Source: 2018 Canadian Open Math Challenge Part B Problem 2 Let ABCD be a square with side length 1. Points $X$ and $Y$ are on sides $BC$ and $CD$ respectively such that the areas of triangels $ABX$, $XCY$, and $YDA$ are equal. Find the ratio of the area of $\triangle AXY$ to the area of $\triangle XCY$.
Problem
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Tags: Comc, 2018 COMC
omriya200
06.12.2018 17:38
$BX=DY=x$ and $YC=CX=1-x$ so $X^2-3x+1=04$ so $x=(3+\sqrt{5})/2$ now easy
Kagebaka
07.12.2018 01:55
$x=(1-x)^2$, as well as $\frac{\triangle AXY}{\triangle XCY}=\frac{1+x}{1-x}=\frac{2-3x}{x}$
, but I couldn't find anything further.
programjames1
07.12.2018 02:03
You can further that to be $3-2x$, but I don't think there is a better (or faster) method.
AlastorMoody
13.12.2018 16:59
Let $CX=CY=x \implies BX=DY=1-x$, Hence, $\frac{1-x}{2}=\frac{x^2}{2} \implies x^2+x-1=0 \implies x=\frac{\sqrt{5}-1}{2}$ and Let $AC \cap XY=Z$ $\frac{[AXY]}{[XCY]} =\frac{AZ}{CZ}=\frac{\sqrt{2}-\tfrac{x}{\sqrt{2}}}{\tfrac{x}{\sqrt{2}}} =\frac{2}{x}-1=\boxed{\sqrt{5}}$