# Given a circle, its center, and a point on the circle find inscribed square

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

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

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

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. $\square$

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.

• Thank you. I was missing that point L. Apr 9, 2018 at 18:40