So Many Intersections

Geometry Level 3

Two non-intersecting circles in the plane Γ 1 \Gamma_1 and Γ 2 \Gamma_2 are drawn centered at points O 1 O_1 and O 2 O_2 respectively. A circle Γ 3 \Gamma_3 intersects Γ 1 \Gamma_1 at points X 1 , X 2 X_1,X_2 and Γ 2 \Gamma_2 at points Y 1 , Y 2 . Y_1,Y_2. Another circle Γ 4 \Gamma_4 intersects Γ 1 \Gamma_1 at points X 3 , X 4 X_3, X_4 and Γ 2 \Gamma_2 at points Y 3 , Y 4 . Y_3, Y_4. Suppose lines X 1 X 2 X_1X_2 and Y 1 Y 2 Y_1Y_2 intersect at K 1 K_1 and X 3 X 4 X_3X_4 and Y 3 Y 4 Y_3Y_4 intersect at K 2 . K_2. Find K 1 O 2 O 1 + O 2 K 1 K 2 \angle K_1O_2O_1 + \angle O_2K_1K_2 in degrees.


The answer is 90.

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2 solutions

Note that X 1 X 2 X_1X_2 is the radical axis of Γ 1 \Gamma_1 and Γ 3 \Gamma_3 and Y 1 Y 2 Y_1Y_2 is the radical axis of Γ 2 \Gamma_2 and Γ 3 . \Gamma_3. Their point of intersection, K 1 , K_1, therefore is the radical center of Γ 1 , Γ 2 , Γ 3 , \Gamma_1, \Gamma_2, \Gamma_3, which implies K 1 K_1 lies on the radical axis of Γ 1 \Gamma_1 and Γ 2 . \Gamma_2. Similarly K 2 K_2 also lies on the radical axis of Γ 1 \Gamma_1 and Γ 2 . \Gamma_2. It follows that K 1 K 2 O 1 O 2 , K_1K_2 \perp O_1O_2, and K O 2 O 1 + O 2 K 1 K 2 = 18 0 9 0 = 9 0 . \angle KO_2O_1 + \angle O_2K_1K_2 = 180^{\circ} - 90^{\circ} = \boxed{90^{\circ}}.

done it the same way sreejato! :)

Sagnik Saha - 7 years, 2 months ago

easy but nice . answer is 90

Care to be a little more elaborate?

Sreejato Bhattacharya - 7 years, 2 months ago

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