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RMO 2008

This is a standard High School Olympiad problem and for an experienced problem solver a quite easy solve. But how was this problem created. To pose a problem, I believe is much harder, than to solve a posed problem.

Here the problem poser had to first make the figure up and then simultaneously realise that $ND$ had the wonderful property of being equal in magnitude to the circumradius. Is there a nifty way to find out these wonderful geometric properties?

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    $\begingroup$ Most olympiad problems I have created come from analyzing the solution to another (preferably hard problem) problem, and looking at some key ideas, and then looking at how we can change the hypothesis so that a similar idea still works. $\endgroup$ – Jorge Fernández Hidalgo Sep 30 '16 at 3:44
  • $\begingroup$ I have also appeared in RMO ,so it is quite interesting to see how the guys at NBHM design these problems :) $\endgroup$ – user369582 Sep 30 '16 at 4:57
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    $\begingroup$ I don't know how this particular example was created, but I found a blog post by Evan Chen which relates how Olympiad geometry problems are created. $\endgroup$ – Ankoganit Sep 30 '16 at 11:34
  • $\begingroup$ @Ankoganit Thanks for that link. What Evan Chen calls "Or shorter yet: build up, then tear down." I'd call even shorter: obfuscation. And this is one of very few contexts where I'd use that word appreciatively ;-) $\endgroup$ – dxiv Oct 1 '16 at 2:49
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Using GeoGebra is a fantastic way to come up with geometry problems such as this. Just screwing around with GeoGebra can give you very interesting and contest-able geometry problems, and I'd bet that this problem was created with GeoGebra or an equivalent program.

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    $\begingroup$ This one appears to come from RMO 2008 so it could have conceivably used GeoGebra in the making (though, as a personal opinion, unlikely). I would point however that challenging and un-obvious geometry problems have been posed from ancient times, long before computers existed. $\endgroup$ – dxiv Sep 30 '16 at 3:50
  • $\begingroup$ @dxiv You're correct, but most geometry problems from recent years (well, those made in the USA at least) are GeoGebra or the like. $\endgroup$ – Carl Schildkraut Sep 30 '16 at 3:52
  • $\begingroup$ I would be honestly curious if you found some evidence to back that assertion up. $\endgroup$ – dxiv Sep 30 '16 at 3:54
  • $\begingroup$ @dxiv I don't have any hard evidence, but I've met about 10 people who write problems for American olympiads and they've all admitted to using GeoGebra to come up with geometry problems. It's not hard evidence (nor would it be conclusive if I had proof of it) but it's a very good indication. In addition, as someone who really dislikes geometry but enjoys writing math olympiad problems, the most interesting problem that I've created (in my opinion) was a geometry problem I made with GeoGebra. It's an amazing tool. $\endgroup$ – Carl Schildkraut Sep 30 '16 at 3:56
  • $\begingroup$ using geogebra sounds more like a way to speed up the process than something essential to the process itself. I don't think the difference between doodling in a notebook and using the computer progrem is THAT big (obviously there's a big difference but I don't think its that essential). $\endgroup$ – Jorge Fernández Hidalgo Sep 30 '16 at 3:57
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Let me show you another solution ;)... Maybe this is how they came up with this problem. Manipulating midpoints, orthogonality and parallel lines.

Draw the perpendicular of edge $AC$ at point $A$ and let it intersect the line $BN$ at point $P$. Since $\angle \, CBP = 90^{\circ} = \angle \, CAP$, it follows that quadrilateral $BCAP$ is inscribed in a circle (the circumcircle of triangle $ABC$) and $PC$ is a diameter of that circle. Since line $FK \equiv NK$ is perpendicular to $AC$, it is parallel to $AP$. As $FK$ passes trough the midpoint $F$ of segment $AB$ and is parallel to $AP$, its intersection point $N$ with segment $BP$ is the midpoint of $BP$ (i.e. $FN$ is the midsegment of triangle $BAP$). Consequently, in triangle $BCP$ the points $N$ and $D$ are midpoints of edges $BP$ and $BC$ respectively, making $ND$ the midsegmetn of triangle $BCP$ parallel to the diameter $CP$ and half of its length, i.e. $ND = \frac{1}{2} CP$ equals the radius of the circumcirlce of triangle $ABC$.

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  • $\begingroup$ +1 Great solution! But I'd still be surprised if this is how they came up with the problem. $\endgroup$ – The Cryptic Cat Sep 30 '16 at 4:50
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This is NOT an answer to your question but is a shorter alternative version to the provided solution.

It should be clear that why the square-marked angles are right angles. Then, we can say that FNBD, and FBDO are cyclic quadrilaterals.

enter image description here

This means B, D, O, F, and N are all on the red dotted circle. Then, $\angle BNO = 90^0$.

Result follows from the fact that DN = OB because they are the diagonals of the rectangle BDON.

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