## Friday, December 30, 2016

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

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below. 1. Problem 1300
According 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

2. https://goo.gl/photos/t4BLKDL5KmEciriP8

Draw 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