سؤال

Задача 1:каждый член последовательности натуральных чисел начиная с третьего равен сумме двух предыдущих. Восьмой член равен 3900. Сколько существует таких последовательностей? Чему равен второй член последовательности если первый равен 13?
2 Задание:
Дан вписанный четырёхугольник ABCD .оказалось что касательная в точке D к описанной окружности параллельна биссектрисе угла ABC. При этом угол ADB равен 12° и угол DBC равен 96° Найдите угол BCA ответ выразите в градусах?

الفكرة

Solving math problems

For the first problem, the sequence follows a Fibonacci-like recurrence. We need to find how many natural number solutions satisfy 8x + 13y = 3900. The second question asks to find y when x = 13. I’ll use the general formula a8 = 8x + 13y to solve both parts. Let's compute the solutions to the equations.

Solving the equation

We need to find solutions for 8x + 13y = 3900, where x and y are natural numbers. After simplifying, we see that x must be a multiple of 13. Setting x = 13k and solving for y, we find that y = 300 - 8k. The condition y ≥ 1 leads to k ≤ 37. So, k ranges from 1 to 37.

Verifying the sequences

The number of solutions from k = 1 to 37 gives us 37 valid sequences. x = 0 is excluded since natural numbers start at 1. For the second question, with x = 13, we find y = 292, which satisfies the condition. I confirm the sequence correctly computes to a8 = 3900. So, the answers are 37 sequences and a2 = 292.

Solving geometry problem

We have a cyclic quadrilateral ABCD, with the tangent at D being parallel to the angle bisector of ∠ABC. Given angle ADB = 12° and angle DBC = 96°, we need to find angle BCA. I'll focus on using the properties of cyclic quadrilaterals and angle relationships to find ∠BCA.

Applying tangent-chord theorem

At point D, the tangent line is parallel to the bisector of angle ABC. The tangent forms equal angles with the chords DA and DC. Using the tangent-chord theorem, we relate those angles to the corresponding inscribed angles. Since the tangent is parallel to the bisector, it creates equal angle relationships, helping us set up equations for angle BCA.

Setting up angle equations

Using the tangent-chord theorem, I express the angle between the tangent at D and the chords DA and DC with respect to ∠DBA and ∠DBC. Through parallelism, I derive relationships between angles with chords and the bisector, aiming to solve for ∠BCA by connecting angles in triangles ABD and DBC.

Verifying angle measures

I'm comparing various angles like ∠BCA and ∠ACB by considering how they intercept arcs AB and BA. Given that ∠BCA intercepts arc BA and the arc between B and A is 24°, I calculate ∠BCA as 12°. Additionally, I confirm my earlier work with Eq1, testing consistency across the angles and arcs.

Clarifying angles at C

Okay, I was mixing up ∠BCA and ∠ACB. Both have vertex C, but they are essentially the same angle despite the different notation (orientation doesn't change the measure). Now that I've straightened that out, both angles at C are 12°. This works out with the angle sum in triangle ABC, where the other angles are yet to be determined.

Verifying angle consistency

In triangle ABC, I finally reached that both ∠BCA and ∠BAC are 12°. This matches the situation where the arcs AB and BC are both 24°. Everything checks out—these angles are consistent with the geometry, and the triangle would indeed be isosceles with sides AB and BC equal.

Thought for 3m 0s

الإجابة

Задача 1

Пусть $a_1=x$ , $a_2=y$ . Тогда по рекуррентной формуле

a_3=x+y,\ a_4=x+2y,\ a_5=2x+3y,\ a_6=3x+5y,\ a_7=5x+8y,\ a_8=8x+13y.

Дано $a_8=3900$ , значит

8x+13y=3900.

Так как $3900\equiv 0\pmod{13}$ , получаем $8x\equiv0\pmod{13}\Rightarrow x\equiv0\pmod{13}$ . Пусть $x=13k$ ( $k\in\mathbb N$ ). Тогда

y=\frac{3900-8\cdot13k}{13}=300-8k.

Требуем $x,y\in\mathbb N\Rightarrow 300-8k\ge1\Rightarrow k\le37$ . Значит $k=1,2,\dots,37$ — 37 вариантов.

Если $a_1=13$ (то есть $k=1$ ), то

a_2=300-8\cdot1=\boxed{292}.

Ответ к Задаче 1: последовательностей 37, при $a_1=13$ имеем $a_2=\boxed{292}$ .

Задача 2

Дано: $ABCD$ — вписанный четырёхугольник, касательная в $D$ к окружности параллельна биссектрисе $\angle ABC$ . Из теоремы о касательной и секущей:

\angle(\text{касат.},\,DA)=\angle DBA,\qquad \angle(\text{касат.},\,DC)=\angle DBC.

Параллельность касательной биссектрисе даёт равенство углов с её «лучами»:

\angle BAD+\angle DBA=\angle DBC+\angle DCB. \tag{1}

В треугольнике $ABD$ : $\angle ADB=12^\circ\Rightarrow \angle BAD+\angle DBA=180^\circ-12^\circ=168^\circ$ . Подставляя в (1) и учитывая $\angle DBC=96^\circ$ , получаем

168^\circ=96^\circ+\angle DCB \Rightarrow \angle DCB=72^\circ.

Тогда в $\triangle DBC$ : $\angle BDC=180^\circ-96^\circ-72^\circ=12^\circ$ .

Отсюда дуга $BC$ равна $2\cdot12^\circ=24^\circ$ и дуга $AB$ тоже $24^\circ$ (по $\angle ADB=12^\circ$ ). Следовательно, вписанные углы, опирающиеся на эти дуги:

\angle BAC=\tfrac12\widehat{BC}=12^\circ,\qquad \angle BCA=\tfrac12\widehat{BA}=12^\circ.

Ответ к Задаче 2: $\boxed{12^\circ}$ .

Задача 1:каждый член последовательности натуральны...