February 2018 Detroit Mercy Math Problem
Top solvers for the February 2018 DMMC Web Problem:
 Mikhail Fedoseev, South Lyon High School
 Kathleen Springer, Huron High School
The original JanuaryFebruary 2018 DMMC Monthly Web Problem follows
Special \(\mathbf{m}\mathop{:}\mathbf{n}\) Triangles.
An acute angled triangle \(ABC\) is called a
special \(m\mathop{:}n\) triangle with vertex \(A\)
if \(AH\mathop{:}BH\mathop{:}CH\) are in the proportion
\(mn\mathop{:}m\mathop{:}n\), where \(H\) is the foot
of the altitude dropped from the vertex \(A\) on to the base \(BC\).
Here is a special \(1\mathop{:}2\) triangle, with
\(AH\mathop{:}BH\mathop{:}CH = 2\mathop{:}1\mathop{:}2\).
Figure 1: Example of a special 1:2 triangle
And here is the February 2018 problem
(PDF)
In the figure below,
\(ABCDE\) is a pentagon such that
(with respect to vertex \(A\)):

\(ABC\) is a special \(3\mathop{:}5\) triangle,

\(ACD\) is a special \(2\mathop{:}7\) triangle, and

\(ADE\) is a special \(8\mathop{:}3\) triangle.
Prove that the pentagon's vertex angle \(\angle A\),
that is \(\angle BAE\), must be a right angle,
using only elementary geometry and/or trigonometry, and without using
a calculator or a computer.
Figure 2: Why must the vertex angle of this pentagon be a right angle?
See the official solution
Back to the
Detroit Mercy Math Competition page.