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The importance of "intuition" in geometry

Hii everyone! Today I will be discussing a few geometry problems in which once you "guess" or "claim" the important things, then the problem can easily be finished using not-so-fancy techniques (e.g. angle chasing, power-of-point etc. Sometimes you would want to use inversion or projective geometry but once you have figured out that some particular synthetic property should hold, the finish shouldn't be that non trivial)

This post stresses more about intuition rather than being rigorous. When I did these problems myself, I used freehand diagrams (not geogebra or ruler/compass) because I feel that gives a lot more freedom to you. By freedom, I mean, the power to guess. To elaborate on this - Suppose you drew a perfect diagram on paper using ruler and compass, then you would be too rigid on what is true in the diagram which you drew. But sometimes that might just be a coincidence. e.g. Let's say a question says $D$ is a random point on segment $BC$, so maybe you took $D$ close to midpoint of $BC$ or close to the feet of altitude from $A$. The thing is, in those cases there are more synthetic properties rather than when $D$ would be any random point. So your diagram would show you some claims to be true which might not be true in a general case.
Moreover, the diagrams on paper are not $100%$ accurate, so suppose something looks cyclic but is not exactly cyclic in a ruler-compass diagram, then you would not really consider it, but in a freehand diagram, since you know your diagram is not exact, if it looks cyclic you'd just assume it to be cyclic and try stuff with that which I think gives you an edge.

Also, I have added the diagrams to the questions I have discussed below, but in general practice, I don't really recommend using geogebra or those diagrams. But these are just for reference for now.

Well, you would be able to understand all the philosophy written above after you try a few questions with freehand diagrams yourself :P so let's just move onto the questions for now.

I will be discussing the following questions :- 
(PUMaC Finals 2017 A3) Let $ABC$ be a triangle with incenter $I$. The line through $I$ perpendicular to $\overline{AI}$
meets the circumcircle of $\triangle ABC$ at points $P$ and $Q$, with $P$ and $B$ on the same side of $\overline{AI}$. Let $X$ be a point such that $\overline{PX} \parallel \overline{CI}$, $\overline{QX} \parallel \overline{BI}$. Prove that $\overline{PB}$, $\overline{QC}$, $\overline{IX}$ are concurrent. 

(Iran TST 2017 P5) Let $ABC$ be a triangle. Points $P$, $Q$ lie on side $BC$ such that $BP=CQ$ and $P$ lies between $B$ and $Q$. The circumcircle of triangle $APQ$ intersects sides $AB$ and $AC$ again at $E$ and $F$ respectively. Point $T$ is the intersection of lines $EP$ and $FQ$. The two lines passing through the midpoint of $\overline{BC}$
and parallel to $\overline{AB}$ and $\overline{AC}$
intersect lines $EP$ and $FQ$ at points $X$, $Y$ respectively. Prove that the circumcircles of triangle $TXY$ and triangle $APQ$ are tangent to each other.

(USA TST 2006 P2) Let $ABC$ be an acute triangle with altitudes $\overline{AD}$, $\overline{BE}$, $\overline{CF}$ concurrent at the orthocenter $H$. A circle $\gamma$ centered at $S$ passing through $A$ and $H$ meets sides $AB$ and $AC$ again at points $P$ and $Q$. Given that the circumcircle of $\triangle SPQ$ is tangent to $\overline{BC}$ at $R$, prove that \[ \frac{BR}{CR} = \frac{DF}{DE}. \]

I will not be giving the solutions/details of the proofs or claims, but rather I'll try to motivate the things and you can finish the proofs of those claims yourself C:

PUMac Finals 2017 A3 : 
At first when you see the problem, you might be tempted to show that $PBIX$ is cyclic. Because then, by symmetry $QCXI$ will also be cyclic and you will be done by radical axis. But after trying a bit you would realize you can't prove this because it is actually not true. I don't see a direct way to disprove it except the observation that the angles in same segment don't look equal at all. e.g. $\angle BIX$ and $\angle BPX$ should be equal for cyclicity to hold but they don't look so. This is what you have to do intuitively.
Okay but what if we can find some other circle involving the points $B,P,I$ and symmetrically a circle with points $C,Q,I$. Then we will have $BP, CQ$ as two radical axis, and we will have to prove that $IX$ is radical axis of those two circles. (Note that all this is thought process behind solving the problem, it's not that easy to spot while actually solving!)

For cyclicity, we try to find the angle we can first. $$\angle PIB = \angle AIB - 90^{\circ} = \frac{C}{2}$$ But we also have $\angle ICB = C/2$. The only thing which we haven't used till now is the parallel condition, try to use that to get a nice cyclic quadrilateral.

Okay so if we introduce intersection of $PX$ and $BC$ as $D$ and intersection of $CX$ and $BC$ as $E$, then you can show that $BPDI, \ CQEI, \ PDEQ$ are cyclic quadrilaterals just by angle chase. And then you can finish using the way mentioned above.

Comments : The thing to prove looked like you need such circles and the angles very "nice" because of the parallel and perpendicular condition. So we basically just angle chase to get nice cyclic quadrilaterals and finish.

Before we start discussing the problem, I would like to state a nice technique which I learnt in a class and is quite helpful in geometry problems :-
Whenever you have a circle and a line and only one intersection of the line with circle is marked, then most of the times you should introduce the second intersection because it might have nice properties.

Iran TST 2017 P5 : 
Ok so now in this problem, you have midpoint of $BC$, say $M$, from which two parallel lines are drawn and you have that $P$ is reflection of $Q$ over $M$. Which means $M$ is quite-heavily involved. If you draw some freehand diagrams, it almost looks true that the tangency point lies on the line $AM$. And this is actually true!! (notice that you had the median and circle $(APQ)$, I just drew it's other intersection using the technique I stated above and that point turned out to be the tangency point which we need, nice, right?)

Okay so now how do we prove all this stuff? Well if you would like to try first then go ahead, because after you notice this it's just angle chase (yay!) 

Anyways, here is what we do - First, let $Z$ as intersection of $AM$ with $\odot{(APQ)}$. Then, in $\triangle PTQ$ you can show that $PMZX$ and $QMZY$ are cyclic quadrilaterals, which by miquel's theorem gives us that $XZYT$ is cyclic. So, we got that $Z$ is one of the intersections of $\odot{(APQ)}, \odot{(XYT)}$.

If you angle chase a bit, you will see that $\angle XMY = \angle BAC$, which let's us think that are these two triangles similar? Notice that we still have not used the fact that $BP = CQ$. Try to use this and the parallel condition to prove $\triangle MXY \sim \triangle ACB$.

Now to prove that $Z$ is the point of tangency, we use a well-known trick to prove two circles tangent: $$\angle PZX = \angle PAZ + \angle ZYX$$ Which you can prove using angle chasing and the above similarity.

Comments : To motivate similarity condition in an other (and probably better?) way, we can work backwards. We know that we have to prove $\angle PZX = \angle PAZ + \angle ZYX$. And we can use the cyclic quadrilaterals which we have already to see what this is reduced to, which is basically : $$\angle ABC = \angle MYX$$
But symmetrically you can get that if the above thing is true, then $\angle ACB = \angle MXY$ should also be true, which means you have to prove the similarity.

USA TST 2006 P2 :

In this question, $\odot{(SPQ)}$ tangent to $BC$ is a bit hard to draw, and doesn't really give anything nice directly. So how do you handle the things here? Well the idea is to see a bigger picture in here. So what we do is basically take any circle passing through $A,H$ with center $S'$ which intersects $AB,AC$ at $P',Q'$. And $\odot{(S'P'Q')}$ intersects $BC$ at two points, say $R_1, R_2$. We forget about the tangency conditions for now. Let's try to find nice things about $R_1$ and $R_2$ if there are any.

The reason we did this is - Suppose $R_1$ has some property say $x$ and $R_2$ has some property say $y$, then in the tangent case, $R$ will have properties of both $R_1$ and $R_2$! This is because, when a line and circle are tangent, the tangency point is basically the case when both the intersections coincide.

Now to see what $x$ and $y$ are you just need intuition. Because, after you guess the things, the proofs of them are just angle chase, so I will state what they are without wasting time xD. (The diagram attached below is for the general case)

One of $R_1,R_2$ lies on the line $S'H$, say $R_1$.
$R_2$ is intersection of perpendicular bisectors of $BP'$ and $CQ'$ (and it lies on $BC$ only :p)

After you prove the above two things, in the tangent case (which the question states) you will have $R$ has both the above properties mentioned, using which you can finish in the following way : $$\triangle QPR \sim \triangle EFD \implies \boxed{\frac{BR}{CR} = \frac{PR}{QR} = \frac{DF}{DE}}$$

And with this, I end this post :D I hope you enjoyed the post and liked the questions. The main takeaway I want you to have is that freehand diagrams are a lot nicer and give you a lot of freedom. Moreover, the better your intuition is with freehand diagrams, the more will be your confidence with geometry problems leading to more improvement :)

Pranav Choudhary


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