Problem:
In hexagon ABCDEF, which is nonconvex but not self-intersecting, no pair of opposite sides are parallel. The internal angles satisfy ∠A=3∠D,∠C=3∠F, and ∠E=3∠B. Furthermore AB=DE,BC=EF, and CD=FA. Prove that diagonals AD,BE, and CF are concurrent.
Solution:
We first give a recipe for constructing hexagons as in the problem statement. Let ACE be a triangle, with all angles less than 2π/3. Let D be the reflection of A across CE; let F be the reflection of C across EA; let B be the reflection of E across AC. Then, ∠BAF=∠BAC+∠CAE+∠EAF=3∠CAE=3∠CDE, and similarly for the other angle equalities. Also, AB=AE=DE, and similarly for the other side equalities. Thus, the hexagon satisfies the equations in the problem statement. The diagonals AD,BE,CF are simply the altitudes of the triangle ACE, so they are concurrent at the orthocenter.
Now we show that the only possible hexagons meeting the conditions of the problem statement are the ones constructed in this manner. This will suffice to complete the solution.
Given the hexagon ABCDEF as in the problem statement, let β,δ,Ï• be the measures of its angles B,D,F. Since 4(∠B+∠D+∠F)=∠A+∠B+∠C+∠D+∠E+∠F=4Ï€, we must have β+δ+Ï•=Ï€. Also, the fact that opposite sides are not parallel implies that Ï€+2β=∠D+∠E+∠Fî€ =2Ï€, so Î²î€ =Ï€/2; likewise δ,Ï•î€ =Ï€/2.
We can construct a hexagon A1​B1​C1​D1​E1​F1​ meeting the angle and side equality conditions, with angles ∠B1​=β,∠D1​=δ,∠F1​=ϕ, by taking A1​C1​E1​ to be a triangle with angles β,δ,ϕ, and reflecting each vertex across the opposite site as above. We wish to show that ABCDEF∼A1​B1​C1​D1​E1​F1​.
Treat the positions of A,B as fixed, and treat β,δ,ϕ as fixed; these are enough to uniquely determine the orientation of each edge of the hexagon, given the known angles. Let x=AB=DE,y=BC=EF,z=CD=FA. Our goal is to show that these lengths are uniquely determined (up to scale) by the given angles.
Let a,b,c,d,e,f be unit vectors in the directions of the edges from A to B,B to C,C to D,D to E,E to F, and F to A, respectively. Then the vector identity
x(a+d)+y(b+e)+z(c+f)=0(1)
holds. Without loss of generality, assume the vertices of ABCDEF are labeled in counterclockwise order. The respective orientations of vectors b,c,d,e,f, measured counterclockwise relative to a, are
b:c:d:e:f:​π−β−β−3ϕ−2ϕπ+2δ−β2δ−ϕ−β​
(These angles are given modulo 2π; we have made liberal use of the identity β+δ+ϕ=π.) Now, whenever two unit vectors point in directions θ and ψ, which do not differ by π, then their sum is a nonzero vector pointing in direction (θ+ψ)/2 or (θ+ψ)/2+π. It follows that vectors a+d,b+e,c+f are all nonzero and point in the following directions (modulo π):
a+d:b+e:c+f:​−ϕδ−βδ−2ϕ−β​
None of the differences between these angles are multiples of Ï€. (This follows from the fact that β,δ,Ï•î€ =Ï€/2.) Thus, a+d,b+e,c+f are not collinear, nonzero vectors. Consequently, the equation (1) determines the coefficients x,y,z uniquely up to scale, as required.
It follows that ABCDEF and A1​B1​C1​D1​E1​F1​ are similar to each other, as required, and this completes the proof.
The problems on this page are the property of the MAA's American Mathematics Competitions