Silly Geometry Solutions

(Drew here.) I've done a post on each of A, C and N, and decided it's finally time to finish off with G. My style of geometry is very unconventional, as people have pointed out throughout the years. If I had a nickel for every time I used inversion after I failed to perform an angle chase... I'd probably have over half a dollar by now.

So here's a compilation of some of my most unique and/or elegant geometry solutions, and comments on my style.

2022 IMO Mock 2 P1

Let $APBCQ$ be a cyclic pentagon. A point $M$ inside triangle $ABC$ is such that $\angle MAB=\angle MCA$, $\angle MAC=\angle MBA$, and $\angle PMB=\angle QMC=90^\circ$. Prove that $AM$, $BP$, and $CQ$ concur.

Being in my inversion addiction phase, after trivial angle chasing didn't work, I immediately attempted inversion. Point $M$ is where all the angle conditions are, so why not invert at $M$?

I'll leave you to check that this just returns the same problem statement. At times even that could produce a useful overlay between the inverted and uninverted diagrams, but there's no relation in this case. Bummer.

It's time to try something more daring -- inversion at $A$. Everything is fine until we have to handle the angle condition $\angle PMB=\angle QMC=90^\circ$, but we'll deal with that later.

The first angle condition translates to $\angle M'AB'=\angle AM'C'$ in the inverted diagram, which means $AB'\parallel M'C'$. Similarly $AC'\parallel M'B'$, and boom we have ourselves a parallelogram $AB'M'C'$.

At this point, I wanted at least some way to construct $P'$. It didn't have to be very tractable, I just needed a theoretical way to do so. $PM\perp MB$ implies that $(AP'M')$ is orthogonal to $(AM'B')$ (meaning the tangents at the intersection are perpendicular), then $P$ lies on $(ABC)$ so $P'$ lies on $B'C'$, and on the ray $C'B'$ past $B'$ because $APBC$ lie on the circle in that order.

In short, these are the steps to construct $P'$:

1. Construct $(AM'B')$.
2. Construct the circle through $A$ and $M'$ orthogonal to $(AM'B')$.
3. This circle intersects $B'C'$ twice, choose the intersection on the ray $C'B'$ past $B'$ to be $P'$.

By symmetry, these are the steps to construct $Q'$:

1. Construct $(AM'C')$.
2. Construct the circle through $A$ and $M'$ orthogonal to $(AM'C')$.
3. This circle intersects $B'C'$ twice, choose the intersection on the ray $B'C'$ past $C'$ to be $Q'$.

They're almost the same, which is to be expected given the nature of the construction. But we need to remember the extra symmetry that we have derived: $AB'M'C'$ is a parallelogram. In particular, there's a reflection through $O$, the center of the parallelogram $AB'M'C'$, which maps $A$ to $M'$ and $B'$ to $C'$.

What happens if we do it to $Q'$?

These are the steps to construct $Q''$, the reflection of $Q'$ through $O$:

1. Construct $(M'AB')$.
2. Construct the circle through $M'$ and $A$ orthogonal to $(M'AB')$.
3. This circle intersects $C'B'$ twice, choose the intersection on the ray $C'B'$ past $B'$ to be $Q''$.

Wait, this is exactly the same as $P'$!

So $O$ is the midpoint of $P'Q'$ and we finish by a simple PoP: $OP'\cdot OB'=OQ'\cdot OC'$ proves that $O$ lies on the radical axis of $(AB'P')$ and $(AC'Q')$, then inverting back gives the desired statement.

"nice" was written alongside the $7$ when I received back my script.

A corollary of this solution is that the $90^\circ$ doesn't actually matter! As long as the two angles $\angle PMB$ and $\angle QMC$ are equal, the construction of the circle (which was orthogonal because of the $90^\circ$ angle) will turn out to be symmetric. In the original problem, we do have another pair of equal angles like this, namely $\angle QMB=\angle PMC$, so $BQ$ and $CP$ concur on $AM$ too, at a point $S$. $S$ turns out to lie on $(BMP)$ and $(CMQ)$ as well, which is used in the "normal" solution. 

USAMO 2016/5

An equilateral pentagon $AMNPQ$ is inscribed in triangle $ABC$ such that $M\in\overline{AB}$, $Q\in\overline{AC}$, and $N,P\in\overline{BC}$. Let $S$ be the intersection of $\overleftrightarrow{MN}$ and $\overleftrightarrow{PQ}$. Denote by $\ell$ the angle bisector of $\angle MSQ$.

Prove that $\overline{OI}$ is parallel to $\ell$, where $O$ is the circumcenter of triangle $ABC$, and $I$ is the incenter of triangle $ABC$.

This was the last problem on a National Team set and I've tried it months ago with no success, so I decided to work on it again on the ride back home. I was doing it on the Sketches app (basically GeoGebra, except the constructions feel more intuitive to me, and what I'm using to create these diagrams) and simply stared at the diagram, not knowing what to do.

Somehow, I had to work with the $OI$ line of the triangle formed by three lines $NP$, $AM$ and $AQ$...? The incenter felt especially out of place. I've always felt that the line $OI$ was difficult to deal with, so things weren't looking good here. Constructing some points trying to relate the equal lengths didn't work for me, either.

After I reached home, I continued thinking about the problem in the shower, with the diagram drawn on the glass. Three vectors along the sides - how does this relate to the line $OI$? How should this relate to the line $OI$, if working backwards? $\ell$ is the angle bisector of the lines formed by the equal length vectors $\overrightarrow{MN}$ and $\overrightarrow{PQ}$, i.e. parallel to $\overrightarrow{MN}+\overrightarrow{QP}$. But these vectors go in the opposite direction along the pentagon $AMNPQ$, so it may be better to say $\ell$ is perpendicular to $\overrightarrow{MN}+\overrightarrow{PQ}=-\big(\overrightarrow{NP}+\overrightarrow{QA}+\overrightarrow{AM}\big)$.

Our goal is now to show that the sum of three equal length vectors going around the sides of $\triangle ABC$ is perpendicular to $OI$. And the last step to reduce this problem to a calculation is to make these vectors have the same length as $BC$ -- the three vectors can then connect up! Our new statement is as follows:

Let $ABC$ be a triangle with circumcenter $O$ and incenter $I$. Let $J$ and $K$ be points on rays $BA$ and $CA$ such that $JB=BC=CK$. Prove that the vector sum $\overrightarrow{JB}+\overrightarrow{BC}+\overrightarrow{CK}=\overrightarrow{JK}$ is perpendicular to $OI$.

 

Miraculously, I've seen and done this problem before, as a headsolve too! It's equivalent to showing that $$\operatorname{Pow}_{(I)}(J)-\operatorname{Pow}_{(O)}(J)=\operatorname{Pow}_{(I)}(K)-\operatorname{Pow}_{(O)}(K)$$where $(I)$ and $(O)$ are the incircle and circumcircle respectively. The calculation is most easily done by subtracting the two sides, which gives

$$\begin{align*}\left[\left(a-\left(\frac{a-b+c}{2}\right)\right)^2-a(a-c)\right]&-\left[\left(a-\left(\frac{a+b-c}{2}\right)\right)^2-a(a-b)\right]\\&=\left(a-\left(\frac{a-b+c}{2}\right)\right)^2-\left(a-\left(\frac{a+b-c}{2}\right)\right)^2+a(c-b)\\&=a(b-c)+a(c-b)=0,\end{align*}$$

so we're done (and I'm still in the shower)!

This solution demonstrates how minimal I can be in my geometry solutions by finding ways to characterise the difficult parts of the diagram. In this case, it's finding how $OI$ relates to the equal lengths along the sides. It also explains my ability to headsolve some geometry problems, this one being my hardest (alongside IMO 2018 P6)!

There's also an easier finish after the reduction to the equal lengths: rotate each of the vectors by $90^\circ$, make one endpoint the circumcenter, and make the length the circumradius. The vector sum is then the orthocenter of the three arc midpoints, which is the incenter.

Again I will mention a corollary of my solution, which is that it only requires $AM=AQ=NP$ and $MP=NQ$. Not as unexpected as the previous one, but it shows that I'm not fully utilising the information in the problem.

Some problem from Mr Wang

Let $G$ be the centroid of $\triangle ABC$ and let the circumcentres of $\triangle GBC$, $\triangle GCA$, $\triangle GAB$ be $D$, $E$, $F$ respectively. Show that the circumcentre of $\triangle ABC$ coincides with the centroid of $\triangle DEF$.

At this point, I've already completed my Math Olympiad journey, and only wanted to crash an internal training session. This problem struck me as bashable, and seeing as nobody in the room was attempting to bash it, I took the task upon myself.

In complex bashing, taking the centroid is really easy, just average the three coordinates. And the circumcentre of $\triangle ABC$ is there for free. So the only nontrivial calculation is the circumcentre of $\triangle GBC$, from which the circumcentres of the other two triangles $\triangle GCA$ and $\triangle GAB$ follow by symmetry.

As usual the complex bash begins with setting $(ABC)$ to be the unit circle, and $g=\frac{a+b+c}{3}$ being the centroid. We can use the circumcentre formula to get $$f=\begin{vmatrix}a&a\overline a&1\\b&b\overline b&1\\g&g\overline g&1\end{vmatrix}\times\begin{vmatrix}a&\overline a&1\\b&\overline b&1\\g&\overline g&1\end{vmatrix}^{-1}.$$

From here, I'll let the solution speak for itself.

For the first matrix, we can create symmetry by calculating $$\begin{vmatrix}a&a\overline a&1\\b&b\overline b&1\\g&g\overline g&1\end{vmatrix}=\frac13\begin{vmatrix}a&1&1\\b&1&1\\3g&3g\overline g&3\end{vmatrix}=\frac13\begin{vmatrix}a&1&1\\b&1&1\\c&3g\overline g-2&1\end{vmatrix}.$$

The second matrix resembles the area formula, and is equal to $$\lambda\coloneqq\frac4i[\triangle GAB]=\frac4{3i}[\triangle ABC]$$

using $[\triangle XYZ]$ to denote the signed area of $\triangle XYZ$.

Finally we can calculate

$$\begin{align*}\frac{d+e+f}{3}&=\frac1{9\lambda}\left(\begin{vmatrix}a&1&1\\b&1&1\\c&3g\overline g-2&1\end{vmatrix}+\begin{vmatrix}a&1&1\\b&3g\overline g-2&1\\c&1&1\end{vmatrix}+\begin{vmatrix}a&3g\overline g-2&1\\b&1&1\\c&1&1\end{vmatrix}\right)\\&=\frac1{9\lambda}\begin{vmatrix}a&3g\overline g&1\\b&3g\overline g&1\\c&3g\overline g&1\end{vmatrix}\\&=0.\end{align*}$$

So their centroid is indeed the circumcentre of $\triangle ABC$.

Scary as determinants may be, absolutely no expansion needs to be done here! I greatly enjoy when I manage to skip calculations, because not only am I prone to making mistakes, but avoiding calculations also saves a great deal of time. Remarkable simplicity remains my signature in these "elegant" geometry solutions, but whether a clean bash counts as elegant depends on you.

Final Remarks

I would definitely love to put more solutions in here, but unfortunately some of my solutions have been lost to time. (Please do write down your solutions especially if you find them remarkable!) From memory, here are four solution outlines:

• Strengthen the statement by allowing the tangency point to be split into two differently defined points. This stronger statement is solved by a generalisation of the Butterfly Theorem.
• Invert, then there are fixed points $A$ and $B$ and a moving point $C$ depending on angle $\theta=\angle ABC$. Show that the length $BC$ is equal to $a\sin\theta+b\cos\theta$ for some constants $a$ and $b$ that don't need to be calculated; this shows that $C$ lies on a fixed circle via polar coordinates centred at $B$.
• Translate the problem to complex numbers, and use the fact that $a+b$ is the direction of the angle bisector if $|a|=|b|$. This resulted in bouncing back between complex numbers and synthetic progress eventually leading to a solve, and this was the very first time I used complex numbers in contest.
• Fix the direction of $\ell$, let it move parallel to a fixed line. Prove that all the circles formed are tangent at one point, and identify this point by choosing the $\ell$ which makes the circle have radius $0$.

My style of geometry has developed to be finding the easiest way to define / identify points, thanks to Euclidea (play it, I dare you) and Sketches / GeoGebra, even if it involves creating odd-looking auxilliary points or bashing. I once called this "how to use GeoGebra without using GeoGebra", and it's pretty useful! Questions such as "is $X$ uniquely determined" and "if $X$ can be defined cleanly, are we done?" have led me to solutions before.

I will admit that my classical synthetic is not very strong, which I tried to patch by learning many different methods. So my geometry toolkit is pretty absurd, which is why I just simply can't get the normal solutions to problems (all the way back in 2021, there was a streak where for 9/10 problems, I either didn't solve it or got a solution vastly different from intended). Over time I learnt to tame this chaos and find consistency in solving, but still found fun solutions at times.

Feel free to take olympiad geometry as a playground and see what's most enjoyable for you! I still find warping the playground via inversion pretty fun :)