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What is the largest possible area of a rectangle(in square units) inscribed in the triangle shown in the picture above?

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@Henry What is this based upon? – ploosu2 May 28 '14 at 9:03
Connect the midpoints of two sides and drop perpendiculars onto the third side. This is a rectangle with half the area of the triangle. Affine transformations mean that if this is optimal for any triangle (e.g. an isosceles right-angled triangle) then it is optimal for any triangle. – Henry May 28 '14 at 9:08
To correct my original comment (since deleted), as this triangle has an obtuse angle, there is one possible rectangle, half the area of the triangle. If all the angles were acute, then there would be three such rectangles. – Henry May 28 '14 at 9:13
@Henry: Affine transformations do not preserve rectangles. – Christian Blatter May 28 '14 at 9:50
Christian: that is true, as affine transformations turn rectangles into parallelograms, but these parallelograms have the same areas as rectangles in the transformed triangle with the same base and perpendicular height. – Henry May 28 '14 at 12:19
up vote 2 down vote accepted

Let $\triangle ABC$ be the given triangle with $AB = 10$, $AC = 17$, and $BC = 21$. Choose an arbitrary point $M$ on the side $AB$, and let $x = AM$. Let $N$ be the point on $AC$ such that $MN \parallel BC$, and points $P$, and $Q$ on $BC$ such that $NP \perp BC$, and $MQ \perp BC$. Thus $MNPQ$ is a rectangle. Let $h_a$ be the length of the altitude from $A$. Note that $h_a$ is constant. The we have the followings proportions:

$\dfrac{BM}{BA} = \dfrac{MQ}{h_a} \to \dfrac{10-x}{10} = \dfrac{MQ}{h_a} \to MQ = \dfrac{h_a}{10}\cdot (10-x)$. Also:

$\dfrac{AM}{AB} = \dfrac{MN}{BC} \to \dfrac{x}{10} = \dfrac{MN}{21} \to MN = \dfrac{21}{10}\cdot x$.

Let $S(x)$ be the area of the rectangle $MNPQ$, then:

$S(x) = MN\cdot MQ = \dfrac{21h_a}{100}\cdot x(10 - x)$. Since:

$x + (10 - x) = 10$, a constant, $S$ achieves a maximum value when $x = 10 - x$, or $x = 5$.

From this we can solve for $MN =\dfrac{21}{2}$, and $MQ = \dfrac{h_a}{2}$. Thus:

$S_{max} = \dfrac{21h_a}{4} = \dfrac{S_{\triangle ABC}}{2}$

Note: $h_a$ can be calculated using Heron formula.

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This is assuming the rectangle that maximize the area has a side parallel to one of the sides of the triangle. This is probably true but I can't see any easy justification of that. – achille hui May 28 '14 at 9:19


You can prove that any such rectangle has the area at most half the triangle area via a few simple cases (of course, there always exists such a rectangle):

  1. Two sides parallel. For any right triangle, and any rectangle with sides parallel to the sides other than hypotenuse: to maximize the area the rectangle has to have proportions of the triangle.
  2. One side parallel. Any triangle and any rectangle with at least one side parallel to some side of the triangle: put a line perpendicular to that side which goes through one of the triangle vertices (it doesn't need to be a height) and use the previous point.two
  3. No sides parallel. Any triangle and any rectangle: take a line that goes through one of the triangle vertices and is parallel to one of the sides of the rectangle, and use previous point.three

I hope this helps $\ddot\smile$

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Beautiful chain of reductions. – zyx May 28 '14 at 12:48
+1 The $3^{rd}$ point is brilliant. – achille hui May 28 '14 at 13:39

Let us generalize the case to obtain the largest rectangle inscribed in any triangle. enter image description here

Let $AB=b$ and $CF=h$. Let $x$ and $y$ be the length and wide of rectangle, respectively. It is easy to notice that $\Delta ABC\sim \Delta CDE$. Hence \begin{align} \frac{x}{b}&=\frac{h-y}{h}\\ x&=\frac{b(h-y)}{h}.\tag1 \end{align} The area of rectangle is $xy$ and by substituting equation $(1)$ to $xy$ we will obtain \begin{align} xy&=\frac{b(h-y)y}{h}\\ &=\frac{-b(y^2-hy)}{h}\\ &=\frac{-b\left(y-\dfrac12h\right)^2+\dfrac14bh^2}{h}.\tag2 \end{align} It can be seen from the equation $(2)$ that the area of rectangle will be maximum when $y=\dfrac12h$, therefore the largest area of rectangle inscribed in any triangle is $\dfrac14bh$ or half of the area of the triangle. Substituting $y=\dfrac12h$ to equation $(1)$ yields $$ x=\frac{b\left(h-\dfrac12h\right)}{h}=\dfrac12b. $$ Thus, the maximum area of rectangle occurs when the midpoints of two sides of the triangle were joined to make a side of the rectangle.

Now, let $AB=21$, $BC=17$, and $AC=10$. Using cosine formula, we will obtain $$ \cos A=\frac{10^2+21^2-17^2}{2\cdot10\cdot21}=\frac35\quad\Rightarrow\quad\sin A=\frac45. $$ The area of $\Delta ABC$ is $$ [\Delta ABC]=\frac12\cdot AB\cdot BC\cdot\sin A=84\text{ square units.} $$ Thus, the maximum area of rectangle is $$ \frac12[\Delta ABC]=\large\color{blue}{42\text{ square units}}. $$

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Don't you have to consider all threes sides as basis $b$? – lhf May 28 '14 at 11:04
@lhf The area of $\Delta ABC$ will always be the same regardless the base we choose. – Tunk-Fey May 28 '14 at 11:15
What if the rectangle isn't parallel to any of the sides of the triangle? – dtldarek May 28 '14 at 11:36
What if $A$ or $B$ is an obtuse angle? – Jeppe Stig Nielsen May 28 '14 at 11:43
@dtldarek I built my argument based on the picture I used. If we construct the rectangle isn't parallel to any of the sides of the triangle (esp. obtuse triangle), then the possible max area is $\dfrac12h^2$, where $h$ is the perpendicular distance between obtuse point to the base. In this case, the area of rectangle will be $32$ square units. – Tunk-Fey May 28 '14 at 11:50

Sharing an elegant approach provided by Dr.Shailesh A Shirali.enter image description here

  • BC = a, CA = b, AB = c
  • PQRS: rectangle with PS ∥ BCand PQ ⊥ BC
  • Triangle APS is similar to Triangle ABC
  • Let AP/AB = t; then AS/AC = t and PS/BC = t
  • Dimensions of Triangle APS: ta,tb, tc
  • AD is an altitude of Triangle ABC
  • AD = h
  • Altitude of Triangle APS: th (by similarity)
  • PQ = h−th = (1−t)h

Following the train of reasoning as shown below figure, we see that:

Area of triangle ABC = (1/2)ah,

Area of rectangle PQRS = t(1−t)ah,

Here 0 <= t <= 1. The maximum value attained by t(1−t) for 0 <= t <= 1 is 1/4 ,

attained when t = 1/2 .

Hence the maximum possible area of PQRS is (1/4)ah, which is half the area of the triangle.

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