Partitioning Many Triangles

Claim: If the larger right triangle is partitioned into $N$ triangles, then the width (say $w$ ) of the larger right triangle is given by, $w=\frac{2^{2N}-1}{2^{N-1}}$ Proof: Define $w(k)$ to be the width of the larger right triangle with $k$ partitions. Now consider a larger right triangle with $N+1$ partitions that has a base length equal to $w(N+1)$ . Notice that it comprises of a large right triangle with $N$ partitions with base length $w(N)$ and 1 smaller triangle (right most triangle) with base length as $w(N+1)-w(N)$ .

Therefore, this smaller triangle has sides $\sqrt{16+w(N)^2}$ , $\sqrt{16+w(N+1)^2}$ and $w(N+1)-w(N)$ . Since the inradius of this triangle is $1$ , $\begin{aligned} 2(w(N+1)-w(N))&=\frac{\sqrt{16+w(N)^2}+\sqrt{16+w(N+1)^2}+w(N+1)-w(N)}{2} \\ \implies 3(w(N+1)-w(N))&=\sqrt{16+w(N+1)^2}+\sqrt{16+w(N)^2} \end{aligned}$ Solving this as a quadratic in terms of $w(N+1)$ , we get the relation $w(N+1)=\frac{3\sqrt{16+w(N)^2}+5w(N)}{4}$ Since $w(1)=3$ , it is easy to see that the first few values of the sequence are $\begin{aligned} w(1)&=\frac{3}{1}=\frac{2^2-1}{2^0} \\ \\ w(2)&=\frac{9}{2}=\frac{2^4-1}{2^1} \\ \\ w(3)&=\frac{63}{4}=\frac{2^6-1}{2^2} \\ \\ w(4)&=\frac{255}{8}=\frac{2^8-1}{2^3} \\ \\ w(5)&=\frac{1023}{16}=\frac{2^{10}-1}{2^4}\end{aligned}$ By induction, we prove that the solution of the above recurrence relation is $w(N)=\frac{2^{2N}-1}{2^{N-1}}$ Hence, our initial claim was correct.

Finally, all we need to do is solve the inequality, $\frac{2^{2N}-1}{2^{N-1}}>10^4$ $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{4x^2-1}{x}>10^4 \;, \; \; \; \; x=2^{N-1}$ $\implies x=2^{N-1}>2500\implies N\ge 13$ Therefore, the minimum value of $N$ that satisfies the above condition is $\boxed{13}$

Define the width $x_N$ is a sequence. We note that $x_1 = 3$ , $x_2 = \dfrac {15}2$ , $x_3 = \dfrac {63}4$ , $\cdots$ . It would appear that $x_N = \dfrac {4^N-1}{2^{N-1}}$ . We can prove this claim using proof by induction as follows:

The area of the last partitioned triangle, $\Delta_N = sr$ , where $s$ is the semiperimeter of the triangle and $r$ is the radius of the incircle, which is $1$ . But $\Delta_N = \dfrac {4(x_N-x_{N-1})}2$ . Therefore,

$\begin{aligned} 2(x_N-x_{N-1}) & = sr = \frac {(x_N-x_{N-1})+\sqrt{x_N^2+4^2}+\sqrt{x_{N-1}^2+4^2}}2 \\ 3(x_N-x_{N-1}) & = \sqrt{x_N^2+16}+\sqrt{x_{N-1}^2+16} \\ 3x_N - \blue{\left(3x_{N-1} + \sqrt{x_{N-1}+16} \right)} & = \sqrt{x_N^2+16} \qquad \qquad \small \blue{\text{Let }a_{N-1} = 3x_{N-1} + \sqrt{x_{N-1}+16}} \\ 3x_N - \blue{a_{N-1}} & = \sqrt{x_N^2+16} \qquad \qquad \small \blue{\text{Squaring both sides}} \\ 8x_N - 6a_{N-1}x_N + a_{N-1}^2 - 16 & = 0 \\ \implies x_N & = \frac {3a_{N-1}+\sqrt{a_{N-1}^2+128}}8 \end{aligned}$

Assuming that the claim $x_N = \dfrac {4^N-1}{2^{N-1}}$ is true for $N$ , then

$\begin{aligned} a_N & = 3x_N + \sqrt{x_N^2+16} = \frac {3(4^N-1)}{2^{N-1}} + \sqrt{\left(\frac {4^N-1}{2^{N-1}}\right)^2+16} \\ & = \frac {3(4^N-1)}{2^{N-1}} + \frac {4^N+1}{2^{N-1}} = \frac {2^{2N+1}-1}{2^{N-2}} \\ \sqrt{a_N^2 + 128} & = \frac {2^{2N+1}+1}{2^{N-2}} \end{aligned}$

And

$\begin{aligned} x_{N+1} & = \frac {3a_N+\sqrt{a_N^2+128}}8 = \frac {2^{2N+3}-2}{2^{N+1}} = \frac {4^{N+1}-1}{2^N} \end{aligned}$

Therefore the claim is also true for $N+1$ , and it is true for all $N \ge 1$ .

Now we need to find the smallest $N$ such that $x_N = \dfrac {4^N-1}{2^{N-1}} > 10^4$ , approximately $2^{N+1} > 10^4 \implies N = \left \lceil \dfrac 4{\log_{10}2} - 1 \right \rceil = \boxed{13}$ .

Starting with the first triangle on the left, its base extends from the origin to $( x(1), 0)$ , where $x(1) = 3$ , and its hypotenuse is $h(1) = 5$ . Now consider the $n$ -th triangle, we want to build the $(n+1)$ -th triangle on its hypotenuse $h(n)$ , so we have to extend the base line to $x(n+1) = x(n) + d$ and we therefore have $h(n+1) = \sqrt{16 + x(n+1)^2}$

Since a unit circle is inscribed in this $(n+1)-th$ triangle, then $2 A = P$ , and therefore,

$4 d = d + h(n) + \sqrt{ 16 + (x(n) + d)^2 }$

Simplifying as follows:

$3 d - h(n) = \sqrt{ 16 + x(n)^2 + 2 x(n) d + d^2 }$

Squaring, and noting that $16 + x(n)^2 = h(n)^2$ , we obtain,

$9 d^2 - 6 d h(n) = 2 x(n) d + d^2$

From which it is straightforward to calculate that

$d = \dfrac{ 3 h(n) + x(n) }{4 }$

Since we have $d$ , then we now have $x(n+1) = x(n) + d$ and $h(n+1) = \sqrt{ 16 + x(n+1)^2 }$

Now, all we have to do is write a tiny code to implement a loop in the index $n$ and compute successive values of $x(n+1), h(n+1)$ , starting with $x(1) = 3$ , and $h(1) = 5$ . We exit the loop when $x(n+1) > 10^4$ . The number of triangles is $N = n + 1$ .

Running this code I found that $n = 12$ , and therefore, $N = \boxed{13}$