Perfect Euler Line
A non-equilateral has centroid , orthocenter , incenter , circumcenter that satisfy . Find the measure of the smallest angle, to the nearest degree.
Note : Figure not necessarily drawn to scale.
The answer is 29.
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1 solution
Relevant wiki: Euler Line
The actual answer is 2 a r c s i n 4 1 , which is approximately 2 8 . 9 5 5 ° .
The first thing to realize is that the triangle is isosceles. O , G , H are always collinear with H G = 2 G O (Head to the wiki for more info). Therefore, in order to make H I = I G = G O , I must be halfway between H and G , indicating that the triangle is isosceles.
Without losing generality, let A be the vertex angle and the distance from H to B C be 1 . Let H I = I G = G O = x . As A G = 2 G O , A G = 4 x + 2 .
Make I K ⊥ A C , then I K = x + 1 , C O = 3 x + 2 . Pythagorean theorem yields C J = 6 x + 3 . Apply Pythagorean theorem another time and get A C = ( 6 x + 3 ) ( 6 x + 4 ) . Then s i n θ = ( 6 x + 3 ) ( 6 x + 4 ) 6 x + 3 = 6 x + 4 1 .
On the other hand, s i n θ = 5 x + 2 x + 1 . Let the equations equal and we get x = 2 . Plugging back yields s i n θ = 4 1 .
Therefore, m ∠ B A C = 2 a r c s i n 4 1 ≈ 2 9 ° < 6 0 ° , so it is the smallest angle thus this is the answer.