In isosceles triangle PQR, angle QPR = 20 degrees and the other two angles are 80 degrees.

In article <gat-130295102032@milo.jpl.nasa.gov>,
Erann Gat <gat@robotics.jpl.nasa.gov> wrote:
>I've corresponded with about half a dozen people so far about the
>so-called "grad-school" puzzle, the one about the isoscoles triangle
>with the two lines drawn at 20 and 30 degrees.  So far, every solution
>I've heard has a flaw.
[...]
>So, my question is: does anyone have a REAL solution to this puzzle?
>That is, does anyone have a closed-form derivation of the answer that
>does not make unwarranted assumptions?

Well, this is probably a bit belated, but here's two versions.  I've
forgotten how things were labelled originally, so here's a new labelling.
(<ABC means the angle at B formed by AB and BC.)

Isosceles triangle PQR, <QPR = 20 degrees, other two 80 degrees.

S is on PQ such that <QRS = 50 degrees.
T is on PR such that <RQT = 60 degrees.

We want to show that <STQ = 30 degrees.  (Or to derive it.  Same thing.)

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1st solution (the trig bash - geometric solution follows after this one)

Call <STQ alpha.  Note that we know the following angles:

<SQT = 20 degrees.
<QTR = 40 degrees.
<QRT = 80 degrees.

Note also that <QSR [Editor: "<SQR" in the original; <SQR = 80] =
<QRS = 50 degrees, hence SQ = QR.

In triangle QST, the sine rule gives: QS/sin(alpha) = ST/sin(20)
(I will omit the degrees at this point.  Take them as read.)

thus (sin(20)/sin(alpha))^2 = ST^2 / QS^2

Using the cosine rule in STQ, ST^2 = QS^2 + QT^2 - 2.QS.QT.cos(20).

In triangle QTR, the sine rule gives: QT/sin(80) = QR/sin(40) = QS/sin(40).
So QT = QS.sin(80)/sin(40) = 2.QS.cos(40).

So:

(sin(20)/sin(alpha))^2 = (QS^2 + 4.QS^2.cos^2(40) - 4.QS^2.cos(40)cos(20))/QS^2
                       = 1 + 4cos^2(40) - 4cos(40)cos(20)

Now, cos^2(x) = (1 + cos(2x))/2
And cos(x)cos(y) = (cos(x+y) + cos(x-y))/2

So:

(sin(20)/sin(alpha))^2 = 1 + 2 + 2cos(80) - 2cos(60) - 2cos(20)
                       = 1 + 2 + 2sin(10) - 1 - 2(1 - 2sin^2(10))
                       = 2(sin(10) + 2 sin^2(10))

So 1/sin^2(alpha) = [2sin(10)(2sin(10) + 1)] / sin^2(20)
                  = [2sin(10)(2sin(10) + 1)] / 4sin^2(10)(1 - sin^2(10))
                  = 2(2sin(10) + 1) / (4sin(10) - 4sin^3(10))

Since sin(3*x) = 3sin(x) - 4sin^3(x):

1/sin^2(alpha) = 2(2sin(10) + 1) / (sin(10) + sin(30))
               = 2(2sin(10) + 1) / (sin(10) + 1/2)
               = 4

So sin(alpha) = +- 1/2.

Since 0 < alpha < 140 degrees (<PTQ = 140 degrees), alpha = 30 degrees.

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And now the geometry (yay!)

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Let X be on PR such that XQ bisects <PQR.  Then we know the following:

QSR is isosceles, with SQ = QR.  Hence SXR is isosceles, with SX = XR,
and QX bisects <SXR.  Many angles can be filled in by angle chasing.
(I'll just refer to the results later.)

We make heavy use of the fact that the angle bisector divides the opposite
side in proportion to the adjacent sides.

Since SX bisects <PXQ, PS/PX = QS/QX = QR/QX  (as QS = QR).
Since XQ bisects <PQR, PX/PQ = RX/RQ.
So PS/PQ = (PS/PX)*(PX/PQ) = (QR/QX)*(RX/RQ) = RX/QX = SX/QX  (as RX = SX)
So PS/XS = PQ/XQ = PT/XT  (as TQ bisects <PQX)

Hence TS bisects <PSX, whence <TSX = 50 degrees, and hence <STQ = 30 degrees.

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-- Geoff Bailey <ftww@cs.su.oz.au>, February 1995

[ Editor: the same puzzle as posted to geometry.puzzles in January
1995, with John Conway's reply. ]

We had the following puzzle in class tonight.  We are to work on it
until next week.  I'll describe the drawing instead of trying to draw
it.

Draw an isosceles triangle ABC where angle B & C are the same (80
degrees).  Place a point somewhere on side AB and call it D.  Place a
point on side AC and call it E.  Connect segment BE, DE, and CD.  We
know this much....
angle EBC = 60     angle EBD = 20     angle DCE = 30     angle DCB = 50
The question is what does angle DEB = ?

The person who gave us this said we would never get it & when we saw
the solution we would say "Of course!"  We want to prove him wrong.

-- Douglas Darfus <ddarfus@nwosun.discovery.k12.oh.us>

This puzzle surfaces quite often (sometimes with other values for the
particular angles).  Here's a quick way to solve it :- draw a regular
enneagon ("9-gon"), and all its diagonals, to get a rather pretty
figure.
   Then you can see a figure similar to the given one somewhere in that
picture.

   This reminds me - the old problem of finding exactly many points
there are in this figure has recently been conclusively solved (not
yet published).  [There was an older attempt by Bol in the 1930s, but
this is now known to have several errors.]

   I remember a paper from about 10 years ago on the problem of finding
other angles (integral when measured in degrees) for which this figure
worked.

    Unfortunately, I don't remember the author(s).

-- John Conway <conway@math.Princeton.EDU>