## sábado, 16 de janeiro de 2016

### Um problema de geometria

(If the square's side measures x, what is the area painted in yellow?)

PCFilho

#### 3 comentários:

1. Before I start, I'm assuming that the green curves are circular arcs? Assuming that, some definitions are in order:

y=area of yellow region
O,P,Q,R=vertices of square, starting with the lower-left vertex and proceeding counterclockwise around the square
A,B,C,D=four inner points, starting with the top one and proceeding clockwise

We then place the square on a coordinate grid so that O is at the origin. That means that:

Coordinates of P=(x,0)
Coordinates of Q=(x,x)
Coordinates of R=(0,x)

Let's start by drawing the line segments OA, OB, PA, AB, BC, CD, and DA.

Since OA, OB, and PA are radii of circles with radius x, we have |OA|=|OB|=|PA|=x. Also, since segment OP is a side of a square with side length x, |OP|=x. We now have |OA|=|PA|=|OP|, so triangle OPA is equilateral. Therefore angle POA has a measure of 60° and the coordinates of A are (x cos 60°,x sin 60°). Since B is the reflection of A in the line y=x, this means that the coordinates of B are (x sin 60°,x cos 60°), or (x cos 30°,x sin 30°).

|AB|=2 sin [|30°-60°|/2]=2x^2 sin 15°=[(√6-√2)/2]x

Since ABCD is a square by the symmetry of the figure, we can now calculate the area of ABCD:

Area of ABCD=([(√6-√2)/2]x)^2=(2-√3)x^2

We now need to find the area of the region bounded by segment AB and arc AB, which is the area of the circular sector AOB minus the area of triangle AOB. Since the coordinates of A are (x cos 60°,x sin 60°) and those of B are (x cos 30°,x sin 30°), it follows that the angular measure of arc AB is 60°-30°=30°. That's 1/12 of a complete circle, so the area of sector AOB is Pi*x^2/12. The measure of angle AOB is the same as the angular measure of arc AB, or 30°. Therefore, the area of triangle AOB is (1/2)*x*x*sin 30°=x^2/4. So, the area of the sector bounded by segment AB and arc AB is (Pi/12-1/4)x^2. Since the yellow region is the union of four of these regions and the square ABCD, its area is calculated thusly:

y=(2-√3)x^2+4(Pi/12-1/4)x^2=(2-√3+Pi/3-1)x^2=(Pi/3+1-√3)x^2

My answer: The area of the yellow region is (Pi/3+1-√3)x^2.

1. Perfect, Jake!! :)

2. For anyone who wants to see an illustrated solution, click here.

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