Proving Area Ratio of Equilateral Triangle Divided by Line

[anemone]

A line divides an equilateral triangle into two parts with the same perimeter and having areas $A_1$ and $A_2$ respectively. Prove that $\dfrac{7}{9} \le \dfrac{A_1}{A_2} \le \dfrac{9}{7}$.

 

[Albert1]

A line divides an equilateral triangle into two parts with the same perimeter and having areas $A_1$ and $A_2$ respectively. Prove that $\dfrac{7}{9} \le \dfrac{A_1}{A_2} \le \dfrac{9}{7}$.

my solution :

Spoiler

View attachment 1735
if DE is not parallel to BC the discussion is similar (x+y=0.5 must hold)

 

[Albert1]

A line divides an equilateral triangle into two parts with the same perimeter and having areas $A_1$ and $A_2$ respectively. Prove that $\dfrac{7}{9} \le \dfrac{A_1}{A_2} \le \dfrac{9}{7}$.

another solution :

Spoiler

View attachment 1737

 

[anemone]

Thanks for participating and well done, Albert! For your second method, I recall vaguely seeing you used pretty much quite similar way to target other type of geometry problem.

Solution proposed by Vishal Lama, Southern Utah University:

Spoiler

Let's name the triangle be triangle ABC. WLOG, let the triangle has the unit side length of 1.The line can divide the triangle into two congruent triangles or into a triangle and a quadrilateral or . If the line cuts the triangle in two congruent triangles, then clearly $\dfrac{A_1}{A_2}=1$.

For the second case, we may assume that the line cuts side $AB$ at $D$ and $AC$ at $E$. Let the area of triangle $ADE=A_1$ and the area of quadrilateral $BDEC=A_2$. Then $A_1+A_2=\dfrac{1}{2}(1)(1)(\sin 60^{\circ})=\dfrac{\sqrt{3}}{4}$.

Let $BD=x$ and $CE=y$. Then $AD=1-x$, $AE=1-y$. Since the regions with areas $A_1$ and $A_2$ have equal perimeter, we must have $BD+BC+CE+AD+AE$.

$\therefore x+1+y=(1-x)+(1-y) \rightarrow x+y=\dfrac{1}{2}$

Now, area of triangle $ADE=A_1=\dfrac{1}{2}(AD)(AE)\sin \angle DAE$,

$A_1=\dfrac{1}{2}(1-x)(1-y)\sin 60^{\circ} \rightarrow A_1=\dfrac{\sqrt{3}}{4}(1-x)(\dfrac{1}{2}+x)$

Denote $k=\dfrac{A_2}{A_1}>0$, we get that

$\dfrac{A_1}{A_1+A_2}=\dfrac{\dfrac{\sqrt{3}}{4}(1-x)(\dfrac{1}{2}+x)}{\dfrac{\sqrt{3}}{4}}=(1-x)(\dfrac{1}{2}+x)=\dfrac{1}{1+k}$

which after simplification yields

$2x^2-x+\dfrac{1-k}{1+k}=0$

The above quadratic equation in $x$ has real roots and the discriminant should be greater than or equal to zero. Thus,

$D=1-4\cdot2\cdot\left( \dfrac{1-k}{1+k} \right)=\dfrac{9k-7}{k+1} \ge 0$

Therefore $k \ge \dfrac{7}{9}$ or $\dfrac{A_2}{A_1} \ge \dfrac{7}{9}$.

Changing the notations, area of triangle $ADE=A_2$ and area of quadrilateral $BDEC=A_1$ we get $\dfrac{A_1}{A_2} \ge \dfrac{7}{9}$.

Thus,

$\dfrac{7}{9} \le \dfrac{A_1}{A_2} \le \dfrac{9}{7}$.

 

[CellCoree]

I would approach this problem by first defining some key terms and concepts. An equilateral triangle is a triangle with three equal sides and three equal angles. The perimeter of a shape is the distance around its outer edge, while the area is the measure of the space inside the shape. In this case, the line dividing the equilateral triangle into two parts creates two smaller triangles with different areas, denoted as $A_1$ and $A_2$.

To prove the given inequality, we can start by using the formula for the area of a triangle, which is given by $A = \frac{1}{2}bh$, where $b$ is the base of the triangle and $h$ is the height. Since the equilateral triangle is divided into two parts with the same perimeter, we can set the base of both smaller triangles as equal to the side length of the original equilateral triangle, denoted as $s$. This means that the height of each smaller triangle will be half of the original height, or $\frac{s}{2}$.

Now, we can calculate the areas of the two smaller triangles. For $A_1$, we have $A_1 = \frac{1}{2}s \cdot \frac{s}{2} = \frac{s^2}{4}$. Similarly, for $A_2$, we have $A_2 = \frac{1}{2}s \cdot \frac{s}{2} = \frac{s^2}{4}$. Since the two triangles have the same perimeter, their side lengths will also be the same, and therefore their areas will be equal.

To prove the given inequality, we can set up the following equation: $\frac{A_1}{A_2} = \frac{\frac{s^2}{4}}{\frac{s^2}{4}} = 1$. This shows that the ratio of the two areas is equal to 1, which falls within the given range of $\frac{7}{9} \le \frac{A_1}{A_2} \le \frac{9}{7}$. Therefore, the inequality is proven.

In conclusion, we can see that the ratio of the areas of the two smaller triangles created by dividing an equilateral triangle with a line is always equal to 1. This means that the two areas are always equal, and therefore, the given inequality holds true for all possible