Problem 3

In a given right triangle

ABC

, the hypotenuse

BC

has length

a

and is divided into

n

equal parts, where

n

is odd. The central part subtends an angle

\alpha

A

. If

h

is the perpendicular distance from

A

BC

, prove that

\tan\alpha=\dfrac{4nh}{a(n^2-1)}

Step 1 of 5: Name the midpoint and the endpoints of the central part

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In plain words

Centering the chosen segment at $M$ makes its endpoints symmetric, even though the altitude foot need not be at $M$ .

M\text{ is the midpoint of }BC,\quad P=M-\dfrac{a}{2n},\quad Q=M+\dfrac{a}{2n}

Detailed analysis

Because $n$ is odd, one of the equal parts is centered at the midpoint $M$ of $BC$ . Its endpoints $P,Q$ lie on $BC$ at signed distances $MP=-a/(2n)$ and $MQ=a/(2n)$ , so $PQ=a/n$ .