## terça-feira, 16 de julho de 2019

(Draw a square with side length 4 and a semicircle whose diameter coincides with the bottom side of the square, as shown in the figure. Draw a line from the top left corner until the right side of the square, tangent to the semicircle. What is the length of this line?)

PCFilho

#### 3 comentários:

1. I am going to use z to represent the length of the line segment, because I plan to use x in a different way.

We start by placing the square in a coordinate plane, but this time, instead of placing it so that the origin is the lower-left hand corner, we shall place it so that the origin is the midpoint of the bottom side. Therefore, the vertices are at (-2,0), (-2,4), (2,4), and (2,0). An equation for the semicircle is x^2+y^2=4, or, solving for y:

y=sqrt(4-x^2)

We know two things about the line segment:

1. It is tangent to the semicircle.
2. It passes through (-2,4).

Since the line is tangent to the semicircle, we need to differentiate y to find an expression that gives the slope of the line segment (we can do this because y is differentiable):

dy/dx = (1/2)[1/sqrt(4-x^2)]*(-2x) = -x/sqrt(4-x^2)

Let c be the x-coordinate of the point where the line segment intersects the semicircle. So the slope of the line is given by the following expression:

m = -c/sqrt(4-c^2)

Now we have an expression for the slope of the line segment and a point that is on the line, so we can now come up with an equation for the line:

y = [-c/sqrt(4-c^2)](x+2)+4

We now have two expressions for y, so we set their right sides equal to each other and let x = c:

sqrt(4-c^2) = [-c/sqrt(4-c^2)](c+2)+4

We now solve this equation for c:

4-c^2 = -c(c+2)+4*sqrt(4-c^2)
4-c^2 = -c^2-2c+4*sqrt(4-c^2)
2c+4 = 4*sqrt(4-c^2)
4c^2+16c+16 = 64-16c^2
20c^2+16c-48 = 0
5c^2+4c-12 = 0
c = [-4+sqrt(4^2-4(5)(-12))]/(2*5)
c = (-4+16)/10
c = 6/5

Therefore:

m = [(-6/5)/sqrt(4-(6/5)^2)] = -3/4

So the equation of the line is y = (-3/4)(x+2)+4 and the coordinates of the line segment's right endpoint are (2,1). Therefore, by the distance formula:

z = sqrt[(2-(-2))^2+(1-4)^2] = sqrt(25) = 5.

I'm sure there's a purely geometric solution, but it seemed to me that it would be easier to use analytic geometry and calculus to solve this.

1. Here's another way to solve for c (this way doesn't introduce an extraneous solution):

sqrt(4-c^2) = -c*sqrt[(2+c)/(2-c)]+4
(2-c)sqrt(2+c) = -c*sqrt(2+c)+4*sqrt(2-c)
2*sqrt(2+c) = 4*sqrt(2-c)
4(2+c) = 16(2-c)
8+4c = 32-16c
20c = 24
c = 6/5

2. Well done, Jake. Awesome work!

My geometric solution uses the fact that the distances of an external point to the circle through the tangent lines are the same. (It can be proved with triangle similarity.)

Let y be the distance along the right side of the square, from the bottom to the intersection with the blue line.

x in the figure can be split in two smaller segments. From the top left corner to the point of tangency, it measures 4. From the point of tangency on, it measures y (this comes from the mentioned fact).

So, x = 4 + y. Applying Pythagoras' Theorem to the right triangle in the upper right portion of the square, we have:

(4 + y)² = 4² + (4 - y)²
16 + 8y + y² = 16 + 16 - 8y + y²
8y = 16 - 8y
y = 1

And then x = 4 + 1 = 5

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