parallelogram and 3 equal circles

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parallelogram and 3 equal circles

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Source: Nordic Mathematical Contest 1987 #2

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#1 h Oct 5, 2017, 5:30 PM • 2 Y

Y by Adventure10, Mango247

Let $ABCD$ be a parallelogram in the plane. We draw two circles of radius $R$ , one through the points $A$ and
$B$ , the other through $B$ and $C$ . Let $E$ be the other intersection point of the circles. We assume that $E$ is not a vertex of the parallelogram. Show that the circle passing through $A, D$ , and $E$ also has radius $R$ .

This post has been edited 1 time. Last edited by parmenides51, Jun 21, 2022, 1:31 AM

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#2 h Oct 5, 2017, 8:30 PM • 1 Y

Y by Adventure10

The proof of the result follows from an immediate application of the Law of Sines (on $\Delta AED$ )after the observation that:

If $E$ is a point inside a parallelogram $ABCD$ then $\angle AED + \angle BEC = 180^o$ .

This follows from $(\overrightarrow{EA} - \overrightarrow{ED}) X (\overrightarrow{EB} - \overrightarrow{{EC}) = \overrightarrow{AD} X \overrightarrow{BC} = 0$ as $AD||BC$ (here we have used the usual vector notation). Alternatively, we may translate $\Delta BEC$ to $\Delta AED$ (with $A$ and $D$ as the images of $B$ and $C$ respectively so that $E$ translates to $E'$ ) and then observe that quadrilateral $AEDE'$ is cyclic (using the fact that, by the constructionn, quadrilaterals $ABEB'$ and $CDE'E$ are parallelograms along with $ABCD$ ).

This post has been edited 4 times. Last edited by ak12sr99, Oct 5, 2017, 8:41 PM
Reason: Standard vector notations not working...

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