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Geometry Difficulty 5.0 AIME Find the answer

1. As shown in Figure 4,ABC4, \triangle A B C has an incircle O1\odot O_{1} that touches side BCB C at point D,O2D, \odot O_{2} is the excircle of ABC\triangle A B C inside A\angle A. If O1B=6,O1C=3,O1D=2O_{1} B=6, O_{1} C=3, O_{1} D=2, then O1O2=O_{1} O_{2}= \qquad

A number or a short expression. Spacing and $ signs are ignored.

Solution

O1BO2=O1CO2=90O1CO2B are concyclic O1CD=O1O2BO1DCO1BO2O1O2=O1BO1CO1D=3×62=9. \begin{aligned} & \angle O_{1} B O_{2}=\angle O_{1} C O_{2}=90^{\circ} \\ \Rightarrow & O_{1} 、 C 、 O_{2} 、 B \text { are concyclic } \\ \Rightarrow & \angle O_{1} C D=\angle O_{1} O_{2} B \\ \Rightarrow & \triangle O_{1} D C \backsim \triangle O_{1} B O_{2} \\ \Rightarrow & O_{1} O_{2}=\frac{O_{1} B \cdot O_{1} C}{O_{1} D}=\frac{3 \times 6}{2}=9 . \end{aligned}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.