midpoint is circumcenter, 2squares on sides of ABC (2017 Polish JMO R1 p2)

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parmenides51

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parmenides51

#1 h Apr 6, 2021, 5:33 PM • 3 Y

Y by samrocksnature, K0NSTANTiNE_hacker, FaThEr-SqUiRrEl

Given an acute triangle $ABC$ with $\angle ACB = 45^o$ . Let $BCED$ and $ACFG$ be squares outside the triangle $ABC$ . Prove that the midpoint of segment $DG$ coincides with the center of the circumscribed circle around the triangle $ABC$ .

https://cdn.artofproblemsolving.com/attachments/b/7/4bc109ddc4182033ba5ddc614e56fbd0b13908.png

This post has been edited 1 time. Last edited by parmenides51, Apr 6, 2021, 5:34 PM

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parmenides51

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parmenides51

#2 h Apr 6, 2021, 5:33 PM • 3 Y

Y by samrocksnature, K0NSTANTiNE_hacker, FaThEr-SqUiRrEl

posted for the image link

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samrocksnature

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samrocksnature

#3 h Apr 6, 2021, 5:57 PM • 1 Y

Y by FaThEr-SqUiRrEl

https://www.geogebra.org/classic/mfqmf8gx

A and B are movable. Both the midpoint and center of the circle is graphed.

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RedFireTruck

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RedFireTruck

#4 h Apr 6, 2021, 6:20 PM • 2 Y

Y by samrocksnature, FaThEr-SqUiRrEl

samrocksnature wrote:

https://www.geogebra.org/classic/mfqmf8gx

A and B are movable. Both the midpoint and center of the circle is graphed.

cool diagram uncle sam

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MortemEtInteritum

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MortemEtInteritum

#5 h Apr 7, 2021, 12:25 AM • 4 Y

Y by FaThEr-SqUiRrEl, samrocksnature, Mango247, Mango247

Complex bash, with the circumcircle of ABC being the unit circle as usual. Then $g=(c-a)i+a, d=-(c-b)i+b$ , so $g+d=a-ai+b+bi$ . Since $\angle ACB=45^{\circ}$ , $\angle AOB=90^\circ$ where O is the circumcenter, so $ai=b, bi=-a$ , and so $g+d=0$ as desired.

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RedFireTruck

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RedFireTruck

#6 h Apr 7, 2021, 1:50 AM • 4 Y

Y by FaThEr-SqUiRrEl, samrocksnature, Mango247, Mango247

MortemEtInteritum wrote:

Complex bash, with the circumcircle of ABC being the unit circle as usual. Then $g=(c-a)i+a, d=-(c-b)i+b$ , so $g+d=a-ai+b+bi$ . Since $\angle ACB=45^{\circ}$ , $\angle AOB=90^\circ$ where O is the circumcenter, so $ai=b, bi=-a$ , and so $g+d=0$ as desired.

wouldn't the midpoint of g and d be $\frac{g+d}2$ and not g+d.

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MortemEtInteritum

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MortemEtInteritum

#7 h Apr 8, 2021, 1:05 AM • 2 Y

Y by FaThEr-SqUiRrEl, samrocksnature

RedFireTruck wrote:

MortemEtInteritum wrote:

Complex bash, with the circumcircle of ABC being the unit circle as usual. Then $g=(c-a)i+a, d=-(c-b)i+b$ , so $g+d=a-ai+b+bi$ . Since $\angle ACB=45^{\circ}$ , $\angle AOB=90^\circ$ where O is the circumcenter, so $ai=b, bi=-a$ , and so $g+d=0$ as desired.

wouldn't the midpoint of g and d be $\frac{g+d}2$ and not g+d.

$\frac{g+d}{2}=0\iff g+d=0$

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RedFireTruck

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RedFireTruck

#8 h Apr 8, 2021, 2:21 AM • 1 Y

Y by samrocksnature

MortemEtInteritum wrote:

RedFireTruck wrote:

MortemEtInteritum wrote:

Complex bash, with the circumcircle of ABC being the unit circle as usual. Then $g=(c-a)i+a, d=-(c-b)i+b$ , so $g+d=a-ai+b+bi$ . Since $\angle ACB=45^{\circ}$ , $\angle AOB=90^\circ$ where O is the circumcenter, so $ai=b, bi=-a$ , and so $g+d=0$ as desired.

wouldn't the midpoint of g and d be $\frac{g+d}2$ and not g+d.

$\frac{g+d}{2}=0\iff g+d=0$

yeah i know, but if it wasnt zero, then would i be right?

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