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Please hold on post until Praneeth addresses some notes re this: Solution
Praneeth, please edit as approp. We don't mind the external links so long as we have the important substance on site, which means problem text, diagrams/pics and solution. One can always look up definitions if such sites go down.
http://www.kalva.demon.co.uk/short/soln/sh883.html
This was the problem text:
The angle bisectors of the triangle ABC meet the circumcircle again at A', B', C'. Show that area A'B'C' ≥ area ABC.
The posted diagram has points A, C’, C”, B, A”, A’, C, B” and B’ as one goes around the circumference of a circle; A is the highest point on the page. A’ is the lowest. Triangles ABC, A”BC and BCH are shown. Chords AA’ and B’C’ are highlighted. The centre of the circle, O, in this diagram, is not mentioned and lays on the other side to the point H.
Solution provided, and modified is:
Let H be the orthocenter. Let AH meet the circumcircle at A". Define B" and C" similarly. ∠A"CB = ∠A"AB = 90o - ∠B = ∠BCH, and A"H is perpendicular to BC. Hence A" is the reflection of H in BC and so triangles A"CB and HCB are congruent. Similarly for C"BA and B"AC. So the hexagon AB"CA"BC" has twice the area of ABC.
Let B'C' meet AA' at X. Then ∠B'XA = ∠B'C'A + ∠XAC' = ∠B'C'A + ∠XAB + ∠C'AB = ∠B'BA + ∠A'AB + ∠C'CB = ∠B/2 + ∠A/2 + ∠C/2 = 90o. Hence A'A is an altitude of the triangle A'B'C'. Hence the hexagon AB'CA'BC' has twice the area of the triangle A'B'C'.
But A' is the midpoint of the arc BC, so area A'BC ≥ area A"BC. Similarly for the other two pairs of triangles, so area AB'CA'BC' ≥ area AB"CA"BC". Hence area A'B'C' ≥ area ABC.
This had its origins with the IMO. Find more at your wish by posting just the site: www.kalva.demon.co.uk |
