MHB What is the area of square ABCD with OQ = OF = 6?

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The area of square ABCD can be determined using coordinate geometry, given that OQ and OF both equal 6. By establishing a coordinate system with O at the origin, the calculations reveal that the midpoint M of line segment HK is crucial for determining the dimensions of the square. The derived relationships between the coordinates lead to the conclusion that the side length of the square is equal to the square root of 12. Consequently, the area of square ABCD is confirmed to be 12. A purely geometric solution is desired but not provided in the discussion.
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Find area of square ABCD if OQ=OF=6.
 

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Here is an outline of a solution using coordinate geometry.

[TIKZ][scale=1.5]
\coordinate [label=above right:{\textcolor{blue}O}] (O) at (0,0) ;
\coordinate [label=above right:{\textcolor{blue}A}] (A) at (0,3.46) ;
\coordinate [label=above right:{\textcolor{blue}B}] (B) at (1.73,0.46) ;
\coordinate [label=above right:{\textcolor{blue}C}] (C) at (4.73,2.19) ;
\coordinate [label=above right:{\textcolor{blue}D}] (D) at (3,5.2) ;
\coordinate [label=above right:{\textcolor{blue}F}] (F) at (6,0) ;
\coordinate [label=above right:{\textcolor{blue}H}] (H) at (5.91,1.02) ;
\coordinate [label=above right:{\textcolor{blue}K}] (K) at (1.02,5.91) ;
\coordinate [label=above right:{\textcolor{blue}Q}] (Q) at (0,6) ;
\coordinate [label=above right:{M}] (M) at (3.46,3.46) ;
\draw [blue, ultra thick] (A) -- (B) -- (C) -- (D) -- cycle ;
\draw [brown, ultra thick] (H) -- (K) ;
\draw [blue, ultra thick] (F) -- (O) -- (Q) ;
\draw [blue, ultra thick] (F) arc (0:90:6) ;
\draw [brown, ultra thick] (K) arc (135:315:3.46) ;
\draw [thin] (A) -- (M) -- (3.46,0) ;
\draw [thin] (K) |- (M) |- (C) ;
\draw (0.9,4.7) node {$s$} ;
\draw (2.2,3.6) node {$s$} ;
\draw (-0.2,1.7) node {$t$} ;
\draw (1.7,-0.2) node {$t$} ;
\draw (3.3,2.8) node {$u$} ;
\draw (4.1,2.1) node {$u$} ;
\draw [thin, dashed] (A) -- (C) ;
\draw [thin, dashed] (B) -- (D) ;
\foreach \point in {O,A,B,C,D,F,H,K,Q} \fill [blue] (\point) circle (2pt) ;
\fill (M) circle (2pt) ;[/TIKZ]
Choose a coordinate system with origin at O, so that F is the point $(6,0)$ and Q is $(0,6)$.

I assume that the brown part of the diagram is meant to be a semicircle and its diameter. Let M be the midpoint of HK, with coordinates $(t,t)$, so that the semicircle has radius $t$. Then the line HK has equation $x+y = 2t$. Let K be the point $(t-s,t+s)$. The distance KM is $t$, from which it follows that $t^2 = 2s^2$. The condition that K lies on the blue circle of radius $6$ is $(t+s)^2 + (t-s)^2 = 36$, from which $3t^2 = 36$ and hence $t = \sqrt{12}$.

Next, let C be the point $(t+u,t-u)$. The centre of the square ABCD is the midpoint of AC, namely $\bigl(\frac12t+ \frac12u, t - \frac12u\bigr)$. You can then calculate that B is the point $\bigl(\frac12t,\frac12t-u\bigr)$ and D is $\bigl(\frac12t + u,\frac32t\bigr)$. The conditions that B lies on the brown semicircle and that D lies on the blue circle both lead to the same equation $2u^2 + 2ut - t^2 = 0$. Therefore $u = \frac12(\sqrt3 - 1)t = 3-\sqrt3$.

Knowing $t$ and $u$, you can then easily check that the distance AB is $t$. So the area of the square ABCD is $t^2 = 12$.

I would much prefer to have a solution using a purely geometric argument, but I do not see how to do that.
 
Hint:
a364ec7d7a6117370504de8ae46ff6f5fb731c (1).jpg
 
Solution pure geometry:
sol1.JPG
 
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