The geometry problem from my previous math puzzle has a nice solution. I particularly like it because it is one of these problems that are pretty hard to solve traditionally – unless you perform the right geometric construction – and at that point, the solution becomes trivial.

We will perform the following trick. From the point D we will draw a segment DE (with E on the longer side – AC in our case). E is chosen such that the angle(ABD) = angle(ADE).

At this point, our puzzle solving strategy is this:

a) Derive as much equations as possible from our new construction, and

b) Eliminate from these euqations any segments involving the point “E”.

First, we immediately observe that the triangle BAD is similar with the triangle DAE, therefore:

AD/AB = AE/AD = DE/BD (1)

From here we can extract AE without involving any other segments containing the point “E”:

AE = AD^2 / BA (2)

Second, because angle(ADE) + angle(EDC) = angle(ADC) = angle(ABD) + angle(BAD) we obtain: angle(BAD) = angle(EDC) = angle(DAE). In conclusion, the triangle ADC is similar with the triangle DEC, so:

DC/AC = EC/DC = DE/AD (3)

From here, we extract EC, again without involving any other segments containing the point “E”:

EC = DC^2 / AC (4)

Third, if we divide the other “unused” parts from equalities from (1) and (3) involving DE (but not involving any other segments containing “E”) we obtain: (AD/AB) / (DC/AC) = (DE/BD) / (DE/AD). AD simplifies, and we obtain:

AB/AC = DB/DC (5)

So, we have now three inequalities (2), (4) and (5), where the first two one involve the segments AE and EC.

Now, how we can get rid of “E” in AE and EC? We observe that AC = AE + EC, and by substituting AE and EC from the equations (2) and (4) above, we finally obtain something that doesn’t contain any “E”s anymore:

AC = AD^2 / AB + DC^2 / AC. (6)

From here, we extract AD^2:

AD^2 = AC * AB – AB/AC * DC^2 = AC * AB – DB/DC * DC^2 = AC * AB – DB/DC * DC^2. (7)

In conclusion, after everyting simplifies nicely, we obtain our initial equality:

AD^2 = AC * AB – DB * DC

P.S. Feeedback summary on the puzzles:

1) MovGP0 posted another possible solution to the geometry problem at http://de.wikipedia.org/wiki/Benutzer:MovGP0/Dreieck but it looks like the proof is apparently incomplete (although I might be wrong).

2) Many readers correctly pointed out that the other sequence puzzle is a classic one, called the Convay sequence (also named the “Look and Say” sequence). As an anecdotic fact, I heard that it was initially done as a scientific study where people with mathematic abilities were tested along with people with literary/linguistic abilities. Interestingly, the persons with linguistic abilities found the solution much quicker than the ones with math abilities.

it was just a try to solve it – the result was a proof for AB/BD=AC/CD; but in the end a one-way calculation witch doesn’t leads to the result that was asked for.

I do not see why it must be the case that

angle(ADC) = angle(ABD) + angle(BAD)

Please elaborate on this.

angle(BAD) + angle (ABD) + angle(BDA) = 180, therefore:

angle(BAD) + angle (ABD) = 180 – angle(BDA) = angle(ADC)

very nice. thanks for the clarification.