Let tanα, tanβ and tanγ; α, β, γ ≠ (2n-1)π/2, n∈N be the slopes of three line segments OA, OB and OC, respectively, where O is origin. If circumcentre of ΔABC coincides with origin and its orthocentre lies on y-axis, then the value of ((cos 3α+cos 3β+cos 3γ)/(cos α cos β cos γ))² is equal to ____________.

Solution

Maths sir · 5 steps

Put the circumcentre at the origin, and the orthocentre is just the sum of the three vertices.

1. Place the vertices. O is the circumcentre, so OA = OB = OC = r. Since OA has slope tan α, take A = (r cos α, r sin α). In the same way, B = (r cos β, r sin β) and C = (r cos γ, r sin γ).
2. Use the orthocentre. When the circumcentre is at the origin, the orthocentre is H = A + B + C. H lies on the y-axis, so its x-coordinate is zero, which gives cos α + cos β + cos γ = 0.
3. Triple angle formula. Write cos 3θ = 4 cos³ θ - 3 cos θ. Adding the three terms gives 4(cos³ α + cos³ β + cos³ γ) - 3(cos α + cos β + cos γ), and the second bracket is zero.
4. Sum of cubes. If a + b + c = 0, then a³ + b³ + c³ = 3abc. So cos³ α + cos³ β + cos³ γ = 3 cos α cos β cos γ. The product is not zero, because none of the angles is an odd multiple of π/2.

   (cos 3α + cos 3β + cos 3γ)/(cos α cos β cos γ) = (4 × 3 cos α cos β cos γ)/(cos α cos β cos γ) = 12

5. So. The required value is 12² = 144.

The common slip: 12. Stopping at the ratio and forgetting to square it gives 12, not 144.

The idea to keep. With the circumcentre at the origin, use H = A + B + C, and from it G = H/3. A condition like the sum of three things being zero should make you reach for a³ + b³ + c³ = 3abc.

Answer from the NTA final answer key. Solution by PrepWiser.