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Quad formed by orthocenters has same area (all 7's!)
v_Enhance   34
N 43 minutes ago by Jupiterballs
Source: USA January TST for the 55th IMO 2014
Let $ABCD$ be a cyclic quadrilateral, and let $E$, $F$, $G$, and $H$ be the midpoints of $AB$, $BC$, $CD$, and $DA$ respectively. Let $W$, $X$, $Y$ and $Z$ be the orthocenters of triangles $AHE$, $BEF$, $CFG$ and $DGH$, respectively. Prove that the quadrilaterals $ABCD$ and $WXYZ$ have the same area.
34 replies
v_Enhance
Apr 28, 2014
Jupiterballs
43 minutes ago
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Incircle problem
AlbertSimonVN   9
N Nov 26, 2017 by tuannghia18.09MNP
Given triangle $ABC$ . $AD,BE,CF$ are altitudes . Let $I,J$ are incenter of triangle $CDF$ and $BCE$ . Prove $C,I,J,B \in $ a circle
9 replies
AlbertSimonVN
Mar 6, 2016
tuannghia18.09MNP
Nov 26, 2017
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Incircle problem
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AlbertSimonVN
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AlbertSimonVN
#1 h  Mar 6, 2016, 12:57 PM • 2 Y
Y by Adventure10, Mango247
Given triangle $ABC$ . $AD,BE,CF$ are altitudes . Let $I,J$ are incenter of triangle $CDF$ and $BCE$ . Prove $C,I,J,B \in $ a circle
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AlbertSimonVN
341 posts
AlbertSimonVN
#2 h  Mar 6, 2016, 2:00 PM • 1 Y
Y by Adventure10
Any one ?
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livetolove212
859 posts
livetolove212
#3 h  Mar 6, 2016, 2:10 PM • 1 Y
Y by Adventure10
See attachment.
Attachments:
This post has been edited 1 time. Last edited by livetolove212, Mar 7, 2016, 2:42 AM
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livetolove212
859 posts
livetolove212
#4 h  Mar 6, 2016, 2:18 PM • 2 Y
Y by Adventure10, Mango247
Generalization.
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TelvCohl
2312 posts
TelvCohl
#5 h  Mar 6, 2016, 2:37 PM • 4 Y
Y by livetolove212, ValidName, enhanced, Adventure10
Proof of the generalization mentioned at post #4 :

Let $ T $ be the incenter of $ \triangle ABC $ and let $ Y, $ $ Z $ be the center of $ \odot (BIE), $ $ \odot (CJD) $, respectively. Clearly, $ P, $ $ Y, $ $ Z $ lie on the external bisector of $ \angle BPC $, so notice the bisector of $ \angle BAC, $ $ \angle BPC $ are parallel we get $ AT $ $ \perp $ $ YZ $ ... $ (\star) $.

On the other hand, from $ AB $ $ \cdot $ $ AE $ $ = $ $ AC $ $ \cdot $ $ AD $ $ \Longrightarrow $ $ A $ lies on the radical axis of $ \odot (Y) $ and $ \odot (Z) $, so combine $ (\star) $ we get $ T $ lies on the radical axis of $ \odot (Y), $ $ \odot (Z) $ $ \Longrightarrow $ $ TB $ $ \cdot $ $ TI $ $ = $ $ TC $ $ \cdot $ $ TJ $, hence we conclude that $ B, $ $ C, $ $ I, $ $ J $ are concyclic.
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LeVietAn
375 posts
LeVietAn
#6 h  Mar 6, 2016, 2:44 PM • 2 Y
Y by Adventure10, Mango247
Proof of the generalization can see: http://www.artofproblemsolving.com/community/c6h1199843 (at post #2)
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lebathanh
464 posts
lebathanh
#7 h  Aug 20, 2016, 10:19 AM • 2 Y
Y by Adventure10, Mango247
AlbertSimonVN wrote:
Given triangle $ABC$ . $AD,BE,CF$ are altitudes . Let $I,J$ are incenter of triangle $CDF$ and $BCE$ . Prove $C,I,J,B \in $ a circle

it must be BF,CE
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lebathanh
464 posts
lebathanh
#8 h  Aug 20, 2016, 10:28 AM • 2 Y
Y by Adventure10, Mango247
hehe I have a solution shorter: written the problem: for ABC is a triangle,AD,BE,CF is altude ,X , Y is incenter CDE and BDF. prove BYXC concylic : DX,DY cut AB,AB at U,V . because D is center spiral rote BF--> EC ==> DX/DU=DY/DV ==> XY//UV . DVU similar DBE ==> 90-C=DBE=DVU=DYX==>BYX=180-C/2 ==>W.E.D
This post has been edited 1 time. Last edited by lebathanh, Aug 20, 2016, 10:29 AM
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jayme
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jayme
#10 h  Aug 20, 2016, 11:38 AM • 1 Y
Y by Adventure10
Dear Mathlinkers,

I have this reference for the initial problem

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=605960

Is this right?

Sincerely
Jean-Louis
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tuannghia18.09MNP
24 posts
tuannghia18.09MNP
#12 h  Nov 26, 2017, 1:47 PM • 3 Y
Y by livetolove212, Adventure10, Mango247
My solution for the generazilation
We rewrite the problem

Let $ABCD$ be the cyclic quadrilateral and $P$ be the intersection of $AC,BD$. $(PAD); (PBC)$
cut $CD$ at $S,T$. Let $I.J$ be the incenter of triangle $SAD; SBC$. Prove that $IJ$ is parallel to
the bisctor of $\angle BPC$.

Let $M,N$ be the mid-points of arc $AD,BC$ of $(PAD); (PBC)$ and $Q$ be the intersection of
$SN$ and $TM$.
We have $\angle MNQ=\angle PDC$ and $\angle NMQ=\angle PCD$.
So $ \triangle PDC \sim \triangle QNM (a.a)$. This means $\frac {QN}{QM}=\frac {PD}{PC}=\frac{AD}{BC}$.
Note that $\angle AND = 180^\circ-\angle APD=180^\circ-\angle BPC=\angle BMC$ and $NA=ND; BM=MC$
so $\triangle AND \sim \triangle BMC$.
This leads to $\frac {NI}{JM}= \frac {AD}{BC}$.
So $\frac {NI}{JM}=\frac {QN}{QM}$ or $MN$ is parallel to $IJ$. (QED)
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