February 2018 Detroit Mercy Math Problem

Problem of the month
PDF version of this problem
Detroit Mercy Math Competition

Submissions closed - See the solution here

Top solvers for the February 2018 DMMC Web Problem:

Mikhail Fedoseev, South Lyon High School
Kathleen Springer, Huron High School

The original January-February 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 \(AH\mathop{:}BH\mathop{:}CH = 2\mathop{:}1\mathop{:}2\).

Example 1:2 triangle
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.

The special pentagon
Figure 2: Why must the vertex angle of this pentagon be a right angle?

See the official solution

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