If we have a circle of radius $R$ around center $O$ and its inscribed triangle $XYZ$ that is acute as well as scalene. $XY$ is the longest side. $XA,YB, ZC$ are the altitudes of the triangle $XYZ$. Let $L$ be the symmetric point of $A$ w.r.t. $BC$ and $M$ be the symmetric point of $B$ in $AC$. $P$ is the intersection of $XL$ and $YM$. $H$ is the orthocenter of triangle $XYZ$.

Then $OP\cdot OH = f(R)$. Find $f(R)$.


  • $\begingroup$ Did you put any thought into this yourself? $\endgroup$ – Daniel R Nov 3 '14 at 10:59

Observe how the whole setup is invariant under scaling. So if you change your unit of scale, say from centimeters to millimeters, then you'd do the same construction but use different numbers. More precisely, for lengths you'd use numbers which are 10 times as large. So the left hand side of your equation, where you multiply two lengths, would increase by a factor of 100. Therefore the right hand side has to do the same, so you know that $f(R)=\alpha R^2$ for some $\alpha\in\mathbb R$.

Now make an educated guess as to the value of $\alpha$, e.g. by doing measurements in your construction. Then you have the answer, although you might still need a proof that this is indeed the correct answer, or that the claimed relation actually holds, i.e. that the product of lengths really only depends on $R$.


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