Problem Post 2-2: A Sampler of `Olympiad' Geometry Concepts

Problem Posts

[Level 4]
Henry

This post is meant to be a `sampler' of sorts, to show the most common tag words one will see in olympiad geometry problems.

I've been fortunate enough to find two remarkable problems, and solve them in ways that form a whirlwind tour of the subject. The first problem demonstrates side chasing, isosceles triangles, some cyclic quadrilaterals and spiral similarity. The second problem demonstrates power theorems, cyclic quadrilaterals, collinearity, and triangle geometry.

Problem 1

This one comes from Andreescu and Gelca's Mathematical Olympiad Challenges.

QUESTION
(Andreescu, Gelca) Let $B$ and $C$ be the endpoints and $A$ the midpoint of a semicircle. Let $M$ be a point on the line segment $AC$, and $P$, $Q$ the feet of the perpendiculars from $A$ and $C$ to the line $BM$, respectively. Prove that $BP=PQ+QC$.

Laconic Solution Sketch This is a side chasing problem. Invoke isoceles triangles to turn it to an angle chasing problem, with a dash of spiral similarity.

Discussion

I think the most obvious clue is the fact that the plus symbol was used with segment lengths. It's highly unnatural in constructions to have that property; midpoints and isosceles triangles are more common. So let's induce that. Since $P$, $Q$, and $B$ are collinear anyway (in the order $B$, $P$, $Q$, no dispute on that) let's make a point $D$ on that line so that $QC=QD$. It's apparent that triangle $QCD$ is isosceles right, which is a good thing, because triangle $ABC$ is too. Now all we have to do is show that $P$ is a midpoint of $BD$ -- that's a more familiar request, isn't it?

Now since we're already seeing isosceles triangles, let's stretch our luck. Iff $P$ were a midpoint of $BD$, then $AB=AD$ by the Pythagorean Theorem. That happens, iff $AC=AD$. One way to show this would be to try to angle chase and show that $\angle ADC=\angle ACD$, but it seems far-fetched that we can successfully access $D$, being artificially introduced. (I haven't really checked if angle chasing won't work, do comment if you've found a way!)

Rather, look at the fact that $CD$ already forms the base of another isosceles triangle. So essentially, if $AC=AD$, we can cleave both triangles $QDC$ and $ADC$ in half by a perpendicular bisector. Essentially, we win if we can show that $A$, $Q$, and the midpoint $N$ of $CD$ are collinear.

The magic is to see how $ANC$ looks a lot like $BQC$ if our assertion is correct; they should be similar. How do we formalize this? Well, for starters, you have to subject $BQC$ to the same spiral similarity that takes $CQ$ to $CN$: rotate by $-\pi/4$ about $C$ (That's if your semicircle goes CAB counterclockwise) and scale by factor $\sqrt{2}/2$ (I'm just giving the numbers but as you probably know that's irrelevant.) Now note that if you do this transformation, $BQ$ should line up perfectly with $QN$. But if you rotate $BQ$ by $-\pi/4$ it lines up perfectly with $AQ$. So our three points are collinear, and we're finished!

Review

Let's look at the important steps in the solution:

• The points were inaccessible to side chasing, so we converted it to an angle chasing problem.
• We evolved things into same-based isosceles triangles, and then into a question of collinearity, and finally spiral similarity.

Problem 2

This next one comes from the 2013 IMO Shortlist.

QUESTION
(IMO SL2013-1) Let $ABC$ be an acute triangle with orthocenter $H$, and let $W$ be a point on the side $BC$, lying strictly between $B$ and $C$. The points $M$ and $N$ are the feet of the altitudes from $B$ and $C$, respectively. Denoteby $\omega_{1}$ the circumcircle of $BWN$, and let $X$ be the point on $\omega_{1}$ such that $WX$ is a diameter of $\omega_{1}$. Analogously, let $\omega_{2}$ be the circumcircle of triangle $CWM$, and let $Y$ be the point so that $WY$ is the diameter of $\omega_{2}$. Prove that $X$,$Y$, and $H$ are collinear. (Diagram is mine; not given)

Discussion

The first thing I did was draw the diagram, and it's a mess, but not overly so. So first things first -- what do we want? To show that $H$ is on line $XY$. Well, the problem is $X$ and $Y$ aren't particularly special points; they're just connected to the rest of the diagram by being diametrically opposite to $W$ on their respective circles.

I had the time to be meticulous about my diagram, so I noticed instantly that $XY$ also hits the other intersection $K$ of $\omega_{1}$ and $\omega_{2}$. Does this always happen? Well, the obvious next step is to look at the centers $O_{1}$ and $O_{2}$ of $\omega_{1}$ and $\omega_{2}$. They're respectively the midpoints of $XW$ and $YW$. So now, $O_{1}O_{2}$ is the midline of triangle $WXY$, so that $W$ is twice as far from $XY$ as it is to $O_{1}O_{2}$. It follows almost naturally to conclude because $K$ is the reflection of $W$ across $O_{1}O_{2}$, it must hit $XY$.

So what have we done here? We've just added another point to the line $XKY$ -- what does that do? Well, $K$ is more related to the original figure than $X$ or $Y$. It's presumably much easier to show that $HK$ is parallel to $XY$ rather than to worry about $HX$ or $HY$.

The next magic observation (again, thanks to my meticulous diagram) is that $K$ appears to be on line $AW$, or rather $A$ appears to be on the line $KW$. But $KW$ is an important construct between the two circles, the radical axis. To say that $A$ is on the radical axis of the circles is equivalent to saying that $AN\cdot AB=AM\cdot AC$.

We pull up our notebook on triangle geometry and we see that \begin{eqnarray*} AN\left(AB\right) & = & \left(b\cos\alpha\right)c\\  & = & \left(c\cos\alpha\right)b\\  & = & \left(AM\right)\left(AC\right) \end{eqnarray*} so our guess is correct: $AKM$ is a line.

This result is another gift from our well-drawn diagram, but we have to ask ourselves, is it even useful? Well, it makes $HK$ a lot more accessible; now all we have to do is to show that $XY$ makes the same angle with $AW$ as $HK$ does. But wait, $O_{1}O_{2}$ is a midline of $WXY$ so it's parallel to $XY$. But $O_{1}O_{2}\bot WK$! So now, we win if we can show that $HK\bot AW$.

But we weren't given any right angles... right? Not really: now's another good time to consult The Codex. It will tell us that orthocenter means cyclic quadrilaterals with lots of right angles. The ones important to our problem are $ANHM$, $BNHP$, and $CMHP$, where $P$ is the foot of the altitude from $A$. More importantly, we have $\angle ANH=\angle AMH=\pi/2$. 

So we do angle chasing (every time we replace an angle with $\pi$ minus another, we are using the relevant cyclic quadrilateral): \begin{eqnarray*} \angle NKM & = & 2\pi-\angle NKW-\angle MKW\\  & = & 2\pi-\left(\pi-\angle NBW\right)-\left(\pi-\angle MCW\right)\\  & = & \angle NBW+\angle MCW\\  & = & \pi-\angle BAC\\  & = & \pi-\angle NAM\\  & = & \angle NHM \end{eqnarray*} which shows us that the points $A,\, N,\, H,\, K,\, M$ are cocyclic, and so $\angle AKH=\angle AMH=\pi/2$. So $KH\bot AK$ and we're done.

Review

As promised, this problem is a sampler of many of the important basic concepts:

• The whole game was to show three points were collinear.
• We were given the two circles and so their radical axis came into play.
• The problem invoked the orthocenter, so we looked up some facts on triangle geometry twice.
• We used five cyclic quadrilaterals to show that five points lie on the same circle (or more professionally, that those five points are cocyclic.)

Moreover, we have lots of `morals' to take away from this problem:

• I had the uncanny feeling that I saw the construct with the two circles before. If you had that in some kind of notebook, you wouldn't have to spend five or ten minutes trying to convince yourself of the collinearity.
• If you had a strong foundation in triangle geometry or notes on triangle centers, the orthocenter would be easy lunch.
• A lot of working backwards. Many, many geometry statements are the equivalence of two statements (they're if-and-only-if statements), so it's far less risky (compared to, say, number theory problems) to start at the endgame.