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Tuesday, May 26, 2015
[OlymPri20150526] The Smallest Angle in a Special Trapezium
challenging because we do not seem to be given much information. For example, we do not know the individual
angles of the trapezium (American: trapezoid). Does this mean that this puzzle cannot be
solved? Are we trapped by this
trapezium? What is the secret key that
unlocks the problem?
First sketch trapezium ABCD.
Introduce point F,
of AD. All contructed lines and points are shown in
grey (American: gray). Then DF = FA
Now draw a line parallel to BC passing through A and intersecting DC at E, say.
Note that ÐAED = ÐBCD (corresponding angles). Then ÐADE + ÐAED = ÐADC + ÐBCD = 120°.
Hence the remaining angle in DDAE, ÐDAE = 180° –
60°. We can deduce this, even though we do not
(currently) know the individual angles ÐADE and ÐAED. This is the key
From here, things get easier. DAFE is an isosceles triangle
with ÐAFE = ÐAEF = (180° – 60°) ¸ 2 = 60°. So DAFE is in fact an equilateral
triangle. That means FE
= FA = FD. So DFDE is an isosceles triangle.
We can now see that actually ÐBCD = ÐAED = 90°, but we do not need to rely on that (or on
accurate drawing) to deduce the answer.
From solving this question, we learn that even though we do not know the
individual angles, by construction and using the sum of angles in a triangle, it
is possible to solve for an important angle, namely ÐAFE.
The rest of the solving uses isosceles triangles and equilateral
triangles, which are part of the common repertoire of tactics.