AoPS - Problem

Source: Romania National Olympiad 2023

Tags: geometry, Metric Relation, perpendicularity

DanDumitrescu

14.04.2023 05:00

We consider triangle $ABC$ with $\angle BAC = 90^{\circ}$ and $\angle ABC = 60^{\circ}.$ Let $D \in (AC) , E \in (AB),$ such that $CD = 2 \cdot DA$ and $DE$ is bisector of $\angle ADB.$ Denote by $M$ the intersection of $CE$ and $BD$ , and by $P$ the intersection of $DE$ and $AM$ . a) Show that $AM \perp BD$ . b) Show that $3 \cdot PB = 2 \cdot CM$ .

lpieleanu

01.07.2023 21:07

Without loss of generality, let

$A=(0,0),$

$B=(0,6),$

$C=(6\sqrt{3},0).$ Then, we have

$D = (2\sqrt{3},0).$ Note that

$\frac{AB}{AD} = \frac{6}{2\sqrt{3}} = \sqrt{3}$ which implies that

$ABD$ is a

$30^\circ-60^\circ-90^\circ$ triangle with right angle at

$A$ and

$30^\circ$ angle at

$B.$ Thus, by the Angle Bisector Theorem, we have

$\frac{AE}{EB} = \frac{AD}{BD} = \frac{1}{2},$ so

$E = (0,2).$ Then, the line going through

$E$ and

$C$ has slope

$\tfrac{2 - 0}{0-6\sqrt{3}} = -\tfrac{1}{3\sqrt{3}} = -\tfrac{\sqrt{3}}{9}$ and

$y$ -intercept

$(0,2),$ so its equation is

$y = -\frac{\sqrt{3}}{9}x +2.$ Similarly, we can derive that the equation of the line containing

$B$ and

$D$ is

$y=-\sqrt{3}x + 6.$ Therefore, the intersection of the lines

$y = -\frac{\sqrt{3}}{9}x +2$ and

$y=-\sqrt{3}x + 6$ is

$M = \left( \frac{3\sqrt{3}}{2}, \frac{3}{2} \right).$ Then, we can find that the slope of the line going through

$A$ and

$M$ is

$\tfrac{\tfrac{3}{2}}{\tfrac{3\sqrt{3}}{2}} = \tfrac{1}{\sqrt{3}} = \tfrac{\sqrt{3}}{3},$ while the slope of the line going through

$B$ and

$D$ is

$\tfrac{0-6}{2\sqrt{3}-0} = -\sqrt{3}.$ Since these slopes are negative reciprocals, we can conclude that

$AM \perp BD.$

$\square$

Let us use the same diagram as for part

$(a).$ Since the slope of the line going through

$A$ and

$M$ is

$\frac{\sqrt{3}}{3}$ and this line goes through the origin, the equation of this line is

$y = \frac{\sqrt{3}}{3}x.$ Recall that

$D=(2\sqrt{3},0),$

$E = (0,2),$ so the slope of the line going through

$D$ and

$E$ is

$\tfrac{2-0}{0-2\sqrt{3}} = -\tfrac{\sqrt{3}}{3}.$ Its

$y$ -intercept is

$(0,2)$ so it follows that the equation of this line is

$y = -\frac{\sqrt{3}}{3}x + 2.$ The intersection of this line and line

$AM$ is

$P=(\sqrt{3}, 1).$ Now, we simply use the distance formula to compute

$PB = \sqrt{(0-\sqrt{3})^2 + (6-1)^2} = \sqrt{28} = 2\sqrt{7},$ and

$CM = \sqrt{\left( 6\sqrt{3} - \frac{3\sqrt{3}}{2} \right)^2 + \left( 0-\frac{3}{2} \right)^2} = \sqrt{\frac{243}{4} + \frac{9}{4}} = \sqrt{63} = 3\sqrt{7} = \frac{3}{2} \cdot PB,$ so we are done.

$\square$