JEE Challenger
More from Conic Sections
MathHardAdv 2009 • Paper 1

Locus of Moving Vertex with Fixed Base in Triangle

In a triangle ABCABC with fixed base BCBC, the vertex AA moves such that
cos⁡B+cos⁡C=4sin⁡2A2.\cos B + \cos C = 4 \sin^2 \frac{A}{2}.
If a,ba, b and cc denote the lengths of the sides of the triangle opposite to the angles A,BA, B and CC, respectively, then

Options

A

b+c=4ab+c = 4a

B

b+c=2ab+c = 2a

Correct
C

locus of point AA is an ellipse

Correct
D

locus of point AA is a pair of straight lines

Topics & Concepts

Conic SectionsProperties of TrianglesEllipse

Step-by-Step Solution

To find the relation between the side lengths and the locus of the vertex AA, we start with the given trigonometric equation:
cos⁡B+cos⁡C=4sin⁡2A2\cos B + \cos C = 4 \sin^2 \frac{A}{2}

Using the sum-to-product identity for the left-hand side:
cos⁡B+cos⁡C=2cos⁡(B+C2)cos⁡(B−C2)\cos B + \cos C = 2 \cos \left(\frac{B+C}{2}\right) \cos \left(\frac{B-C}{2}\right)

In any triangle ABCABC, A+B+C=πA + B + C = \pi, which implies B+C2=π2−A2\frac{B+C}{2} = \frac{\pi}{2} - \frac{A}{2}. Therefore:
cos⁡(B+C2)=sin⁡A2\cos \left(\frac{B+C}{2}\right) = \sin \frac{A}{2}

Substituting this into the identity gives:
2sin⁡A2cos⁡(B−C2)=4sin⁡2A22 \sin \frac{A}{2} \cos \left(\frac{B-C}{2}\right) = 4 \sin^2 \frac{A}{2}

Since AA is an angle of a non-degenerate triangle, sin⁡A2≠0\sin \frac{A}{2} \neq 0. Dividing both sides by 2sin⁡A22 \sin \frac{A}{2}:
cos⁡(B−C2)=2sin⁡A2\cos \left(\frac{B-C}{2}\right) = 2 \sin \frac{A}{2}

Now, using the sine rule asin⁡A=bsin⁡B=csin⁡C=2R\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R:
b+ca=sin⁡B+sin⁡Csin⁡A\frac{b+c}{a} = \frac{\sin B + \sin C}{\sin A}

Applying the sum-to-product formula to the numerator and the double-angle formula to the denominator:
b+ca=2sin⁡(B+C2)cos⁡(B−C2)2sin⁡A2cos⁡A2=2cos⁡A2cos⁡(B−C2)2sin⁡A2cos⁡A2=cos⁡(B−C2)sin⁡A2\frac{b+c}{a} = \frac{2 \sin \left(\frac{B+C}{2}\right) \cos \left(\frac{B-C}{2}\right)}{2 \sin \frac{A}{2} \cos \frac{A}{2}} = \frac{2 \cos \frac{A}{2} \cos \left(\frac{B-C}{2}\right)}{2 \sin \frac{A}{2} \cos \frac{A}{2}} = \frac{\cos \left(\frac{B-C}{2}\right)}{\sin \frac{A}{2}}

Substituting cos⁡(B−C2)=2sin⁡A2\cos \left(\frac{B-C}{2}\right) = 2 \sin \frac{A}{2}:
b+ca=2sin⁡A2sin⁡A2=2  ⟹  b+c=2a\frac{b+c}{a} = \frac{2 \sin \frac{A}{2}}{\sin \frac{A}{2}} = 2 \implies b+c = 2a

Thus, we have:
b+c=2ab + c = 2a

Geometrically, the side BCBC is a fixed base, meaning the points BB and CC are fixed in the plane and the distance between them is BC=a=constantBC = a = \text{constant}.

The position of vertex AA satisfies:
AB+AC=c+b=2a=constantAB + AC = c + b = 2a = \text{constant}

Since the sum of the distances from point AA to the two fixed points BB and CC is a constant greater than the distance between the two fixed points (2a>a2a > a), by definition, the locus of the point AA is an ellipse with foci at BB and CC.

Therefore, the correct options are B and C.

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