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

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

Click the figure below to enlarge the problem 950.

## Saturday, December 21, 2013

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

Labels:
circumcenter,
concyclic,
intersecting circles,
secant

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Let S(B, k, θ) be the homothetic rotation transformation,

ReplyDeletecentered 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.

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

ReplyDelete1. 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

<BE1A=<BD1A=<BC1A. But

ReplyDelete<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

Problem 950

ReplyDeleteIn 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