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

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Hyper-Brilliant!

@Zakir Husain

Yajat Shamji - 11 months, 2 weeks ago

Easy Math Editor

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Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
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`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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

Zakir Husain - 11 months, 2 weeks ago

@Yajat Shamji, @Kumudesh Ghosh, @Kriti Kamal, @Aryan Sanghi, @Vinayak Srivastava, @Mahdi Raza, @Alak Bhattacharya, @Marvin Kalngan

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.

Justin Travers - 11 months, 2 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.

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

A Former Brilliant Member - 11 months, 2 weeks ago

@A Former Brilliant Member – I myself don't use any book for geometry, what I will recommend is to practice yourself. It will help a lot

@Zakir Husain – Thanks

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 $Q$ .

X X - 11 months, 2 weeks ago

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

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

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

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.

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 $S$ .

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

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

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.

@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.

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$