# What fraction is shaded?

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

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Solution:

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

• Definitely the colored circles are not tangent to the large one Dec 6, 2019 at 14:30
• You correctly identified the error in your solution. Try to find the large radius as a function of the smaller radius. Dec 6, 2019 at 14:30
• Revise your drawing by constructing the circle that will be tangent to $5/6$ of the smaller circles. Dec 6, 2019 at 14:48
• 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. Dec 6, 2019 at 17:44

## 3 Answers

We need to find the radius of the large circle and the rest is straight forward:

• It is not clear why A, B, C are on the same line. Dec 6, 2019 at 16:54
• @Vasya, $A$, $B$ and $C$ are on the diameter of the red circle and clearly they have to be collinear. Dec 6, 2019 at 16:59
• A is a point of tangency of blue and red circles, it does not have to be collinear with B Dec 6, 2019 at 17:01
• @Vasya, The center of small circles are on an equilateral triangle and $\frac{5}{6}$ of the white part of small circles are symmetry. Dec 6, 2019 at 17:09

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 $$\frac{5 \over 2}{49 \over 12}=\frac{30}{49}$$

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 $$15/26$$.

• You might have a typo here, the correct answer is $\dfrac{15}{26}$. Dec 6, 2019 at 17:16
• Yep, a typo. Thanks. Dec 6, 2019 at 17:36