terça-feira, 28 de abril de 2020

Geometria – Quadrilátero e hexágono


Na figura acima, o hexágono e ambos os triângulos são regulares. Qual é a razão de áreas entre o quadrilátero rosa formado e o hexágono regular?

(In the figure above, the hexagon and both triangles are regular. What is the ratio of areas between the pink quadrilateral formed and the regular hexagon?)

PCFilho
(do Twitter de Ed Southall)

3 comentários:

  1. Labeling some of the points:
    A, B, C, D, E, F: vertices of the hexagon, starting with the upper left and going clockwise
    G: third vertex of the small equilateral triangle (the other two are points E and F)
    H: third vertex of the large equilateral triangle (the other two are points C and G)
    I: point of intersection of segments AB and CH

    Variable definitions:
    x = length of segment CG
    y = length of segment CI
    z = length of segment AI
    s = semiperimeter of quadrilateral AICG (the pink quadrilateral)

    We will take the side length of the hexagon to be 1 to simplify the math.

    Given: ABCDEF is a regular hexagon, △EFG and △CGH are equilateral
    Find: ratio of the areas of quadrilateral AICG and hexagon ABCDEF

    The first observation is that both △EFG and hexagon ABCDEF have side length 1. Therefore, AG (the length of one of quadrilateral AICG’s sides) is AF + FG = 1 + 1 = 2. Also, DG = DE + EG = 1 + 1 = 2. Shifting our focus to △CDG, we see that CD = 1 (since that segment is one of the hexagon’s sides). Now we know the lengths of two sides of △CDG—segments CD and DG. The angle between these sides (∠CDG) is an interior angle of the hexagon, so its measure is 2π/3 radians. So, by the Law of Cosines:

    x² = 1² + 2² – 2(1)(2) cos (2π/3)
    x² = 1 + 4 – 2(-1/2)
    x² = 7
    x = √7

    Now we look at △IBC. We see that ∠IBC is another interior angle of the hexagon. Since the hexagon is regular, ∠IBC is congruent to ∠CDG and its measure is also 2π/3 radians. The hexagon interior angle between these two angles (∠BCD) is divided into ∠BCI, ∠ICG, and ∠DCG. ∠BCD is an interior angle of the hexagon, so its measure is 2π/3 radians, and ∠ICG is an interior angle of the large equilateral triangle, so its measure is π/3 radians. This means that m∠BCI + m∠DCG = 2π/3 – π/3 = π/3. Also, since ∠DCG and ∠DGC are two angles of a triangle whose third angle measures 2π/3 radians, m∠DCG + m∠DGC = π – 2π/3 = π/3. This means that m∠BCI = m∠DGC; that is, ∠BCI and ∠DGC are congruent. A similar argument can be used to show that ∠BIC and ∠DCG are congruent. Since all three pairs of corresponding angles of △IBC and △CDG are congruent, △IBC ~ △CDG. The segment on △IBC that corresponds to the segment DG is BC, and DG = 2(BC), so that also means that x = 2y. Since x = √7, that means that y = √7/2. Also, since segments IB and CD correspond, 1 = 2z; that is, z = 1/2. Now we know all four side lengths of quadrilateral AICG, so we can compute s:

    s = (2 + x + y + z)/2 = (2 + 1/2 + √7/2 + √7)/2 = 5/4 + (3/4)√7

    We know that m∠ICG = π/3. The angle that is opposite this angle in quadrilateral AICG is ∠GAI, which is an interior angle of the hexagon, so m∠GAI = 2π/3. Since opposite angles of the quadrilateral sum to π radians, that means the quadrilateral is cyclic; in other words, it can be inscribed in a circle. There is a neat little formula for computing the area of such a quadrilateral. It’s known as Brahmagupta’s formula and it looks like this:

    Area = √[(s – a)(s – b)(s – c)(s – d)]

    where a,b,c,d are the side lengths of a cyclic quadrilateral and s is its semiperimeter. (You’ll notice that this formula looks very much like Heron’s formula. In fact, Heron’s formula is a special case of Brahmagupta’s formula where d = 0.) Using this formula with quadrilateral AICG, we obtain the following:

    Area of AICG = √[(3/4 + (3/4)√7)(-3/4 + (3/4)√7)(5/4 + (1/4)√7)(5/4 – (1/4)√7)]
    = √[(63/16 – 9/16)(25/16 – 7/16)]
    = √[(27/8)(9/8)]
    = √(243/64)
    = (9/8)√3

    The area of the hexagon is 6 times the area of △EFG; that is:

    Area of ABCDEF = 6(√3/4) = (3/2)√3

    Therefore, (area of AICG)/(area of ABCDEF) = [(9/8)√3]/[(3/2)√3] = 3/4; that is, the ratio of the area of the pink quadrilateral to the area of the hexagon is 3:4.

    My answer → 3:4

    ResponderExcluir
    Respostas
    1. Beautiful demonstration, Jake!! Brahmagupta's formula is amazing!

      Excluir
    2. It sure is.

      BTW, my previous post has a small math error. The Law of Cosines calculation should have 4(-1/2) where 2(-1/2) appears. But, the rest of the calculation is correct.

      Excluir

Regras para postar comentários:

I. Os comentários devem se ater ao assunto do post, preferencialmente. Pense duas vezes antes de publicar um comentário fora do contexto.

II. Os comentários devem ser relevantes, isto é, devem acrescentar informação útil ao post ou ao debate em questão.

III. Os comentários devem ser sempre respeitosos. É terminantemente proibido debochar, ofender, insultar e/ou caluniar quaisquer pessoas e instituições.

IV. Os nomes dos clubes devem ser escritos sempre da maneira correta. Não serão tolerados apelidos pejorativos para as instituições, sejam quais forem.

V. Não é permitido pedir ou publicar números de telefone/Whatsapp, e-mails, redes sociais, etc.

VI. Respeitem a nossa bela Língua Portuguesa, e evitem escrever em CAIXA ALTA.

Os comentários que não respeitem as regras acima poderão ser excluídos ou não, a critério dos moderadores do blog.