Question

Ram places three identical cones of radius on a table in such a way that each cone base touches the other two, also the altitude of the cone is perpendicular to the table. The radius of the circle drawn through the cone vertices would be?

• A. Smaller than r
• B. Equal to r
• C. Larger than r
• D. Cannot be determined

Answer 1

The Correct Option is C

To find the radius of the circle drawn through the vertices of the cones, consider the arrangement of the cones. Each cone is touching the other two at their bases, forming an equilateral triangle when viewed from above.
1. Arrangement and Geometry:
- The centers of the three cones' bases form an equilateral triangle with side length \(2r\) (since the distance between the centers of two touching cones is twice the radius, \(r\)).
  - The vertices of the cones are directly above these centers.
2. Circumcircle of the Equilateral Triangle:
  - The radius \(R\) of the circumcircle of an equilateral triangle with side length \(2r\) can be found using the formula \(R = \frac{a}{\sqrt{3}}\), where \(a\) is the side length of the triangle.
3. Calculation:
  - For our equilateral triangle with side length \(2r\), the circumradius \(R\) is:
    \[ R = \frac{2r}{\sqrt{3}} = \frac{2r \sqrt{3}}{3}\]
4. Comparison:
  - Simplifying, \( \frac{2r \sqrt{3}}{3} \) is clearly larger than \(r\), because \( \sqrt{3} \) is approximately 1.732, making \( \frac{2r \times 1.732}{3} \approx 1.154r \).
Thus, the radius of the circle drawn through the vertices of the cones is larger than \(r\).
Explanation: Since the radius of the circumcircle of the equilateral triangle formed by the centers of the cones' bases is larger than the radius of the cones' bases, the radius of the circle drawn through the vertices of the cones will be larger than \(r\).

Therefore, the correct answer is  C Larger than r.

Answer 2

Three identical cones, each of base radius \( r \), placed so that every pair of bases just touches, have their centers forming an equilateral triangle. Since each pair of circles of radius \( r \) touches externally, the distance between any two centers is \( r + r = 2r \), so the triangle joining the three centers has every side equal to \( 2r \).

The vertices, or tips, of the cones sit directly above these centers, so the circle passing through the three vertices, when viewed from above, is exactly the circumscribed circle of this equilateral triangle of side \( 2r \).

Using the law of sines for a triangle, side over sine of the opposite angle equals \( 2R \). In an equilateral triangle every angle is \( 60 \degree \), so:

\( 2R = \dfrac{2r}{\sin 60 \degree} = \dfrac{2r}{\frac{\sqrt{3}}{2}} = \dfrac{4r}{\sqrt{3}} \)

\( R = \dfrac{2r}{\sqrt{3}} \approx 1.155r \)

1. Option A (Smaller than r): Since \( R \approx 1.155r \) is larger than \( r \), not smaller, this is incorrect.
2. Option B (Equal to r): \( R \) works out to about \( 1.155r \), not exactly \( r \), so this is incorrect.
3. Option C (Larger than r): This matches the computed value \( R \approx 1.155r \), which is indeed larger than \( r \), so this is correct.
4. Option D (Cannot be determined): The value of \( R \) can in fact be pinned down exactly using the law of sines, so this is incorrect.

The circle through the cone vertices has radius \( \frac{2r}{\sqrt{3}} \), which is always larger than \( r \).

the correct answer is Option C: Larger than r.