# Four points in a square!

Given arbitrary $3$ points $A,B,C$ , can we construct (only using compass and straightedge) an equilateral triangle $\triangle PQR$ such that $A, B, C$ lie on the sides of $\triangle PQR$?

Yes, we can! Try it for yourself before reading on.

We use the following sequence of steps:

• Draw a arbitrary line from point $A$ not touching point $B$ and $C$

• Take an arbitrary point $P'$ on the line and draw another line from $P'$ till $K$ such that $\angle AP'K=60^\degree$ (see how to draw $60^\degree$ with compass here)

• Join $K$ and $B$

• Extend a line from $B$ till $K'$ such that $\angle P'KB=\angle KBK'$ (see how to draw)

• Extend $\overline{BK'}$ intersecting $\overline{AP'}$ at $P$

• As in step $2$ draw angle $\angle PQ'L=60^\degree$

• As in step $3$ join $L$ and $C$

• As in step $4$ Draw line $\overline{CQ}$

• Extend $\overline{PB}$ and $\overline{QC}$ and let them intersect at $R$

• Now points $A,B$ and $C$ lie on the sides of the equilateral triangle $\triangle PQR$

Justification:

After step $4$ $\angle P'KB=\angle KBK'$ $\Rightarrow \overline{P'K} \parallel \overline{PR}$ $\Rightarrow \angle QPR=\angle QP'K$ As from step $2$ $\angle QP'K=60^\degree$ $\Rightarrow QPR=60^\degree........[1]$ Similarly we can prove that $\angle PQR=60^\degree.......[2]$ From angle sum property of triangle $\angle PRQ=60^\degree........[3]$ From $[1],[2]$ and $[3]$ it is proved that $\triangle PQR$ is an equilateral triangle

Now a bonus : Given arbitrary points $A,B,C$ and $D$ can you draw a square $PQRS$ such that points $A,B,C$ and $D$ lies on the sides $\overline{PQ},\overline{QR},\overline{RS},\overline{SP}$ - A problem given by Jeff Giff

Note by Zakir Husain
4 months, 4 weeks ago

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@Jeff Giff - I didn't write my construction for rectangle (it will take more time and lines which I don't have this time! )

- 4 months, 4 weeks ago

- 4 months, 4 weeks ago

# Hyper-Brilliant!

@Zakir Husain

- 4 months, 4 weeks ago

Are you allowed to extend a side of the square beyond the segment between the vertices? If not, then if A,B,C are the vertices of an equilateral triangle and D is the centre of the triangle, no square can be constructed.

- 4 months, 3 weeks ago

@Justin Travers - If $A,B,C$ and $D$ forms a concave quadrilateral then no rectangle can be constructed, and as a square is a rectangle also therefore if $A,B,C$ and $D$ forms a concave quadrilateral then the square we desired is not possible. Therefore $A,B,C$ and $D$ must form a convex quadrilateral together.

- 4 months, 3 weeks ago

@Zakir Husain Yes,for concave quadrilateral no rectangle can be drawn.Can you recommend me some books for geometry??Thanks in advance

- 4 months, 3 weeks ago

I myself don't use any book for geometry, what I will recommend is to practice yourself. It will help a lot

- 4 months, 3 weeks ago

Thanks

- 4 months, 3 weeks ago

I will try to give a method of the Bonus Question, though I think it can be simplfied.

Construct circles with diameter $AC$, $BD$, and call them $\omega_A$, $\omega_B$, respectively.

Construct circles with the same radius $r$ centered at A and B, call them $O_A,O_B$, respectively.

The radical axes of $O_A$ and $\omega_A$ is $L_A$, and define $L_B$ similarly.

Construct a circle with diameter $AB$, call it $\Omega$, and its center $M$.

Choose $K$ on $\Omega$ such that $KA=KB$.

$KA$ intersects $L_A$ at $E$, $F$ is on $KM$ such that $FE\perp EK$.

The circle with diameter $KF$ intersects $L_A$ at point $G$ other than $E$.

$H$ is on $L_B$ such that $HG\perp GE$, and $BH$ intersect $\Omega$ at a point $I$ other than $B$.

Reflect $I$ about $M$ and we get $Q$.

- 4 months, 3 weeks ago

About what radius $r$ do you mean in the second step?

- 4 months, 3 weeks ago

You can choose any length to be the radius, just make sure the two circles are the same size.

- 4 months, 3 weeks ago

Can you also give a justification or proof that how these steps work so that the answer becomes mathematically acceptable

- 4 months, 3 weeks ago

I used inversion, Steiner Conic, and spiral similarity to come up with this. I will try to figure out if there is a simpler proof for this construction.

- 4 months, 3 weeks ago

Second attempt:) (Much simpler)

Call the circle with diameter $AB$ $\Omega$. Choose $K$ on $\Omega$ such that $KA=KB$.

$AC$ intersect $\Omega$ at point $E$ other than $A$, and construct the circumcircle of $CEK$, call it $\omega$.

$\omega$ intersects the circle with diameter $CD$ at a point other than $C$, which is the desired $S$.

- 4 months, 3 weeks ago

There are two choices of $K$, so there are two squares that satisfies the condition.

- 4 months, 3 weeks ago

Can you please give a proof or justification also for your steps of construction?

- 4 months, 3 weeks ago

I will give an idea of how I come up with this construction.

First of all, let the foot of $A$, $B$ on $RS$, $SP$ be $A'$, $B'$, respectively.

We know that $A',B'$ lie on the circle with diameter $AC,BD$, respectively.

Also, we have $AA'=BB'$, and $AA'\perp BB'$, thus the intersection of $AA'$ and $BB'$ lies on $\Omega$.

We now know that the Miquel point($K$) of complete quadrilateral $AA'B'B$ is on $\Omega$, and since $AA'=BB'$, we have $KA=KB$ (thus the location of K), and $KA'=KB'$.

Hence the goal is to find two points $A',B'$, which lie on the circle with diameter $AC,BD$, respectively, and satisfies $KA'=KB'$ and $KA'\perp KB'$.

So actually $B'$ is the point obtained by rotating $A'$ around $K$ for $90^{\circ}$, hence we rotate all the possible location of $A'$, which is the circle with diameter $AC$, around $K$ for $90^{\circ}$.

Let $C'$ be the point obtained by rotating $C$ around $K$ for $90^{\circ}$, then the circle we obtained on the previous step has diameter $BC'$, so $B'$ lies on both the circle with diameter $BC'$ and the circle with diameter $BD$.

So it is the matter of how to find $C'$. Since the Miquel point of $ACC'B$ is also $K$, the intersection of $AC$ and $BC'$ lies on $\Omega$, which is $E$.

According to the above, $C'$ lies on the circumcircle of $CEK(\omega)$. Also $CC'$ is the diameter of $\omega$ since $\angle CKC'$ is a right angle.

Since $B'$ lies on both the circle with diameter $BC'$ and the circle with diameter $BD$, $BB'$ is the radical axes of the two circles, so $BB'$ is perpendicular to $C'D$.

Also, $SC// BB'$, so $SC\perp C'D$. We have $CS\perp DS$, so $C',D,S$ are collinear, hence $\angle CSD=\angle CSC'=90^{\circ}$, which leads to $S$ lying on $\omega$.

We have that $S$ lies on $\omega$, and also $S$ lies on the circle of diameter $CD$, hence the construction.

- 4 months, 3 weeks ago

@X X This is how I thought of the construction, a bit long, but perhaps there is a simpler explanation or proof of why this construction works.

- 4 months, 3 weeks ago