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# Beautifully Crafted Circles

Two circles Γ and Σ in the plane intersect at two distinct points A and B and the centre of Σ lies on Γ . Let points C and D be on Γ and Σ, respectively, such that C, B and D are collinear. Let point E on Σ be such that DE is parallel to AC. Show that AE = AB.

Note by Shashwat Gokhe
4 months, 2 weeks ago

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Let $$O$$ be the centre of Γ, and let $$P$$ be the centre of Σ. Let the point $$F$$ be on Γ, such that $$F,O,P$$ are collinear. Let $$\angle ACB = x$$.

From the picture, we can see that there are two similar possibilities for how the diagram turns out. By properties of parallel lines, $$\angle BDE$$ either equals $$x$$ or $$180^{\circ} - x$$. In both cases, since $$ABDE$$ is a cyclic quadrilateral, this implies that $$\angle BAE = x = \angle ACB$$. [1]

Since $$FCBA$$ is a cyclic quadrilateral, we have $$\angle ACB = \angle AFB$$. Combining this with [1] gives us $$\angle BAE = \angle BFA$$. [2]

Since $$F,O,P$$ are collinear, then $$FP$$ is a diameter of Γ. Therefore $$\angle FAP = \angle FBP = 90^{\circ}$$. [3]

Since $$PA$$ and $$PB$$ are radii of Σ, we have $$PA = PB$$. Therefore the triangles $$\triangle FPA$$ and $$\triangle FPB$$ have two pairs of sides equal ($$PA = PB$$ and $$FP = FP$$) and one pair of right angles. Hence $$\triangle FPA$$ and $$\triangle FPB$$ are congruent. Hence $$FA = FB$$. [4]

Since $$P$$ is the center of Σ and $$E$$ is a point on Σ on the same side of $$AB$$ as $$P$$, therefore $$\angle BPA = 2\angle BEA$$. Now since $$\triangle BPA$$ is an isosceles triangle, and since the angles in a triangle sum to $$180^{\circ}$$, we have $$\angle BAP = \dfrac{180^{\circ} - \angle BPA}{2} = 90^{\circ} - \angle BEA$$. But thanks to [3], we know that $$\angle FAB = 90^{\circ} - \angle BAP$$. Hence, $$\angle FAB = \angle BEA$$. [5]

Therefore, by [2] and [5], the triangles $$\triangle BEA$$ and $$\triangle FAB$$ have two pairs of equal angles, which means they have three pairs of equal angles. Therefore $$\triangle BEA$$ and $$\triangle FAB$$ are similar triangles.

Therefore, by [4], since $$FA = FB$$, then $$AE = AB$$ also. QED

- 4 months ago

thats a very good solution you are the best

- 4 months ago

Thanks! This was a very interesting problem.

- 4 months ago