SAMO 2012 Problem 2

[Fahim Shahriar]

Let $ABCD$ be a square and $X$ a point such that $A$ and $X$ are on opposite sides of $CD$ . The lines $AX$ and $BX$ intersect $CD$ in $X$ and $Y$ respectively. If the area of $ABCD$ is $1$ and the area of $XYZ$ is $\frac {2} {3}$ , Determine the length of $YZ$.

[Fahim Shahriar]

Structure

ABCD=1.jpg (28.99KiB)Viewed 3103 times

[SANZEED]

Let us assume that $YZ=a$ and the altitude of $\triangle XYZ$ is $h$. Then the altitude of $\triangle XAB$ is $1+h$. Now $AB^{2}=1\Rightarrow AB=1$. So by similarity between $\triangle XYZ$ and $\triangle XAB$, we have $\frac{a}{AB}=\frac{a}{1}=\frac{h}{1+h}$. Again we are given that $\frac{1}{2}ah=\triangle XYZ=\frac{2}{3}$. So $a=\frac{4}{3h}$. Substituting this we have, $\frac{4}{3h}=\frac{h}{1+h}$ i.e. $3h^{2}-4h-4=0\Rightarrow (h-2)(3h+2)=0$. Clearly $h=2$, since $h$ is positive, and thus $YZ=a=\frac{4}{6}=\frac{2}{3}$.

[photon]

let,YZ=a.height of trapizoid $AYZB$ is $1$. Area($AYZB$)=$\frac{1}{2}.1.(YZ+AB)=\frac{a+1}{2}$
Area($AXB$)=$XYZ+AYZB=\frac{2}{3}+\frac{a+1}{2}=\frac{3a+7}{6}$ . in similar triangle $XYZ,AXB$
$\frac{a^2}{1}=\frac{XYZ}{AXB}=\frac{\frac{2}{3}}{\frac{3a+7}{6}}$ , $\Rightarrow 3a^3+7a^2-4=0 \Rightarrow (a+1)(3a^2+4a-4)=0 $ . $a=-1$ and $3a^2+4a-4=0$ gives $a=\frac{2}{3},-2$ . so $YZ=\frac{2}{3}$

by the way what is SAMO?

[Fahim Shahriar]

by the way what is SAMO?

SAMO means "South African Math Olympiad". It's a problem of SAMO, Senior Round 3.