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Langley’s Adventitious Angles is a puzzle in which one must infer an angle in a geometric diagram from other given angles. It was posed by
Edward Mann Langley Edward Mann Langley (22 January 1851 – 9 June 1933) was a British mathematician, author of mathematical textbooks and founder of the Mathematical Gazette. He created the mathematical problem known as Langley’s Adventitious Angles. Biography ...
in ''
The Mathematical Gazette ''The Mathematical Gazette'' is an academic journal of mathematics education, published three times yearly, that publishes "articles about the teaching and learning of mathematics with a focus on the 15–20 age range and expositions of attractive ...
'' in 1922..


The problem

In its original form the problem was as follows: ::ABC is an
isosceles triangle In geometry, an isosceles triangle () is a triangle that has two sides of equal length. Sometimes it is specified as having ''exactly'' two sides of equal length, and sometimes as having ''at least'' two sides of equal length, the latter versio ...
with \angle=\angle=80^\circ. ::CF at 30^\circ to AC cuts AB in F. ::BE at 20^\circ to AB cuts AC in E. ::Prove \angle=30^\circ. .


Solution

The problem of calculating angle \angle is a standard application of Hansen's resection. Such calculations can establish that \angle is within any desired precision of 30^\circ, but being of only finite precision, always leave doubt about the exact value. A direct proof using classical geometry was developed by James Mercer in 1923. This solution involves drawing one additional line, and then making repeated use of the fact that the internal angles of a triangle add up to 180° to prove that several triangles drawn within the large triangle are all isosceles. ::Draw BG at 20^\circ to BC intersecting AC at G and draw FG. (See figure on the lower right.) ::Since \angle=80^\circ and \angle=20^\circ then \angle=80^\circ and triangle BCG is isosceles with BC=BG. ::Since \angle=50^\circ and \angle=80^\circ then \angle=50^\circ and triangle BCF is isosceles with BC=BF. ::Since \angle=60^\circ and BF=BG then triangle BGF is
equilateral In geometry, an equilateral triangle is a triangle in which all three sides have the same length. In the familiar Euclidean geometry, an equilateral triangle is also equiangular; that is, all three internal angles are also congruent to each oth ...
. ::Since \angle=100^\circ and \angle=40^\circ then \angle=40^\circ and triangle BGE is isosceles with GB=GE. ::Therefore all the red lines in the figure are equal. ::Since GE=GF triangle EFG is isosceles with \angle=70^\circ. ::Therefore \angle=30^\circ. Many other solutions are possible. Cut the Knot list twelve different solutions and several alternative problems with the same 80-80-20 triangle but different internal angles.


Generalization

A quadrilateral such as BCEF is called an adventitious quadrangle when the angles between its diagonals and sides are all rational angles, angles that give
rational number In mathematics, a rational number is a number that can be expressed as the quotient or fraction of two integers, a numerator and a non-zero denominator . For example, is a rational number, as is every integer (e.g. ). The set of all ra ...
s when measured in degrees or other units for which the whole circle is a rational number. Numerous adventitious quadrangles beyond the one appearing in Langley's puzzle have been constructed. They form several infinite families and an additional set of sporadic examples.. Classifying the adventitious quadrangles (which need not be convex) turns out to be equivalent to classifying all triple intersections of diagonals in regular polygons. This was solved by
Gerrit Bol Gerrit Bol (May 29, 1906 in Amsterdam – February 21, 1989 in Freiburg) was a Dutch mathematician who specialized in geometry. He is known for introducing Bol loops in 1937, and Bol’s conjecture on sextactic points. Life Bol earned his PhD ...
in 1936 (Beantwoording van prijsvraag # 17, Nieuw-Archief voor Wiskunde 18, pages 14-66). He in fact classified (though with a few errors) all multiple intersections of diagonals in regular polygons. His results (all done by hand) were confirmed with computer, and the errors corrected, by Bjorn Poonen and Michael Rubinstein in 1998. The article contains a history of the problem and a picture featuring the regular
triacontagon In geometry, a triacontagon or 30-gon is a thirty-sided polygon. The sum of any triacontagon's interior angles is 5040 degrees. Regular triacontagon The ''regular triacontagon'' is a constructible polygon, by an edge- bisection of a regular ...
and its diagonals. In 2015, an anonymous Japanese woman using the pen name "aerile re" published the first known method (the method of 3 circumcenters) to construct a proof in elementary geometry for a special class of adventitious quadrangles problem. - English translation of the article from ''Gendaisūgaku'' (現代数学). This work solves the first of the three unsolved problems listed by Rigby in his 1978 paper.


References


External links


Angular Angst
MathPages {{DEFAULTSORT:Langley's Adventitious Angles Triangle problems Articles containing proofs