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Straight Lines and Pair of Straight Lines question

2021 · 17 Mar · Shift 2 · Q36
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  5. /2021 · 17 Mar · Shift 2 · Q36

Straight Lines and Pair of Straight Lines question

2021 · 17 Mar · Shift 2 · Q36

JEE MainMathematicsStraight Lines and Pair of Straight LinesNumerical+4 / −1
Let tan α\alphaα, tan β\betaβ and tan γ\gammaγ; α\alphaα, β\betaβ, γe(2n−1)π2\gamma e{{(2n - 1)\pi } \over 2}γe2(2n−1)π​, n ∈\in∈ N be the slopes of three line segments OA, OB and OC, respectively, where O is origin. If circumcentre of Δ\DeltaΔ ABC coincides with origin and its orthocentre lies on y-axis, then the value of (cos⁡3α+cos⁡3β+cos⁡3γcos⁡αcos⁡βcos⁡γ)2{\left( {{{\cos 3\alpha + \cos 3\beta + \cos 3\gamma } \over {\cos \alpha \cos \beta \cos \gamma }}} \right)^2}(cosαcosβcosγcos3α+cos3β+cos3γ​)2 is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 144

  1. Interpret the geometry

The points A,B,CA,B,CA,B,C lie on rays from the origin with arguments α,β,γ\alpha,\beta,\gammaα,β,γ, so we may write

A=R(cos⁡α,sin⁡α),B=R(cos⁡β,sin⁡β),C=R(cos⁡γ,sin⁡γ)A=R(\cos\alpha,\sin\alpha),\quad B=R(\cos\beta,\sin\beta),\quad C=R(\cos\gamma,\sin\gamma)A=R(cosα,sinα),B=R(cosβ,sinβ),C=R(cosγ,sinγ)

for some common radius RRR, because the circumcentre of △ABC\triangle ABC△ABC is the origin. Hence A,B,CA,B,CA,B,C all lie on the circle centered at the origin.

So the orthocentre HHH of a triangle inscribed in a circle of radius RRR centered at origin is

H⃗=A⃗+B⃗+C⃗.\vec H = \vec A+\vec B+\vec C.H=A+B+C.

Thus

H=R(cos⁡α+cos⁡β+cos⁡γ, sin⁡α+sin⁡β+sin⁡γ).H=R(\cos\alpha+\cos\beta+\cos\gamma,\ \sin\alpha+\sin\beta+\sin\gamma).H=R(cosα+cosβ+cosγ, sinα+sinβ+sinγ).
  1. Use the condition that orthocentre lies on the yyy-axis

If HHH lies on the yyy-axis, then its xxx-coordinate is zero:

cos⁡α+cos⁡β+cos⁡γ=0.\cos\alpha+\cos\beta+\cos\gamma=0.cosα+cosβ+cosγ=0.

Let

S1=cos⁡α+cos⁡β+cos⁡γ=0.S_1=\cos\alpha+\cos\beta+\cos\gamma=0.S1​=cosα+cosβ+cosγ=0.
  1. Use the fact that the slopes are finite

Since slopes are tan⁡α,tan⁡β,tan⁡γ\tan\alpha,\tan\beta,\tan\gammatanα,tanβ,tanγ, none of the lines is vertical, so

cos⁡αcos⁡βcos⁡γ≠0.\cos\alpha\cos\beta\cos\gamma\neq 0.cosαcosβcosγ=0.

This ensures the required expression is well-defined.

  1. Evaluate the numerator

We need

cos⁡3α+cos⁡3β+cos⁡3γ.\cos 3\alpha+\cos 3\beta+\cos 3\gamma.cos3α+cos3β+cos3γ.

Using

cos⁡3θ=4cos⁡3θ−3cos⁡θ,\cos 3\theta = 4\cos^3\theta-3\cos\theta,cos3θ=4cos3θ−3cosθ,

we get

cos⁡3α+cos⁡3β+cos⁡3γ=4(cos⁡3α+cos⁡3β+cos⁡3γ)−3(cos⁡α+cos⁡β+cos⁡γ).\cos 3\alpha+\cos 3\beta+\cos 3\gamma =4(\cos^3\alpha+\cos^3\beta+\cos^3\gamma)-3(\cos\alpha+\cos\beta+\cos\gamma).cos3α+cos3β+cos3γ=4(cos3α+cos3β+cos3γ)−3(cosα+cosβ+cosγ).

Since S1=0S_1=0S1​=0, this becomes

cos⁡3α+cos⁡3β+cos⁡3γ=4(cos⁡3α+cos⁡3β+cos⁡3γ).\cos 3\alpha+\cos 3\beta+\cos 3\gamma=4(\cos^3\alpha+\cos^3\beta+\cos^3\gamma).cos3α+cos3β+cos3γ=4(cos3α+cos3β+cos3γ).

Now use the identity

a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca).a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca).a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca).

When a+b+c=0a+b+c=0a+b+c=0, this gives

a3+b3+c3=3abc.a^3+b^3+c^3=3abc.a3+b3+c3=3abc.

Taking

a=cos⁡α,b=cos⁡β,c=cos⁡γ,a=\cos\alpha,\quad b=\cos\beta,\quad c=\cos\gamma,a=cosα,b=cosβ,c=cosγ,

we obtain

cos⁡3α+cos⁡3β+cos⁡3γ=3cos⁡αcos⁡βcos⁡γ.\cos^3\alpha+\cos^3\beta+\cos^3\gamma=3\cos\alpha\cos\beta\cos\gamma.cos3α+cos3β+cos3γ=3cosαcosβcosγ.

Hence

cos⁡3α+cos⁡3β+cos⁡3γ=12cos⁡αcos⁡βcos⁡γ.\cos 3\alpha+\cos 3\beta+\cos 3\gamma =12\cos\alpha\cos\beta\cos\gamma.cos3α+cos3β+cos3γ=12cosαcosβcosγ.
  1. Compute the required expression

Therefore,

cos⁡3α+cos⁡3β+cos⁡3γcos⁡αcos⁡βcos⁡γ=12.\frac{\cos 3\alpha+\cos 3\beta+\cos 3\gamma}{\cos\alpha\cos\beta\cos\gamma}=12.cosαcosβcosγcos3α+cos3β+cos3γ​=12.

So

(cos⁡3α+cos⁡3β+cos⁡3γcos⁡αcos⁡βcos⁡γ)2=122=144.\left(\frac{\cos 3\alpha+\cos 3\beta+\cos 3\gamma}{\cos\alpha\cos\beta\cos\gamma}\right)^2=12^2=144.(cosαcosβcosγcos3α+cos3β+cos3γ​)2=122=144.
  1. Final answer

The required integer is

144.\boxed{144}.144​.
  1. Comparison with stored answer

Stored correct answer = 144144144, which matches the derived result.

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