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In equilateral triangle $ABC$ with side length $1$ inscribed another triangle $DEF$, so that vertices of $DEF$ belong to different sides of $ABC$ and vertices of $DEF$ are not closer to vertices of $ABC$ than given distance $L, L<1/2$. Prove that minimal area of triangle $DEF$ is achieved when two vertices of $DEF$ are at distance $L$ from the same vertex of $ABC$ or when two vertices of $DEF$ are at distance $1-L$ from the same vertex of $ABC$. On picture below minimal area triangle $DEF$ cases are shown in green and blue and general case (with bigger area) is in purple.enter image description here

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  • $\begingroup$ Suppose to slide point D of the pink triangle on side BC, where do you attain the smallest area? At that point you can slide either E or F on their side and repeat the reasoning. $\endgroup$ May 15 at 16:23
  • $\begingroup$ did not quite get what you mean, because D is located on side AB and should stay there in any case, but I did realize when I was putting my lengthy proof online that all that can be proved without derivatives on "h", just with "sliding" argument, so if that what you mean I totally agree... and of course I mean that minimal area green and blue triangles can be rotated 120 degrees both clock- and anti-clock-wise, so that we have 6 minimal triangles in total, and thanks for comment! $\endgroup$
    – Vladimir_U
    May 15 at 17:10
  • $\begingroup$ Yes, D slides on AB, sorry for the typo. But I see you understood what I meant. $\endgroup$ May 15 at 17:12

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To prove the statement above we first consider triangle $DEF$ symmetric with respect to one of the three axes of symmetry of triangle $ABC$ as shown on $picture 1$ below. If we denote height of general symmetric triangle $DEF$ (purple) as $h$ and height of $ABC$ as $1$, we have the following formulas for area of $DEF$ as a function of $h$ and its derivative with respect to $h$:

$$S(DEF) = S(h) = \frac{h(1-h) }{\sqrt{3}}$$ $$S´(h) = \frac{1-2h }{\sqrt{3}}$$

Clearly area of $DEF$ monotonically increases with $h$ until $h$ reaches half of height of triangle $ABC$ and $E,F$ become middle points of sides $BC,AC$, and after that point area of $DEF$ monotonically decreases. Blue and green triangles are both minimal cases allowed by parameter $L$, because their areas are equal, this can be seen if we slide green $D$ along side $AB$ in direction of vertex $A$, until it is at distance $L$ from $A$, this move doesn’t change area of green $DEF$ because $EF||AB$, and after that we slide green $F$ along side $AC$ to middle of this side, this also doesn’t change area of green $DEF$ because we already have in it $ED||AC$, after these two moves green $DEF$ becomes equal to blue $DEF$ up to translation and rotation.

enter image description here

Now we will prove that in general case of triangle $DEF$ we also have its area larger than or equal to areas of minimal blue and green triangles cases. Suppose the opposite. Note that none of the sides of $DEF$ can be parallel to any side of $ABC$, because in that case, if we slide the vertex of $DEF$ which is opposite to the side which is parallel to some side of $ABC$ into middle point of the side of $ABC$ where it is located (we can do that because side parallel to side of $ABC$ serves as base and height is not changed by that move), we will transform $DEF$ into symmetric triangle that was considered above and we already found that it is larger than or equal to in area to blue and green minimal triangles cases. So, without loss of generality we can assume that $AD>AF$ as in $picture 2$ below. Point $E$ (see $picture 2$) cannot be above point $F$ in height relative to side $AB$ because in that case we can slide it alone side $BC$ into position with the same height with point $F$ (rose sides transformed into grey sides in $picture 2$), this move will decrease area of $DEF$ because we assumed that lines $DF$ and $BC$ intersect in opposite half-plane from point $C$, if we consider two half-planes defined by line $AB$, and thus height of point $E$ relative to side $DF$ of triangle $DEF$ is decreasing during such move, and we already showed that triangle $DEF$ with some side parallel to some side of $ABC$ is larger than or equal to our minimal cases in area. This means that point $E$ is below $F$ relative to $AB$ as shown in $picture 2$ in red triangle $DEF$. Now we need to consider two cases, first case is when angle $∠BDE$ is less than 60 degrees, it is shown in $picture 2$. In that case we have lines $AC$ and $DE$ intersecting in opposite half-plane from point $C$, if we consider two half-planes defined by line $AB$, and thus, we can slide point $F$ along side $AC$ in direction of $A$ until we have $EF||AB$ (purple $F$), while decreasing area of $DEF$. After that we can slide $D$ into middle point of side $AB$, while not changing area of $DEF$, and we obtain final purple triangle $DEF$ which is symmetric and less in area than initial red $DEF$, which leads us to contradiction to our assumption.

enter image description here

Second case is when angle $∠BDE$ is higher than 60 degrees, it is shown in $picture 3$. Because lines $AC$ and $DE$ intersect in the same half-plane with point $C$, if we consider two half-planes defined by line $AB$, we can slide $F$ along side $AC$ in direction of $C$ until $DF||BC$, while decreasing area of $DEF$, and after that slide $E$ into middle point of side $BC$, not changing area of $DEF$, creating symmetric rose triangle $DEF$, this is contradiction, meaning that our assumption is false and in general case of triangle $DEF$ its area is larger than or equal to areas of minimal blue and green triangles cases.

enter image description here

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