Solution to February 2018 DMMC Web Problem

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Official solution to the February 2018 DMMC problem

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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.

The special pentagon

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.

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