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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.

enter image description here


I heard this puzzle from Solomon Golomb, a brilliant mathematician and lover of puzzles who recently passed away.

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    $\begingroup$ Thanks for sharing -- didn't realize he passed away. I vividly remember learning about Golomb rulers as a kid from a Martin Gardner book. $\endgroup$ – Tyler Seacrest May 10 '16 at 8:17
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    $\begingroup$ Penguino's solution is both clever and elegant showing this is definitely a puzzle and not a math-textbook style problem. I'm voting to reopen. $\endgroup$ – Tyler Seacrest May 12 '16 at 3:24
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    $\begingroup$ How many "textbook-style" math problems have a solution that only involves folding the paper and no calculations at all? $\endgroup$ – ffao May 12 '16 at 16:55
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Graphically... enter image description here

...where H is the height of the triangle

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  • $\begingroup$ The A>=H/2 case doesn't necessarily have AxB+Overlap as the total size, it may also have extra. $\endgroup$ – hvd May 10 '16 at 5:09
  • $\begingroup$ Should add an example where A = H/2, you have no extra or overlap, but the areas are equal. $\endgroup$ – Darrel Hoffman May 10 '16 at 14:10
  • $\begingroup$ All I can think of is i.imgur.com/r6yHlo5.gif $\endgroup$ – corsiKa May 10 '16 at 15:07
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    $\begingroup$ While this is pretty I don't see how it's a proof (as requested) without showing why the triangles must fit as shown in all cases. Do they? $\endgroup$ – Jonathan Allan May 10 '16 at 15:31
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    $\begingroup$ @Jonathan Allan In my opinion the only other thing I believe is needed is that the the two internal angles of the two triangles that meet at each pivot point sum to 90 degrees. That assures the fact that the constructed 'pink' shape is rectangular (and we know its height is A and width is B the same as the green square. $\endgroup$ – Penguino May 10 '16 at 20:37
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Any such arrangement may be rotated to have the adjoining edge at the base (as pictured in the question).

enter image description here 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.

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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.

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  • $\begingroup$ Good and simple answer. $\endgroup$ – cst1992 May 10 '16 at 9:26
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A basic not so clever answer.

enter image description here

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.

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    $\begingroup$ I don't think you need to mark this as spoiler. It's hard to understand as it is. :) $\endgroup$ – Marius May 10 '16 at 10:30
  • $\begingroup$ but still understandable, isn't it? $\endgroup$ – clem steredenn May 10 '16 at 11:32
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    $\begingroup$ Yep it makes sense, but only if you follow it closely. $\endgroup$ – Marius May 10 '16 at 11:34
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I threw the kitchen sink at it. Please inform of any mistakes.

Step 1: Consider the triangle in a convenient coordinate system.

enter image description here

Step 2: Find the area of the triangle.

It's one half the base width multiplied by the height.

$$\frac{1}{2}ac$$

Step 3: Find an equation for the line passing through points (0,0) and (b,c) in slope-intercept form.

I won't go into the details unless someone asks me to. It's:

$$y=\frac{c}{b}x$$

Step 4: Find an equation for the line passing through points (b,c) and (a,0) in slope-intercept form

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): $$c=\frac{-c}{a-b}b+i$$ $$i=\frac{c}{a-b}b+c$$ $$y=\frac{-c}{a-b}x+\frac{c}{a-b}b+c$$ $$y=\frac{cb-cx}{a-b}+c$$

Step 5: Find a relationship between x1 and x2.

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.

$$\frac{c}{b}x_1=\frac{cb-cx_2}{a-b}+c$$ $$\frac{c}{b}x_1-c=\frac{cb-cx_2}{a-b}$$ $$\frac{cx_1(a-b)}{b}-c(a-b)=cb-cx_2$$ $$-\frac{cx_1(a-b)}{b}+c(a-b)=-cb+cx_2$$ $$-\frac{cx_1(a-b)}{b}+c(a-b)+cb=cx_2$$ $$-\frac{x_1(a-b)}{b}+(a-b)+b=x_2$$ $$-\frac{x_1(a-b)}{b}+a=x_2$$ $$x_2=-\frac{x_1(a-b)}{b}+a$$

Step 6: Find the area of the rectangle.

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:

$$area=(-\frac{x_1(a-b)}{b}+a - x_1)\frac{cx_1}{b}$$

Step 7: Find the maximum of the area function between x = 0 and x = a.

Let's just use WolframAlpha

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.

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    $\begingroup$ Hey, welcome to Puzzling! This is a fantastic answer - I hope to see a lot more from you later! c: $\endgroup$ – Deusovi May 11 '16 at 4:02
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enter image description here 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.

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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 $\frac12$).

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