Mar 2002
We received quite a few correct solutions for this month's puzzle. In case it defeated you, the key was to look for symmetries in the information you are given, and to use
similar triangles.
Part I
With
and
as in the diagram,
, so it is sufficient to find
. By similar triangles,
as
, so 
By Pythagoras' Theorem,
But

, so
Solving this equation for

, and taking only the positive solution, yields
Now we find

.
So
and
Therefore
Part II
There are a number of approaches to this problem, but the one presented here follows from the observation that the information given is left-right symmetric - in other words, finding the two vertical heights will be equally easy (or hard!). We will label these two heights
and
, as in the diagram.
By similar triangles, and dividing
up into two segments
and
, we see that
as
and
as
; therefore
Adding gives

and multiplying gives

Combining these two results gives
Leaving this equation to one side for a moment, we can use Pythagoras' Theorem to find each of

and

in terms of our unknown

, as shown in the diagram above. Combining these two expressions gives that

.
Using Pythagoras' Theorem, we draw a triangle to represent this relationship:
On its own, the information in this diagram is not enough to solve the triangle - for that, we would need one more bit of information. But now we can use the identity
All we have to do is find some combination of trig functions of
that takes the form
. Since
and
, it is easy to see that
Using a calculator gives

and

. Therefore