Algebra (Olympiad)hardFree

Solve the Symmetric System for All Triples (x,y,z) | IOQM

JEE Maths question with a full step-by-step solution.

Question

The total number of all ordered triples

(x, y, z)

of real numbers such that

x^2 - yz + xy + zx = 82, \qquad y^2 - zx + xy + yz = -18, \qquad z^2 - xy + yz + zx = 18.

Solution

Answer: 4

Label the equations (i), (ii), (iii) in order. Step 1: Add (ii) and (iii). The cross terms cancel, leaving

y^2 + z^2 + 2yz = (y + z)^2 = -18 + 18 = 0 \implies z = -y. \quad \text{...(iv)}

Step 2: Add (i) and (iii):

x^2 + z^2 + 2zx = (x + z)^2 = 82 + 18 = 100 \implies x + z = \pm 10. \quad \text{...(v)}

Step 3: Add (i) and (ii):

x^2 + y^2 + 2xy = (x + y)^2 = 82 - 18 = 64 \implies x + y = \pm 8. \quad \text{...(vi)}

Step 4: Using

z = -y

from (iv), equation (v) becomes

x - y = \pm 10

and (vi) is

x + y = \pm 8

. Solving the four sign combinations gives

(x, y, z) = (9, -1, 1),\ (1, -9, 9),\ (-1, 9, -9),\ (-9, 1, -1).

Answer:

(x, y, z) = (9, -1, 1),\ (1, -9, 9),\ (-1, 9, -9),\ (-9, 1, -1)

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