Geometry Problem. Post your solution in the comment box below.

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

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## Saturday, December 31, 2016

### Geometry Problem 1301: Arbelos, Semicircles, Diameters, Circle, Incircle, Incenter, Square, Midpoint, Concurrency

Labels:
arbelos,
circle,
concurrent,
diameter,
incenter,
incircle,
midpoint,
semicircle,
square

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Problem 1301

ReplyDeleteLet K medium the arc AB and L medium the arc BC, then K is medium M1O1 and L is medium M2O2.Draw IH perpendicular in AC ,

intersecting the circle (Ι,ΙΤ) in P.So from problem 1298 is IP=PH.

But IP/PH=M1K/KO1=M2L/LO2 =1 so the points M1,I and C are collinear and points A,I and M2 are collinear.

APOSTOLIS MANOLOUDIS KORYDALLOS PIRAEUS GREECE

To Apostolis

DeleteReferring to last line of your solution " But IP/PH=M1K/KO1=M2L/LO2 =1 so the points M1,I and C are collinear and points A,I and M2 are collinear"

In my opinion , we need to show that points C, P and K are collinear before conclude that M1, I and C are collinear.

Yes to peter.

DeleteThe points C,P and T1 are collinear from problem 1300 , <BT1C+<AT1B+<KT2A=45+90+45=180

so the points K,T1,P and C are collinear. Thenks

Relation between three inradius (for all three circles inscribed in arbelos)

ReplyDeleteRi=[R1R2(R1+R2)]/(R1²+R1R2+R2²)

PN perpendicular to AC, P intersection CM1 and AM2, tr PNA ~ tr M2O2A

ReplyDeleteand tr PNC ~ tr M1O1C + Theorem pythagore for tr O1IN and O2IN

then just to prove PN = 2Ri