Weird transformations WT999999P2

Geometry Level 5

There is a weird transformation known as $WT999999P2$ , also known as The W9 transformation . Little is known about it. However, it is known that it is certainly used to transform a square into a smaller square, following few steps and other wierd transformations. Some weird manual indicates the following about the transformation:

Step 1: Inscribe a regular octagon in the square, using method I8-v2 .
Step 2: Find isosceles triangle by rule T8 in octagon from step 1
Step 3: Inscribe a circle in the triangle from step 2
Step 4: Inscribe a square in the circle from step 3.
The initial square is transformed into the square from step 4

Luckily, some images explains what method I8-v2 and rule T8 should be. The pictures are provided from the manual (just click on the name of method/rule).

And now, the questions will be asked. If we have a square whose side length is $10^5$ , after applying transformation three times, what will be the new side length. If that new length is $L$ , answer shall be entered as $\left\lfloor L \right\rfloor$ .

If you just grasped: "Oh dear"

Because within you, lies a fear,

That something is left unclear,

It can be asked right over here .

The answer is 749.

1 solution

Milan Milanic
Jul 10, 2016

Solution:

If initial side length is $a$ , after the transformation $WT999999P2$ , new side length can be shown as $k \times a$ . If the number $k$ is found, then the answer is $\left\lfloor 10^5 \times k^3 \right\rfloor$ .

x is length of PQ segment

(Labels are used from picture above)

The angle at point $O$ of triangle $T$ is $\frac{360°}{8} = 45°$

$h = \frac{a}{2}, \space \tan{22.5°} = \sqrt{2} - 1 = \frac{\frac{x}{2}}{h}, \space x = a \times (\sqrt{2} - 1)$

The area of triangle $T$ can be calculated with formula: $\frac{h \times x}{2}$ and it is $\frac{a^2}{4}\times (\sqrt{2} - 1)$ . [1]

The same area can be calculated with formula: $r \times (c + \frac{x}{2})$ [2], where $r$ is radius of inscribed circle in the triangle.

Using Pythagoras theorem, $c = \frac{a}{2} \sqrt{4 - 2\sqrt{2}}$ . With [1] and [2], $r = \frac{a}{2} \times \frac{\sqrt{2} - 1}{\sqrt{2} - 1 + \sqrt{4 - 2\sqrt{2}}}$

The new side length $a'$ , will be equal to $r \sqrt{2}$ , which is $a \times \frac{\sqrt{2} - 1}{2 - \sqrt{2} + 2\sqrt{2 - \sqrt{2}}} = a \times k$ .

Carefully calculating, $k = 0.19570499200257185213999335635743$ .

So: $10^5 \times k^3 = 749.55880... \approx \boxed{749}$

Weird transformations WT999999P2

The answer is 749.

1 solution

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