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I'm trying to form similar arcs (like the blue one) to join points X-Z, and A-B at both ends of the line BX in the following picture.

XY and YZ are unknown but equal.

How do I find the radius of the arc and the exact position of BX relative to the rest of the shape?

enter image description here

Kais Hasan
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Coronos W
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  • Is this what you want? Trace the line through $Z$ perpendicular to $YZ$ and the line through $X$ perpendicular to $XY$. Consider the intersection of these two lines call it $O$. Then draw the circle with center at $O$ and radius $OX=OZ$. – Julian Mejia May 22 '20 at 21:38
  • That is close, but assumes I know the absolute position of BX, which I don't. – Coronos W May 22 '20 at 22:24

1 Answers1

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Let $t = XY = YZ = AC=BC$ (where $C$ is the intersection of the vertical and slanted lines near $A$ and $B$). Let $D$ be the intersection of the horizontal and vertical lines at top left. Thus $CDY$ is a $45$ degree right triangle with hypotenuse $30+2t$ and legs $30-t$, so $30+2t = \sqrt{2} (30 - t)$. Solve for $t$.

The centre of arc $AB$ is then a point $P$ such that $ACP$ and $BCP$ are congruent right triangles, with angle $APC$ of $22.5^\circ $ . Thus $AP = t \cot(22.5 ^\circ)$.

Robert Israel
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  • I think this might be what I'm looking for. A question though: Are ACP and BCP actually right triangles? Wouldn't P be slightly closer to C than the line AB needed to form right triangles? (Either way will be close enough for my purposes.) – Coronos W May 22 '20 at 22:36