Largest Parallelogram Contained in Triangle

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Theorem

Let $T$ be a triangle.

Let $P$ be a parallelogram contained within $T$.

Let $P$ have the largest area possible for the conditions given.


Then:

$(1): \quad$ One side of $P$ is coincident with part of one side of $T$, and hence two vertices lie on that side of $T$
$(2): \quad$ The other two vertices of $P$ bisect the other two sides of $T$
$(3): \quad$ The area of $P$ is equal to half the area of $T$.


Proof

We will first find the maximum area of $P$ when $(1)$ is satisfied, that is, when $P$ is inscribed in $T$.

Proof of $(2)$

Consider the diagram below.

Largest-parallelogram-in-triangle-1.png

Here $DEGF$ is our inscribed parallelogram $P$.


Since $FG \parallel BC$, by Equiangular Triangles are Similar:

$\triangle AFG \sim \triangle ABC$


Let $FG : BC = 1 : r = \text {Height of } \triangle AFG : \text {Height of } \triangle ABC$.

The area of $T$, which is fixed, is given by:

$\dfrac {BC \times \text {Height of } \triangle ABC} 2$

The area of $P$ is given by:

\(\ds FG \times \text {Height of } P\) \(=\) \(\ds \paren {BC \times \frac 1 r} \times \paren {\text {Height of } \triangle ABC \times \frac {r - 1} r}\)
\(\ds \) \(=\) \(\ds \frac {BC \times \text {Height of } \triangle ABC} 2 \times \frac {2 \paren {r - 1} } {r^2}\)

which is $\dfrac {2 \paren {r - 1} } {r^2}$ of the area of $T$.


Notice that:

\(\ds 0\) \(\le\) \(\ds \paren {r - 2}^2\) Square of Real Number is Non-Negative
\(\ds \leadstoandfrom \ \ \) \(\ds 0\) \(\le\) \(\ds r^2 - 4 r + 4\) Square of Difference
\(\ds \leadstoandfrom \ \ \) \(\ds 4 r - 4\) \(\le\) \(\ds r^2\)
\(\ds \leadstoandfrom \ \ \) \(\ds \frac {2 \paren {r - 1} } {r^2}\) \(\le\) \(\ds \frac 1 2\)

Equality holds if and only if $r = 2$ for the first inequality.

Therefore the maximum of $\dfrac {2 \paren {r - 1} } {r^2}$ occurs at $r = 2$.

This implies $AF : AB = AG : AC = 1 : 2$.

Hence both sides $AB$ and $AC$ are both bisected by vertices of $P$.

$\Box$


Proof of $(3)$

At $r = 2$:

$\dfrac {2 \paren {r - 1} } {r^2} = \dfrac {2 \paren {2 - 1} } {2^2} = \dfrac 1 2$

therefore the maximum area of $P$ is equal to half the area of $T$.

$\Box$


Proof of $(1)$

Now we consider the cases where $P$ is not inscribed in $T$.

Suppose less than $3$ vertices of $P$ lie on the sides of $T$.

By constructing parallel lines to the sides of $T$, we can find a smaller triangle $T'$ that is similar to $T$, and the sides of $T'$ touches at least $3$ vertices of $P$.

Largest-parallelogram-in-triangle-2.png

In the above figure, $\triangle A'B'C'$ is our $T'$.


The case where all $4$ vertices of $P$ lie on the sides of $T'$ has been covered above.

Suppose, then, that only $3$ vertices of $P$ lie on the sides of $T'$.

We can split $T'$ using a line parallel to one of the sides of $P$ that passes through a vertex of $T'$.

We connect the base of that line to the vertex of $P$ that does not lie on the sides of $T'$.

Largest-parallelogram-in-triangle-3.png

In the figure above, we can see that $P$ has also been split into two parallelograms, both are inscribed in the two smaller triangles $\triangle XB'C'$ and $\triangle XYC'$.


By our results above:

\(\ds \text {Area of } P\) \(\le\) \(\ds \frac 1 2 \text {Area of } \triangle XB'C' + \frac 1 2 \text {Area of } \triangle XYC'\)
\(\ds \) \(\le\) \(\ds \frac 1 2 \text {Area of } \triangle XB'C' + \frac 1 2 \text {Area of } \triangle XA'C'\)
\(\ds \) \(=\) \(\ds \frac 1 2 \text {Area of } T'\)
\(\ds \) \(\le\) \(\ds \frac 1 2 \text {Area of } T\)

so the area of $P$ cannot exceed half the area of $T$.

Hence we see that half the area of $T$ is indeed the maximum area of $P$, and occurs when all three conditions above are satisfied.

$\blacksquare$


Sources

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