(related to Problem: The Compasses Puzzle)

Let $AB$ in the following diagram be the given straight line With the centers $A$ and $B$ and radius $AB$ describe the two circles. Mark off $DE$ and $EF$ equal to $AD.$ With the centers $A$ and $F$ and radius $DF$ describe arcs intersecting at $G.$ With the centers $A$ and $B$ and distance $BG$ describe arcs $GHK$ and $N.$ Make $HK$ equal to $AB$ and $HL$ equal to $HB.$ Then with centres $K$ and $L$ and radius $AB$ describe arcs intersecting at $I.$ Make $BM$ equal to $BI$. Finally, with the center $M$ and radius $MB$ cut the line in $C,$ and the point $C$ is the required middle of the line $AB.$ For greater exactitude you can mark off $R$ from $A$ (as you did $M$ from $B$), and from $R$ describe another arc at $C.$ This also solves the problem, to find a point midway between two given points without the straight line.


I will put the young geometer in the way of a rigid proof. First prove that twice the square of the line $AB$ equals the square of the distance BG, from which it follows that $HABN$ are the four corners of a square. To prove that $I$ is the center of this square, draw a line from $H$ to $P$ through $QIB$ and continue the arc $HK$ to $P.$ Then, conceiving the necessary lines to be drawn, the angle $HKP,$ being in a semicircle, is a right angle. Let fall the perpendicular $KQ,$ and by similar triangles, and from the fact that $HKI$ is an isosceles triangle by the construction, it can be proved that $HI$ is half of $HB.$ We can similarly prove that $C$ is the center of the square of which $AIB$ are three corners.

I am aware that this is not the simplest possible solution.

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Project Gutenberg

  1. Dudeney, H. E.: "Amusements in Mathematics", The Authors' Club, 1917

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