Na figura acima, os dois hexágonos são regulares. Qual é o comprimento do segmento rosa?
(In the figure above, both hexagons are regular. How long is the pink segment?)
PCFilho
(mais um problema pescado no Twitter de Catriona Shearer)
domingo, 29 de março de 2020
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That is impossible to determine with the given information. How much bigger is the light blue hexagon than the dark blue hexagon?
ResponderExcluirIn fact, it does not matter. Whatever the proportion between the hexagons is, the pink line is always the same length (provided that the other line measures 6).
ExcluirI think I have the solution.
ExcluirFirst, there is a nifty little formula that will figure in solving this problem. It's called the Law of Cosines and it goes a little something like this: if a, b, and c are the side lengths of a triangle, and θ is the measure of the angle opposite the side with length c, then c² = a² + b² - 2ab cos θ.
Now that we've gotten that out of the way, let's get to solving.
We start with some variable definitions:
a = side length of larger hexagon
b = side length of smaller hexagon
θ = measure of angle opposite the 6-unit-long segment
p = width of larger hexagon
q = width of smaller hexagon
x = length of pink segment
Let's label a couple points, too:
P,Q = left and right endpoints, respectively, of the pink segment
We focus first on the triangle at the top of the figure, which is formed by the 6-unit-long segment and one side from each hexagon. Observe that the angle opposite the 6-unit-long segment has a measure of 120°; that is, θ = 120°. By the Law of Cosines, we have:
6² = a² + b² - 2ab cos 120°
36 = a² + b² - 2ab(-1/2)
a² + ab + b² = 36 (Equation 1)
(I labeled this equation because we'll use it later.)
Now we turn our attention to the bottom of the figure. We locate a point on the rightmost side of the larger hexagon where a line drawn perpendicular to that segment goes through P. Let's call this new point R. Therefore, triangle PQR is a right triangle. Some additional variable definitions are in order here:
y = PR = p + q
z = QR = a - b
So, by the Pythagorean Theorem (which is a special case of the Law of Cosines, BTW), x² = y² + z².
All right, we already have z defined in terms of a and b, so let's define y the same way. Observe that a short diagonal of the large hexagon--whose length is equal to the width of the hexagon--forms a triangle with two of the sides. So, by the Law of Cosines:
p² = a² + a² - 2(a)(a) cos 120°
p² = 3a²
p = √3a
Using similar reasoning with the smaller hexagon reveals that q = √3b. This means that y = √3(a + b). Therefore:
x² = [√3(a + b)]² + (a - b)²
x² = 3(a + b)² + (a - b)²
x² = 3(a² + 2ab + b²) + a² - 2ab + b²
x² = 4a² + 4ab + 4b²
x² = 4(a² + ab + b²)
Using Equation 1:
x² = 4(36)
x² = 144
x = 12
My answer: The pink segment is 12 units long.
Excellent work, Jake. What a beautiful explanation!
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