MHB A Guide to Finding the Perfect Vacation Spot

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The discussion focuses on a geometric proof involving the rotation of a diagram through 180 degrees about the midpoint of segment MN, resulting in two new circles that intersect at points A' and B'. It establishes that the angles MBM and MAN sum to 180 degrees using the alternate segment theorem, thereby demonstrating that MA'NB forms a cyclic quadrilateral. The circumcircles of triangles MAN and MBN are shown to have the same radius, as they transform into each other upon rotation. This geometric relationship highlights the properties of cyclic quadrilaterals and their circumcircles. The proof effectively illustrates the interconnectedness of angles and circles in geometric configurations.
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Rotate the diagram through $180^\circ$ about the midpoint of $MN$ to get two new circles, intersecting at say $A'$ and $B'$, so that $MANA'$ and $MBNB'$ are parallelograms. Use the alternate segment theorem in triangles $MAB$ and $NAB$ to show that angles $MBN$ and $MAN$ add up to $180^\circ$. Hence so do the angles $MBN$ and $MA'N$, and it follows that $MA'NB$ is a cyclic quadrilateral. So in the diagram below, the green circle (circumcircle of triangle $MAN$) and the red circle (circumcircle of triangle $MBN$) have the same radius: each of them goes to the other one when the diagram is rotated through $180^\circ$ about the midpoint of $MN$.
[TIKZ]%preamble \usetikzlibrary{calc,through}
\coordinate [label=above:$M$] (M) at (0,0) ;
\coordinate [label=above:$N$] (N) at (7,0) ;
\draw [thick] (-5,0) -- (10,0) ;
\node (P) [draw, thick, circle through=(M)] at (0,-5) {} ;
\node (Q) [draw,thick, circle through=(N)] at (7,-4.25) {} ;
\node (R) [draw, thin, circle through=(N)] at (7,5) {} ;
\node (S) [draw, thin, circle through=(M)] at (0,4.25) {} ;
\node [draw, red, circle through=(M)] at (3.5,-3) {} ;
\node [draw, green, circle through=(M)] at (3.5,3) {} ;
\coordinate [label=below: $A$] (A) at (intersection 2 of P and Q) ;
\coordinate [label=below: $B$] (B) at (intersection 1 of P and Q) ;
\coordinate [label=below: $A'$] (C) at (intersection 2 of R and S) ;
\coordinate [label=above: $B'$] (D) at (intersection 1 of R and S) ;
\draw [thick] (A) -- (M) -- (B) -- (N) -- cycle ;
\draw [thin] (C) -- (M) -- (D) -- (N) -- cycle ;[/TIKZ]
 
Thank you so much!
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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