# Draw a line that cuts a triangle with minimal area

I'm solving problems on extrema of multivariable functions and Lagrange multipliers. The problem is:

"Through a fixed point M inside a fixed angle draw a line, that cuts a triangle with minimal area from the angle".

I know the way of solving this problem using a function of one variable, i.e. writing equation of the line $y=k(x-a)+b$ and then minimizing area of the triangle this line cuts from the given angle using $k$ as a variable. But we need to minimize a function of at least two variables. Maybe we could denote length of line segment from the given point $M$ to the intersection with one side of the angle as $x$, and to the intersection with another side of the angle as $y$? But then how do we write the area of the triangle as a function of $x,y$? Or maybe there is another way of solving this problem using multivariable function minimization?

• Why should this be a two-variable problem? Given the angle and the point $M$ there is just one degree of freedom under any accounts. Oct 21, 2017 at 14:36
• Because one-variable optimization problems are long way before this problem in the book, which I took the problem from. They are under "Applications of derivatives" section. And the section this problem is from deals with multivariable functions, Lagrange multipliers and alike. Oct 21, 2017 at 17:41

We can solve this problem without using multivariable function minimization.

Indeed, let $\angle BAC$ be our angle, where $B$ and $C$ be chose such that $M$ is a midpoint of $BC$.

Easy to see that $\Delta ABC$ has a minimal area.

Indeed, let $B_1$ and $C_1$ be placed on rays $AB$ and $AC$ respectively such that $M\in B_1C_1$.

Now, let $B_1M>C_1M$, $B_2\in B_1M$ such that $B_2M=C_1M$.

Hence, $$\Delta MBB_2\cong\Delta MCC_1,$$ which says that $$S_{\Delta BB_1M}>S_{\Delta CC_1M}$$ and from here $$S_{\Delta AB_1C_1}=S_{ABMC_1}+S_{BB_1M}>S_{ABMC_1}+S_{CC_1M}=S_{\Delta ABC}.$$

The rest is easy:

Let $D$ be a point on $AM$ such that $M$ is a midpoint of $AD$.

Now, we can get the parallelogram $ABDC$ and we are done!

• In fact, this is the answer, that I saw in the very beginning in the book, the problem is from. The point is, that I don't understand, how to prove, that when $M$ is a midpoint of $BC$, we get the minimal area. I don't see this. Oct 21, 2017 at 17:36
• Andrei, I added something. See now. Oct 21, 2017 at 17:49
• Thank you for this geometrical solution, I understood it and added +1. However, I don't see any expressions for the area, which I could minimize using derivatives and can't tie your solution with the theme I'm studying now (multivariable optimization). Oct 21, 2017 at 20:40