Geometry

[MathStudent1999]

An infinite plane is completely filled with congruent circles in the pattern shown. What percent of the plane is filled with circles?


I put the circles' radius as one. I found a circle's area would be pi. The area of the shape between the circles is (pi/2) - √3.
I found that for the first three circles there would be a shape between the circles, but after the first three for every new circle there would be another shape. But since the plane is infinite to i just disregard the first part and just do the area of a circle over the area of the circle plus the shape?

 

[pappus]

An infinite plane is completely filled with congruent circles in the pattern shown. What percent of the plane is filled with circles?

View attachment 2676
I put the circles' radius as one. I found a circle's area would be pi. The area of the shape between the circles is (pi/2) - √3.
I found that for the first three circles there would be a shape between the circles, but after the first three for every new circle there would be another shape. But since the plane is infinite to i just disregard the first part and just do the area of a circle over the area of the circle plus the shape?

I've modified your sketch a little bit:

1. The white triangle is equilateral with side-length 2r.

2. The area of the white triangle contains half of a circle and the white gap.

3. The area of the white gap is the difference of the area of an equilateral triangle and a half circle.

 

[MathStudent1999]

I already got the area of the white gap: (pi/2) - √3.

But then I found that for the first three circles there would be one white gap, but after the first three for every new circle there would be another white gap. Since the plane is infinite to i just disregard the first part and just do the area of a circle over the area of the circle plus the white gap?

 

[pappus]

I already got the area of the white gap: (pi/2) - √3.

But then I found that for the first three circles there would be one white gap, but after the first three for every new circle there would be another white gap. Since the plane is infinite to i just disregard the first part and just do the area of a circle over the area of the circle plus the white gap?

Use the equilateral triangles as tiles to cover the plane: Each triangle contains exactly one white gap.

So the ratio of $\displaystyle \displaystyle{\frac{white}{blue} = \frac{\sqrt{3} - \frac12 \cdot \pi}{\frac12 \cdot \pi}=\frac{2 \sqrt{3} - \pi}{\pi} \approx 0.1026...}$

 

[lookagain]

I already got the area of the white gap: (pi/2) - √3.

But then I found that for the first three circles there would be one white gap, but after the first three
for every new circle there would be another white gap. Since the plane is infinite to i just disregard
the first part and just do the area of a circle over the area of the circle plus the white gap?

Look at that white triangle that was drawn.

We need the ratio of the three parts of the circle (which are shaded here) to the area of the whole triangle.

The white triangle is an equilateral triangle, and its

area formula is $\displaystyle \dfrac{s^2\sqrt{3}}{4}.$

Each side of that triangle is the sum of two radii of each circle.

Each radius =1, so each side of the triangle = 2.

With that formula, the area of the triangle is $\displaystyle \sqrt{3.}$

Use the equilateral triangles as tiles to cover the plane: Each triangle contains exactly one white gap.

So the ratio of $\displaystyle \displaystyle{\frac{white}{blue} = \frac{\sqrt{3} - \frac12 \cdot \pi}{\frac12 \cdot \pi}=\frac{2 \sqrt{3} - \pi}{\pi} \approx 0.1026...}$

The result in the quote box is not the answer (yet) to the OP's question.

That question asks what percent do the areas of the circles have to the plane.




Percent of the areas of the circles to the whole plane $\displaystyle \ = \ \ \dfrac{Area \ \ of \ \ the \ \ triangle \ - \ area \ \ of \ \ the \ \ white \ \ region}{Area \ \ of \ \ the \ \ triangle}$



$\displaystyle \dfrac{\sqrt{3}}{\sqrt{3} - \bigg(\dfrac{\pi}{2} - \sqrt{3}\bigg)} \ =$



$\displaystyle \dfrac{\sqrt{3}}{2\sqrt{3} - \dfrac{\pi}{2}} \ =$


$\displaystyle \dfrac{2\sqrt{3}}{4\sqrt{3} - \pi} \ \approx \ 91.5 \%$



So, we have calculated $\displaystyle \ \dfrac{area \ \ of \ \ blue}{area \ \ of \ \ blue \ \ + \ \ area \ \ of \ \ white}$Edit: I used your value for the area of the white region. But it is actually the negative of that.

 

[MathStudent1999]

I got something different. Using the triangles to tile the plane, I got about 90.6%

The area of the triangle is √3. The the radius of the circle is pi. Since there is 1/2 of a circle inside the triangle, wouldn't the answer be (pi/2)/(√3)?

 

[lookagain]

I got something different. Using the triangles to tile the plane, I got about > 90.6% <

The area of the triangle is √3. The the radius of the circle is pi. Since there is 1/2 of a circle inside the triangle, wouldn't the answer be (pi/2)/(√3)?

MathStudent1999, when the correct area of the white area is used (which is negative one multiplied by what you gave in your first post), then you are correct.
$\displaystyle (\dfrac{\pi}{2})/(\sqrt{3}) \ = \dfrac{\pi}{2 \sqrt{3}} \ \approx \ \ 90.7\%$