# Making 7 congruent triangles from the pieces of a triangle dissection

I got this challenging geometrical conundrum from a Russian geometrical magazine. It states:

(A. Soifer) Use six lines to cut a triangle into parts such that it is possible to compose seven congruent triangles from them.

In other words, given an arbitrary triangle, how can you use six straight cuts to dissect the triangle into some number of pieces, such that the pieces can be combined to form seven congruent triangles?

The solution must work for any given triangle. And, the six cuts must be made all at once (i.e. You can't make one cut, move the pieces around, then make another cut), though I wouldn't mind if anyone shared a solution with this methodology.

I found this problem extremely fun and rewarding to crack. Hope you guys enjoy it too!

• Do the seven triangles have to use all different pieces, or can we use the same piece in two different triangles? – Ankoganit Aug 19 '20 at 4:12
• Must all the pieces be used? Because rot13( gurer vf n fvzcyr jnl gb phg vg vagb avar pbatehrag gevnatyrf jvgu fvk phgf, naq vs lbh bzvg bar phg lbh trg frira gevnatyrf naq na rkgen aba-gevnatyr cvrpr. ) – Jaap Scherphuis Aug 19 '20 at 5:34
• All pieces must be used, no pieces may be reused for different triangles. – greenturtle3141 Aug 19 '20 at 5:52
• Feel free to post such a dissection. – greenturtle3141 Aug 19 '20 at 6:42

• @PaulPanzer Yes. The 5 times smaller square worked because $5$ can be written as the the sum of two squares $5=a^2+b^2$. For this triangle it works because $7$ can be written in the form $7=a^2+ab+b^2$ with integers $a$, $b$. – Jaap Scherphuis Aug 19 '20 at 8:05
• @PaulPanzer In the 5-times-smaller-square problem, the larger square splits into a central square and four right-triangles. If the triangles have legs $a$ and $b$ units in length, then the area of the whole square is $(a-b)^2+4(ab/2) = a^2+b^2$ square units. In this problem we have a central triangle surrounded by 3 triangles with legs of $a$ units along one axis and $b$ along another. These each have an area of $ab$ unit triangles, the centre triangle has an area of $(a-b)^2$ unit triangles, for a total of $a^2+ab+b^2$ unit triangles. – Jaap Scherphuis Aug 20 '20 at 14:59