ABCD cyclic --> incenters of BCD, etc. form rectangle

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ABCD cyclic --> incenters of BCD, etc. form rectangle

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cyclic quadrilateral

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Source: Kedlaya; also: APMO 1996, Problem 3

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3660 posts

Arne

#1 h Jan 23, 2005, 5:15 PM • 2 Y

Y by Adventure10, Mango247

If $ABCD$ is a cyclic quadrilateral, then prove that the incenters of the triangles $ABC$ , $BCD$ , $CDA$ , $DAB$ are the vertices of a rectangle.

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kueh

#2 h Jan 23, 2005, 5:58 PM • 6 Y

Y by Pluto1708, Adventure10, anonman, Mango247, Stuffybear, and 1 other user

By angle chasing.

You can show that the incentres of ABC (K), BCD (L), B and C are cyclic because angle CLB = 180 - DCB/2 - DBC/2 = 90 + BDC/2, and similarly, angle CKB = 90 + BAC/2. But BAC = BDC. Thus CLB = CKB and so B,C,K,L are concyclic. Also if we call the incentre of ADC (M), D,C,L,M is cyclic. angle KLC = 180 - angle LBC, angle MLC = 180 - MDC, thus angle KLM = angle KBC + angle MDC = (ADC+ABC)/2 = 90, which is what we want.

PS: Can someone give me a crashcourse on latex? or point to me where?

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6555 posts

#3 h Jan 23, 2005, 6:01 PM • 2 Y

Y by Adventure10, Mango247

It can be proved by complex numbers... See also

Paul Yiu, Notes on Euclidean Geometry, paragraph 10.2.6 (page 154);
Barry Wolk, Hyacinthos message #8154,
Alexey A.Zaslavsky, Hyacinthos message #8178.

darij

This post has been edited 1 time. Last edited by darij grinberg, Jun 15, 2006, 3:54 PM

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#4 h Mar 1, 2006, 8:06 PM • 1 Y

Y by Adventure10

There is the converse of this fact (from Russian Summer TST 2005):
Let $ABCD$ be a quadrilateral. The incentres of the triangles $ABC, BCD, CDA, DAB$ form a rectangle. Proove, that $ABCD$ is cyclic.
It's very interesting and rather difficult problem.

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9775 posts

jayme

#5 h Mar 20, 2018, 7:51 AM • 3 Y

Y by govind7701, Adventure10, Mango247

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Le%20rectangle%20de%20Ryokan%20Maruyama.pdf

Sincerely
Jean-Louis

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299 posts

#6 h May 10, 2019, 7:08 AM • 3 Y

Y by Pluto1708, Adventure10, Mango247

We use complex numbers. Plot the points on the unit circle. Then we can choose $a,b,c,d$ such that $A=a^2,B=b^2,C=c^2,D=d^2$ , and $I_{ABC}=-bc-ca-ab,I_{BCD}=-cd-db-bc,I_{CDA}=da+ac-cd,I_{DAB}=-ab+bd+da.$ Then we can compute $I_{ABC}+I_{CDA}=-ab-bc-cd+da=I_{BCD}+I_{DAB}$ and
$\begin{align*} \left|I_{ABC}-I_{CDA}\right|^2&=\left|-bc-2ca-ab-da+cd\right|^2=\frac{1}{abcd}\left(-bc-2ca-ab-da+cd\right)\left(-ad-2bd-cd-bc+ab\right)\\ &=\frac{1}{abcd}\left(-cd-2db-bc-da+ab\right)\left(-ab-2ac-ad-bc+cd\right)=\left|-cd-2db-bc-da+ab\right|^2=\left|I_{BCD}-I_{DAB}\right|^2 \end{align*}$ so the incenters form a rectangle.

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520 posts

zuss77

#7 h Oct 13, 2020, 4:28 PM • 1 Y

Y by yellowhanoi

Labels are on the diagram.
By Fact 5 $AMNB$ is cyclic. And so is $AMQD$ .
$\angle NMQ=\angle B/2 +\angle D/2=90^\circ$ . And similarly the others.

Attachments:

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denery

#8 h Dec 15, 2020, 6:19 AM

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i only could find those right angles on the rectangle by drawing external tangents
and since it is a quadrilateral
it must be a square or rectangle
and by length chasing interestingly we could say that that square/rectangle is definitely composed of rectangles

This post has been edited 1 time. Last edited by denery, Dec 15, 2020, 6:21 AM

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primesarespecial

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primesarespecial

#9 h Jul 12, 2021, 1:19 PM

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We use complex numbers on this one.
We just easily prove this a parallelogram and then just show perpendicularity once or prove diagonals equal in length. :blush:

:blush:

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4379 posts

#10 h Jul 13, 2021, 4:50 PM

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My solution at https://artofproblemsolving.com/community/c6t48f6h24483_abcd_cyclic_gt_incenters_of_bcd_etc_form_rectangle
I hope you will enjoy it!

Best regards,
sunken rock

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#11 h Jul 13, 2021, 9:42 PM

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https://groups.io/g/euclid/topic/81006484#1476

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#12 h Jul 13, 2021, 9:45 PM

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Another property.
https://geometry-diary.blogspot.com/2020/11/1521.html

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ismayilzadei1387

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ismayilzadei1387

#13 h Oct 23, 2023, 7:20 AM • 1 Y

Y by FriIzi

Also China 1986 TsT

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