Nesting some triangles

Geometry Level 3

Given that right $\Delta A_0B_0C_0$ has legs $a_0$ and $b_0$ such that $a_0=b_0=2$ . Another right triangle $\Delta A_1B_0C_1$ is drawn from the hypotenuse of $\Delta A_0B_0C_0$ such that its non-adjacent perpendicular side is drawn $\frac{2}{3} c_0$ from $B_0$ . Then, another right triangle $\Delta A_2B_0C_2$ is drawn from the hypotenuse of $\Delta A_1B_0C_1$ such that its non-adjacent perpendicular side is drawn $\frac{2}{3} c_1$ from $B_0$ and so on, as is depicted by the image above.

If all the triangles drawn are similar to one another, what is the sum of areas of all these triangles drawn up to infinity?

Clarification : Uppercase letters denote vertices; lowercase letters denote side lengths.

The answer is 18.

2 solutions

Efren Medallo
Aug 16, 2015

This may be a fast and not-so-thorough solution. Firstly, we derive an explicit formula for finding the area of $\Delta A_nB_nC_n$ . By iteration, we will find out that

$A_{\Delta n} = \frac{9}{16}b^{2} (\frac{8}{9})^{n+1}$

where $b$ is the original side length. Let us rearrange this relation as

$\frac{16}{9b^{2}} A_{\Delta n} = (\frac{8}{9})^{n+1}$

Let $\large S_{\Delta n} = \sum_{0}^{\infty} A_{\Delta n}$ .

From that, we can get

$\large \frac{16}{9b^{2}} S_{\Delta n} = \frac{\frac{8}{9}}{1 - \frac{8}{9}}$

$\large \frac{16}{9b^{2}} S_{\Delta n} = 8$ .

$\large S_{\Delta n} = 8 \cdot \frac{9b^{2}}{16}$

$\large S_{\Delta n} = \frac{9}{2} b^{2}$

substituting $b = 2$ gives us $\large{ \boxed {18}}$ .

Vincent Miller Moral
Aug 16, 2015

Infinite geometric

A1 = 2
common ratio = 8/9

Sum of areas = 2 / ( 1 - 8/9 )
Sum of  areas = 18 sq. units

Nesting some triangles

The answer is 18.

2 solutions

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