1991 OIM Problems/Problem 2

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Problem

Two perpendicular lines divide a square into four parts, three of which each have an area equal to 1. Show that the area of the square is four.

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

1991 OIM p2c.png

First let's find the area of $A_1$:

$A_1=x\frac{y-x.tan(\theta)+y}{2}+\frac{y^2tan(\theta)}{2}$

$A_1=\frac{y^2-x^2}{2}tan(\theta)+xy=1$

  • Note. I actually competed at this event in Argentina when I was in High School representing Puerto Rico. I got partial points because I couldn't prove this but had somewhat of an approach to get there.

~Tomas Diaz. orders@tomasdiaz.com

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.


See also

https://www.oma.org.ar/enunciados/ibe6.htm