordinates will be determined from the isosceles condition. The goal is to find length CK, which depends on side lengths and angle using geometry.

Computing CK squared using geometry

I’m working out the distance CK by breaking it down into coordinates and simplifying the equation for CK^2. The formula incorporates the side lengths, and I see that some terms cancel out. I end up simplifying to CK^2 = 21 [5 - 4d/a], where d = √21, and this tells me more about the geometry of the situation.

Verifying the square of BC

I’m confirming that the computed side length a=3212a = \frac{3\sqrt{21}}{2} is valid, and with it, a2=1894a^2 = \frac{189}{4}. This works geometrically as well. The tangents from point C to the incircle are indeed 21\sqrt{21}, confirming that the power of C is 21. It appears that the final answer for BC2BC^2 is 1894\frac{189}{4}.

Thought for 59s

الإجابة

Обозначим BC=AB=aBC=AB=a, AC=221AC=2\sqrt{21}. Тогда полупериметр

s=a+a+2212=a+21.s=\frac{a+a+2\sqrt{21}}2=a+\sqrt{21}.

1) Степень точки CC относительно вписанной окружности

Точка касания окружности со стороной ACAC делит её на отрезки длиной s−as-a и s−cs-c. Так как c=AB=ac=AB=a, то отрезок от CC до точки касания на ACAC равен

s−c=s−a=21.s-c=s-a=\sqrt{21}.

Значит, длина касательной из CC к вписанной окружности равна 21\sqrt{21}, и степень точки CC:

Pow⁡ω(C)=21.\operatorname{Pow}_\omega(C)=21.

Прямая CKCK пересекает окружность в точках MM и KK, причём порядок C−M−KC-M-K. Тогда

CM⋅CK=21.CM\cdot CK = 21.

Но MK=4MK=4, значит CK=CM+4CK=CM+4. Пусть CM=xCM=x. Тогда

x(x+4)=21  ⇒  x2+4x−21=0  ⇒  x=3.x(x+4)=21 \;\Rightarrow\; x^2+4x-21=0 \;\Rightarrow\; x=3.

Следовательно,

CK=3+4=7.CK=3+4=7.

2) Выразим CKCK через aa

По свойству касательных на стороне ABAB:

AK=s−a=21.AK=s-a=\sqrt{21}.

Поставим координаты: A(−21,0)A(-\sqrt{21},0), C(21,0)C(\sqrt{21},0), B(0,h)B(0,h). Тогда

AB2=21+h2=a2  ⇒  h2=a2−21.AB^2 = 21+h^2 = a^2 \;\Rightarrow\; h^2=a^2-21.

Точка KK лежит на ABAB и AK=21AK=\sqrt{21}, значит доля отрезка

AKAB=21a.\frac{AK}{AB}=\frac{\sqrt{21}}a.

Отсюда координаты KK:

K=A+21a(B−A).K=A+\frac{\sqrt{21}}a(B-A).

После вычисления получается

CK2=21(5−421a).CK^2 = 21\Bigl(5-\frac{4\sqrt{21}}a\Bigr).

Но CK=7CK=7, значит

49=21(5−421a)  ⇒  4921=73=5−421a  ⇒  421a=83  ⇒  a=3212.49=21\Bigl(5-\frac{4\sqrt{21}}a\Bigr) \;\Rightarrow\; \frac{49}{21}=\frac73=5-\frac{4\sqrt{21}}a \;\Rightarrow\; \frac{4\sqrt{21}}a=\frac83 \;\Rightarrow\; a=\frac{3\sqrt{21}}2.

Требуется a2a^2:

a2=(3212)2=9⋅214=1894.a^2=\left(\frac{3\sqrt{21}}2\right)^2=\frac{9\cdot 21}{4}=\frac{189}{4}.

Ответ: 1894\displaystyle \frac{189}{4}.

سؤال

Среди 15 человек некоторые − друзья. Но любые два человека, имеющие среди других одинаковое число друзей, между собой не дружат. Найдите максимальное возможное число разных пар друзей среди этих 15 человек.

Python

python
import pkgutil, math, itertools, collections, random, time, functools, statistics, sys
any(m.name=='networkx' for m in pkgutil.iter_modules())


True

Python

python
any(m.name=='pulp' for m in pkgutil.iter_modules()), any(m.name=='ortools' for m in pkgutil.iter_modules())


(False, False)

Python

python
any(m.name=='cvxpy' for m in pkgutil.iter_modules())


False

Python

python
any(m.name=='pyscipopt' for m in pkgutil.iter_modules()), any(m.name=='mip' for m in pkgutil.iter_modules())


(False, False)

Python

python
import itertools, math
def max_edges_complete_multipartite(n=15):
    best=None
    best_part=None
    # partitions into distinct positive ints, order descending
    # brute force subsets of 1..n
    for r in range(2,n+1):
        for parts in itertools.combinations(range(1,n+1), r):
            if sum(parts)==n:
                # distinct already
                # edges
                s=sum(parts)
                edges=(s*s - sum(p*p for p in parts))//2
                if best is None or edges>best:
                    best=edges
                    best_part=parts
    return best,best_part
max_edges_complete_multipartite()


(85, (1, 2, 3, 4, 5))

Python

python
any(m.name in ('pysat','python_sat','pysat.solvers') for m in pkgutil.iter_modules())


False

Python

python
any(m.name=='z3' or m.name=='z3solver' or m.name=='z3-solver' for m in pkgutil.iter_modules())


False

Python

python
from functools import lru_cache
def partitions(n, max_part=None):
    if max_part is None or max_part>n: max_part=n
    if n==0: 
        yield []
        return
    for first in range(min(max_part,n),1-1,-1):
        for rest in partitions(n-first, first):
            yield [first]+rest

# generate all integer partitions of 15
parts_list=list(partitions(15))
len(parts_list), parts_list[:5], parts_list[-1]


(176,
 [[15], [14, 1], [13, 2], [13, 1, 1], [12, 3]],
 [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1])

Python

python
N=15
def max_sum_for_partition(sizes):
    sizes=tuple(sizes)
    L=[N - s for s in sizes]
    # order by maybe size descending; but assignment best do recursion on parts
    # use DP on index and used degrees bitmask (15 bits for 0..14)
    @lru_cache(None)
    def dp(i, used_mask):
        if i==len(sizes):
            return 0
        best=-1
        for d in range(L[i], -1, -1):  # high to low
            if used_mask>>d &1: 
                continue
            val = sizes[i]*d + dp(i+1, used_mask | (1<<d))
            if val>best: best=val
        return best
    return dp(0,0)

best=0
best_parts=None
best_sum=None
for p in parts_list:
    if len(p)==1: continue
    p_sorted=sorted(p, reverse=True)
    ms=max_sum_for_partition(p_sorted)
    if ms>best:
        best=ms
        best_parts=p_sorted
        best_sum=ms
best, best_parts, best_sum/2


(170, [5, 4, 3, 2, 1], 85.0)

الفكرة

Confirming upper bound and degrees assignment

Our upper bound of 170 is reached with partition sizes 5, 4, 3, 2, and 1, with corresponding degrees of 10, 11, 12, 13, and 14. This suggests a maximum of 85 edges. Since this configuration achieves the upper bound through a complete multipartite graph, the maximum number of edges is 85.

Clarifying degree constraints and confirming upper bound

We allowed distinct degrees, but didn't impose that sizes correspond to degrees. The constraint d_i ≤ N - s_i holds for all graphs, which gives us the upper bound sum of degrees as 170. Since degrees are distinct across classes, the inequality d_i ≤ N - s_i properly bounds the sum.

Formulating proof for maximum sum of degrees

If we sort classes by decreasing degree, our dynamic programming computation showed that the maximum sum of degrees is 170, so edges will be ≤ 85. To prove this, we can analytically show that for distinct class sizes, taking each degree as 15 - s_i maximizes the sum, leading to the result 225 - Σ s_i^2.

Maximizing sum degrees and proving optimal partition

If sizes aren't distinct, classes would have equal upper bounds which means we'd need to reduce the degree for some class, lowering the sum. To maximize the degree sum, we choose distinct sizes. By minimizing the sum of squares (1, 2, 3, 4, 5), we find the maximum sum and edges to be 170 and 85, respectively. Now, I need to prove no other partition can achieve a higher sum.

Maximizing sum with distinct sizes and degrees

I’m looking at how classes are paired with missing neighbors within the class. For a class of size s, we want to minimize missing degrees. By ordering classes by increasing size, we can match the smallest size with the largest degree, ensuring distinct degrees while maximizing the degree sum. This shows choosing distinct sizes and degrees is optimal since larger degrees will correspond to smaller sizes to satisfy constraints. Numerical adjustments or transformations come into play for ensuring optimality.

Maximizing sum of degrees with constraints

Since ti+si≤15t_i + s_i \leq 15, and degrees tit_i are distinct, we want to maximize the sum ∑siti\sum s_i t_i. The key is balancing the classes' sizes sis_i and degrees tit_i by making ti=15−sit_i = 15 - s_i where possible, ensuring distinct and feasible degree values. This leads to an optimal solution when the classes have distinct sizes summing to 15. Finally, the assumption is that k=5k=5 for five distinct degree classes.

Maximizing edges with constraints

For k=4k = 4, the maximum degree values are 11–14, but sizes must sum to 15. The degree constraints suggest that achieving degree 14 requires size 1. After testing, it looks like the optimum comes when k=5k = 5. We aim to minimize nonedges across classes and total nonedges, leading to an inequality showing that edges are maximized up to a bound of 85.

Maximizing degrees with distinct sizes

We need to list degree values in decreasing order and consider the constraints: ti+si≤15t_i + s_i \leq 15 with distinct tit_is. The sum Σtisi\Sigma t_i s_i is maximized when tit_i is as large as possible. To do this efficiently, we need distinct size values, as equal sizes would lower the degree sum. Hence, we aim for distinct sizes to ensure all constraints hold and maximum degrees are achieved.

Maximizing sum degrees with distinct sizes

The sum of tisit_i s_i is bounded by 225−Σsi2225 - \Sigma s_i^2, and we want to minimize the sum of squares of distinct sizes sis_i that add up to 15. The minimum occurs when si=1,2,3,4,5s_i = 1, 2, 3, 4, 5, giving a sum of squares of 55. This leads to a maximum sum degree of 170 and a total degree E≤85E \leq 85.

Maximizing degree sum with distinct sizes

To maximize the upper bound, I need to minimize Σsi2\Sigma s_i^2. If sizes aren't distinct, there's a penalty due to repeated sizes, lowering the potential degree sum. To avoid this penalty, we can use distinct sizes. With distinct sizes minimizing squares gives an upper bound of 85. If degree classes repeat, the sum degrees won’t exceed 170. This can be shown with a concrete case analysis of degree distributions. For instance, if k≥6k \geq 6, we get 14, 13, 12, etc. degrees, constraining the sum.

Maximizing edges in graph with degree constraints

To maximize the sum of degrees, we aim to take as many high degrees as possible, with limitations on the number of vertices per degree. With the distinctness condition, the distribution results in m14=1m_{14}=1, m13=2m_{13}=2, m12=3m_{12}=3, m11=4m_{11}=4, and m10=5m_{10}=5, totaling 170 degrees. Dividing by 2 gives the maximal number of edges: 85.

Thought for 6m 46s

الإجابة

Рассмотрим граф дружбы на 15 вершинах, степень вершины — число друзей.

Пусть ada_d — число людей, имеющих ровно dd друзей. По условию любые два человека с одинаковым числом друзей не дружат, значит внутри множества из ada_d вершин степени dd рёбер нет. Тогда человек со степенью dd может дружить только с людьми вне этого множества, то есть максимум с 15−ad15-a_d людьми. Поэтому

d≤15−ad⟹ad≤15−d.d \le 15-a_d \quad\Longrightarrow\quad a_d \le 15-d.

Отсюда:

a14≤1, a13≤2, a12≤3, a11≤4, a10≤5,…a_{14}\le 1,\ a_{13}\le 2,\ a_{12}\le 3,\ a_{11}\le 4,\ a_{10}\le 5,\ldots

Хотим максимизировать число рёбер E=12∑d adE=\frac12\sum d\,a_d, то есть сумму степеней. Заметим, что

1+2+3+4+5=15,1+2+3+4+5=15,

то есть можно «заполнить» всех 15 людей степенями 14,13,12,11,1014,13,12,11,10 в максимально возможных количествах:

a14=1, a13=2, a12=3, a11=4, a10=5.a_{14}=1,\ a_{13}=2,\ a_{12}=3,\ a_{11}=4,\ a_{10}=5.

Любое появление степени ≤9\le 9 означает, что кого-то пришлось “снять” с степени ≥10\ge 10, а это уменьшит сумму степеней, значит максимум достигается именно так.

Тогда

∑d ad=14⋅1+13⋅2+12⋅3+11⋅4+10⋅5=14+26+36+44+50=170,\sum d\,a_d=14\cdot 1+13\cdot 2+12\cdot 3+11\cdot 4+10\cdot 5 =14+26+36+44+50=170, Emax⁡=1702=85.E_{\max}=\frac{170}{2}=85.

Пример, где это достигается: разбить людей на группы размеров 1,2,3,4,51,2,3,4,5, дружить всем между разными группами и не дружить внутри групп (полный 5-дольный граф). Тогда степени будут 14,13,12,11,1014,13,12,11,10 с нужными кратностями.

Ответ: 85.

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