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?

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    $\begingroup$ 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. $\endgroup$ Oct 21, 2017 at 14:36
  • $\begingroup$ 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. $\endgroup$ Oct 21, 2017 at 17:41

1 Answer 1


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!

  • $\begingroup$ 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. $\endgroup$ Oct 21, 2017 at 17:36
  • $\begingroup$ Andrei, I added something. See now. $\endgroup$ Oct 21, 2017 at 17:49
  • $\begingroup$ 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). $\endgroup$ Oct 21, 2017 at 20:40

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