Geometry Problem

Level: Mathematics Education, High School, Honors Geometry, College.

This entry contributed by Ajit Athle.

Click the figure below to see the problem 769 details.

## Wednesday, June 20, 2012

### Problem 769: Triangle, Circumcircle, Circumcenter, Orthocenter, Midpoint of arc, Parallel to chord, Congruence

Labels:
arc,
circumcenter,
circumcircle,
congruence,
midpoint,
orthocenter,
parallel,
triangle

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Partial Solution:

ReplyDeleteTriangles AOM, COM are congruent,

OE // AM, and OF // MC.

So Triangles AEM,CFM have equal area.

M lies on the bisector of angle ABC.

So altitudes from M on bases

AE and FC are of equal length.

So AE = FC

total solution was sent to the email

ReplyDeleteby

Mixalis Tsourakakis from Greece

Problem 769: Solution by Mixalis Tsourakakis from Greece.

DeleteThanks Mixalis.

To Mixalis Tsourakakis

ReplyDeleteI appreciate if you can translate your solution of problem 769 to English so that we can understand it

Peter Tran

Because i don't speak english very fluently i only send you the text of the solution that Antonio Gutierrez attached translated in english without any symbols. The text is in the same order. My solution is very detailed and this is why is so lengthy. I hope this was helpful.

DeletePage 1

PROOF

USEFUL ELEMENTS FOR THE MAIN PROOF

As known (proof is very simple) it applies that ΝΒΜ= Let’s assume that Α>C in that case ΝΒΜ= [If angle Α< angle C then ΝΒΜ= ]

ΗΝC+HXC=180 thus ΑΗΝ=C

ΑΗD=C+ΝΗD=C+ΝΒΜ (HD//BM)= C+ =

S is the midpoint AC .MS mediator of ΑC (since ΑΜ=ΜC) and ΒCMS=Q .Then QA=QC and QAC=C.Also BAL=A-C and due to symmetry LCB=A.In addition MQ,AB,CL intersect at G.

Page 2

I define yellow triangles ΗDC ,OFC,OEG and blue triangles HAD,QOF,EOA. All yellow triangles are similar and all blue triangles are similar. Indeed:

ΑΗD=AHN+NHD=C+NHD=C+NBM (HD//BM)= C+ =

DHC=CHN-HDN=A- = (CHN=A because UHNA inscribable) and so HD is the bisector of ΑΗC .

HAD=NBC(NABX inscribable)=(BN//QM)= OQF

Since OF//MC, then

Therefore OF is the bisector of triangle QOC. But QOF=OMC= (ABCM is inscribable and so ΑΜC=A+C). Thus also FOC=

Since ΕΟ//ΑΜ,OF//MC then ΕΟF=AMC=A+C and since A+C+B=180 BEOF will be inscribable therefore ΟFC=BEO=EAM=A+ (CAM=ABM= )

It applies …………………………. Therefore OE is the bisector of triangle GOA.

Page 3

However GOE= thus also ΕΟΑ=

Also it is obvious that G= HCD (G+A= A+HCD=90 G= HCD)

Finally ΑΕΟ= ΒFO= (OF//MC) and ΗDC=BIC=A+

Therefore

triangle ΑΗD triangle QOF since ΑΗD=QOF= and

HAD= OQF

Triangle QOF triangle EOA since EOA= QOF= and AEO=QFO=

Thus all blue triangles are similar

Triangle HDC triangle OFC since FOC= FOC=CHD and ΟFC=ΗDC=A+

Triangle HDC triangle GEO since G=UCA and DHC=GOE=

Thus all yellow triangles are similar.

MAIN PROOF

Since ΗD is the bisector of triangle AHC it will be:

Howerver triangle AHC triangle OQF and therefore

triangle HDC triangle OFC and therefore

Thus triangle FDC is isosceles, therefore DF=FC (1)

triangle ΕΟΑ triangle HAD therefore

triangle EOG triangle HDC thus

Therefore

Page 4

Therefore triangle ΑΕD is isosceles which means that ΕΑ=ΕD (2)

For triangle ΕΖΑ applies ΑΕΖ=Γ+ , ΕΑΖ=Α-Γ therefore ΕΖΑ= Γ+ and hence triangle ΑΕΖ is isosceles and therefore ΕΑ=ΖΑ. But obviously ΖΑ=FC (3)

From (1),(2),(3) DF=FC =ΕΑ=ΕD

http://netload.in/dateiTp8iGM7rtV/Translation.rar.htm

Michael Tsourakakis from Greece