perpencicular bisector, angle bisector, moving points (2009 VWO Flanders MO p3)

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parmenides51

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parmenides51

#1 h Aug 7, 2020, 11:59 AM

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Consider a line segment $[AB]$ with midpoint $M$ and perpendicular bisector $m$ . For each point $X \ne M$ on m consider we are the intersection point $Y$ of the line $BX$ with the bisector from the angle $\angle BAX$ . As $X$ approaches $M$ , then approaches $Y$ to a point of $[AB]$ . Which?

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original wording

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parmenides51

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parmenides51

#2 h Aug 7, 2020, 11:59 AM

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posted for the image link

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Possible

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Possible

#3 h Aug 7, 2020, 4:21 PM • 1 Y

Y by Mango247

parmenides51 wrote:

Consider a line segment $[AB]$ with midpoint $M$ and perpendicular bisector $m$ . For each point $X \ne M$ on m consider we are the intersection point $Y$ of the line $BX$ with the bisector from the angle $\angle BAX$ . As $X$ approaches $M$ , then approaches $Y$ to a point of $[AB]$ . Which?

original wording

Which what?

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superagh

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superagh

#4 h Aug 7, 2020, 4:37 PM

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which point I think

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PROA200

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PROA200

#5 h Aug 7, 2020, 4:51 PM

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By the angle bisector theorem for degenerate triangle $MAB$ , $MY/YB=AM/AB=1/2$ . So $Y$ lies one-third of the way from $M$ to $B$ and two-thirds of the way from $A$ to $B$ .

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Possible

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Possible

#6 h Aug 7, 2020, 5:47 PM • 2 Y

Y by Mango247, Mango247

superagh wrote:

which point I think

Yeah, thats correct. Parmenides51 messaged me

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vanstraelen

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vanstraelen

#7 h Aug 7, 2020, 6:19 PM

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Given $A(-a,0),M(0,0),B(a,0)$ , choose $X(0,\lambda)$ .

Point $Y(\frac{a\sqrt{a^{2}+\lambda^{2}}}{2a+\sqrt{a^{2}+\lambda^{2}}},/)$ .

$\lim_{\lambda \to 0} \frac{a\sqrt{a^{2}+\lambda^{2}}}{2a+\sqrt{a^{2}+\lambda^{2}}}=\frac{a}{3}$ .

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