Source: Mathcenter Contest / Oly - Thai Forum 2008 R2 p4 https://artofproblemsolving.com/community/c3196914_mathcenter_contest
Tags: geometry, trapezoid
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The trapezoid $ABCD$ has sides $AB$ and $CD$ that are parallel $\hat{DAB} = 6^{\circ}$ and $\hat{ABC} = 42^{\circ}$. Point $X$ lies on the side $AB$ , such that $\hat{AXD} = 78^{\circ}$ and $\hat{CXB} = 66^{\circ}$. The distance between $AB$ and $CD$ is $1$ unit . Prove that $AD + DX - (BC + CX) = 8$ units.
(Heir of Ramanujan)
$\sin 6^{\circ}=\frac{1}{AD}$ and $\sin 78^{\circ}=\frac{1}{DX}$ and $\sin 42^{\circ}=\frac{1}{BC}$ and $\sin 66^{\circ}=\frac{1}{CX}$.
$T=AD + DX - (BC + CX) = \frac{1}{\sin 6^{\circ}}+\frac{1}{\sin 78^{\circ}}-(\frac{1}{\sin 42^{\circ}}+\frac{1}{\sin 66^{\circ}})$.
$T=(\frac{1}{\sin 6^{\circ}}-\frac{1}{\sin 66^{\circ}})+(\frac{1}{\sin 78^{\circ}}-\frac{1}{\sin 42^{\circ}})$.
$T=(\frac{\sin 66^{\circ}-\sin 6^{\circ}}{\sin 6^{\circ}\sin 66^{\circ}})+(\frac{\sin 42^{\circ}-\sin 78^{\circ}}{\sin 42^{\circ}\sin 78^{\circ}})$.
$T=\frac{2\sin 30^{\circ}\cos 36^{\circ}}{\sin 6^{\circ}\sin 66^{\circ}}+\frac{-2\sin 18^{\circ}\cos 60^{\circ}}{\sin 42^{\circ}\sin 78^{\circ}}$.
$T=\frac{2\cos 36^{\circ}}{2\sin 6^{\circ}\sin 66^{\circ}}-\frac{2\sin 18^{\circ}}{2\sin 42^{\circ}\sin 78^{\circ}}$.
$T=\frac{2\cos 36^{\circ}}{\cos 60^{\circ}-\cos 72^{\circ}}-\frac{2\sin 18^{\circ}}{\cos 36^{\circ}-\cos 120^{\circ}}$.
$T=\frac{2\cos^{2}36^{\circ}+\cos 36^{\circ}-\sin 18^{\circ}+2\sin 18^{\circ}\cos 72^{\circ}}{(\frac{1}{2}-\cos 72^{\circ})(\cos 36^{\circ}+\frac{1}{2})}$.
$T=4 \cdot \frac{\cos 72^{\circ}+1+\cos 36^{\circ}-\sin 18^{\circ}+\sin 90^{\circ}-\sin 54^{\circ}}{(1-2\cos 72^{\circ})(2\cos 36^{\circ}+1)}$.
$T=4 \cdot \frac{1+1}{2\cos 36^{\circ}+1-2\cos 72^{\circ}-4\cos 36^{\circ}\cos 72^{\circ}}$.
$T=8 \cdot \frac{1}{2\cos 36^{\circ}+1-2\cos 72^{\circ}-2(\cos 108^{\circ}+\cos 36^{\circ})}=8$.