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Given a circle, its center, and a point on the circle find inscribed square using only straightedge and compass.

The easy way is to draw a line through the two points to find another vertex of the square, then a perpendicular to this line through the center of the circle and find the two remaining vertices.

But (*) there's another solution using two circles to find "helper" points (I cheated for this solution).

Why does this solution work? What properties of circles, triangles, or squares are at play here?

(*) problem 1.7 of 'euclidea' app ( https://www.euclidea.xyz/ )

======== edit to add gif of the construction: https://makeagif.com/i/-F6Gdu

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  • $\begingroup$ Hint: if $\;r\;$ is the circle's radius, the side length of any inscribed square in that circle is $\;r{\sqrt2}\;$ ... $\endgroup$
    – DonAntonio
    Apr 8, 2018 at 13:35

2 Answers 2

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From @Blue's answer to my Mathematics StackExchange question:


enter image description here

  1. Construct $\bigcirc A$ through $O$.

    • Let $P_1$ and $P_2$ be the points where $\bigcirc A$ meets $\bigcirc O$.
  2. Construct $\bigcirc P_1$ through $P_2$.

    • Let $C$ be the (other) point where $\bigcirc P_1$ meets $\bigcirc O$.
  3. Construct $\overleftrightarrow{OP_1}$.

    • Let $Q_1$ and $Q_2$ be the points where $\overleftrightarrow{OP_1}$ meets $\bigcirc P_1$.
  4. Construct $\overleftrightarrow{CQ_1}$.

    • Let $B$ be the point where $\overleftrightarrow{CQ_1}$ meets $\bigcirc O$.
    • Note that we have constructed $\overleftrightarrow{BC}$.
  5. Construct $\overleftrightarrow{CQ_2}$.

    • Let $D$ be the point where $\overleftrightarrow{CQ_2}$ meets $\bigcirc O$.
    • Note that we have constructed $\overleftrightarrow{CD}$.
  6. Construct $\overleftrightarrow{AB}$.

  7. Construct $\overleftrightarrow{AD}$.

Square $\square ABCD$, with constructed edge-lines, is inscribed in $\bigcirc O$. (Proof that the quadrilateral is, in fact, a square, is left as an exercise to the reader.)


Edit. Having been asked to elaborate on the square ...

  • As of Step 2, we know $\triangle P_1 P_2 C$ is equilateral and that $\overline{OA}$ is on the perpendicualr bisector of side $\overline{P_1 P_2}$. Therefore, $\overline{AC}$ is a diameter of $\bigcirc O$, and we have that $\angle ABC$ and $\angle ADC$ (for point $D$ constructed later) are right angles by Thales' Theorem.

  • As of Step 4, as observed by Jan and Tristan in the comments, $\overline{Q_1 Q_2}$ is a diameter of $\bigcirc{P_1}$, so $\angle O_1 C Q_2$ is a right angle. Therefore, $\square ABCD$ is at least a rectangle.

  • Now, define $a := |\overline{OA}|$, so that $|\overline{P_1P_2}| = a\sqrt{3}$ and $|\overline{OQ_1}| = a( 1 + \sqrt{3})$. Since $\angle AOQ_1 = 60^\circ$, if we let $R$ be the foot of the perpendicular from $Q_1$ to $\overleftrightarrow{OA}$, then $|\overline{OR}| = \frac{a}{2}( 1 + \sqrt{3})$ and $$|\overline{Q_1R}| = \frac{a \sqrt{3}}{2}(1+\sqrt{3}) = \frac{a}{2}(3 + \sqrt{3}) = a + |\overline{OR}| = |\overline{CR}|$$ Thus, $\angle Q_1 C R = 45^\circ$ and we may conclude that $\square ABCD$ is a square. $\square$

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If $AO=1$ then $BC=CE=\sqrt3$ and it is easy to show (by the similarity of triangles $COM$ and $EOL$) that $EL=DL=(3+\sqrt3)/2$. It follows that $\angle EDA=45°$. Moreover $\angle EDG=90°$ because it is inscribed in a half-circle.

enter image description here

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  • $\begingroup$ Thank you. I was missing that point L. $\endgroup$
    – maraguida
    Apr 9, 2018 at 18:40

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