1
$\begingroup$

Consider any $\bigtriangleup PQR$ in the $x-y$ plane. Let $f(x,y)=ax+by+c$ , where $a,b,c\in\mathbb{R}$. Let $A\in\mathbb{R^2}$ be any point in the interior or on the $\bigtriangleup PQR$. Prove that $f(A)\leq max(f(P),f(Q),f(R))$ where $A,P,Q,R\in \mathbb{R^2}$.
Source- Indian Statistical Institute Entrance Exam, 2014
N.B.- Intuitively it seems correct but I don't know how to give a rigorous argument. Also please check if the "tag" and "title" of the question are appropriate.Thanks

$\endgroup$
1
$\begingroup$

It is easy to see that $f(x,y)=c_0$ for a constant $c_0$ defines a line. Particularly, $f(x,y)=f(A)$ defines a line. This line divides the plane in two parts: $f(x,y)<f(A)$ an $f(x,y)\geq f(A)$. It is easy to see, that the part $f(x,y)\geq f(A)$ contains at least one vertex of the $\bigtriangleup PQR$. QED.

$\endgroup$
3
  • $\begingroup$ This is all good. Rather this was also in my mind. I wanted to know how do we prove that a line divides a plane into 2 parts such that $f(x,y)<f(A)$ in one part and $f(x,y)\geq f(A)$ in other. I mean how to prove that on one side of the line $f(x,y)>0$ and keeps on increasing in magnitude as we move away from the line? $\endgroup$ – idpd15 Jun 15 '14 at 13:37
  • $\begingroup$ Firstly we should that points with different "signs" belong to different half-planes. Due to Intermediate value theorem for two points $(x_1,y_1)$ and $(x_2,y_2)$ with $f(x_1,y_1)>c_0$ and $f(x_2,y_2)<c_0$ there is a point $(x,y)$ on line segment $[(x_1,y_1),(x_2,y_2)]$ with $f(x_1,y_1)=c_0$, i.e, $(x,y)$ is on the line and hence $(x_1,y_1)$ and $(x_2,y_2)$ are in different half-planes. $\endgroup$ – IBazhov Jun 15 '14 at 18:05
  • $\begingroup$ Secondly, we show that there are a half-plane with "<" and a half-plane with ">". Check that function $f(x,y)$ increases in direction $(a,b)$ and decreases in direction $-(a,b)$. Now if for a point $(x,y)$ we have $f(x,y)=c_0$ then for $(x+a,y+b)$ we have $f(x+a,y+b)>c_0$ and for $(x-a,y-b)$ we have $f(x-a,y-b)<c_0$. $\endgroup$ – IBazhov Jun 15 '14 at 18:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.