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 2 of 5: Apply the tangent subtraction formula with signed distances

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\tan\alpha=\frac{AH\cdot PQ}{AH^2+(QH)(PH)}=\frac{h(a/n)}{h^2+(QH)(PH)}

Detailed analysis

Let $H$ be the foot of the perpendicular from $A$ to $BC$ , so $AH=h$ . The angles made by $AP$ and $AQ$ with $AH$ have signed tangents $PH/AH$ and $QH/AH$ . Their difference is $\alpha$ , hence the tangent subtraction formula gives $\tan\alpha=AH\cdot(QH-PH)/(AH^2+QH\cdot PH)=AH\cdot PQ/(AH^2+QH\cdot PH)$ .