## Saturday, December 21, 2013

### Geometry Problem 950: Intersecting Circles, Secant, Circumcenter, Cyclic Quadrilateral, Concyclic Points

Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to enlarge the problem 950.

1. Let S(B, k, θ) be the homothetic rotation transformation,
centered at B, such that O₁→O₂, C₁→C₂, D₁→D₂, E₁→E₂.

By Problem 948, BC₁FC₂ are concyclic with circumcenter O₃.

Thus, ∠O₁O₃B = 1/2 ∠C₁O₃B = ∠C₁C₂B = ∠O₁O₂B.
Hence, O₁BO₂O₃ are concyclic.

By similar argument, O₁BO₂O₄ and O₁BO₂O₅ are concyclic.
Hence, O₁BO₂O₃O₄O₅ are concyclic.

2. http://img607.imageshack.us/img607/3732/m2gu.png

1. Observe that ∠ (C1BC2)= ∠ (D1BD2)= ∠ (E1BE2)= ∠ (O1BO2)
2. Since O2O3 and O1O3 are perpendicular bisectors of BC2 and BC1 so ∠ (O1O3O2) supplement to ∠ (C1BC2)
And ∠ (O1O3O2) supplement to ∠ (O1BO2) …..(See blue color)
So O1, O2, B , O3 are concyclic
Similarly ∠ (O1O4O2) supplement to ∠ (O1BO2)… ( see red color)
And ∠ (O1O5O2) supplement to ∠ (O1BO2)… ( see green color)
So O1, B, O2, O3,O4,O5 are concyclic

3. <BE1A=<BD1A=<BC1A. But
<BE1A=<BO1O2 (in circle O1 and O1O2 perpendicular to AB),
<BE1E2=<BO5O2 (in circle O5 and O5O2 perpendicular to BE2),
<BD1D2=<BO4O2 (in circle O4 and O4O2 perpendicular to BD2)
and BC1C2=<BO3O2 (in circle O3 and O3O2 perpendicular to BC2),
so <BO1O2=<BO5O2=<BO4O2=<BO3O2

4. Problem 950
In complete quadrilateral FC_1AD_2 D_1 C_2 the centers O_1,O_2,O_3,O_4 and the point B (Miquel ) they belong to the same circle of the Miquel.
Also in complete quadrilateral GD_1AE_2 E_1 D_2 the centers O_1,O_2,O_5 and the point B (Miquel) they belong to the same circle of the Miquel.
Therefore the points O_1,O_2,O_3,O_4,O_(5,)B are concyclic.
APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE