“The end,” as @alison_kiddle is fond of saying, “is not the end.”

When you’ve finished solving a problem, it’s almost always going back and figuring out what extra value you can get from it. Good questions to ask include:

• “How do I feel now?”
• “What went wrong?”
• “What held me up?”
• “Was there a better method that I missed?”
• “What would have helped me get the solution more quickly?”
• “What did I learn?”
• “Did I make any interesting errors?”

So, when my student and I struggled with a BMO question – question 5 here – it struck me that going back to do a post-mortem would be good practice.

There are spoilers below the line, so if you want to try it yourself first, do it before you read on.

### My (eventual) solution

The first thing was getting a diagram that satisfied all of the conditions. Here’s how I got mine to look OK:

• I set up my diagram with line BAP horizontal.
• If DP = DA, then D lies on the perpendicular bisector of AP,
• Circle $\Delta$ passes through P, A and D.
• I added a tangent to this circle at A
• Then circle $\Gamma$ passes through A, B and D
• The line and $\Gamma$ intersect at C

The angles at the base of the isosceles triangle, DAP and DPA, are equal.

The alternating segment theorem says that DAC is also equal to DPA.

Here is the nugget of the solution.

I added segments DC and DB; because DCA and DBA subtend the same chord, they are equal (red angles).

Then BDP and CDA are equal, because angles in a triangle sum to the same total; therefore triangles BDP and CDA are similar.

Because DP = DA, these triangles are in fact congruent, so BP = CA as required.

### How do I feel now?

Three parts pretty good – it’s a really pleasing solution! – to one part slightly annoyed I didn’t think to try building those triangles sooner.

### What went wrong?

I didn’t get any sort of solution until long after class had finished!

### What held me up?

I think I focussed too closely on the triangle to begin with, chasing angles in the circle that matched those there.

I also struggled to reconcile that the two lines purported to be the same length didn’t start from the same point – I felt like I needed to make a circle out of them, which is obviously not the way forward.

I also tried splitting AC into smaller parts, without any success.

### Was there a better method that I missed?

Unclear. The two congruent triangles are rotations of each other about D; being able to show that DC = DB might make it drop out with even less work.

### What would have helped me get the solution more quickly?

Running through a checklist of circle theorems.

### Did I make any interesting errors?

I almost fell into a trap of assuming that DC was a tangent! Luckily I caught that one in time.

### What did I learn?

That nemesis follows hubris? The solution came to me after I reflected the circle ABC in the line L so that PC’ corresponded to AC. That led to a correct (but messy) solution that refined itself in a couple of drafts to the version I gave here.

Possible recommendations for future questions:

• Go through a checklist of theorems
• Draw bigger and clearer
• Use symmetry, but discard it if needed.
• Watch out for false assumptions!

So that’s how I do a puzzle post-mortem. Are there other questions you would ask? Anything you would do differently? Let me know!