Os dois círculos pequenos têm raio 4. Qual é o raio do semicírculo grande?
PCFilho
(pescado no perfil @tsatie do Twitter)
sexta-feira, 4 de setembro de 2020
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Given info: The circle on the left is tangent to the large semicircle and the top side of the triangle, the circle on the right is tangent to all three sides of the triangle, and the radius of both circles is 4.
ResponderExcluirTask: Find the radius of the large semicircle.
Labeling some points:
O = center of circle
A = left endpoint of the triangle’s longest side
B = right endpoint of the triangle’s longest side
C = triangle’s third vertex
M = point of tangency between the circle on the left and the semicircle
N = point of tangency between the circle on the left and segment AC
Defining some variables:
r = radius of the semicircle (whose value we are tasked with finding)
θ = angle between segments AB and AC (note that 0 < θ < π/2)
Since segment AB is a diameter of a circle, and point C is a third point on that circle, this means that △ABC is a right triangle. So, we can come up with some expressions for the lengths of the triangle’s sides:
AB = 2r, AC = 2r cos θ, BC = 2r sin θ
Let’s now consider points M, N, and O. O is the midpoint of segment AB and M is the midpoint of segment AC, so segment MN is parallel to segment AB and its length is half that of segment AB. In other words, OM = r sin θ. Segment MN is a diameter of the circle on the left, so its length must be 8 (= 2*4). Segment ON is a radius of the semicircle, so its length is r. Therefore:
r = r sin θ + 8
r(1 – sin θ) = 8
r = 8/(1 – sin θ)
Now we consider the circle on the right. It is the incircle of △ABC, so the line segments from its center to points A, B, and C divide △ABC into three triangles, all of which have height equal to the incircle’s radius, which is 4. The bases of these triangles have lengths 2r, 2r cos θ, and 2r sin θ, so the area of the triangle can be given thusly:
A = 4(r + r cos θ + r sin θ) = 4r(1 + cos θ + sin θ)
But, since △ABC is a right triangle, its area can also be given thusly:
A = (1/2)(2r cos θ)(2r sin θ) = 2r² sin θ cos θ
Setting the right sides of these two equations equal to each other, we obtain:
2r² sin θ cos θ = 4r(1 + cos θ + sin θ)
r = 2(1 + cos θ + sin θ)/(sin θ cos θ)
Now we have two expressions that define r in terms of θ. We can now solve for θ:
8/(1 – sin θ) = 2(1 + cos θ + sin θ)/(sin θ cos θ)
8 sin θ cos θ = 2(1 + cos θ + sin θ)(1 – sin θ)
8 sin θ cos θ = 2 cos θ (1 – sin θ) + 2(1 + sin θ)(1 – sin θ)
8 sin θ cos θ = 2 cos θ – 2 cos θ sin θ + 2(1 – sin² θ)
10 sin θ cos θ = 2 cos θ + 2 cos² θ
10 sin θ = 2 + 2 cos θ (since we know that cos θ ≠ 0)
5 sin θ – cos θ = 1
(5/√26) sin θ – (1/√26) cos θ = 1/√26
sin [θ – arcsin (1/√26)] = 1/√26
θ – arcsin (1/√26) = arcsin (1/√26)
θ = 2 arcsin (1/√26) = arcsin [2(1/√26)(5/√26)] = arcsin (5/13)
We then can determine r:
r = 8/(1 – 5/13) = 8/(8/13) = 13
My answer: 13 units