Equilateral Triangle Division

Source: William Wu Puzzle Page

Problem: Draw an equilateral triangle (all sides same length). Divide it into four identical shapes. remove the bottom left hand shape. now divide the resulting shape into four identical shapes.

Update (18 Oct 2010):
Solution: Identical solutions posted independently by Siddhant Agarwal (Senior Undergraduate, EE, IITB) and Shaunak Chapparia (Senior Undergraduate, CSE, IITB) in comments.


  1. divide the triangle into 16 congruent equilateral triangles (by the obvious method). Now number them 1-16.. as 1 for the top one, 2 3 4 from left to right of the next three and so on

    So we are removing the bottom left part i.e 5,10,11,12..
    the 4 congruent parts are: (1,2,3),(6,7,13),(14,15,16) and (4,8,9)

  2. Draw a triangle ABC with P, Q and R as mid-points of AB, BC and CA respectively. Remove the triangle PQB (as asked in the question).

    Let mid-points of AR, PR, QR and RC be X, Y, Z and W respectively. Join XY, YZ and ZW.

    Thus, the four identical shapes are:
    AXYP, PYZQ, XYZW and ZWCQ :)

  3. @Siddhant, Shaunak..
    Both the solutions are correct (and identical!)

    @Shaunak.. Nice explanation. Thanks!

  4. P is midpoint of AB, Q is midpoint of BC, R is midpoint of CA. Remove triangle PQB. Divide segment PQ into 4 parts of ratio(1:5:1:1)by points P1 PQ1 and Q1 . Divide AC into 4 segments of ratio(5:1:5:5) by points P2 PQ2 and Q2 [Note that PQ:AC=1:2]. 4equal parts are A P P1 P2, P1 P2 PQ1 PQ2, PQ1 PQ2 Q1 Q2 , Q1 Q2 C Q


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