Let's say $AM$ and $AN$ are tangent lines to a circle centered at $O$. $L$ is a point on arc $MN$. Line $ML$ and $NL$ intersect with the line passing $A$ parallel to $MN$, at $P$ and $Q$. If $\angle POQ=45°$, prove that the area of circle $O$ is $2\pi$ times the area of $\triangle OPQ$.

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I have discovered that points $ALMQ$ are concyclic, as well as points $ALNP$, but I cannot connect them with the asked area. I believe the problem can be solved using power of a point.

  • $\begingroup$ Very good, that construction takes more than just a compass and a straight edge. There is a fair amount of figuring to make sure that line QP passes through the intersections of tangents, radii and chord extensions, and is parallel with the chord spanning the tangent points. $\endgroup$ – Phil H May 6 '18 at 18:26

From power of $P$ with respect to circles $ALMQ$ and $MLN$ one gets: $$ QA\cdot QP=QL\cdot QN=(OQ+r)(OQ-r), $$ where $r$ is the radius of circle $MLN$. From power of $Q$ with respect to circles $ALNP$ and $MLN$ one gets: $$ PA\cdot PQ=PL\cdot PM=(OP+r)(OP-r). $$ Adding these two equations together one gets $$ PQ^2=OP^2+OQ^2-2r^2, \quad\hbox{that is:}\quad OP^2+OQ^2-PQ^2=2r^2. $$ On the other hand, from the cosine rule applied to triangle $OPQ$ one has: $$ OP^2+OQ^2-PQ^2=2OP\cdot OQ{\sqrt2\over2}=4\,\text{Area}_{OPQ}. $$ It follows that $$ \text{Area}_{OPQ}={r^2\over2}. $$


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