BX = CY if AX = XY, equilateral ABC (2016 AESC MSU Internet Olympiad 8.4)

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BX = CY if AX = XY, equilateral ABC (2016 AESC MSU Internet Olympiad 8.4)

parmenides51 2

N Feb 19, 2021 by CT17

The point $X$ is marked on the side $BC$ of an equilateral triangle $ABC$ , and point $Y$ on the extension of side $AC$ beyond point $C$ such that $AX = XY$ . Prove that $BX = CY$ .

2 replies

parmenides51

Feb 19, 2021

CT17

Feb 19, 2021

BX = CY if AX = XY, equilateral ABC (2016 AESC MSU Internet Olympiad 8.4)

G H J

Reveal topic tags

geometry

equal segments

Equilateral

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parmenides51

30630 posts

parmenides51

#1 h Feb 19, 2021, 5:59 AM • 3 Y

Y by Mango247, Mango247, Mango247

The point $X$ is marked on the side $BC$ of an equilateral triangle $ABC$ , and point $Y$ on the extension of side $AC$ beyond point $C$ such that $AX = XY$ . Prove that $BX = CY$ .

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donian9265

541 posts

donian9265

#2 h Feb 19, 2021, 1:46 PM • 3 Y

Y by Mango247, Mango247, Mango247

sol

Let $\angle{YAX}=\angle{AYX}=\alpha$ . Notice that $\angle{YCX}=120$ so $\angle{YXC}=\angle{XAB}=60-\alpha$ . By law of sines on $\triangle{AXB}$ we know $\frac{\sin(XAB)}{BX}=\frac{\sin(60)}{AX}$ $\frac{\sin(60-\alpha)}{BX}=\frac{\sin(120)}{XY}$ $BX=\frac{\sin(60-\alpha)(XY)}{\sin(120)}$ By law of sines on $\triangle{CXY}$ $\frac{\sin(120)}{XY}=\frac{\sin(60-\alpha)}{CY}$ $CY=\frac{\sin(60-\alpha)(XY)}{\sin(120)}=BX$ as desired.

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CT17

1481 posts

CT17

#3 h Feb 19, 2021, 2:25 PM • 1 Y

Y by tenebrine

Synthetic

Let $P$ be the foot of the altitude from $X$ to $AC$ . Since $\triangle AXY$ is isosceles, $PA = PY$ . Hence,

$CY = PY - PC = PA - PC = AC - 2PC$ .

Since $\triangle PXC$ is $30$ - $60$ - $90$ ,

$BX = BC - XC = AC - 2PC = CY$

and we are done.

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