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# Point in a triangle (ISI B.MATH 2014 SUBJECTIVE SOLUTION)

Let PQR be a triangle. Take a point A on or inside the triangle. Let f(x, y) = ax + by + c. Show that $$\mathbf { f(A) \le \max { f(P), f(Q) , f(R)} }$$

Discussion:

Basic idea is this: First we take A on a side, say PQ. We show $$\mathbf { f(A) \le \max { f(P), f(Q) } \implies f(A) \le \max { f(P), f(Q) , f(R)} }$$
Next we take A in the interior. Lets join RA and produce it to meet PQ at T. Using the previous argument we show $$\mathbf { f(A) \le \max { f(T), f(R) } \text{but} f(T) \le max { f(P), f(Q) } \implies f(A) \le \max { f(P), f(Q) , f(R)} }$$

Hence it is sufficient to show for a point A on the line PQ (and the rest will follow):

Let $$\mathbf { A = (x, y), P = (x_p , y_p) , Q = (x_q, y_q) }$$ Since A is on PQ, it is possible to write $$\mathbf { x = \frac{m x_p + n x_q }{ m+n} , y = \frac{ m y_p + n y_q} {m+n}, m,n in \mathbb{R}^{+} }$$

Suppose f(A) is larger than both f(P) and f(Q). Then f(A) – f(P) and f(A) – f(Q) are both positive.

Let $$\mathbf { \lambda_1 = \frac{m}{m+n} , \lambda_2 = \frac{n}{m+n} implies \lambda_1 + \lambda_2 = 1 }$$

$$\mathbf { f(A) – f(P) = ax+by+c – ax_p – by_p -c = a(x-x_p) + b(y – y_p) = \newline a(\lambda_1 x_p + \lambda_2 x_q – x_p) + b (\lambda_1 y_p + \lambda_2 y_q – y_p) }$$
$$\mathbf {\implies f(A) – f(P) = a(\lambda_2 x_q – (1-\lambda_1) x_p) + b(\lambda_2 y_q – (1-\lambda_1) y_p) = a \lambda_2 (x_q – x_p) + b \lambda_ 2 (y_q – y_p) }$$
Similarly we can show $$\mathbf { f(A) – f(Q) = a \lambda_1 (x_p – x_q) + b \lambda_1 (y_p – y_q) }$$
$$\mathbf { f(A) – f(Q) = \frac{-\lambda_2}{\lambda_1} (f(A) – f(P) ) }$$
Hence f(A) – f(Q) and f(A) – f(P) are opposite signs. The conclusion follows.

May 13, 2014

1. THIS CAN BE SOLVABLE USING PARTIAL DIFFERENTIATION

2. Can’t we do this by using the theorem of Linear Programming. When we subject the pt A to the required constraints, and maximise ax+by+c, by the theorem we get the maximum value at hte corner pts i.e P,Q,R

• That might be possible. Can you send a detailed solution? We may publish it (with your name as author), if it is good.