Solution to February 2018 DMMC Web Problem
Top solvers for the February 2018 DMMC Web Problem!
- Mikhail Fedoseev, South Lyon High School
- Kathleen Springer, Huron High School
Official solution to the February 2018 DMMC problem
View the original problem
A key fact behind the solution is the following
If \(x\) and \(y\) are acute angles
with \(\tan x = \dfrac{1}{2}\) and \(\tan y = \dfrac{1}{3}\),
then \(x + y = 45^\circ\).
(Stated differently, the lemma says that the vertex angle
of a special \(2\mathop{:}3\) triangle is \(45^\circ\).)
Proof. The conclusion of the lemma follows immediately from the following observation:
\begin{align*}
\tan(x+y) &= \frac{\tan x + \tan y}{1 - \tan x \tan y}\\
&= \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2}\cdot\frac{1}{3}} \;=\; 1,
\end{align*}
QED.
Drop perpendiculars from the vertex \(A\) onto the base of
each of the three triangles \(ABC\), \(ACD\), and \(ADE\), producing
six right triangles with angles \(p\), \(q\), \(r\), \(s\), \(t\),
and \(u\), as shown in the figure below.
By the given conditions we get:
\[
\tan p = \frac{1}{5},\;
\tan q = \frac{1}{3},\;
\tan r = \frac{1}{7},\;
\tan s = \frac{1}{2},\;
\tan t = \frac{1}{3},\;
\tan u = \frac{1}{8}.
\]
We need to show that \(p + q + r + s + t + u = 90^\circ\).
Since \(\tan s = 1/2\) and \(\tan q = 1/3\), it follows from the lemma that
\begin{equation}
q + s = 45^\circ.
\label{eq:first}
\end{equation}
Next, note that
\[
\tan (r + t) = \frac{\tan r + \tan t}{1 - \tan r \tan t}
= \frac{\frac{1}{7} + \frac{1}{3}}{1 - \frac{1}{7}\cdot\frac{1}{3}}
= \frac{1}{2},
\]
and
\[
\tan (p + u) = \frac{\tan p + \tan u}{1 - \tan p \tan u}
= \frac{\frac{1}{5} + \frac{1}{8}}{1 - \frac{1}{5}\cdot\frac{1}{8}}
= \frac{1}{3},
\]
hence again by the lemma we get
\begin{equation}
(r + t) + (p + u) = 45^\circ.
\label{eq:second}
\end{equation}
Combining the two results (\(\ref{eq:first}\)) and (\(\ref{eq:second}\)), we get
\[
q + s + r + t + p + u = 45^\circ + 45^\circ = 90^\circ,
\]
QED.
Top solvers for the February 2018 DMMC Web Problem!
- Mikhail Fedoseev, South Lyon High School
- Kathleen Springer, Huron High School
Go to the original problem page
Back to the Detroit Mercy Math Competition page.