# February 2018 Detroit Mercy Math Problem

### 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 $$AH\mathop{:}BH\mathop{:}CH = 2\mathop{:}1\mathop{:}2$$.

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 $$A$$):
1. $$ABC$$ is a special $$3\mathop{:}5$$ triangle,
2. $$ACD$$ is a special $$2\mathop{:}7$$ triangle, and
3. $$ADE$$ 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?

Back to the Detroit Mercy Math Competition page.