Geometry Kings, Solve This!

Geometry Level 5

Let the excircle of the triangle A B C ABC lying opposite to A A touch its side B C BC at the point A 1 A_{1} . Define the points B 1 B_{1} and C 1 C_{1} analogously. Suppose that the circumcentre of the triangle A 1 B 1 C 1 A_{1}B_{1}C_{1} lies on the circumcircle of the triangle A B C ABC . Find the measure of the largest angle in triangle A B C ABC in degrees.


The answer is 90.

Department 8 by Department 8 , 27, Israel

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3 solutions

It is not logical that there would be three different angles for the Δ A B C . There must be some symmetry. So, the triangle must be an isosceles. Let BC be the base. ∴ A 1 A ⊥ B C . \text{It is not logical that there would be three different angles for the } \Delta ABC. \\ \text{There must be some symmetry.}\\ \text{ So, the triangle must be an isosceles. Let BC be the base.}~\therefore ~A_1A\perp BC. \\ The line of symmetry is thus A A 1 . Plotting the triangle ABC in X-Y, plane, C(0,0), and BC along X-asis, A above BC with ∠ A = 2 α . \text{The line of symmetry is thus }AA_1. ~~~~~ \text{Plotting the triangle ABC in X-Y, plane,}\\ \text{C(0,0), and BC along X-asis, A above BC with } \angle A= 2\alpha.\\ L e t O , O 1 b e t h e c i r c u m c e n t e r s o f Δ s A B C , A 1 B 1 , C 1 . O B t h e e x − c e n t e r , e x − c i r c l e t o u c h i n g A C a t B 1 . Let ~ O,~ O_1 ~be~ the ~ circumcenters ~ of ~ \Delta s ABC, ~A_1B_1,C_1.\\ O_B ~ the ~ ex-center,~~ex-circle ~ touching ~ AC ~ at B_1.\\ C O B , a n d A O B are external angle bisectors. ∴ all angles as shown on the sketch. O_1 must be above BC on the side of A. CO_B,~ and ~AO_B ~\text{ are external angle bisectors. } \therefore ~ \text{all angles as shown on the sketch.} \\ \text{O\_1 must be above BC on the side of A.}\\ B e c a u s e o f s y m m e t r y , A 1 , A , O 1 , O m u s t b e o n A A 1 , t h e l i n e o f s y m m e t r y . S o , O 1 m u s t b e a t A . W i t h o u t l o s s o f g e n e r a l i t y l e t A C = 1. B U T i n Δ O B A C , b a s e a n g l e s a r e e q u a l , ∴ A O B = A C = 1. Because~ of ~ symmetry, ~ A_1, ~ A, ~ O_1, ~ O ~ must ~ be ~ on ~AA_1,~ the ~ line ~ of ~ symmetry.\\ So, O_1 ~ must ~ be ~ at ~ A. \\ Without ~ loss ~ of ~ generality ~ let ~ AC=1. ~ BUT ~ in ~ \Delta O_BAC,~ base ~ angles ~ are ~ equal, ~ \therefore~AO_B=AC=1.\\ A i s ( − S i n α , C o s α ) . ∵ ∠ O B A C = ∠ A C B , A O B ∣ ∣ B C . O B X = A O B − S i n α = 1 − S i n α , O B Y = A Y = C o s α . A ~ is ~ (-Sin\alpha,Cos\alpha). ~ \because ~ \angle O_BAC=\angle ACB, AO_B || BC.\\ \color{#3D99F6}{O_{B_X}=AO_B - Sin\alpha=1 - Sin\alpha, ~~ O_{B_Y}=A_Y=Cos\alpha.}\\ R e m e m b e r , A ≡ O 1 . C i r c u m r a d i u s o f A 1 B 1 C 1 , R 1 = O 1 A 1 = O 1 B 1 . O 1 A 1 = 1 ∗ C o s α . ∴ O 1 B 1 = C o s α . ∴ B 1 C O 1 C = 1 − C o s α 1 . Remember, ~ A\equiv O_1. ~~~Circumradius of A_1B_1C_1, ~ R_1=O_1A_1=O_1B_1. ~~ O_1A_1=1*Cos\alpha.\\ \therefore ~O_1B_1=Cos\alpha. ~~ \therefore ~ \dfrac{B_1C}{O_1C}=\dfrac{1 - Cos\alpha}{ 1}.\\ ⟹ B 1 i s ( ( 1 − C o s α ) ∗ ( − S i n α ) , ( 1 − C o s α ) ∗ C o s α ) . ∴ S l o p B 1 O B = O B Y − B 1 Y O B X − B 1 X \implies~\color{#3D99F6}{ B_1 ~ is ~\left ( (1 - Cos\alpha)*( - Sin\alpha),~ (1 - Cos\alpha)*Cos\alpha \right ).}\\ \therefore ~Slop ~ B_1O_B=\dfrac{O_{B_Y} - B_{1_Y} }{ O_{B_X} - B_{1_X} }\\ = C o s α − ( 1 − C o s α ) ∗ C o s α 1 − S i n α − ( 1 − C o s α ) ∗ ( − S i n α . ) ∴ s l o p B 1 O B = C o s 2 α 1 − C o s α ∗ S i n α . =\dfrac{ Cos\alpha - (1 - Cos\alpha)*Cos\alpha}{ 1 - Sin\alpha - (1 - Cos\alpha)*( - Sin\alpha.)}\\ \therefore ~\color{#3D99F6}{ slop ~ ~ B_1O_B =\dfrac{ Cos^2\alpha}{1 - Cos\alpha*Sin\alpha}.}\\ B 1 O B ⊥ A C . S l o p A C = − C o t α . ∴ s l o p B 1 O B = T a n α . S o l v i n g = C o s 2 α 1 − C o s α ∗ S i n α − T a n / a l p h a = 0 , i n g r a p h i n g c a l c u l a t o r , α = 4 5 o . ⟹ g r e a t e s t a n g l e A = 2 ∗ 45 = 9 0 o . B_1O_B ~ \perp ~AC. ~~Slop ~ AC = - Cot\alpha. ~~ \therefore ~~\color{#3D99F6}{ slop ~ ~ B_1O_B =Tan\alpha.}\\ Solving =\dfrac{ Cos^2\alpha}{1 - Cos\alpha*Sin\alpha} - Tan/alpha= 0, ~ in ~graphing ~ calculator,\\ \alpha=45^o. ~~~~~~~~\implies ~ greatest ~ angle ~A=2*45=90^o.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

This is problem No.3 of IMO 2013.

The official solution is attached below.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Uhh... How is 2 ∠ A 0 B 0 C 0 = 18 0 ∘ − ∠ A B C 2 \angle A_0B_0C_0 = 180^{\circ}- \angle ABC ?

Deeparaj Bhat - 5 years ago
Trevor Arashiro
Oct 30, 2015

How are we supposed to know which angle is angle B? I knew it was a 90-45-45 but I didn't know to answer 90 or 45 so I just guessed 90

1 Helpful 0 Interesting 0 Brilliant 0 Confused

"I knew it was a 90-45-45 ".How did you know this ? Please let us know. Thanks.

Niranjan Khanderia - 5 years, 6 months ago

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