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There is a rectangle inside a triangle, with one of the rectangle's sides up against one of the triangle's sides, and the other two corners touching the triangle's other two sides.
Prove that the area of the rectangle is at most $1/2$ the area of the triangle.
In other words, prove that the blue region is no bigger than the red region.
I heard this puzzle from Solomon Golomb, a brilliant mathematician and lover of puzzles who recently passed away.
Graphically...
...where H is the height of the triangle
A>=H/2 case doesn't necessarily have AxB+Overlap as the total size, it may also have extra.
$\endgroup$
Any such arrangement may be rotated to have the adjoining edge at the base (as pictured in the question).
Say that the rectangle is $n$ high and $m$ wide and the large triangle is $p$ high and $q$ wide.
The large triangle will have area $A=\frac{pq}2$.
The upper of the smaller triangles is a similar triangle to the large triangle and as such:
$m=q\frac{p-n}p$,
so the area of the rectangle is:
$nm=qn\frac{p-n}p=q\frac{(p/2)^2-(n-p/2)^2}{p}$
which may be maximised by letting $(n-p/2)^2=0\rightarrow n=p/2$,
whereupon
$nm=q\frac{(p/2)^2}{p}=\frac{pq}{4}=\frac12A$
Alternatively
one could take the derivative of $n(p-n)$ with respect to $n$ and set it to zero for the same result.
I'm not sure if this needs spoiler marking, but I'll play it safe:
Look at it as three separate triangles. Flip each of the three red triangles on its edge with the blue rectangle. If the rectangle's height is exactly half the total height, the triangles perfectly cover it. If the rectangle's height is less than or greater than half the total height, the triangles more than cover it.
A basic not so clever answer.
We have area of the three triangles as
$$\frac{AC\times BD}{2} + \frac{EF\times FB}{2} + \frac{DH\times HI}{2}$$
if we write $a=BF$ and $b=BD$, we get that
$$area=\frac{1}{2}[b.AC + a.(EF+HI)]$$
From Thales, we get that
$$\frac{b}{EI}=\frac{AB}{AE}=\frac{AD}{AI}=\alpha$$
Thus,
$$EF+HI=(\frac{1}{\alpha}-1)b$$
We have $AC=AD sin(\theta)$, which leads, combining with above
$$AC = \frac{\alpha}{1-\alpha}a$$
and thus
$$area=\frac{a.b}{2}(\frac{\alpha}{1-\alpha}+\frac{1-\alpha}{\alpha})$$
We know that $\alpha\in[0;1]$, so we can write $\alpha=0.5+\epsilon$, where $|\epsilon|\in[0;0.5]$. Thus, it becomes
$$area=a.b.(\frac{0.5^2+\epsilon^2}{0.5^2-\epsilon^2})\ge a.b.(\frac{0.5^2}{0.5^2})=a.b$$
QED.
PS. I did not know how to mark the whole text as spoiler and still have a reasonable formatting.
I threw the kitchen sink at it. Please inform of any mistakes.
It's one half the base width multiplied by the height.
$$\frac{1}{2}ac$$
I won't go into the details unless someone asks me to. It's:
$$y=\frac{c}{b}x$$
The slope of that line is: $$\frac{-c}{a-b}$$
So the equation will be (i is the x intercept... I'm already using b...): $$y=\frac{-c}{a-b}x+i$$
To find i, plug in the point (b,c): $$c=\frac{-c}{a-b}b+i$$ $$i=\frac{c}{a-b}b+c$$ $$y=\frac{-c}{a-b}x+\frac{c}{a-b}b+c$$ $$y=\frac{cb-cx}{a-b}+c$$
Looking at the diagram, notice that the top left corner of the square is on the line addressed in step 3, and the top right corner of the square is on the line addressed in step 4. Moreover, their Y coordinates are the same (a consequence of how we chose our coordinate system). Given this, we should be able to find a relationship between x1 and x2 by setting their line equations to be equal.
$$\frac{c}{b}x_1=\frac{cb-cx_2}{a-b}+c$$ $$\frac{c}{b}x_1-c=\frac{cb-cx_2}{a-b}$$ $$\frac{cx_1(a-b)}{b}-c(a-b)=cb-cx_2$$ $$-\frac{cx_1(a-b)}{b}+c(a-b)=-cb+cx_2$$ $$-\frac{cx_1(a-b)}{b}+c(a-b)+cb=cx_2$$ $$-\frac{x_1(a-b)}{b}+(a-b)+b=x_2$$ $$-\frac{x_1(a-b)}{b}+a=x_2$$ $$x_2=-\frac{x_1(a-b)}{b}+a$$
The area of the rectangle is the base times the height. The base is x2 - x1, the height can be found by plugging x1 into the equation in step 3. Since we have a relationship between x1 and x2, we can find the rectangle's size now as a function of x1 and only x1.
Height:
$$\frac{cx_1}{b}$$
Base:
$$-\frac{x_1(a-b)}{b}+a - x_1$$
Put them together as base * height:
$$area=(-\frac{x_1(a-b)}{b}+a - x_1)\frac{cx_1}{b}$$
Who kindly tells us that the maximum is:
$$\frac{1}{4}ac$$
If you look back at step 1, you will see that that's half the area of the triangle.
Let's prove a somewhat stronger version of the problem: the blue area is no bigger than the white area ($r \le p+q$).
The proof relies on the fact that the ratio of areas between two similar triangle is the ratio of the lengths squared.
Let $t = AE/AB$ and $s = p + q + r$. It's easy to see that $1-t = EB/AB$
Top white triangle ($p$) is similar to ABD ($s$) with ratio $t$, so $p/s = t^2 => p = s * t^2$
Left white triangle ($q$) is similar to ABD ($s$) with ratio $1-t$, so $q/s = (1-t)^2 => q = s * (1-t)^2$
The difference ($d$) between white and blue areas is: $$ d = p + q - r = p + q - (s - p - q) = 2 * p + 2 * q - s $$ $$ d = 2 * s* t^2 + 2 * s * (1-t)^2 - s = s * ( 4 * t^2 - 4*t + 1) = $$ $$ d = s * (2*t - 1)^2 \ge 0 $$
Coming back to the big picture, this means that the area of the left piece of the big rectangle is smaller (or equal) than half the area of the left piece of the big triangle. A mirrored proof shows that the area of the right piece of the big rectangle is smaller (or equal) than half the area of the right piece of the big triangle.
In conclusion the area of the big rectangle is no bigger than half the area of the big triangle, qed.
PS: As a fun fact, it looks like any rectangle inscribed in a triangle has its area at most half of the area of the triangle. Looking forward to a proof of that.
Any (triangle, rectangle) pair as described in the question may be split (by cutting perpendicular to the base, through the opposite vertex) into two smaller cases where triangle is right-angled and rectangle shares a vertex with the right angle. Thanks to the additive property of inequalities we need only prove it in this special case. We can also scale any such instance of the special case so that the triangle is isosceles, because such scaling doesn't change the proportions of the rectangle and triangle.
Say that the non-hypotenuse side length of our triangle is 1 and the rectangle has dimensions $a \times (1 - a)$. It is easy to show by differentiating that this has a maximum where $a = \frac12$, and the value there is $\frac14$ (half of the triangle's area which is $\frac12$).
