The Concurrent Lines
In acute angled △ A B C , ∠ A B C = 6 0 ∘ .
Circles ω 1 , ω 2 , ω 3 are constructed having diameters B C , C A , A B respectively.
The two tangents from A to ω 1 touch ω 1 at points A 2 and A 3 , the two tangents from B to ω 2 touch ω 2 at points B 2 and B 3 , and the two tangents from C to ω 3 touch ω 3 at points C 2 and C 3 . It turns out that the lines A 2 A 3 , B 2 B 3 , C 2 C 3 concur at a point T within the triangle. Find ∠ B A T in degrees.
Details and assumptions
- You might refer to the following figure to see how points A 2 and A 3 are constructed. Points B 2 , B 3 , C 2 , C 3 are defined analogously.
This problem is adapted from the Proofathon Geometry contest, and was originally posed by Nicolae Shapoval.
The answer is 30.
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4 solutions
Solution without using poles and polar -- let N be the mid-point of BC and AL be foot of perpendicular to BC , H=orthocentre => A,A1,L,N,A2 to be concyclic with AN as diameter.Let P= {AN U A1A2}, Consider AA1M ; AP X AN = AA1^2 = AN^2 - A1N^2 (its trivial) ; AP X AN=AN^2-BN^2 (as A1N=BN) (R1) ; Let CH U AB = K & BH U AC= M => considering cyclic quad KBLH => AK X AB=AH X AL (R2) ; quad KBCM => AK X AB= AN^2-BN^2 ( as BC is diameter and N as midpoint)(R3) => combining the results (R1 R2 R3) => AP X AN=AH X AL (R4) LET {A1A2 U AL} = H1 . Quad PH1LN cyclic ( as A1A2 perpendicular to AN and H1L to BC ) => AP X AN= AH1 X AL (R5) comb ( R4 R5) => AH=AH1 => H1=H => H belongs to A1A2 similarly for others => T=H.
Well done!
WELL....I considered that the points A2 and A3 correspond to B and C (in the diagram...just assumed)...Then i assumed the incentre of triangle of ABC as the circles w1 and w2(The same circle has tangents from B and C)...the point T came out to be midpoint of B and C.....so angle BAT is 30!!!
Doesn't make any sense!
each of A1A2,B1B2,C1C2,passes through the orthocenter of the triangle.so T=H(orthocenter of ABC)hence <BAT=90-60=30
Why do they pass through the orthocenter of the triangle?
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A2A3 is the polar of A w.r.t the circle with diameter BC.now consider the duality principle of POLE and POLARS
Could you explain how u used the concept of pole-polar to solve it??
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Let H be the orthocenter of △ A B C . By Brokard's theorem, A lies on the polar of H (with respect to ω 1 ). By La Hire's theorem, H lies on the polar on A . Thus H lies on A 2 A 3 . Similarly H also lies on B 2 B 3 and C 2 C 3 , which implies that A 2 A 3 , B 2 B 3 , C 2 C 3 are concurrent at H .
I hate to say this, but since such a point of concurrency T appears to exist for any acute triangle with at least one 60 degree vertex, then it has to be true for the equilateral triangle, and so it's easy to find ∠BAT = 30, by reason of symmetry.