Conrguent $\frac{5}{6}$ circles in a circle. What fraction is shaded?

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Solution: enter image description here

Let $r$ be the radius of the small circles and $R$ the radius of the big one.

The colored section is three times five sixths of the area of one of the small circles.

Colored Section area= $3\times\dfrac{5}{6}\times\pi r^2=\dfrac{5\pi r^2}{2}$

The radius $R$ of the big circle is equal to $r$ plus the radius of the circumscribed circle of equailateral triangle ABC, whose side is $2r$. The radius of the circumscribed circle of an equilateral triangle is the length of the sides divided by $\sqrt{3}$. Since the side here measures $2r$, the radius of the circumscribed circle is $\dfrac{2r}{\sqrt{3}}$.

So we have $R = r+\dfrac{2r}{\sqrt{3}}$

The area of the big circle is $\pi \times R^2$, which here is equal to $(r+\dfrac{2r}{\sqrt{3}})^2$

which, when expanded, gives

Big Circle area = $\dfrac{\pi r^2(7+4\sqrt{3})}{3}$

To obtain the shaded fraction, we need to divide the area of the colored region by the area of the big circle:

Shaded fraction = $\dfrac{\dfrac{5\pi r^2}{2}}{\dfrac{\pi r^2(7+4\sqrt{3})}{3}}$

Shaded fraction = $\dfrac{5\pi r^2}{2} \times \dfrac{3}{\pi r^2(7+4\sqrt{3})}$

Shaded fraction = $\dfrac{15}{2(7+4\sqrt{3})} \simeq 53.847 \% $

I think it's wrong. In the drawing the smaller circles are not tangent to the largest

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    $\begingroup$ Definitely the colored circles are not tangent to the large one $\endgroup$ – Vasya Dec 6 '19 at 14:30
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    $\begingroup$ You correctly identified the error in your solution. Try to find the large radius as a function of the smaller radius. $\endgroup$ – Calvin Lin Dec 6 '19 at 14:30
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    $\begingroup$ Revise your drawing by constructing the circle that will be tangent to $5/6$ of the smaller circles. $\endgroup$ – Vasya Dec 6 '19 at 14:48
  • $\begingroup$ Good catch. But the small circles are tangent to each other. But A,B,C,D can be calculated independant of the outer circle. Then E can be figured in terms of triangles, not circles. $\endgroup$ – fleablood Dec 6 '19 at 17:44

We need to find the radius of the large circle and the rest is straight forward: enter image description here

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  • $\begingroup$ It is not clear why A, B, C are on the same line. $\endgroup$ – Vasya Dec 6 '19 at 16:54
  • $\begingroup$ @Vasya, $A$, $B$ and $C$ are on the diameter of the red circle and clearly they have to be collinear. $\endgroup$ – Seyed Dec 6 '19 at 16:59
  • $\begingroup$ A is a point of tangency of blue and red circles, it does not have to be collinear with B $\endgroup$ – Vasya Dec 6 '19 at 17:01
  • $\begingroup$ @Vasya, The center of small circles are on an equilateral triangle and $\frac{5}{6}$ of the white part of small circles are symmetry. $\endgroup$ – Seyed Dec 6 '19 at 17:09

enter image description here

Using your original drawing. Let the radius of the smaller circle $r=1$. The radius of the big circle is:
$AD+AG=\frac{1}{\cos 30°}+\cos 30°=\frac{7\sqrt 3}{6}$.

Thus, the ratio of the colored area to the area of the big circle is $$\frac{5 \over 2}{49 \over 12}=\frac{30}{49}$$

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Take the upper right point $A$ of the red circle, the center $B$ of the yellow circle and the center $C$ of the big circle as vertices of a triangle. Call the radius of the smaller circles $r$. Then $AB=3r$, $BC=r\sqrt3$ and $\angle CBA=30^\circ$. Law of cosine gives $R=AC$, the radius of the big circle.

My results: $R^2=13r^2/3$ and the ratio is $15/26$.

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  • $\begingroup$ You might have a typo here, the correct answer is $\dfrac{15}{26}$. $\endgroup$ – Seyed Dec 6 '19 at 17:16
  • $\begingroup$ Yep, a typo. Thanks. $\endgroup$ – Michael Hoppe Dec 6 '19 at 17:36

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