A geometry problem by Rahil Sehgal

Let $AX=x$ and let $YC=y.$ Then $XB=2x$ and $AY=2y.$ Let $m\angle BAC=\theta.$

The area of $\triangle AXY$ can be expressed in terms of $x,$ $y,$ and $\theta :$

$A_{\triangle AXY}=\frac{1}{2}(x)(2y)\sin{\theta}=xy\sin{\theta}.$

The area of $\triangle ABC$ can also be expressed in terms of $x,$ $y,$ and $\theta :$

$A_{\triangle ABC}=\frac{1}{2}(3x)(3y)\sin{\theta}=\frac{9}{2}xy\sin{\theta}.$

Thus, $A_{\triangle ABC}=\frac{9}{2}A_{\triangle AXY}=\boxed{45}.$

$Given \space \triangle{ABC} \space in \space which \space X \space and \space Y \space are \space points \space on \space AB \space and \space AC \space respectively \space such \space that \space {AX\space \colon\space XB}\space={1 \space\colon\space 2} \space and \space{AY \space \colon\space YC} \space = \space {2 \space\colon\space 1} \space and \space ar\triangle{AXY}\space=\space 10$ $\rightarrow Let \space there \space be \space a \space point \space *D* \space on \space AC\space such \space that \space {AD \space\colon\space DC}\space=\space{1\space\colon\space2}$ $\Rightarrow By \space SAS \space property \space of \space similarity \space , \space \boxed{\triangle{AXD}\space\sim\space\triangle{ABC}}$

$\rightarrow \space{AD\space \colon\space AC}\space =\space {1 \space \colon \space 3} \space \therefore AD\space = \space \frac{1}{3}AC$

$\rightarrow \space {AY\space \colon\space AC}\space=\space{2\space \colon\space 3} \space \therefore AY\space=\space \frac{2}{3}AC$

$\rightarrow \space DY\space=\space {AY-AD} \space \Rightarrow DY \space=\space \frac{1}{3}AC \space=\space AD$

$In \space \triangle{AXY}\space ,\space XD \space is \space median\space \boxed{\therefore \space ar\triangle{AXD}\space=\space \frac{10}{2}\space=\space 5}$

$Now\space,\space {ar\triangle{AXD}\space \colon\space ar\triangle{ABC}}\space=\space [1\space \colon\space 3]^{2}\space=\space{1\space \colon\space 9}$

$\Rightarrow \space ar\triangle{ABC} \space=\space 9ar\triangle{AXD}\space=\space 9 \space\times\space 5 \space=\space 45$

$\boxed{\therefore \space ar\triangle{ABC}\space=\space 45}$