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Given a triangle ABC with fixed point A(5,2) and also other points lie on lines y=x and y=0 respectively . How can I minimize the perimeter of the triangle ie **

|AB+BC+CA|

** should get minimum value ? I tried to assume the points as (p,p) and (q,0) but cannot reach anywhere with the distance formula.

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2 Answers 2

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Reflect $A$ with respect to line $\ell_1 : y = x$ to get $B' = (2,5)$.

Reflect $A$ with respect to line $\ell_2 : y = 0$ to get $C' = (5,-2)$.

For any point $B$ on $\ell_1$ and point $C$ on $\ell_2$, we have

$$AB + BC + CA = B'B + BC + CC'$$ Notice $B'BCC'$ is a path joining $B'$ and $C'$. Since the shortest path joining two points is the straight line between them, we have

$$AB + BC + CA \ge B'C'$$

Since the intersections of $\ell_1$ and $\ell_2$ with line $B'C'$ lies between $B'$ and $C'$ (see illustration below), the perimeter is minimized when $B$, $C$ are these two intersections. The minimum perimeter equals to $B'C'$ = $\sqrt{(2-5)^2 + (5+2)^2} = \sqrt{58}$.

$\hspace 1in$ minimize the perimeter

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  • $\begingroup$ i have differetiate the function $$f(p,q)$$ and solved the partial derivatives and i have got another solution $$p=p,q=\frac{3p}{p-2}$$ $\endgroup$
    – Dr. Sonnhard Graubner
    Sep 18, 2017 at 9:11
  • $\begingroup$ Why is B'B +BC+CC'>B'C' ? $\endgroup$
    – user345710
    Sep 18, 2017 at 9:39
  • $\begingroup$ @AnmolAgarwal the shortest path joining two points is the straight line! $\endgroup$
    – achille hui
    Sep 18, 2017 at 10:29
  • $\begingroup$ @Dr.SonnhardGraubner I don't understand your comment. $\endgroup$
    – achille hui
    Sep 18, 2017 at 11:04
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we have $$AB=\sqrt{(5-p)^2+(2-p)^2},AC=\sqrt{(q-5)^2+4},BC=\sqrt{(p-q)^2+p^2}$$ then $$|AB+BC+CA|=\sqrt{(5-p)^2+(2-p)^2}+\sqrt{(p-q)^2+p^2}+\sqrt{(q-5)^2+4}$$ can you reach this? solving $$f_p=0,f_q=0$$ i got $$p=\frac{29}{10},q=\frac{29}{7}$$ Setting the Solutions in our function we get $$\sqrt{58}$$

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