Library / /12 of 28

, 2024

Geometry Difficulty 5.2 AIME, harder Prove it United States

Problem:
Let $A_{1} A_{2} \ldots A_{19}$ be a regular nonadecagon. Lines $A_{1} A_{5}$ and $A_{3} A_{4}$ meet at $X$ . Compute $\angle A_{7} X A_{5}$ .
Proposed by: Nithid Anchaleenukoon

Solution

Solution:

Inscribing the nonadecagon in a circle, note that
$\angle A_{3} X A_{5}=\frac{1}{2}\left(\widehat{A_{1} A_{3}}-\widehat{A_{4} A_{5}}\right)=\frac{1}{2} \widehat{A_{5} A_{3} A_{4}}=\angle A_{5} A_{3} X$
Thus $A_{5} X=A_{5} A_{3}=A_{5} A_{7}$ , so
$\begin{aligned} \angle A_{7} X A_{5} & =90^{\circ}-\frac{1}{2} \angle X A_{5} A_{7}=\frac{1}{2} \angle A_{1} A_{5} A_{7} \\ & =\frac{1}{4} \widehat{A_{1} A_{8} A_{7}}=\frac{1}{4} \cdot \frac{13}{19} \cdot 360^{\circ}=\frac{1170^{\circ}}{19} \end{aligned}$

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.