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There is a square $ABCD$, line $EF$ and point $G$ on a plain. Can you construct the foot of perpendicular to line $EF$ through point $G$, using only a straightedge (in traditional Euclidean constructions)? Is there any construction shorter than mine?

I'd appreciate it if you posted a construction or provided any kinds of reference.

Here is my construction.


Complexity: 42

  • Create a point $H$ on line $AC$.
  • Create 2x2 grid: ACBDI BHCDJ BCDHK AJBCL AKCDM ABHLN ADHMO CDINP BCIOQ
  • Create 2x4 grid: ADNQR BCNOS BCOPT ADPQU NPRSV NPTUW
  • Create line $Id\bot EF$: IOEFX ABVXY CDWXZ EFYZa ABIab AIObc ADNcd
  • Create rectangle $IhGl$: GNIOe GPIOf NfPeg GgIOh GOINi GQINj OjQik GkINl
  • Create $Gp\bot EF$: IGhlm GkIdn IOmno EFGop

Geogebra Screenshot

  • The complexity of a construction can be described by the quantity of points.
  • XYZUV is short for "creating point $V$ which is the intersection of line $XY$ and $ZU$".
  • Notice that points can be in either capital and lowercase letters.
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  • $\begingroup$ That's a lot of points. It isn't easy to see what you did, but if there is such a construction not using compasses I'd think it was shorter. $\endgroup$
    – coffeemath
    Jun 23 at 10:33

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