Four congruent circles are tangent to each other and tangent to the edges of a sector as shown. If the straight edges are joined to form a right circular cone with the vertex at P, the radius of the base would be 2/3 the slant height of the cone. Compute the ratio of the radius of the sector to the radius of each circle.

Here's a drawing:

Sorry for the poor artwork. Assume that all four of the small circles are tangent to each other. Assume that everything that seems tangent is tangent.

enter image description here

I'm thinking about using angle/360 = radius/slant or maybe area of sector = pi * r * l

Hints or solutions are both appreciated, thank you!

  • $\begingroup$ 4 circles are identical, if you connect P and three tangent points, you divide you cone into 4 identical parts. You just deal with one of them $\endgroup$
    – Yimin
    May 20, 2015 at 16:56
  • $\begingroup$ But since it involves the sector, so don't we have to look at the whole thing? $\endgroup$
    – Rex
    May 20, 2015 at 17:00
  • 1
    $\begingroup$ Clarification : The radius of the sector is the same as the slant height of the cone ? $\endgroup$
    – WW1
    May 20, 2015 at 17:41

2 Answers 2


The discussion of the cone convinces me that each of the four sectors containing a single circle will have central angle at $P$ of $\frac{\pi}{3}$

For one sector, the angle bisector at $P$ cut the sector into 2 conguent sectors, each with central angle $\frac{\pi}{6}$ and will pass through the centre of the circle.

Let $x$ represent the radius of the circle , $y$ the distance from $P$ to the centre of the circle and $r$ the slant height of the cone.

clearly $$x+y=r$$

Construct a radius that meets one of the straight sides at the point of tangency. You will have constructed a right angled triangle with hypotenuse $y$ and one side of length $x$ opposite to the central angle of $\frac{\pi}{6}$

so $$ \sin \frac{\pi}{6} = \frac xy \Rightarrow y=2x $$

putting the two equations together gives $x=\frac r3$

  • $\begingroup$ The slant height of the cone is the radius of the sector. With the vertex angles of the four small triangles at $\frac\pi 3$, the base angle of the cone is $\frac{2\pi}{3}$ and the radius of the base is $\frac{\sqrt{3}}{2}$ times the slant height of the cone. This is not the same as $\frac23$. $\endgroup$ May 21, 2015 at 19:21
  • $\begingroup$ I don't agree that the base angle of the cone is $\frac{2 \pi}{3}$ I think it should be $ \sin^{-1}(\frac 23)$ , or maybe double that, depending on your definition of base angle. $\endgroup$
    – WW1
    May 21, 2015 at 21:49

Without loss of generality for theanswer, we can take the slant height (which is the radius of the sector as well) to be $1$.

Step 0: Let $A$ be the intersection of the sector circumference with the line $PA$ on the let; let $BB$ be the corresponding point on the right. Since the cone is a right circular cone, $\overline{AB}$ is a diameter of the base of the cone so the radius of the cone will then be $\frac12 \overline{AB}$.

Let $M$ be the midpoint of $\overline{AB}$; $\overline{AM}$ is a radius of the cone: $\overline{AM}=\frac{2}{3}$. But if angle $APB = 2\phi$, in triangle $AMP$ we see that $$ \phi = \sin^{-1}\left(\frac23\right)$$

Step 1: Each of the small circles are in a triangular wedge of radius $1$ with central angle $2\theta$. Since for of these angles, plus the $2\phi$ in angle $APB$ gives a full circle, $$ \phi + 4 \theta = \pi$$ Step 2: Within that wedge of radius $r=1$ and angle $2\theta$, is inscribed a smaller circle, of radius $s$. (We wish to find $\frac{1}{s}$ which is the answer required.) What is $s$ in terms of $\theta$?

Label the corners of the wedge $XPY$ where $P$ is the same $P$ as in the main diagram. Inscribe the circle, and label the center of the circle $O$. Let the tangent intersection points of circle $O$ with sides $PX$ and and $PY$ be $U$ and $V$ respectively. And let $PO$ meat circle $P$ at $N$.

Angle $PUO$ is a right angle because the circle is tangent to the line. In right triangle $PUO$, side $\overline{OU} = s$ and side $\overline{OP} = s\csc \theta$. Since $\overline{PN} = 1$ (the radius of circle $P$), $$ s + s\csc\theta = 1 \implies \frac{1}{s} = 1 + \csc \theta $$ Step 3: We can find $\sin (4\theta)$ easily enough:

$$ 4\theta = \pi - \sin^{-1}\left(\frac23\right) \\ \sin(4\theta) = \frac23 $$
and we have to remember later that $4\theta$ is in the second quadrant, so $$\cos(4\theta) = -\frac{\sqrt{5}}{9}$$

Step 4: We now work with half-angle formulas: $$\cos (2\theta) = \sqrt{\frac12(1+\cos(4\theta))} = \sqrt{\frac12\left(1-\frac{\sqrt{5}}{9}\right)} $$ $$ \sin\theta = \sqrt{\frac12(1-\cos(2\theta))} = \sqrt{\frac12\left(1-\sqrt{\frac12\left(1-\frac{\sqrt{5}}{9}\right)}\right)} $$

Step 5: Now we must take the reciprocal of $\sin \theta$, add $1$ and simplify. The answer comes out numerically to about 4.8766.


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