 |
 |
Problem Posts |  |
 |
QUESTION
(Canadian Mathematical Olympiad 2013, Problem 3) Let $G$ be the centroid of a right-angled triangle $ABC$ with $\angle BCA = 90^{\circ}$. Let $P$ be the point on ray $AG$ such that $\angle CPA = \angle CAB$, and let $Q$ be the point on ray $BG$ such that $\angle CQB = \angle ABC$. Prove that the circumcircles of triangles $AQG$ and $BPG$ meet at a point on side $AB$.
(Canadian Mathematical Olympiad 2013, Problem 3) Let $G$ be the centroid of a right-angled triangle $ABC$ with $\angle BCA = 90^{\circ}$. Let $P$ be the point on ray $AG$ such that $\angle CPA = \angle CAB$, and let $Q$ be the point on ray $BG$ such that $\angle CQB = \angle ABC$. Prove that the circumcircles of triangles $AQG$ and $BPG$ meet at a point on side $AB$.
Laconic Solution Sketch.
Using similar triangles, we can show that this common point is the foot of the altitude on $AB$ from $C$.
Discussion
This is perhaps a tad more difficult than the usual post, but I decided to include it in the lineup to demonstrate how simple ideas used correctly can fell a formidable problem.
A great deal of geometry problems can be done through two "arts":
- Manipulating angles. Some know it as angle chasing. Tools of the trade (and things we look for) are
- Cyclic quadrilaterals
- Isosceles triangles
- Parallel lines
- Angle bisectors
- Manipulating sides and sometimes areas. Tools of the trade include
- "Same base" theorem for triangles, and mass points to some extent.
- Midpoints
But some of the nastier problems (like this one) have markers for both angle and side chasing. A really good way to start is to find a conduit that translates side data to angle data or vice versa. Even the simple $AB=AC\Leftrightarrow \angle ABC = \angle BCA$ could be powerful if used correctly. Examples include:
- Power Theorems
- Trigonometry
- Ptolemy's Theorem
- Angle bisector theorem
- Menelaus and Ceva's Theorems (angles can imply concurrency/collinearity, which has consequences about side lengths)
- Isosceles Triangles
- Similar Triangles
- Reflections
- Spiral Similarities (see section 2 of the linked document)
But now to our problem.
For the entirety of this problem, we'll be using directed angles mod $180^\circ$, which basically means all angles are measured in the same counterclockwise direction, hence $\angle ABC = - \angle CBA$, and $\angle A_1BA_n = \angle A_1BA_2 + \angle A_2BA_3+\cdots+\angle A_{n-1}BA_n$. This has several advantages, because it is not configuration-specific: for example, using these angles, you don't need to know the clockwise order of four points to show they're on the same circle. Hence concyclic points, not cyclic quadrilateral.
Aside from the fact that it has both side and angle data, what's so ugly about the problem is the fact that the angles $\angle CPA$ and $\angle CQB$ are so inaccessible. The most sensible thing to do is equate it with something nicer. Drop altitude $CD$ to $AB$, and reflect $A$ across $CD$ to get $A^{\prime}$. Now $\angle AA^{\prime}C =\angle CAA^{\prime}=\angle APC$ implying that $C$, $A$, $A^{\prime}$, and $P$ are concyclic. (Look at that portentous isosceles triangle $CAA^\prime$!) Now that we can construct $P$ easily (as the intersection of the circumcircle of $CAA^{\prime}$ with ray $AG$, it's not too hard to draw the figure to scale:
Again, it's not too hard to guess that the common point could be the projection $D$ of $C$ onto $AB$. (there are only so many candidate points along $AB$, right? This is a logical choice to gun for.) Now, we want to show that $P,G,D,B$ are concyclic points. (It stands to reason that if you can do this, you can do exactly the same for $Q, G, D, A$.) To to that, we just have to show that $\angle DPA = \angle ABG$, or $\bigtriangleup GAB \sim \bigtriangleup DAP$. We know that $\bigtriangleup GAB \sim \bigtriangleup G^{\prime}AC$, because they're corresponding parts of similar triangles. Now for a master move:
That's easily provable - just use SAS. But how did we come up with this? We see it because a spiral similarity centered at $A$ takes $\bigtriangleup G^{\prime}AC$ to $\sim\bigtriangleup DAP$. A familiarity with spiral similarities extends far beyond applying the techniques; it tells you to expect things like this coming.For the entirety of this problem, we'll be using directed angles mod $180^\circ$, which basically means all angles are measured in the same counterclockwise direction, hence $\angle ABC = - \angle CBA$, and $\angle A_1BA_n = \angle A_1BA_2 + \angle A_2BA_3+\cdots+\angle A_{n-1}BA_n$. This has several advantages, because it is not configuration-specific: for example, using these angles, you don't need to know the clockwise order of four points to show they're on the same circle. Hence concyclic points, not cyclic quadrilateral.
Aside from the fact that it has both side and angle data, what's so ugly about the problem is the fact that the angles $\angle CPA$ and $\angle CQB$ are so inaccessible. The most sensible thing to do is equate it with something nicer. Drop altitude $CD$ to $AB$, and reflect $A$ across $CD$ to get $A^{\prime}$. Now $\angle AA^{\prime}C =\angle CAA^{\prime}=\angle APC$ implying that $C$, $A$, $A^{\prime}$, and $P$ are concyclic. (Look at that portentous isosceles triangle $CAA^\prime$!) Now that we can construct $P$ easily (as the intersection of the circumcircle of $CAA^{\prime}$ with ray $AG$, it's not too hard to draw the figure to scale:
Again, it's not too hard to guess that the common point could be the projection $D$ of $C$ onto $AB$. (there are only so many candidate points along $AB$, right? This is a logical choice to gun for.) Now, we want to show that $P,G,D,B$ are concyclic points. (It stands to reason that if you can do this, you can do exactly the same for $Q, G, D, A$.) To to that, we just have to show that $\angle DPA = \angle ABG$, or $\bigtriangleup GAB \sim \bigtriangleup DAP$. We know that $\bigtriangleup GAB \sim \bigtriangleup G^{\prime}AC$, because they're corresponding parts of similar triangles. Now for a master move:
If $\bigtriangleup G^\prime AD \sim\bigtriangleup CAP$, then $\bigtriangleup G^{\prime}AC\sim\bigtriangleup DAP$.
It's then pretty easy to prove $\bigtriangleup G^\prime AD \sim\bigtriangleup CAP$ by AA: $\angle PAD \equiv \angle GAB = \angle CAG^{\prime}$, hence $\angle G^\prime AD = \angle PAC$. Next, $\angle DG^\prime A = \angle G^\prime DC + \angle DCA + \angle CAG^\prime$. Whew! Now, for two magical moves:
First, extend ray $DG^\prime$ and you'll hit the midpoint $M$ of $AC$. Because $ACD$ is right, we have $DM=MC$. Isosceles triangle! There, we have $ \angle G^\prime DC = \angle DCA = \angle A^\prime CD$.
Next, consider $\angle CAG^\prime = \angle PAA^\prime = \angle PCA^\prime$.
Now, we have $\angle DG^\prime A = \angle PCA^\prime + \angle A^\prime CD + \angle DCA = \angle PCA$.
And we know that's all we need to prove the whole shebang, so we're done.
Solution
We will be using directed angles modulo $180^\circ$.
We will be using directed angles modulo $180^\circ$.
Let $D$ be the foot of the altitude from $C$ to $AB$ and $G^{\prime}$ the centroid of triangle $ADC$.
Reflect $A$ across $CD$ to get $A^{\prime}$. Now $\angle AA^{\prime}C =\angle CAA^{\prime}=\angle APC$ implying that $C$, $A$, $A^{\prime}$, and $P$ are concyclic.
We show by AA that $\bigtriangleup G^\prime AD \sim\bigtriangleup CAP$.
Reflect $A$ across $CD$ to get $A^{\prime}$. Now $\angle AA^{\prime}C =\angle CAA^{\prime}=\angle APC$ implying that $C$, $A$, $A^{\prime}$, and $P$ are concyclic.
We show by AA that $\bigtriangleup G^\prime AD \sim\bigtriangleup CAP$.
- First, $\angle PAD \equiv \angle GAB = \angle CAG^{\prime}$, hence $\angle G^\prime AD = \angle PAC$.
- Next, we prove $\angle DG^\prime A = \angle PCA$.
- Extend ray $DG^\prime$ and to midpoint $M$ of $AC$. We find $\angle G^\prime DC = \angle DCA = \angle A^\prime CD$.
- Moreover, $\angle CAG^\prime = \angle PAA^\prime = \angle PCA^\prime$.
- Now, $\angle DG^\prime A \angle G^\prime DC + \angle DCA + \angle CAG^\prime = \angle PCA^\prime + \angle A^\prime CD + \angle DCA = \angle PCA$
$\bigtriangleup G^\prime AD \sim\bigtriangleup CAP$ implies $\angle CAG^\prime = \angle PAD$ and also $CA/AP =AG^\prime/AD \Leftrightarrow CA/G^\prime = AP/AD$. Hence $$\bigtriangleup G^{\prime}AC\sim\bigtriangleup DAP \sim \bigtriangleup GAB$$ Then $\angle GPD \equiv \angle APD = \angle GBA \equiv \angle GBD$, implying that $P,G,D,B$ are concyclic.
We can similarly show that $Q, G, D, A$ are concyclic.
In conclusion, the circumcircles of triangles $AQG$ and $BPG$ meet at $D$, which is on side $AB$.
We can similarly show that $Q, G, D, A$ are concyclic.
In conclusion, the circumcircles of triangles $AQG$ and $BPG$ meet at $D$, which is on side $AB$.
All that thinking, and the solution only spans half a page! And to think even this is horrendously long for a solution: There's wonderfully brief solution using power theorems here.
No comments:
Post a Comment