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The evenly spaced lines are drawn parallel to the base of triangle. What percentage of the triangle is grey?

enter image description here

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12 Answers 12

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A visual approach...

Draw nine lines parallel to each of the other 2 edges.

enter image description here

Now there are 100 congruent triangles, 45 of which are grey.

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    $\begingroup$ Now this is what I call a perfect answer. $\endgroup$
    – Simd
    Oct 19, 2022 at 23:48
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I believe the answer to be

45%

Justification:

Observe that all of the triangles are similar to each other (base angles are equal by alternate angles). Further the $n$th triangle ($n$ corresponds to how many bands the triangle contains) has side length ratio $n:1$ with the top triangle, using the fact that the bands are evenly spaced. We can use the fact that the ratio of areas of similar triangles is equal to the square of the ratio of side lengths to obtain:$$\frac{T_n}{T} = \frac{n^2}{1^2}$$ Where $T$ is the area of the top triangle and $T_n$ is the area of the $n$th triangle. Rearranging yields:$$T_n = n^2T$$ To find the area of the $n$th band from the top we can take the difference of $T_n$ and $T_{n-1}$:$$A_n = T_n - T_{n-1} = n^2T - (n-1)^2A = (2n-1)A$$That is, the area of the $n$th band is equal to $2n-1$ times the area of the top triangle. We can then compute the area of the shaded parts as they are simply the odd bands:$$A_{shaded} = \sum_{i \in \{1, 3, 5, 7, 9\}}A_{i} = \sum_{i \in \{1, 3, 5, 7, 9\}}(2i-1)T = 45T$$We can also compute the area of all regions by summing all the shaded areas:$$A_{all} = \sum_{i = 1}^{10}A_{i} = \sum_{i = 1}^{10}(2i-1)T = 100T$$Then taking the ratio gives:$$\frac{A_{shaded}}{A_{all}} = \frac{45T}{100T} = 0.45$$Giving the answer of 45%

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  • $\begingroup$ Obviously, the limit as the number of row-pairs grows large has to be 50%. (I.e. as the triangle grows larger and larer or, alternatively, as we slice a fixed size triangle into thinner and thinner grey-white pairs.) $\endgroup$
    – Kaz
    Oct 21, 2022 at 21:45
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$(1+5+9+13+17)/10^2 = 5*9\%=45\%$.

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    $\begingroup$ I can see the first equation from the divisions in the visual answer, but don't understand their derivation here. Or how to get to the equivalency in the center. More detail / explanation would be appreciated. $\endgroup$
    – brichins
    Oct 22, 2022 at 22:03
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An approach that avoids having to count lots of tiny triangles:

enter image description here Cutting the triangle into 4 subtriangles of identical shape we find that triangles marked 2,3,4 also have the same shading, whereas 1 inverts grey and white bits. Now looking at 2 and 3 (or 3 and 4) together we see that the white to grey ratio is 3:2 in triangles 2,3,4 and 2:3 in triangle 1. Total ratio is therefore (3+3+3+2):(2+2+2+3) = 11:9, i.e. 55% white, 45% grey.

Variation:

enter image description here Starting with two copies of the triangle cut off subtriangle 4 from both and rearrange as shown in the bottom panel. Count white and grey stripes and take the percentage.

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The parallel lines can be seen as the bottom edges of 10 similar triangles of alternating color, all sharing a common upper corner and stacked one atop another. Since the bottom edges are all evenly spaced, if the largest triangle has height h and base b, the total area is that of the largest triangle ($n$ = 10, counting smallest to largest): $$A_{10} =\frac{bh}{2}$$ and the area of each $n$th triangle is proportional to its fractional height and width of the largest triangle: $$A_n = \frac{(\frac{n}{10}b)(\frac{n}{10}h)}{2} = (\frac{n}{10})^2 *\frac{bh}{2} = \frac{n^2}{100}*A_{10} = n^2 \%$$ The shaded areas then are the exposed bottom strip of only the gray (odd-numbered) triangles, and that strip's area is the difference between the total area of its gray triangle and the area of the next-smaller white (even-numbered) neighbor obscuring it: $$A_{n_{exposed}} = n^2-(n-1)^2$$ Therefore the total shaded percentage is the sum of the exposed gray strips, or $$A_{exposed} = \sum_{n \in \{1, 3, 5, 7, 9\}}n^2-(n-1)^2 = 45$$

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This visual solution is similar to @cap's but generalises with ease to any number n of stripes. n can be even or odd in either case the longest stripe shall be white.

The example has angles 45°,45°,90° for visual clarity but obviously other angles work just as well:

enter image description here

Using two copies of the original triangle joined at their white side we get a parallelogram that can be subdivided into smaller parallelograms as shown. Except for the all-white diagonal these smaller parallelograms are half white, half grey.

As the diagonal comprises 1/n of the total area, the grey fraction is 1/2 x (1 - 1/n) for any number of stripes n.

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A simple[citation-needed] answer:

Let the height of each row be $h$ (so the height of the triangle is $10h$).
Knowing the triangles are similar (AAA) and equally spaced, let the lengths of the bottom sides be $1w, 2w, 3w, \dots, 10w$.

Using the formula for the area of a trapezoid $A = \frac{a + b}{2} \times h$:
The area of the whole triangle is $\frac{0w + 10w}{2} \times 10h = 50wh$.
The area of the first grey row is $\frac{0w + 1w}{2} \times h = 0.5wh$.
The area of the second grey row is $\frac{2w + 3w}{2} \times h = 2.5wh$.
Continuing, we get:
$\frac{\textrm{area in grey}}{\textrm{total area}} = \frac{0.5wh + 2.5wh + 4.5wh + 6.5wh + 8.5wh}{50wh} = \frac{22.5wh}{50wh} = \frac{45}{100} = 45\%$

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Another approach which starts the same way as loopy walt's answer. Divide the whole triangle into triangular quarters as in loopy walt's diagram.

Consider triangles 2 and 3. They have half the total area, and $2/5$ of it is grey, which is thus $1/5$ of the total area.

Consider triangles 1 and 4. Those two triangles have half the total area, and half of it is grey, which is thus $1/4$ of the total area.

So the total grey area $1/5+1/4=(4+5)/20=9/20=45\%$ of the total area."

My point is that you don't need to establish the white/grey proportions in each triangle, merely to see that 1 and 4 have the same division but opposite shading, so they collectively have as much grey as white. Once triangles 2 and 3 are considered together, the resulting stripes are congruent, so the $3:2$ ratio is easier to see there.

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Here is a little Scala program for arbitrary n.

def puzzle(n: Int): Double =
  if (n == 2) {
    1.0/4.0
  } else if (n % 2 == 1) {
    (puzzle(n - 1) * (n - 1) * (n - 1) + (n + (n - 1))) / (n * n)
  } else {
    (puzzle(n - 1) * (n - 1) * (n - 1)) / (n * n)
  }

@main def main: Unit =
  val height = 10
  val percentage = puzzle(height)
  println(s"$percentage")

which prints

> run
[info] running main 
0.45
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I am not great with calculus and whatnot, but I think this puzzle was within my reach-- Here's what I was able to figure out, and I hope it helps anybody who struggled to understand the math lingo in the other answers:

The answer is

45%.

This can be seen as

.. a nested stack of triangles, each one 10% narrower and 10% shorter than the last. This is because if all angles are the same, a 10% reduction in height must create a 10% reduction in base width, or else the triangles would no longer be similar.

The long form, which I'm sure other answers have written in a more compact mathematical format, is as follows:

Following from the top triangle down-
((.1b*.1h)/2) -- Top triangle, 1st grey area
+(((.3b*.3h)/2)-((.2b*.2h)/2)) --2nd grey area
+(((.5b*.5h)/2)-((.4b*.4h)/2)) --3rd grey area
+(((.7b*.7h)/2)-((.6b*.6h)/2)) --4th grey area
+(((.9b*.9h)/2)-((.8b*.8h)/2)) --5th grey area

We can strip out the b, h, and even the "/2" portion because these are all in proportion to each other anyway, leaving us with .1² + (.3²-.2²) + (.5²-.4²) - (.7²-.6²) + (.9²-.8²) = .45

enter image description here

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FYI

Given @cap's clever visualization solution, following recursive formulas:

  • number $g(n)$ of grey triangles

$g(n)=n==1$ $?$ $1:g(n-1)+(2*n-1)*(n\%2)$

  • total number $t(n)$ of triangles

$t(n)=n==1$ $?$ $1:t(n-1)+(2*n-1)$

confirm $45\%$ for $n=10$

I found no OEIS® entry for grey, white or total series.

The highest amount $n$ of rows with an integer percentage $p$ of grey ones seems:

$n=50$ with $p=49\%$

Also, if we write $w(n)$ for number of white triangles, then following non-recursive equations hold

$t(n) = g(n) + w(n) = n^2$

and

$w(n) - g(n) = (-1)^n * n$

And, if we write $x(n)$ and $y(n)$ for the alternating smallest and biggest one of $w(n)$ and $g(n)$ then

$x(n) = n * (n - 1) / 2$

and

$y(n) = n * (n + 1) / 2$

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One other visual approach independent of $n$ ($n=10$ here)

enter image description here

The biggest one of both areas (white in OP case) can be seen as sum of $n$ areas (colored red green alternating top down here) each one equal to $k$ $(1<=k<=n)$ units. One unit being the 'top' triangle area.

But then that area equals (by Gauss formula): $n*(n+1)/2$, in this case: $55$

Similarly smallest one (grey in OP case, but not given a color here) has area: $n*(n-1)/2$, in this case: $45$

As a consequence this also confirms:

Total area equals $n^2$ units, in this case: $100$ giving both percentages $45$ and $55$.

Required percentages for any $n$ now are easy to express in terms of above for any $n$.

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