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

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

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## Friday, December 30, 2016

### Geometry Problem 1300: Arbelos, Semicircles, Diameters, Circle, Incircle, Tangent, Perpendicular, Concurrent Lines

Labels:
arbelos,
circle,
concurrent,
incircle,
perpendicular,
semicircle,
tangent

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Problem 1300

ReplyDeleteAccording to the problems of 1168 and 1298, K medium the arc AB and L medium the arc BC. Is K the center the circle passing through the points A,B,T_2 and T, and L is center ( T,T_1,B,C).

The TB intersects IH in P in the circle (Ι,ΙΤ).Ιf AP intersects the circle

(I,IT) and in T’, then <PTT’=<PIT’/2=90-<IPT’=90-<APB=<BAT’ so the points T,T’ ,B and A are concyclic.Then the points T’ and T_2 coincide.

So the points A,P and T_2 are collinear.Similar the points T_1,P and C are

collinear.

APOSTOLIS MANOLOUDIS KORYDALLOS PIRAEUS GREECE

https://goo.gl/photos/t4BLKDL5KmEciriP8

ReplyDeleteDraw points D, E , F as per sketch

Per the result of problem 638 quadrilaterals ATT2B and CTT1B are cyclic with centers at D and E.

Per the result of problem 1298 T, P and B are collinear

Perform geometry inversion center C , power of inversion= CB.CA circle E will become line AF

T1 is the intersection of circles O1 and E so the image of T1 is the intersection of circle O1 and line AF which is point D.

So C, T1 and D are collinear

Similarly we also have A, T2, E are collinear.

Let DT1 cut circle I at P’

Triangles O1DT1 and IT1P’ are isosceles and similar ( case AA)

So ∠DO1T1=∠T1IP’ => IP ‘⊥AC => P’ coincide to P

With similar way ET2 will cut circle I at P

So TB, IH, CT1 and AT2 are concurrent