8
$\begingroup$

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

enter image description here]1

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

$\endgroup$
  • 5
    $\begingroup$ Definitely the colored circles are not tangent to the large one $\endgroup$ – Vasya Dec 6 '19 at 14:30
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
2
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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 $15/26$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.