Difference between revisions of "1991 OIM Problems/Problem 2"

Line 6: Line 6:
 
== Solution ==
 
== Solution ==
 
[[File:1991_OIM_p2c.png|700px]]
 
[[File:1991_OIM_p2c.png|700px]]
 +
 +
First let's find the area of <math>A_1</math>:
 +
 +
<math>A_1=\frac{y-x.tan(\theta)+y}{2}+\frac{y}{y.tan(\theta)}{2}</math>
  
 
* 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.
 
* 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.

Revision as of 00:20, 23 December 2023

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 $A_1$:

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

  • 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