KVPY 2016 SA Question 5

Geometry Level 3

Let P be a point inside the triangle ABC such that Angle A B C = 9 0 o \ ABC = 90^{o} . Let P 1 P_1 and P 2 P_2 are two images of P under reflection of AB and BC. Distance Between the circumcenters of traingles A B C ABC and P 1 P P 2 P_1PP_2 is

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A P + B P + C P 3 \dfrac{AP+BP+CP}{3} B P 2 \dfrac{BP}{2} A B 2 \dfrac{AB}{2} A P 2 \dfrac{AP}{2} A C 2 \dfrac{AC}{2} B C 2 \dfrac{BC}{2}

Md Zuhair by Md Zuhair , 20, India

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2 solutions

Clearly P 1 P 2 P_1 P_2 is passing through the origin (obtain the equation for line P 1 P 2 P_1 P_2 from its coordinates and you will find that the line is of the form y = m x y=mx ), then clearly the required distance is M B MB .

We know that when a right angled triangle is circumscribed by a circle, it's hypotenuse is the diameter of the circle.

Then, M B = A M = M C = A C 2 MB = AM =MC = \boxed{\frac{AC}{2}}

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S i n c e P 1 i s r e f l e c t i o n o f P o n B A , B A i s ⊥ b i s e c t o r o f P 1 P . S i n c e P 2 i s r e f l e c t i o n o f P o n B C , B C i s ⊥ b i s e c t o r o f P 2 P . B u t s i n c e a c i r c l e p a s s e s t h r o u g h t h r e e p o i n t s , P , P 1 , P 2 , a n d P P 1 i s ⊥ B A , P P 2 , ⊥ B C , Δ P P 1 P 2 i s a r i g h t Δ , a t P . ⊥ b i s e c t o r s o f i t s t w o c h o r d s , P P 1 , P P 2 m e e t a t c e n t e r B . ∴ B i s i t s c i r c u m c e n t e r . M i d p o i n t o f h y p o t e n u s e A C i s t h e c i r c u m c e n t e r o f Δ A B C . S o B t o t h i s m i d p o i n t , w h i c h i s a l s o c i r c u m r a d i u s o f Δ A B C . ∴ c i r c u m r a d i u s = 1 2 ∗ A C . Since~P_1~is ~reflection~ of~ P~ on~BA, ~BA~is~\bot ~bisector ~of~P_1P.\\ Since~P_2~is ~reflection~ of~ P~ on~BC, ~BC~is~\bot ~bisector ~of~P_2P.\\ But~since~a ~circle~passes~through~three~points, P, P_1, P_2 ~,\\ and~PP_1~is~\bot~BA, ~~PP_2, \bot~BC, ~\Delta~P P_1 P_2 ~ is~a~right~\Delta,~at~P.\\ \bot ~bisectors ~of~its~two~chords, PP_1,~PP_2~meet~at~center~B.\\ \therefore~B~is~its~circumcenter. \\ Midpoint~of~hypotenuse~AC~is~the~circumcenter~of~\Delta~ABC.\\ So~B~to~this~midpoint, ~which~is~also~circumradius~of~\Delta~ABC.\\ \therefore~circumradius=\frac 1 2*AC.

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