Page 1 of 1

IGO 2016 Medium/3

Posted: Tue Jan 10, 2017 3:59 pm
3. Find all positive integers \$N\$ such that there exists a triangle which can be dissected into \$N\$ similar quadrilaterals.

Re: IGO 2016 Mediam/3

Posted: Wed Jan 11, 2017 4:06 pm
It is possible for all integer \$N \ge 3\$

It is obvious that a triangle can't be partitioned in \$1\$ quadrilateral.

Also if we divide a triangle into \$2\$ quadrilaterals, one is convex but other one is not.

For \$N=3\$, take a equilateral triangle. And divide the triangle into \$3\$ congruent quadrilaterals. Like the figure. \$O\$ is the circumcenter. We draw \$\angle OFB\$ such that it is equal to \$60^o\$. Also same thing for \$ \angle ODC \$ and \$ \angle OEA\$.

It is easy to prove that the quadrilaterals are congruent.

Now we can make a quadrilateral \$BCYX\$ such that it is similar to \$EOFB\$. And \$\angle ABX = \angle ACY = 60^o+120^o=180^o\$. So that means \$\triangle AXY\$ is a new triangle with \$5\$ similar quadrilaterals. We can make this process over and over. So it is possible for all \$N \ge 3\$