### Reim

(KY here.)

In this blog we look at two problems. If you have not done them before, I would highly recommend trying them first before continuing with the rest of the blog.

- (IMO 2017) Let $R$ and $S$ be different points on a circle $\Omega$ such that $RS$ is not a diameter. Let $\ell$ be the tangent line to $\Omega$ at $R$. Point $T$ is such that $S$ is the midpoint of the line segment $RT$. Point $J$ is chosen on the shorter arc $RS$ of $\Omega$ so that the circumcircle $\Gamma$ of triangle $JST$ intersects $\ell$ at two distinct points. Let $A$ be the common point of $\Gamma$ and $\ell$ that is closer to $R$. Line $AJ$ meets $\Omega$ again at $K$. Prove that the line $KT$ is tangent to $\Gamma$.
- (IMO 2018) Let $\Gamma$ be the circumcircle of acute triangle $ABC$. Points $D$ and $E$ are on segments $AB$ and $AC$ respectively such that $AD = AE$. The perpendicular bisectors of $BD$ and $CE$ intersect minor arcs $AB$ and $AC$ of $\Gamma$ at points $F$ and $G$ respectively. Prove that lines $DE$ and $FG$ are either parallel or they are the same line.

# Reim Theorem

The idea is very simple. Take two circles, say $\omega_1, \omega_2$ that intersect at two points $A,B$. Through $A$ draw a line, say intersecting $\omega_1, \omega_2$ at $C,E$. Through $B$ draw another, intersecting $\omega_1,\omega_2$ at $D,F$. Then $CD\parallel EF$.

Check that this holds through for as many configurations as you can draw - no matter how you draw the circles (as long as they intersect), or no matter how you draw the lines!

Check that the converse also holds: let $ABCD$ be a trapezium, $AB\parallel CD$. A circle through $A,B$ meets lines $AD, BC$ at $E,F$. Then $CDEF$ is also cyclic.

Two simple examples:

The keywords I like to tell myself when it comes to using this theorem is a **common chord** or **parallel lines**. The common chord makes sense - more often than not you use Reim when the circles intersect at two points, in which case those points form the common chord.

Sidenote - you don't actually need the circle to intersect at two points! The parallel lines still occur if the circles only intersect at one point(i.e. tangent), in which case we have the usual homothety argument.

It is worth noting that the entire proof of this theorem is a (very simple) angle chase - so all information you have about the diagram is encoded in **angle conditions**. Hence it should make sense that it does not really matter if the circles intersect at one point or two (or if the lines you draw through the intersections is tangent or not).

Now let's talk about the problems.

## IMO 2017 P4

Let $R$ and $S$ be different points on a circle $\Omega$ such that $RS$ is not a diameter. Let $\ell$ be the tangent line to $\Omega$ at $R$. Point $T$ is such that $S$ is the midpoint of the line segment $RT$. Point $J$ is chosen on the shorter arc $RS$ of $\Omega$ so that the circumcircle $\Gamma$ of triangle $JST$ intersects $\ell$ at two distinct points. Let $A$ be the common point of $\Gamma$ and $\ell$ that is closer to $R$. Line $AJ$ meets $\Omega$ again at $K$. Prove that the line $KT$ is tangent to $\Gamma$.

Diagram:

First thoughts: the entire problem is phrased in terms of just intersection of circles and tangent lines, i.e. angle conditions, except for that one pesky midpoint condition. So, the problem difficulty is to translate that midpoint into some angle conditions, so that we can angle chase and prove tangency.

Instantly, Reim's theorem tells us that $RK\parallel AT$. This motivates the construction of a *parallelogram*, and it is important to note that one common way of removing the midpoint from a picture is to form a parallelogram (One similar place this appears is for example, reflecting the orthocenter of some triangle across some midpoint).

So, we construct the point $X$, which is $A$ reflected across $S$. Then, we have a parallelogram, and quite nicely $K$ lies on $RX$.

It turns out that the difficult part of the problem is solved - the rest is a simple angle chase, with an intermediate step being showing that $XKST$ is cyclic.

Note: most solutions are mainly motivated by trying to deal with the midpoint. Reflecting $K$ across $S$ is also possible. Another way is to draw the midpoint of $KR$, then have some similar triangles and midpoints.

## IMO 2018 P1

Let $\Gamma$ be the circumcircle of acute triangle $ABC$. Points $D$ and $E$ are on segments $AB$ and $AC$ respectively such that $AD = AE$. The perpendicular bisectors of $BD$ and $CE$ intersect minor arcs $AB$ and $AC$ of $\Gamma$ at points $F$ and $G$ respectively. Prove that lines $DE$ and $FG$ are either parallel or they are the same line.

I was here!!! Let's start with the diagram:

While technically the diagram does not require it, I have also drawn in the segments $BF,DF$ and same for $CG,EG$. This should be natural - they are kind of the only way to link the perpendicular bisector to the diagram.

First thoughts: there should not be some very easy angle chase that gets us the result. Although the "perpendicular bisector" length condition can somewhat be changed read as an angle condition due to reflection (e.g. $\triangle BDF$ isoceles), the line $FG$ looks way to out of place and it does not seem possible to chase any angles related to $FG$ directly.

There are very few points in the diagram. Bashers can probably get this quite quickly.

Unfortunately I have no experience in bashing, so let's look for other ways (I had a slight regret at this point during the contest but quickly moved on). Since $FG$ is completely out of place, we likely have to construct new points to be able to solve the problem.

Honestly I recall being stuck at this point for awhile, not being sure what to construct, probably a combination of lack of sleep (about 4.5-5 hours the previous night) and the very first problem in contest and all.

After calming a little, my thought process was: let's try a degenerate case. What happens when $B=D$? Then $F$ is also the same point, and we must have $B,E,G$ collinear.

In this case the problem is clearly true - say we extend $BE$ to meet circle again at $G$, then if $\triangle ABE$ is isoceles then so is $\triangle CGE$.

But what is the difference from the normal case? It is easy here because we have some "butterfly" structure in a circle, so that gives us similar triangles. More specifically, we have the "extension" of $GE$ in this diagram!

So let's extend $FD$ and $GE$ in the original diagram, meeting the circle again at $X,Y$.

Now the problem's quickly done. Because $BF=FD$, we have $AX=XD$, and similarly $AY=AE$, but problem condition tells us that $AD=AE$, so $XYDE$ is cyclic with center $A$ and $DE\parallel FG$ by Reim.

Embarrassingly, this took me 45 minutes in the contest, but at least it's still done. In hindsight, a crucial idea that I definitely should have been aware of (probably said by Jeck before) is that for any chord in a circle, chances are you want to extend it - to meet the circle at 2 points.

After the contest Yu Peng said that he took 10 minutes because he had Reim theorem in mind, because I kept talking about it over IMO training. Glad it has helped yay.

## Comments + thoughts

Reim theorem is definitely something that is not very well-known. For instance none of my trainers over my 5 years of training have talked about it. Even now, I don't think I've seen any article specifically talking about Reim, say compared to the number written for LTE or maybe Chicken Feet Lemma(Fact 5).

Yet, it is in my toolbox - I recall constantly having the words **common chord** in my head when seeing any problems that have circles that intersect, or parallel lines (or both).

Maybe people don't talk about it too much because it's too easy. But I think there's definitely value in knowing this exists (or other easy lemmas) and adding it to your toolbox.

This is because it makes **observations more quick** - e.g. in 2017 P4 we can instantly tell that the lines are parallel. Sure, someone without knowledge of Reim can find those parallel lines after short time, but when the diagram gets more complicated, it might be less clear.

Also, it **motivates constructions**, and lets you be more confident in constructing points. E.g. in 2018 P1, the constructions of $X,Y$ make sense because of the parallel lines we should get at the end.

Which leads to my next point - *there are many useful ideas that don't appear in the form of a theorem or a training set*, but these are what you should keep personalizing and have in mind when solving problems. For example, in the two problems, some ideas we covered:

- Reim theorem (parallel lines, common chord)
- Changing midpoint condition to an angle condition by reflection
- Realizing that solving the problem probably requires construction of new points
- Checking degenerate cases
- A chord in a circle should be extended to meet the circle twice

So, an important takeaway: after solving any problem, try not to only think of what theorem you used to solve them. Try to recall the more specific ideas you have used/how the problem motivates you to use that idea and internalize it. This way there stands to be much more value from each problem.

## Practice problems

I will just link to the contest page to prevent spoilers. In approximate order of difficulty:

I am actually quite surprised by how many IMO problems allow the use of Reim. And they're all recent!

## Script

David(tall) dug up my script. Nostalgia purposes I guess.

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