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High School Math Grades 9-12, Ages 13-18

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High School Math Grades 9-12, Ages 13-18

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<DAE=? <ADC=<ADE, CD > 2CM - 2016 Rusanovsky Lyceum Olympiad 7.8

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#1 h Jul 4, 2021, 9:15 PM • 1 Y

Y by centslordm

In an equilateral triangle $ABC$ , the point $M$ is the midpoint of the side $AB$ . On the ray $CM$ , select point $D$ is such that $CD > 2CM$ . On the extension of the altitude drawn on the side $AC$ of the triangle $ABC$ , the point $E$ is taken so that $\angle ADC =\angle ADE$ . Find the value of the angle $DAE$ .

(D. Ershov)

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#2 h Jul 6, 2021, 4:14 PM • 1 Y

Y by centslordm

Let $A(-a,0),B(a,0),C(0,\sqrt{3}\ a)$ . Choose $G(0,d)$ .

Point $E(\frac{2ad(a-\sqrt{3}d)}{\sqrt{3}(d^{2}-a^{2}+2ad},\frac{a(3d^{2}-a^{2})}{\sqrt{3}(d^{2}-a^{2}+2ad})$ .

Slope of the line $AE\ :\ m_{AE}=\frac{3d^{2}-a^{2}}{4ad-\sqrt{3}(a^{2}+d^{2})}$ .
Slope of the line $AD\ :\ m_{AD}=\frac{d}{a}$ .

$\tan \angle DAE = \frac{m_{AE}-m_{AD}}{1+m_{AE} \cdot m_{AD}}=\frac{1}{\sqrt{3}}$ , so $\angle DAE=30^{\circ}$ .

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