Origin of the Puzzle: It was shown on TV in 1966 in Saigon, South Vietnam, when I was 5 years old. My 10-year-old brother could not solve it then, but our dad came up with a trigonometric solution after 3 days. Two years ago, we visited our dad in Seattle. My brother brought it up again and gave the problem to me. Below is my third solution from October 3, 2008, using elementary Euclidean geometry only. (My first two solutions used analytic geometry on the x-y plane and trigonometry. They are "brute force," not very elegant. I have a fourth solution, which is also not very elegant because it needs the square root of minus one.) When my dad passed away in his sleep at age 90 last November, my brother said that surrounding him were many pieces of paper, on which he had written proofs of various geometric theorems.
If you have worked on this problem, whether or not you found a solution, I hope it gave you some pleasure and reminded you of the time you spent at the National Youth Science Camp in Bartow, West Virginia, and the friendship you made there.
I will send Puzzle #2 to Andy Blackwood soon, and it is also a geometric one. Draw on and cut out of stiff cardboard an ellipse and hold it in direct sunlight. The shadow of this ellipse appears to be another ellipse, regardless of how you hold it. This is not true for, say a cardboard square, which casts not only square but also rectangular, paralellogram, or rhomboid shadows, nor for, say a cardboard circle, which casts not only circular but also elliptical shadows. Is it always true that the shadow of any (2-dimensional) ellipse cast on any (2-dimensional) surface is another (2-dimensional) ellipse, or does it only appear that way most of the time but not true in general, in which case, when is it true?
(In case you are curious, the above fact about an ellipse is possibly the key to a solution to a mathematics problem that underlies absolutely unbreakable un-eavedroppable quantum encryption, transmission, and decryption of information.)
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Solution to Puzzle #1
D. B. Zoon Nguyen
On AB, pick E such that the angle ECB is 60 degrees.
Join E to C, which intersects MB at P.
Join E to M and N to P.
Consider triangle NBC. The sum of interior angles being 180 degrees
implies that angle BNC is 50 degrees. So, triangle NBC is isoceles,
and, therefore,
(1) BN = BC.
By construction, triangle PBC is equilateral; with both base angles being 60 degrees, the remaing angle must also be 60 degrees for the sum to be 180. So,
(2) BP = BC.
(1) and (2) imply that BN = BP, and therefore triangle NBP is isoceles and so,
(3) Angle NPB = 80 degrees (since angle NBP, which is the same as angle ABM, is 20 degrees by the construction of M).
As established above, triangle PBC is equilateral, and, therefore,
(4) angle BPC = 60 degrees.
(3) and (4) imply that
(5) angle NPE = 40 degrees.
Consider triangle BEC. That the sum of interior angles must be 180
degrees implies that
(6) angle BEC = 40 degrees.
(5) and (6) show that triangle ENP is isoceles, and therefore,
(7) NE = NP.
The two triangles MBC and ECB are congruent because they have the same
side BC and equal side angles (60 and 80 degrees adjacent
to BC). So,
(8) EB = MC.
Because triangle ABC is isoceles, AB = AC. But AB = AE + EB, and AC =
AM + MC, therefore, (8) implies that AE = AM making triangle AEM
isoceles. So,
(9) angle AME = 80 degrees.
In triangle MCB, the sum of interior angles being 180 degrees means that
(10) angle CMB = 40 degrees.
(9) and (10) imply that
(11) angle PME = 60 degrees.
By construction,
(12) angle EPM = angle BPC , which, from (4), is equal to 60 degrees.
Consider triangle EPM. (11) and (12) show that the remaining angle
PEM must also be 60 degrees. Hence, triangle EPM is equilateral. Therefore,
(13) Angle PME = 60 degrees, and
(14) ME = MP.
Compare triangle NEM to triangle NPM. They share one side (NM), and
the other two sides are equal because of (7) and (14). So, triangle
NEM is congruent to triangle NPM, and therefore,
(15) angle PMN = angle EMN.
But angle PMN + angle EMN = angle PME, which is 60 degrees by (13).
So, angle PMN, which is the same as angle NMB, equals half of 60 degrees, or 30 degrees.
QED
