Question

How many times is the area of the outermost triangle compared to the area of the innermost triangle (triangle \(⇒\) incircle \(⇒\) inscribed triangle \(⇒\) incircle \(⇒\) inscribed triangle)? 

• A. 16
• B. 9
• C. 9.42
• D. 12

Hint:

For similar shapes, areas scale with the square of the linear scale. Each triangle-from-incircle step halves the side of an equilateral triangle \(⇒\) area becomes one-fourth.

Answer 1

The Correct Option is A

Step 1: Relations for an equilateral triangle of side \(a\).
Inradius: \(r=\dfrac{\sqrt{3}}{6}a\).
Circumradius: \(R=\dfrac{a}{\sqrt{3}}\).
Area: \(A=\dfrac{\sqrt{3}}{4}a^{2}\).
Step 2: From triangle to next inscribed triangle via its incircle.
Outer triangle side \(a_0\). Its incircle radius \(r_0=\dfrac{\sqrt{3}}{6}a_0\).
The next (inscribed) triangle uses this circle as circumcircle, so
\(a_1=\sqrt{3}\,r_0=\sqrt{3}\left(\dfrac{\sqrt{3}}{6}a_0\right)=\dfrac{a_0}{2}\).
Step 3: Repeat once more.
For triangle \(a_1\), incircle radius \(r_1=\dfrac{\sqrt{3}}{6}a_1=\dfrac{\sqrt{3}}{6}.\dfrac{a_0}{2}=\dfrac{\sqrt{3}}{12}a_0\).
Innermost triangle side \(a_2=\sqrt{3}\,r_1=\sqrt{3}\left(\dfrac{\sqrt{3}}{12}a_0\right)=\dfrac{a_0}{4}\).
Step 4: Area ratio.
\[ \frac{A_{\text{outer}}}{A_{\text{inner}}} =\frac{\frac{\sqrt{3}}{4}a_0^{2}}{\frac{\sqrt{3}}{4}\left(\frac{a_0}{4}\right)^{2}} =\frac{a_0^{2}}{\left(\frac{a_0^{2}}{16}\right)} =\boxed{16}. \] Final Answer: \[ \boxed{\text{(A) }16} \]