MHB Proving Area Ratio of Equilateral Triangle Divided by Line

AI Thread Summary
The discussion centers on proving that when a line divides an equilateral triangle into two regions with equal perimeters, the ratio of their areas, A1 and A2, satisfies the inequality 7/9 ≤ A1/A2 ≤ 9/7. Participants share their approaches to the proof, emphasizing the geometric properties of the triangle and the implications of equal perimeter. Acknowledgment is given to contributors for their solutions and methods. The conversation highlights the interplay between geometry and area ratios in equilateral triangles. The focus remains on the mathematical proof and its implications within the context of triangle geometry.
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A line divides an equilateral triangle into two parts with the same perimeter and having areas $A_1$ and $A_2$ respectively. Prove that $\dfrac{7}{9} \le \dfrac{A_1}{A_2} \le \dfrac{9}{7}$.
 
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anemone said:
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 :
View attachment 1735
if DE is not parallel to BC the discussion is similar (x+y=0.5 must hold)
 

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anemone said:
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 :
 

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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.:cool:

Solution proposed by Vishal Lama, Southern Utah University:
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,

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