MathLabs

Problem 3

In a given right triangle ABCABC, the hypotenuse BCBC has length aa and is divided into nn equal parts, where nn is odd. The central part subtends an angle α\alpha at AA. If hh is the perpendicular distance from AA to BCBC, prove that tan⁡α=4nha(n2−1)\tan\alpha=\dfrac{4nh}{a(n^2-1)}.
Step 3 of 5: Use the symmetry of P and Q around M
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(QH)(PH)=MH2−(a2n)2(QH)(PH)=MH^2-\left(\dfrac{a}{2n}\right)^2
Detailed analysis

Along the oriented line BCBC, write PH=MH−a/(2n)PH=MH-a/(2n) and QH=MH+a/(2n)QH=MH+a/(2n). Their product is therefore MH2−a2/(4n2)MH^2-a^2/(4n^2).