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$ABCD$, $AEFG$, $GHIJ$ are squares; $ABCD$ is inscribed in the semicircle, and $E$ is the middle of $[AD]$. Attention! $JG+GA+AO \neq r$. We have to find the areas of the squares.

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It's easy to find that $Area(ABCD)=\frac{4r^2}{5}$ and $Area(AEFG)=\frac{r^2}{5}$, but I can't figure out how to find the area of the last square. Can you help me? Thanks!

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  • $\begingroup$ y coord of I = h; x coordinate is OA+AG+h; Pythagoras can be now applied. $\endgroup$ – Narasimham Jun 5 '18 at 0:58
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HINT

The first square can be found by the condition

  • $\sqrt{r^2-x^2}=2x\implies r^2-x^2=4x^2 \implies x_D=\frac r {\sqrt 5}\implies A_1=(2x_D)^2=\frac45r^2$

For the second we have

  • $\sqrt{r^2-x^2}=x-x_D$ from which we can obtain $x_F$ and then $A_2=(x_F-x_D)^2$

For the third we have

  • $\sqrt{r^2-x^2}=x-x_F$ from which we can obtain $x_I$ and then $A_3=(x_I-x_F)^2$
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  • $\begingroup$ I'm not sure what you mean by $\implies$ in the second and third statements. $\endgroup$ – saulspatz Jun 4 '18 at 20:23
  • $\begingroup$ @saulspatz it stands for "from which we can obtain..." $\endgroup$ – user Jun 4 '18 at 20:23
  • $\begingroup$ So obvious... I did not notice that $OI=r$. Thanks! $\endgroup$ – Iulian Oleniuc Jun 4 '18 at 20:24
  • $\begingroup$ @IulianOleniuc You are welcome! Bye $\endgroup$ – user Jun 4 '18 at 20:27
  • $\begingroup$ @saulspatz You are welcome! I've changed the notation to be more clear. Thanks $\endgroup$ – user Jun 4 '18 at 20:27
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Let's let the radius of the circle be $1$ for a while and account for $r$ at the end. I'm also flipping your circle so that all of the interesting stuff happens in the first quadrant.

enter image description here

Consider the vertical line $x= x_i$. To build a square to the left of that with sides of length $s_i$, the upper right corner of that square will have coordinates of $(x_i+s_i, s_i)$ and will be on the unit circle. Hence

\begin{align} (x_i + s_i)^2 + s_i^2 &= 1 \\ 2s_i^2 + 2 x_i s_i - (1-x_i^2) &= 0 \\ s_i &= \dfrac{-2x_i + \sqrt{4 x_i^2 + 8 - 8 x_i^2}}{4} \\ s_i &= \dfrac{\sqrt{2 - x_i^2} - x_i}{2} \\ x_{i+1} &= \dfrac{\sqrt{2 - x_i^2} + x_i}{2} \\ \end{align}

If we start with $x_1 = \dfrac{1}{\sqrt 5}$

Then $s_2= \dfrac{1}{\sqrt 5}$ and $x_2 = \dfrac{2}{\sqrt 5}$

$s_3= \dfrac{\sqrt 6 - 2}{2\sqrt 5}$ and $x_3 = \dfrac{\sqrt 6 + 2}{2\sqrt 5}$

$s_4= \dfrac{2+\sqrt 6 - \sqrt{30-4 \sqrt 6}}{4\sqrt 5}$ and $x_4 = \dfrac{2+\sqrt 6 + \sqrt{30-4 \sqrt 6}}{4\sqrt 5}$

and so on

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