## domingo, 19 de abril de 2020

### Geometria – O eneágono

Na figura acima, o eneágono é regular e o quadrilátero compartilha quatro vértices com ele. Qual fração do eneágono está pintada de rosa?

(In the figure above, the eneagon is regular and the quadrilateral shares four vertices with it. What fraction of the eneagon is painted pink?)

PCFilho

#### 5 comentários:

1. Is it supposed to come out as a nice and neat fraction?

1. Yes, it is. :)

2. First, let’s take the length of the enneagon’s sides to be 1 to simplify the math. Then, let’s define some variables:

F = fraction of the enneagon’s area represented by the pink quadrilateral
A₁ = area of white triangle on the lower left
A₂ = area of the pink quadrilateral
A₃ = area of the enneagon
t = the second-longest side of the pink quadrilateral
u = the longest side of the pink quadrilateral

This problem asks us to find F, which can be expressed thus:

F = A₂/A₃ (Equation 1)

Let’s start by calculating A₁, the area of the white triangle. The white triangle’s an isosceles triangle whose congruent sides have length 1 (since they’re two of the enneagon’s sides). The angle between them is one of the enneagon’s interior angles, so its measure is π(9 – 2)/9 = 7π/9. Therefore:

A₁ = (1/2)(1)(1) sin (7π/9)
A₁ = (1/2) sin (2π/9)
A₁ = sin (π/9) cos (π/9) (Equation 2)

Now we calculate the length t:

t² = 1² + 1² – 2(1)(1) cos (7π/9)
t² = 2[1 – cos (7π/9)]
t² = 2[1 + cos (2π/9)]
t² = 4 cos² (π/9)
t = 2 cos (π/9)

To find u, we draw segments from the top vertex of the enneagon to the upper left and lower right vertices of the pink quadrilateral. This construction divides the pink quadrilateral into a smaller quadrilateral and an equilateral triangle with side length 1, as well as dividing the white hexagon into a triangle congruent to the white triangle whose area is A₁, a pentagon, and an equilateral triangle with side length t. But wait! It also divides the segment with length u into two segments that are each part of one of the two equilateral triangles just constructed. Therefore:

u = t + 1 = 2 cos (π/9) + 1

Neat, huh?

Next, since we’ve already drawn two diagonals from one of the enneagon’s vertices, let’s draw enough of them to make the diagram symmetrical, shall we? This divides the enneagon into five regions. Two are triangles with area A₁, two are quadrilaterals (one pink and one white) with area A₂, and the fifth is a triangle whose angles have measures 4π/9, 4π/9, and π/9 and whose sides have length u, u, and 1. So:

A₃ = 2 sin (π/9) cos (π/9) + 2A₂ + (1/2)[2 cos (π/9) + 1]² sin (π/9)
A₃ = 2 sin (π/9) cos (π/9) + 2A₂ + [2 cos² (π/9) + 2 cos (π/9) + 1/2] sin (π/9)
A₃ = 4 sin (π/9) cos (π/9) + 2A₂ + [2 cos² (π/9) + 1/2] sin (π/9)
A₃ = 4 sin (π/9) cos (π/9) + 2A₂ + 2 sin (π/9) cos² (π/9) + 1/2 sin (π/9)

Next, we draw a diagonal that divides the pink quadrilateral into two triangular regions. The smaller triangle has two sides with lengths 1 and t, and the angle between them has a measure of 2π/3. The larger triangle has two sides with lengths 1 and u, and the angle between them has a measure of π/3. We calculate A₂ as follows:

A₂ = (1/2)(1)(t) sin (2π/3) + (1/2)(1)(u) sin (π/3)
A₂ = (√3/4)(t + u)
A₂ = (√3/4)(t + t + 1)
A₂ = (√3/4)(2t + 1)
A₂ = (√3/4)[4 cos (π/9) + 1]
A₂ = √3 cos (π/9) + √3/4

Thus 2A₂ = 2√3 cos (π/9) + √3/2. So:

A₃ = 4 sin (π/9) cos (π/9) + 2√3 cos (π/9) + √3/2 + 2 sin (π/9) cos² (π/9) + (1/2) sin (π/9)
A₃ = 4 cos (7π/18) cos (π/9) + 2√3 cos (π/9) + √3/2 + 2 cos (7π/18) cos² (π/9) + (1/2) cos (7π/18)
A₃ = 2 cos (5π/18) + 2√3 cos (π/9) + √3/2 + cos (5π/18) cos (π/9) + (1/2) cos (7π/18)
A₃ = 2 cos (5π/18) + 2√3 cos (π/9) + √3/2 + (1/2)[cos (7π/18) + cos (π/6)] + (1/2) cos (7π/18)
A₃ = 2 cos (5π/18) + 2√3 cos (π/9) + 3√3/4 + cos (7π/18)
A₃ = 2 cos (π/6 + π/9) + 2√3 cos (π/9) + 3√3/4 + sin (π/9)
A₃ = √3 cos (π/9) – sin (π/9) + 2√3 cos (π/9) + 3√3/4 + sin (π/9)
A₃ = 3√3 cos (π/9) + 3√3/4

Therefore:

F = [√3 cos (π/9) + √3/4]/[3√3 cos (π/9) + 3√3/4] = 1/3

My answer: The pink quadrilateral’s area is 1/3 the area of the enneagon.

1. I worked on this all afternoon. The hardest part was getting the expression for A₃ to condense down nicely.

2. Amazing demonstration, Jake!! Thank you so much for this. Realizing that u = t + 1 was outstanding!

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