Find the minimum perimeter of the triangle.

Question

Consider point $A(5, 2)$ and variable points $B(a, a)$ and $C(b,0),\, a\in R, \, b\in R$. If the perimeter of $\triangle ABC$ is minimum, find $a$ and $b$.

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My attempt:

$\begin{align} P&=AB+BC+CA\\ &=\sqrt{(a-5)^2+(a-2)^2}+\sqrt{(a-b)^2+(a)^2}+\sqrt{(5-b)^2+(2)^2} \end{align}$

I tried to partial differentiate the equation w.r.t. $a$ and $b$ and solve for $a$ and $b$ and obviously I didn't get the answer as the derivative of square root into its argument is getting much more complicated. I didn't show the calculations here as it is very tedious.

I also know that this is not the expected method to solve this problem. It has something to do with "For minimum length, path followed by light ray is to be traced". But I don't know how to apply it here.

Any hints are appreciated!

Answer 1

Suppose we draw $\triangle ABC$. Reflect $A$ about the line $y = x$, to the point $A' = (2,5)$. Also, reflect $A$ about the line $y = 0$ to the point $A'' = (5,-2)$. Then the side $AB$, when reflected about $y = x$, becomes $A'B$, and the side $AC$, when reflected about $y = 0$, becomes $A''C$, and the path $A'BCA''$ has the same length as the perimeter of $\triangle ABC$ because $A'B = AB$ and $A''C = AC$. For what choice of $B$ and $C$ do we have the shortest possible distance between $A'$ and $A''$, and therefore the least possible perimeter?

Answer 2

This may not be what you are going for, but here is a solution without calculus.

Let $l_1$ and $l_2$ be the lines $y = x$ and $y = 0$, respectively. The problem is to choose $B$ and $C$ on lines $l_1$ and $l_2$ so as to minimize $AB + BC + CA$.

Let $A_1 = (2,5)$ be the reflection of $A$ with respect to $l_1$ and let $A_2 = (5,-2)$ be the reflection of $A$ with respect to $l_2$. Then since $B \in l_1$, we have $A_1 B = AB$. Similarly, $CA_2 = CA$.

Thus the problem now becomes to select $B$ and $C$ so as to minimize the length of the broken line $A_1 B C A_2$. This length is minimal when the broken line is in fact a segment. Therefore $B$ and $C$ must be chosen as the intersections of segment $A_1 A_2$ with lines $l_1$ and $l_2$.

Answer: $B = (29/10,29/10), C = (29/7,0)$.

Answer 3

Using derivatives.

If you compute $$\frac{\partial P}{\partial a}=\frac{2 (a-b)+2 a}{2 \sqrt{a^2+(a-b)^2}}+\frac{2 (a-5)+2 (a-2)}{2\sqrt{(a-5)^2+(a-2)^2}}=0$$ you can solve for $b$ (just a quadratic equation i $(a-b)$. This gives $$b_1=\frac{3 a}{5-a} \qquad \text{and} \qquad b_2=\frac{3 a}{a-2}$$ Now, use $$\frac{\partial P}{\partial b}=\frac{b-5}{\sqrt{(b-5)^2+4}}-\frac{a-b}{\sqrt{a^2+(a-b)^2}}=0$$

Replace $b$ by $b_1$ and plot $\frac{\partial P}{\partial b}$ as a function of $a$ and discard any solution corresponding to $a=2$ or $a=5$. Do the same with $b_2$. You just have one solution which is very close to $a=3$ for $b_1 \sim \frac 92$ which makes, as an approximation, $$P_{min}\sim \frac{1}{2} \left(5 \sqrt{5}+\sqrt{17}\right)\approx 7.65172$$

Now, we can polish the solution for $a$ using Newton method. It would give the following iterates $$\left( \begin{array}{cc} n & a_n \\ 0 & 3.00000 \\ 1 & 2.90898 \\ 2 & 2.90008 \\ 3 & 2.90000 \end{array} \right)$$ So, $a=\frac {29}{10}$, $b=\frac {29}{7}$ give $$P_{min}=\sqrt{58} \approx 7.61577$$