Triangles $OAB,OBC,OCD$ are isoceles triangles with $\angle OAB=\angle OBC=\angle OCD=\angle 90^o$. Find the area of the triangle $OAB$ if the area of the triangle $OCD$ is 12.
Problem
Source: Malaysia National Olympiad 2010 Muda Category Problem 1
Tags: geometry, geometry unsolved
cobbler
24.02.2014 21:56
MathSolver94 wrote: Triangles $OAB,OBC,OCD$ are isoceles triangles with $\angle OAB=\angle OBC=\angle OCD=\angle 90^o$. Find the area of the triangle $OAB$ if the area of the triangle $OCD$ is 12. Notice that the 3 said triangles are all 45-45-90. Let $OC=CD=x$. We have $OB=BC=\frac{x}{\sqrt{2}}$ and $OA=AB=\frac{x}{2}$. Moreoever, $[OCD]=12\implies \frac{x^2}{2}=12\implies x=2\sqrt{6}$. It follows $[OAB]=\frac{\left(\frac{x}{2}\right)^2}{2}=\frac{\left(\frac{2\sqrt{6}}{2}\right)^2}{2}=\boxed{3}$.
sunken rock
24.02.2014 23:38
Take $M,N$ midpoints of $OD, OC$ respectively, see that $\triangle OCM\cong\triangle OCB,\ \triangle OBN\cong\triangle OBA$, hence $[OCD]=2[OCB]=4[OBA]$. Best regards, sunken rock