Langley's Adventitious Angles

In its original form the problem was as follows:

${\displaystyle ABC}$ is an isosceles triangle.

${\displaystyle \angle {B}=\angle {C}=80^{\circ }.}$

${\displaystyle CF}$ at ${\displaystyle 30^{\circ }}$ to ${\displaystyle AC}$ cuts ${\displaystyle AB}$ in ${\displaystyle F.}$

${\displaystyle BE}$ at ${\displaystyle 20^{\circ }}$ to ${\displaystyle AB}$ cuts ${\displaystyle AC}$ in ${\displaystyle E.}$

Prove ${\displaystyle \angle {BEF}=30^{\circ }.}$ [1][2][3]

A solution was developed by James Mercer in 1923.[2] 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 ${\displaystyle BG}$ at ${\displaystyle 20^{\circ }}$ to ${\displaystyle BC}$ intersecting ${\displaystyle AC}$ at ${\displaystyle G}$ and draw ${\displaystyle FG.}$ (See figure on the lower right.)

Since ${\displaystyle \angle {BCG}=80^{\circ }}$ and ${\displaystyle \angle {CBG}=20^{\circ }}$ then ${\displaystyle \angle {BGC}=80^{\circ }}$ and triangle ${\displaystyle BCG}$ is isosceles with ${\displaystyle BC=BG.}$

Since ${\displaystyle \angle {BCF}=50^{\circ }}$ and ${\displaystyle \angle {CBF}=80^{\circ }}$ then ${\displaystyle \angle {BFC}=50^{\circ }}$ and triangle ${\displaystyle BCF}$ is isosceles with ${\displaystyle BC=BF.}$

Since ${\displaystyle \angle {FBG}=60^{\circ }}$ and ${\displaystyle BF=BG}$ then triangle ${\displaystyle BGF}$ is equilateral.

Since ${\displaystyle \angle {BGE}=100^{\circ }}$ and ${\displaystyle \angle {GBE}=40^{\circ }}$ then ${\displaystyle \angle {GEB}=40^{\circ }}$ and triangle ${\displaystyle BGE}$ is isosceles with ${\displaystyle GB=GE.}$

Therefore all the red lines in the figure are equal.

Since ${\displaystyle GE=GF}$ triangle ${\displaystyle EFG}$ is isosceles with ${\displaystyle \angle {GEF}=70^{\circ }.}$

Therefore ${\displaystyle \angle {BEF}=30^{\circ }.}$

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.[4]

adventitious quadrangles problem

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 numbers 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.[5]

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 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.[6] The article contains a history of the problem and a picture featuring the regular triacontagon 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.[7][8][9] This work solves the first of the three unsolved problems listed by Rigby in his 1978 paper.[5]

• Angular Angst, MathPages