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

Revision as of 23:35, 22 December 2023

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

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

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

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

Now lets find the area of $A_2$ :

$A_2=\frac{L-x}{2}((L-x)tan(\theta)+y+y)-\frac{y^2}{2}tan(\theta)$

$A_2=\frac{(L-x)^2-y^2}{2}tan(\theta)+(L-x)y=1$

Now lets find the area of $A_3$ :

$A_3=\frac{L-y}{2}(x+x-(L-y)tan(\theta))+\frac{x^2}{2}tan(\theta)$

$A_3=\frac{x^2-(L-y)^2}{2}tan(\theta)+x(L-y)=1$

Now we need to solve for $x$ , $y$ and $L$ in the following system of equations:

$\begin{cases} \frac{y^2-x^2}{2}tan(\theta)+xy=1\\ \frac{(L-x)^2-y^2}{2}tan(\theta)+(L-x)y=1\\ \frac{x^2-(L-y)^2}{2}tan(\theta)+x(L-y)=1 \end{cases}$

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.

@@ Line 27: / Line 27: @@
 Now we need to solve for <math>x</math>, <math>y</math> and <math>L</math> in the following system of equations:
-\begin{cases}
+<math>\begin{cases}
 \frac{y^2-x^2}{2}tan(\theta)+xy=1\\
 \frac{(L-x)^2-y^2}{2}tan(\theta)+(L-x)y=1\\
 \frac{x^2-(L-y)^2}{2}tan(\theta)+x(L-y)=1
-\end{cases}
+\end{cases}</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.