Question

Let 'P' be any point inside the triangle OAB whose vertices are O(0, 0), A(3, 0) & B(3, 4). Symbol d(P, AB) represents distance of point P from the side AB. If P satisfies d(P, OB) $\le$ d(P, AB) and d(P, OB) $\le$ d(P,OA), then number of integral points inside the region in which P lies, is (where a point is said to be an integral point whose both co-ordinates are integers)

Answer 1

3

Answer 2

The vertices of the triangle OAB are O(0, 0), A(3, 0), and B(3, 4).

The equations of the sides are: OA: y = 0 OB: The line passing through O(0,0) and B(3,4). Slope = (4-0)/(3-0) = 4/3. Equation is y - 0 = (4/3)(x - 0), which is 4x - 3y = 0. AB: The line passing through A(3,0) and B(3,4). This is a vertical line with x-coordinate 3. Equation is x = 3.

A point P(x, y) is inside the triangle OAB if $0 < x < 3$, $y > 0$, and P is below the line OB. For a point below 4x - 3y = 0, 4x - 3y > 0.

The distance of P(x, y) from a line ax + by + c = 0 is given by $\frac{|ax + by + c|}{\sqrt{a^2 + b^2}}$. $d(P, OA) = d(P, y=0) = \frac{|y|}{\sqrt{0^2+1^2}} = |y|$. Since P is inside the triangle, y > 0, so $d(P, OA) = y$. $d(P, OB) = d(P, 4x-3y=0) = \frac{|4x-3y|}{\sqrt{4^2+(-3)^2}} = \frac{|4x-3y|}{5}$. Since P is inside the triangle, 4x - 3y > 0, so $d(P, OB) = \frac{4x-3y}{5}$. $d(P, AB) = d(P, x=3) = d(P, x-3=0) = \frac{|x-3|}{\sqrt{1^2+0^2}} = |x-3|$. Since P is inside the triangle, $0 < x < 3$, so x - 3 < 0, thus $d(P, AB) = -(x-3) = 3-x$.

The point P satisfies the conditions $d(P, OB) \le d(P, AB)$ and $d(P, OB) \le d(P, OA)$.

1. $\frac{4x-3y}{5} \le 3-x$ $4x - 3y \le 15 - 5x$ $9x - 3y \le 15$ $3x - y \le 5 \implies y \ge 3x - 5$.

2. $\frac{4x-3y}{5} \le y$ $4x - 3y \le 5y$ $4x \le 8y$ $x \le 2y \implies y \ge x/2$.

So, P(x, y) must be an integral point (x, y are integers) satisfying: (i) $0 < x < 3$ (from inside the triangle) (ii) $y > 0$ (from inside the triangle) (iii) $4x - 3y > 0 \implies y < 4x/3$ (from inside the triangle) (iv) $y \ge 3x - 5$ (v) $y \ge x/2$

Combining the conditions on y: $\max(y_{min}, y_{min\_triangle}) \le y < y_{max\_triangle}$. The lower bound for y is $\max(0, x/2, 3x-5)$. The upper bound for y is $4x/3$. So, we need integral points (x, y) such that: $x \in \{1, 2\}$ (since x is an integer and $0 < x < 3$) $y$ is an integer $\max(0, x/2, 3x-5) \le y < 4x/3$.

Case 1: x = 1 The conditions become: $\max(0, 1/2, 3(1)-5) \le y < 4(1)/3$ $\max(0, 0.5, -2) \le y < 4/3$ $0.5 \le y < 1.33...$ The only integer value for y in this range is y = 1. For (1, 1): $0 < 1 < 3$ (True) $1 > 0$ (True) $4(1) - 3(1) = 1 > 0$ (True, inside triangle) $1 \ge 3(1) - 5 \implies 1 \ge -2$ (True) $1 \ge 1/2$ (True) So, (1, 1) is an integral point in the region.

Case 2: x = 2 The conditions become: $\max(0, 2/2, 3(2)-5) \le y < 4(2)/3$ $\max(0, 1, 1) \le y < 8/3$ $1 \le y < 2.66...$ The integer values for y in this range are y = 1 and y = 2. Let's check these points: For (2, 1): $0 < 2 < 3$ (True) $1 > 0$ (True) $4(2) - 3(1) = 8 - 3 = 5 > 0$ (True, inside triangle) $1 \ge 3(2) - 5 \implies 1 \ge 1$ (True) $1 \ge 2/2 \implies 1 \ge 1$ (True) So, (2, 1) is an integral point in the region.

For (2, 2): $0 < 2 < 3$ (True) $2 > 0$ (True) $4(2) - 3(2) = 8 - 6 = 2 > 0$ (True, inside triangle) $2 \ge 3(2) - 5 \implies 2 \ge 1$ (True) $2 \ge 2/2 \implies 2 \ge 1$ (True) So, (2, 2) is an integral point in the region.

The integral points inside the region are (1, 1), (2, 1), and (2, 2). There are 3 such integral points.