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I have a fixed size square on which I need to draw a regular pentagon to use as a classroom aid. This pentagon should take up as much of the square as possible, and does not need to have any of its sides aligned with those of the square (though that would be considered a bonus).

I’ve found instructions online for inscribing a regular pentagon inside a circle (mathopenref.com/printpentagon.html), but I need to draw inside a square. Is there a construction method for doing so?

Alternatively, if a regular pentagon is inscribed in a circle and the smallest possible square is drawn around the pentagon, is there a way to find the location of the center of the circle relative to that of the square and the ratio of the length of a side of the square to the circle’s radius?

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  • $\begingroup$ Is it a REGULAR pentagon? $\endgroup$
    – Intelligenti pauca
    Commented Oct 21, 2021 at 19:55
  • $\begingroup$ Yes. Edit to make that clear. $\endgroup$
    – rpspringuel
    Commented Oct 21, 2021 at 19:58
  • $\begingroup$ inscribing a circle in a square is easy, the center is the intersection of diagonals. However, it may not be the optimal solution $\endgroup$
    – Vasili
    Commented Oct 21, 2021 at 19:58
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    $\begingroup$ You can fit a regular pentagon centered at origin with one vertex at $(\cos \frac{\pi}{20},\sin \frac{\pi}{20} )$ inside an axis-aligned square $[-\cos\frac{3\pi}{20}, \cos \frac{\pi}{20}]^2$. $\cos\frac{\pi}{20} + \cos\frac{3\pi}{20}$ is the minimum side length for a square to completely cover a pentagon w/ circumradius $1$. $\endgroup$
    – achille hui
    Commented Oct 21, 2021 at 23:23
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    $\begingroup$ @IvanKaznacheyeu That looks like an answer. $\endgroup$
    – rpspringuel
    Commented Oct 22, 2021 at 18:43

1 Answer 1

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Since no one is posting an answer, here’s the method I settled on for drawing one which is symmetrical on the diagonal of the square (which comments indicated would result in the largest possible pentagon).

  1. Using the information provided by @IvanKaznacheyeu, we set a compass to 0.44246 of the length of a side of the square.
  2. Placing the compass center on one corner of the square, trace the arc that intersects the two adjacent sides.
  3. Connect the two intersections, this will be the base of your pentagon $\overline{AB}$.
  4. (From here on out we’re following the method given on Wikipedia for constructing a regular pentagon when a side length is given. I reproduce it here for completeness and in case that page changes.) Draw an arc of a circle, center point $B$, with the radius $\overline{AB}$.
  5. Draw an arc of a circle, center point $A$, with the radius $\overline{AB}$; there arises the intersection $F$.
  6. Construct a perpendicular to the segment $\overline{ AB}$ through the point $F$; there arises the intersection $G$.
  7. Draw a line parallel to the segment $\overline{ FG}$ from the point $A$ to the circular arc about point $A$; there arises the intersection $H$.
  8. Draw an arc of a circle, center point $G$ with the radius $\overline{ GH}$ to the extension of the segment $\overline{ AB}$; there arises the intersection $J$.
  9. Draw an arc of a circle, center point $B$ with the radius $\overline{ BJ}$ to the perpendicular at point $G$; there arises the intersection D on the perpendicular, and the intersection $E$ with the circular arc that was created about the point $A$.
  10. Draw an arc of a circle, center point $D$, with the radius $\overline{ BA }$ until this circular arc cuts the other circular arc about point $B$; there arises the intersection $C$.
  11. Connect the points $BCDEA$. This results in the pentagon.

This method requires only one measurement using a ruler (step 1), the rest is classical construction techniques using compass and straightedge.

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