Title: Openai/69679cde-35e4-8007-ac39-bb5264fd84d2 - freem

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## gpt-5-2[[edit](https://freemwiki.com/index.php?title=Openai/69679cde-35e4-8007-ac39-bb5264fd84d2&veaction=edit&section=1 "Edit section: gpt-5-2") | [edit source](https://freemwiki.com/index.php?title=Openai/69679cde-35e4-8007-ac39-bb5264fd84d2&action=edit&section=1 "Edit section's source code: gpt-5-2")]

### User: Por favor, resolva e mostre a resolução do seguinte problema de geometria:[[edit](https://freemwiki.com/index.php?title=Openai/69679cde-35e4-8007-ac39-bb5264fd84d2&veaction=edit&section=2 "Edit section: User: Por favor, resolva e mostre a resolução do seguinte problema de geometria:") | [edit source](https://freemwiki.com/index.php?title=Openai/69679cde-35e4-8007-ac39-bb5264fd84d2&action=edit&section=2 "Edit section's source code: User: Por favor, resolva e mostre a resolução do seguinte problema de geometria:")]

Por favor, resolva e mostre a resolução do seguinte problema de geometria:

Um aluno apaixonado por geometria plana, desenha com régua e compasso um triângulo ABC. Ele notou que os três lados do triângulo ABC são números naturais consecutivos e que o maior ângulo era igual ao dobro do menor. Se os lados do triângulo ABC são medidos em centímetros, é CORRETO afirmar que a área do triângulo ABC é igual a?

### Assistant: Vamos resolver passo a passo.[[edit](https://freemwiki.com/index.php?title=Openai/69679cde-35e4-8007-ac39-bb5264fd84d2&veaction=edit&section=3 "Edit section: Assistant: Vamos resolver passo a passo.") | [edit source](https://freemwiki.com/index.php?title=Openai/69679cde-35e4-8007-ac39-bb5264fd84d2&action=edit&section=3 "Edit section's source code: Assistant: Vamos resolver passo a passo.")]

Vamos resolver passo a passo.

#### 1. Interpretação do problema[[edit](https://freemwiki.com/index.php?title=Openai/69679cde-35e4-8007-ac39-bb5264fd84d2&veaction=edit&section=4 "Edit section: 1. Interpretação do problema") | [edit source](https://freemwiki.com/index.php?title=Openai/69679cde-35e4-8007-ac39-bb5264fd84d2&action=edit&section=4 "Edit section's source code: 1. Interpretação do problema")]

Sejam os lados do triângulo ABCABCABC três números naturais consecutivos. Denotemos:

a=n−1,b=n,c=n+1a = n-1,\quad b = n,\quad c = n+1a=n−1,b=n,c=n+1 onde ccc é o maior lado. Sejam os ângulos opostos a esses lados, respectivamente:

α≤β≤γ\alpha \le \beta \le \gammaα≤β≤γ O enunciado informa que o maior ângulo é o dobro do menor, isto é,

γ=2α\gamma = 2\alphaγ=2α

#### 2. Relação entre os ângulos[[edit](https://freemwiki.com/index.php?title=Openai/69679cde-35e4-8007-ac39-bb5264fd84d2&veaction=edit&section=5 "Edit section: 2. Relação entre os ângulos") | [edit source](https://freemwiki.com/index.php?title=Openai/69679cde-35e4-8007-ac39-bb5264fd84d2&action=edit&section=5 "Edit section's source code: 2. Relação entre os ângulos")]

A soma dos ângulos internos de um triângulo é:

α+β+γ=π\alpha + \beta + \gamma = \piα+β+γ=π Substituindo γ=2α\gamma = 2\alphaγ=2α:

α+β+2α=π⇒β=π−3α\alpha + \beta + 2\alpha = \pi \quad\Rightarrow\quad \beta = \pi - 3\alphaα+β+2α=π⇒β=π−3α

#### 3. Aplicação da Lei dos Senos[[edit](https://freemwiki.com/index.php?title=Openai/69679cde-35e4-8007-ac39-bb5264fd84d2&veaction=edit&section=6 "Edit section: 3. Aplicação da Lei dos Senos") | [edit source](https://freemwiki.com/index.php?title=Openai/69679cde-35e4-8007-ac39-bb5264fd84d2&action=edit&section=6 "Edit section's source code: 3. Aplicação da Lei dos Senos")]

Pela Lei dos Senos:

asin⁡α=bsin⁡β=csin⁡γ\frac{a}{\sin\alpha} = \frac{b}{\sin\beta} = \frac{c}{\sin\gamma}sinαa​=sinβb​=sinγc​ Substituindo os valores:

n−1sin⁡α=nsin⁡(3α)=n+1sin⁡(2α)\frac{n-1}{\sin\alpha} = \frac{n}{\sin(3\alpha)} = \frac{n+1}{\sin(2\alpha)}sinαn−1​=sin(3α)n​=sin(2α)n+1​

##### 3.1 Usando[[edit](https://freemwiki.com/index.php?title=Openai/69679cde-35e4-8007-ac39-bb5264fd84d2&veaction=edit&section=7 "Edit section: 3.1 Usando") | [edit source](https://freemwiki.com/index.php?title=Openai/69679cde-35e4-8007-ac39-bb5264fd84d2&action=edit&section=7 "Edit section's source code: 3.1 Usando")]

sin ⁡ ( 2 𝛼 ) = 2 sin ⁡ 𝛼 cos ⁡ 𝛼 sin(2α)=2sinαcosα

Comparando o primeiro e o terceiro termos:

n−1sin⁡α=n+12sin⁡αcos⁡α\frac{n-1}{\sin\alpha} = \frac{n+1}{2\sin\alpha\cos\alpha}sinαn−1​=2sinαcosαn+1​ Cancelando sin⁡α\sin\alphasinα:

2(n−1)cos⁡α=n+12(n-1)\cos\alpha = n+12(n−1)cosα=n+1 ⇒cos⁡α=n+12(n−1)(1)\Rightarrow\quad \cos\alpha = \frac{n+1}{2(n-1)} \tag{1}⇒cosα=2(n−1)n+1​(1)

##### 3.2 Usando[[edit](https://freemwiki.com/index.php?title=Openai/69679cde-35e4-8007-ac39-bb5264fd84d2&veaction=edit&section=8 "Edit section: 3.2 Usando") | [edit source](https://freemwiki.com/index.php?title=Openai/69679cde-35e4-8007-ac39-bb5264fd84d2&action=edit&section=8 "Edit section's source code: 3.2 Usando")]

sin ⁡ ( 3 𝛼 ) = 3 sin ⁡ 𝛼 − 4 sin ⁡ 3 𝛼 sin(3α)=3sinα−4sin 3 α

Comparando o primeiro e o segundo termos:

n−1sin⁡α=n3sin⁡α−4sin⁡3α\frac{n-1}{\sin\alpha} = \frac{n}{3\sin\alpha - 4\sin^3\alpha}sinαn−1​=3sinα−4sin3αn​ Simplificando:

(n−1)(3−4sin⁡2α)=n(n-1)(3 - 4\sin^2\alpha) = n(n−1)(3−4sin2α)=n Como sin⁡2α=1−cos⁡2α\sin^2\alpha = 1 - \cos^2\alphasin2α=1−cos2α, temos:

3−4(1−cos⁡2α)=4cos⁡2α−13 - 4(1 - \cos^2\alpha) = 4\cos^2\alpha - 13−4(1−cos2α)=4cos2α−1 Logo:

(n−1)(4cos⁡2α−1)=n(2)(n-1)(4\cos^2\alpha - 1) = n \tag{2}(n−1)(4cos2α−1)=n(2)

#### 4. Determinação de[[edit](https://freemwiki.com/index.php?title=Openai/69679cde-35e4-8007-ac39-bb5264fd84d2&veaction=edit&section=9 "Edit section: 4. Determinação de") | [edit source](https://freemwiki.com/index.php?title=Openai/69679cde-35e4-8007-ac39-bb5264fd84d2&action=edit&section=9 "Edit section's source code: 4. Determinação de")]

𝑛 n

Substituindo (1)(1)(1) em (2)(2)(2):

cos⁡2α=(n+1)24(n−1)2\cos^2\alpha = \frac{(n+1)^2}{4(n-1)^2}cos2α=4(n−1)2(n+1)2​ 4cos⁡2α−1=(n+1)2−(n−1)2(n−1)2=4n(n−1)24\cos^2\alpha - 1 = \frac{(n+1)^2 - (n-1)^2}{(n-1)^2} = \frac{4n}{(n-1)^2}4cos2α−1=(n−1)2(n+1)2−(n−1)2​=(n−1)24n​ Então:

(n−1)⋅4n(n−1)2=n(n-1)\cdot \frac{4n}{(n-1)^2} = n(n−1)⋅(n−1)24n​=n ⇒4nn−1=n\Rightarrow\quad \frac{4n}{n-1} = n⇒n−14n​=n ⇒n(n−1)=4n⇒n=5\Rightarrow\quad n(n-1) = 4n \quad\Rightarrow\quad n = 5⇒n(n−1)=4n⇒n=5 Portanto, os lados do triângulo são:

4, 5, 6\boxed{4,\;5,\;6}4,5,6​

#### 5. Cálculo da área (Fórmula de Heron)[[edit](https://freemwiki.com/index.php?title=Openai/69679cde-35e4-8007-ac39-bb5264fd84d2&veaction=edit&section=10 "Edit section: 5. Cálculo da área (Fórmula de Heron)") | [edit source](https://freemwiki.com/index.php?title=Openai/69679cde-35e4-8007-ac39-bb5264fd84d2&action=edit&section=10 "Edit section's source code: 5. Cálculo da área (Fórmula de Heron)")]

O semiperímetro é:

s=4+5+62=152s = \frac{4+5+6}{2} = \frac{15}{2}s=24+5+6​=215​ A área é:

A=s(s−a)(s−b)(s−c)A = \sqrt{s(s-a)(s-b)(s-c)}A=s(s−a)(s−b)(s−c)​ A=152⋅72⋅52⋅32=157516A = \sqrt{\frac{15}{2}\cdot\frac{7}{2}\cdot\frac{5}{2}\cdot\frac{3}{2}} = \sqrt{\frac{1575}{16}}A=215​⋅27​⋅25​⋅23​​=161575​​ Como 1575=225⋅71575 = 225\cdot 71575=225⋅7:

A=1574A = \frac{15\sqrt{7}}{4}A=4157​​

#### 6. Resposta final[[edit](https://freemwiki.com/index.php?title=Openai/69679cde-35e4-8007-ac39-bb5264fd84d2&veaction=edit&section=11 "Edit section: 6. Resposta final") | [edit source](https://freemwiki.com/index.php?title=Openai/69679cde-35e4-8007-ac39-bb5264fd84d2&action=edit&section=11 "Edit section's source code: 6. Resposta final")]

É correto afirmar que a área do triângulo ABCABCABC é:

1574 cm2\boxed{\displaystyle \frac{15\sqrt{7}}{4}\ \text{cm}^2}4157​​cm2​

## References[[edit](https://freemwiki.com/index.php?title=Openai/69679cde-35e4-8007-ac39-bb5264fd84d2&veaction=edit&section=12 "Edit section: References") | [edit source](https://freemwiki.com/index.php?title=Openai/69679cde-35e4-8007-ac39-bb5264fd84d2&action=edit&section=12 "Edit section's source code: References")]

1.   [↑](https://freemwiki.com/index.php?oldid=595621&title=Openai%2F69679cde-35e4-8007-ac39-bb5264fd84d2#cite_ref-1 "Jump up")["Problema Triângulo Geometria"](https://chatgpt.com/share/69679cde-35e4-8007-ac39-bb5264fd84d2). ChatGPT. Retrieved 2026-01-15.