# Projective geometry

Is using projective, inversive and advanced Euclidean geometry allowed on STEP? An example of what I mean:

10 S3 Q5
The vertices $A$, $B$, $C$ and $D$ of a square have coordinates $(0,0)$, $(a,0)$, $(a,a)$ and $(0,a)$, respectively. The points $P$ and $Q$ have coordinates $(an,0)$ and $(0,am)$ respectively, where $0 < m < n < 1$. The line $CP$ produced meets $DA$ produced at $R$ and the line $CQ$ produced meets $BA$ produced at $S$. The line $PQ$ produced meets the line $RS$ produced at $T$. Show that $TA$ is perpendicular to $AC$.

Solution
Let $X=TA\cap SC$. The complete quadrilateral $SQPR$ induces the harmonic bundle $(S,Q;X,C)$, and since $\angle QAS=90^\circ$, it follows that $AQ$ bisects $\angle CAX$; whence we obtain the result.

### Try again...

It should read like this: Is using projective, inversive and advanced Euclidean geometry allowed on STEP? An example of what I mean:

10 S3 Q5
The vertices $A$, $B$, $C$ and $D$ of a square have coordinates $(0,0)$, $(a,0)$, $(a,a)$ and $(0,a)$, respectively. The points $P$ and $Q$ have coordinates $(an,0)$ and $(0,am)$ respectively, where $0 < m < n < 1$. The line $CP$ produced meets $DA$ produced at $R$ and the line $CQ$ produced meets $BA$ produced at $S$. The line $PQ$ produced meets the line $RS$ produced at $T$. Show that $TA$ is perpendicular to $AC$.

Solution
Let $X=TA\cap SC$. The complete quadrilateral $SQPR$ induces the harmonic bundle $(S,Q;X,C)$, and since $\angle QAS=90^\circ$, it follows that $AQ$ bisects $\angle CAX$; whence we obtain the result.

### The general rule of thumb is

The general rule of thumb is that you shouldn't really be using knowledge beyond the specification that makes the question trivial. Since your solution is very short, I'd argue that perhaps it's not permissible. I could be wrong though.

### Results

I suppose that if you proved any and all projective, inversive and advanced Euclidean geometry results that you used starting from A-level/GCSE results then it might be ok. Might be a lot more work though!

It is hard to see how your solution could be worth 20 marks :-)

### I see

I admit that the "solution" above is very concise, and by no means morally correct. However, proofs like that usually get full marks at maths olympiads (like the BMO and IMO), so thought it better to ask before I went Cartesian, and hopefully got my signs correct :)

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