Four points in a square!

Given arbitrary 33 points A,B,CA,B,C , can we construct (only using compass and straightedge) an equilateral triangle PQR \triangle PQR such that A,B,CA, B, C lie on the sides of PQR \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 AA not touching point BB and CC

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

  • Join KK and BB

  • Extend a line from BB till KK' such that PKB=KBK\angle P'KB=\angle KBK' (see how to draw)

  • Extend BK\overline{BK'} intersecting AP\overline{AP'} at PP

  • As in step 22 draw angle PQL=60°\angle PQ'L=60^\degree

  • As in step 33 join LL and CC

  • As in step 44 Draw line CQ\overline{CQ}

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

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

Justification:

After step 44 PKB=KBK\angle P'KB=\angle KBK' PKPR\Rightarrow \overline{P'K} \parallel \overline{PR} QPR=QPK\Rightarrow \angle QPR=\angle QP'K As from step 22 QPK=60°\angle QP'K=60^\degree QPR=60°........[1]\Rightarrow QPR=60^\degree........[1] Similarly we can prove that PQR=60°.......[2]\angle PQR=60^\degree.......[2] From angle sum property of triangle PRQ=60°........[3]\angle PRQ=60^\degree........[3] From [1],[2][1],[2] and [3][3] it is proved that PQR\triangle PQR is an equilateral triangle

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

Note by Zakir Husain
4 months, 3 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! )

Zakir Husain - 4 months, 3 weeks ago

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Hyper-Brilliant!

@Zakir Husain

Yajat Shamji - 4 months, 3 weeks ago

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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 - 4 months, 3 weeks ago

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

Zakir Husain - 4 months, 3 weeks ago

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

Kriti Kamal - 4 months, 3 weeks ago

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

Zakir Husain - 4 months, 3 weeks ago

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@Zakir Husain Thanks

Kriti Kamal - 4 months, 3 weeks ago

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

Construct circles with diameter ACAC, BDBD, and call them ωA\omega_A, ωB\omega_B, respectively.

Construct circles with the same radius rr centered at A and B, call them OA,OBO_A,O_B, respectively.

The radical axes of OAO_A and ωA\omega_A is LAL_A, and define LBL_B similarly.

Construct a circle with diameter ABAB, call it Ω\Omega, and its center MM.

Choose KK on Ω\Omega such that KA=KBKA=KB.

KAKA intersects LAL_A at EE, FF is on KMKM such that FEEKFE\perp EK.

The circle with diameter KFKF intersects LAL_A at point GG other than EE.

HH is on LBL_B such that HGGEHG\perp GE, and BHBH intersect Ω\Omega at a point II other than BB.

Reflect II about MM and we get QQ.

X X - 4 months, 3 weeks ago

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About what radius rr do you mean in the second step?

Zakir Husain - 4 months, 3 weeks ago

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

X X - 4 months, 3 weeks ago

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

Zakir Husain - 4 months, 3 weeks ago

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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.

X X - 4 months, 3 weeks ago

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Second attempt:) (Much simpler)

Call the circle with diameter ABAB Ω\Omega. Choose KK on Ω\Omega such that KA=KBKA=KB.

ACAC intersect Ω\Omega at point EE other than AA, and construct the circumcircle of CEKCEK, call it ω\omega.

ω\omega intersects the circle with diameter CDCD at a point other than CC, which is the desired SS.

X X - 4 months, 3 weeks ago

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

X X - 4 months, 3 weeks ago

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

Zakir Husain - 4 months, 3 weeks ago

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

First of all, let the foot of AA, BB on RSRS, SPSP be AA', BB', respectively.

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

Also, we have AA=BBAA'=BB', and AABBAA'\perp BB', thus the intersection of AAAA' and BBBB' lies on Ω\Omega.

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

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

So actually BB' is the point obtained by rotating AA' around KK for 9090^{\circ}, hence we rotate all the possible location of AA', which is the circle with diameter ACAC, around KK for 9090^{\circ}.

Let CC' be the point obtained by rotating CC around KK for 9090^{\circ}, then the circle we obtained on the previous step has diameter BCBC', so BB' lies on both the circle with diameter BCBC' and the circle with diameter BDBD.

So it is the matter of how to find CC'. Since the Miquel point of ACCBACC'B is also KK, the intersection of ACAC and BCBC' lies on Ω\Omega, which is EE.

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

Since BB' lies on both the circle with diameter BCBC' and the circle with diameter BDBD, BBBB' is the radical axes of the two circles, so BBBB' is perpendicular to CDC'D.

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

We have that SS lies on ω\omega, and also SS lies on the circle of diameter CDCD, hence the construction.

X X - 4 months, 3 weeks ago

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@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.

X X - 4 months, 3 weeks ago

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