## 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

#### 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

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.

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

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.

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