Consider N circles stacked on top of each other inside an isosceles triangle, as shown in below for the case of N=4. Let \(A_C\) be the sum of the areas og the N circles, and let \(A_T\) be the area of the triangle. In terms of N, what should the vertex angle \(\alpha\) be so that the ratio \(A_C / A_T\) is maximized? Assume that N is large, and ignore terms in your answer that are of sublieading order in N.

Find a ratio as a function of \(\alpha\) and N.

In the context of a triangle, the relationship between the areas and the radii of the circles is represented by the equation:

\[S_{\Delta ABC}=\frac{1}{2} C_{\Delta ABC} \times r\]

where r is proportional to C, or the length of the side segment.

Assuming the radius of the bottom circle is 1, let the ratio of the radii of the first circle to that of the second circle be denoted as b, which is also a function of \(\alpha\).

\[h=2+2b+2b^2+2b^3+...=\frac{2}{1-b}\] \[\beta=\frac{\alpha}{2}\] \[C_{\Delta ABC}=2h \tan \beta=2\frac{2}{1-b}\frac{1-b}{2\sqrt{b}}=\frac{2}{\sqrt{b}}\] \[S_{\Delta ABC}=\frac{2}{\sqrt{b}(1-b)}\] \[k=\frac{\pi (1+b^2+b^4+...+b^{2(N-1)})}{S_{\Delta ABC}}=\frac{\pi}{2}\frac{\sqrt{b}(1-b^{2N})}{1+r}\]

Calculate the derivative of the function to determine the maximum value.

\[1-b-(4N+1)b^{2N}-(4N-1)b^{2N+1}=0\]

We can address the scenario when N=1, b=1/3, \(\alpha=60^{\circ}\)

for large N,

\[1-b-4Nb^{2N}-4Nb^{2N+1}=0\] \[1-b-8Nb^{2N}=0\]

when N is large, let \(1-b=\epsilon, (1-\epsilon)^{2N}\approx e^{-2N\epsilon}\)

\[e^{-2N\epsilon}=\frac{\epsilon}{8N}\]

solve to get \(\epsilon\approx\frac{\ln N}{N}\)

\[\alpha=2\beta\approx 2 \sin \beta=2*\frac{1-b}{1+b}\approx=\epsilon\]

consider the case \(N=1\), assume \(S=1\), find the largest ratio = the largest r = the smallest \(C\)

\[C=\sqrt{\frac{2}{\sin \alpha}}(2+2\sin(\frac{\alpha}{2}))\]

plot the function

the minimum happens when \(\alpha=\pi/3\) , an equilateral triangle

when \(N=2\), the radii ratio \(b=\frac{h-2r_1}{h}=1-\frac{\frac{2}{l+l\sin \frac{\alpha}{2}}}{l \cos \frac{\alpha}{2}}\)

find the minimum \(r^2(1+b^2)\)

in Mathematica

When N = 2

In[18]:= siDe = Sqrt[2/(Sin[anGle])]

Out[18]= Sqrt[2] Sqrt[Csc[anGle]]

In[19]:= boTTom = 2*siDe*Sin[anGle/2]

Out[19]= 2 Sqrt[2] Sqrt[Csc[anGle]] Sin[anGle/2]

In[20]:= raD = 2/(2*siDe + boTTom)

Out[20]= 2/(2 Sqrt[2] Sqrt[Csc[anGle]] + 2 Sqrt[2] Sqrt[Csc[anGle]] Sin[anGle/2])

In[21]:= heiG = siDe*Cos[anGle/2]

Out[21]= Sqrt[2] Cos[anGle/2] Sqrt[Csc[anGle]]

In[22]:= raTio = (heiG - 2*raD)/heiG

Out[22]= (1/(Sqrt[2] Sqrt[Csc[anGle]]))
Sec[anGle/2] (Sqrt[2] Cos[anGle/2] Sqrt[Csc[anGle]] - 4/(
   2 Sqrt[2] Sqrt[Csc[anGle]] + 2 Sqrt[2] Sqrt[Csc[anGle]] Sin[anGle/2]))

In[23]:= raTioS = Pi*raD^2*(1 + raTio^2)

Out[23]= (4 \[Pi] (1 + 
     1/2 Sec[anGle/
       2]^2 (Sqrt[2] Cos[anGle/2] Sqrt[Csc[anGle]] - 4/(
        2 Sqrt[2] Sqrt[Csc[anGle]] + 
         2 Sqrt[2] Sqrt[Csc[anGle]] Sin[anGle/2]))^2 Sin[anGle]))/(2 Sqrt[2]
     Sqrt[Csc[anGle]] + 2 Sqrt[2] Sqrt[Csc[anGle]] Sin[anGle/2])^2

In[29]:= Plot[raTioS, {anGle, 0, Pi}]