arthham

JEE Advanced 2009 · Linear Inequalities

Let (x, y, z) be points with integer coordinates satisfying the system of homogeneous equations 3x−y−z=0,−3x+z=0,−3x+2y+z=0.3x - y - z = 0, -3x + z = 0, -3x + 2y + z = 0. Then the number of such points for which x2+y2+z2≤100x^{2} + y^{2} + z^{2} \le 100 is ______.

Answer: 77

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Written solution

3x−y−z=0,−3x+z=0,−3x+2y+z=03x - y - z = 0,\quad -3x + z = 0,\quad -3x + 2y + z = 0

x,y,z∈Zx, y, z \in \mathbb{Z}

To find: number of points with x2+y2+z2≤100\text{To find: number of points with } x^2 + y^2 + z^2 \le 100

Step 1: solve the system⇒all integer solutions (x,y,z)\text{Step 1: solve the system} \Rightarrow \text{all integer solutions } (x, y, z)

Step 2: keep those with x2+y2+z2≤100\text{Step 2: keep those with } x^2 + y^2 + z^2 \le 100

−3x+z=0⇒z=3x-3x + z = 0 \Rightarrow z = 3x

3x−y−z=0⇒3x−y−3x=03x - y - z = 0 \Rightarrow 3x - y - 3x = 0

⇒y=0\Rightarrow y = 0

−3x+2y+z=−3x+0+3x=0 ✓-3x + 2y + z = -3x + 0 + 3x = 0\ \checkmark

(x,y,z)=(t,0,3t),t∈Z(x, y, z) = (t, 0, 3t),\quad t \in \mathbb{Z}

line through the origin, direction (1,0,3)\text{line through the origin, direction } (1, 0, 3)

x2+y2+z2≤100x^2 + y^2 + z^2 \le 100

t2+0+(3t)2=10t2≤100t^2 + 0 + (3t)^2 = 10t^2 \le 100

t2≤10⇒∣t∣≤3t^2 \le 10 \Rightarrow |t| \le 3

t∈{−3,−2,−1,0,1,2,3}t \in \{-3, -2, -1, 0, 1, 2, 3\}

Number of values of t=7\text{Number of values of } t = 7

(t,0,3t): t=−3,…,3⇒7 points(t, 0, 3t):\ t = -3,\dots,3 \Rightarrow 7 \text{ points}

7\boxed{7}

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