Competition problem — Geometry — Maths Olympiad Prep

Geometry Difficulty 4.4 AIME Prove it Greece

A triangle $AB\Gamma$ is given with $\hat{A} = 105°$ and $\hat{\Gamma} = \frac{\hat{B}}{4}$ .

α. Determine the measures of the angles $\hat{B}$ and $\hat{\Gamma}$ .

β. If $O$ is the center of the circumcircle of the triangle $AB\Gamma$ and $\Delta$ is the antipodal of $B$ , prove that the distance of $\Gamma$ from $B\Delta$ is $\frac{B\Delta}{4}$ .

Solution

α. Since $\hat{A} + \hat{B} + \hat{\Gamma} = 180°$ and $\hat{A} = 105°$ , $\hat{\Gamma} = \frac{\hat{B}}{4}$ , we have
$105° + \hat{B} + \frac{\hat{B}}{4} = 180° \Leftrightarrow \hat{B} = 60°, \text{ and hence } \hat{\Gamma} = 15°.$

β. Since $OB = O\Delta = O\Gamma$ , it follows that $\widehat{B\Gamma\Delta} = 90°$ . Moreover we have
$\widehat{BO\Gamma} = 360° - 2 \cdot 105° = 150°.$
Therefore
$\widehat{O\Gamma B} = \frac{180° - 150°}{2} = 15°.$
$\text{Hence } \Gamma O\Delta = 30° \text{ and } \Gamma E = \frac{O\Gamma}{2} = \frac{B\Delta}{4}.$

Figure 4

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