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What is the height of the red bar?

the problem

My try: with respect to the picture, it seems for the green bar $\frac{h}{H}=\frac{2}{3}$. So, I think that ratio is the same for the red bar, and the height of the red bar is $$\frac{h}{6+4}=\frac 23\qquad\to\qquad h_{red}=\frac{20}{3}$$

Is this correct?

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    $\begingroup$ A sanity check you can make for yourself: are all the lengths that you are adding together (namely, the 6m and 4m lengths) pointing in the same direction? That is the property they must have for directly adding them to be the right operation. $\endgroup$
    – Daniel Wagner
    Oct 14, 2017 at 6:39
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    $\begingroup$ 3 m of horizontal shadow corresponds to 2 m in height. So the 6 m of horizontal shadow from the red bar corresponds to 4 m in height. The 4 m of vertical shadow however corresponds to 4 m in height (just a 1:1 correspondence), so the total height of the red bar is 4 m + 4 m = 8 m. $\endgroup$
    – md2perpe
    Oct 14, 2017 at 6:39
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    $\begingroup$ For the dupe voter, I hope they don't close it as a dupe of that question because it has considerably lower quality than this one... don't close as dupe just because there's an older one, but close as dupe if there's a better alternative. (if possible, that question should be closed as a dupe to this one) $\endgroup$
    – Andrew T.
    Oct 14, 2017 at 12:12
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    $\begingroup$ I do not want to sound disrespectful, but is this a real question by the OP? It seems improbable to me that a PhD candidate in a Mathematics dept would be duped by this question. I am honestly puzzled. Am I missing something? $\endgroup$
    – Thanassis
    Oct 15, 2017 at 3:16
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    $\begingroup$ Why are we deleting this question? This question is better than the duplicate target. If anything we should close the other question as a duplicate of this one. $\endgroup$
    – Matt Samuel
    Jun 6, 2020 at 21:15

12 Answers 12

Reset to default
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Here is a different visualization. Shadow visualization

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    $\begingroup$ +1. A picture is worth a thousand words (or $140$ characters of MathJax, at current exchange rates). $\endgroup$
    – Brian Tung
    Oct 13, 2017 at 21:59
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    $\begingroup$ Is there an implicit assumption that the light source is far away? It looks like that is the assumption that would make these slanted lines parallel. $\endgroup$
    – Zach Boyd
    Oct 14, 2017 at 6:36
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    $\begingroup$ @ZachBoyd a) It is the only assumption that makes the problem statement tractable. b) There is also the unstated assumption that the bars and the walls are vertical, and that the floor is horizontal, and that the dark areas are shadows and not some painting, and much more $\endgroup$
    – Hagen von Eitzen
    Oct 14, 2017 at 9:17
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    $\begingroup$ So the picture isn't drawn to scale? The red bar doesn't look 4 times the height of the green bar. $\endgroup$
    – camden_kid
    Oct 14, 2017 at 10:12
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    $\begingroup$ @ctrl-alt-delor All questions can only be answered by first assuming things that make the question make sense. For starters, you assume that the question is asked in English, that it has intended meaning, etc. $\endgroup$
    – Joker_vD
    Oct 15, 2017 at 1:16
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3D histograms are evil according to Edward Tufte. Here, they are used to obfuscate information and make this geometry problem harder than it is. Also, as mentioned by @CandiedOrange and @LamarLatrell, the original drawing isn't to scale.

Here's a 3D render with correct heights:

enter image description here

By playing with perspective and point of view, you can seamlessly merge lengths that appear on distinct axes. It might give you the wrong impression that you could simply add those lengths.

from above

But if you select the correct perspective, the problem becomes much clearer.

enter image description here

enter image description here

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  • $\begingroup$ This is now the kickass answer. +1 $\endgroup$
    – Lamar Latrell
    Oct 18, 2017 at 9:19
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    $\begingroup$ This is some impressive visualization work. Any of these would improve the test over using the original image. $\endgroup$
    – candied_orange
    Oct 18, 2017 at 16:52
  • $\begingroup$ Late comment: I think 3D histograms aren't inherently evil, and have their place, but it's certainly true that their use is fraught with peril. $\endgroup$
    – Brian Tung
    Jun 20, 2020 at 18:24
  • $\begingroup$ @BrianTung. Do you have a good example? It'd be interesting to see if there are better alternatives. $\endgroup$
    – Eric Duminil
    Jun 20, 2020 at 20:06
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    $\begingroup$ @JimmyYang SketchUp. I described a similar animation here : math.stackexchange.com/questions/2550096/… $\endgroup$
    – Eric Duminil
    Nov 29, 2020 at 6:25
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There are three similar right triangles here and the ratios of their catheti (or legs) are all equal: $$\frac 23=\frac{4}{x}=\frac{h}{6+x} \implies x=6,\quad h=\frac{2(6+x)}{3}=8$$ ($x$ is the an imaginary cathetus which goes beyond the blue wall).

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  • $\begingroup$ You might want to motivate what $x$ is for the OP. $\endgroup$
    – Brian Tung
    Oct 13, 2017 at 19:07
  • $\begingroup$ @BrianTung Is it better now? $\endgroup$
    – Robert Z
    Oct 13, 2017 at 19:21
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    $\begingroup$ Yes, although "cathetus" is a pretty uncommon word (in American English, at least). I tend to hear "leg." $\endgroup$
    – Brian Tung
    Oct 14, 2017 at 6:25
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    $\begingroup$ @BrianTung I see. I re-edited my answer by adding the term "leg" (but I prefer the old word ;-)). $\endgroup$
    – Robert Z
    Oct 14, 2017 at 6:33
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The floor shadow is twice as long, and anything above that is reflective of the actual height (the bar and the wall are parallel, so their angle with the light source is the same). So, the red bar is twice the green bar $+ \ 4$ meters, which is $2\cdot2+4=8$ meters.

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    $\begingroup$ +1 the accepted answer is a visual representation of the same logic. $\endgroup$
    – Gaurang Tandon
    Oct 15, 2017 at 13:06
  • $\begingroup$ +1 for actually explaining what the accepted answer depicts (specifically the reasoning for why the vertical components have identical length). This is the key point and a vital missing link for anyone who didn't already understand it. $\endgroup$
    – brichins
    Oct 16, 2017 at 18:55
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Basic approach. The $2$-meter green bar has a shadow of $3$ meters. It is $6$ meters from the red bar to the wall—how tall would it have to be for its shadow to extend right up to the foot of the wall? Call that height $h$.

How much taller would a bar have to be for the shadow to go up, vertically, another $4$ meters? That plus $h$ is your answer.

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  • $\begingroup$ @EricDuminil: That's a fair point; I'll edit. $\endgroup$
    – Brian Tung
    Oct 15, 2017 at 19:44
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Suppose we had a third bar $2$m high that was at a distance $0.01$m from the back wall. All but $0.01$m of the shadow of that third bar would be on the wall, and the shadow on the wall could not be higher than the bar (because the light is shining at a downward angle), so it must be less than $2$m tall.

So the total length of shadow of the third bar is less than $2.01$m.

Your claim is that a bar's height is $\frac23$ the total length of its shadow on the ground and back wall combined. So you predict a height of less than $\frac23 (2.01)\mathrm m = 1.34\mathrm m.$

That is clearly not the height of the third bar.

Think again about your model. It may help to ask where you should put a notch in the taller bar so that the shadow of the notch is exactly at the corner where the ground shadow meets the back-wall shadow. Then figure how far is that notch from the top of the bar, and how far from the bottom.

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Using similar triangles, we have $\dfrac{h_\text{red}-4}6=\dfrac23$, yielding $h_\text{red}=8$.

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As Zach Boyd mentions, this problem can only be solved if we assume that the light source is sufficiently distant (eg the sun) so that the light rays are parallel. miracle173 mentions that it looks like the light source is far away because the shadows of the two bars are parallel, but of course that could simply be a diagram artifact.

We also need to assume that the bars and back wall are vertical, and hence perpendicular to the floor (or if they aren't vertical they are at least parallel to each other)

So let's assume that the light rays are parallel, that the floor is horizontal, and that the bars and back wall are vertical. In which case, the lines connecting any object point to its corresponding shadow point are parallel. In particular, the lines from the tops of the bars to their shadow are parallel, so we have two key similar right triangles.

The key triangle of the green bar has a base of 3m and height of 2m. The key triangle of the red bar has a base of 6m and since it's similar to the green key triangle it has a height of 4m since $\frac{3}{2} = \frac{6}{4}$. That height is measured above the top of the shadow of the red bar (as illustrated in the diagram in Doug M's excellent answer). The top of the shadow of the red bar is 4m above floor level, thus the total height of the red bar is 8m.

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Without doing any real math it's fairly simple to determine that the red bar is 8 meters tall.

Rationale: The 2m green bar casts a 3m shadow, so it would take a 4m bar to cast the 6m shadow between the red bar and the wall, and thus the lower portion of the red bar is 4m tall. The red bar is parallel to the wall, so the 4m shadow on the wall is cast by the upper 4m of the bar. 4m + 4m = 8m, so the bar is 8 meters tall.

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The vertical bar height is$ \frac23$ length of its uninturrupted shadow length. So $ \dfrac23 \cdot 6 =4.$

Rest of the red bar casts s a direct $1:1$ vertical projection/translation of $4$ length onto the wall. So red height total height of red bar is $4+4=8.$

Else symbolically

$$ h-4 = \frac23 \cdot 6 \, \rightarrow h= 8. $$

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This image is flawed. At least if those black bars are shadows some one has been playing with the lighting.

If the red bar is meant to be 8m (and I think it is) it should look like this:

enter image description here

Instead it looks like this:

enter image description here

Which I can prove is not 8m by arguing that the 4m mark is halfway up the blue wall. That means the blue wall is 8m and a 8m red bar should be no taller than the blue wall. Yet it is taller, as the yellow box shows.

The red bar given is somewhere between 11-12m. Sure, that isn't what the person who drew this was thinking. But this is a terrible inconsistency to confront students with. This isn't math. This is "guess what I'm thinking".

Now sure, this assumes the image is drawn in a particular 3D style. But it's one that mathematics students are familiar with. Can you guess which one?

enter image description here

The essential assumption is that lines that are parallel in 2D are parallel in 3D.

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    $\begingroup$ The drawing is only here to show the concept and the assumptions. It isn't to scale probably to prevent the reader from simply guessing or measuring the height. It's actually pretty common in math : The alpha angle in this triangle could be any value between 0° and 180°. There's no need to take every drawing literally. $\endgroup$
    – Eric Duminil
    Oct 17, 2017 at 17:56
  • $\begingroup$ @EricDuminil sorry but that's just "guess what I'm thinking" talk. Real math doesn't care how you solve it. If you only want it solved one way say what way. $\endgroup$
    – candied_orange
    Oct 18, 2017 at 2:14
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    $\begingroup$ Since it's a math problem, you might want to trust and use the numbers ;) $\endgroup$
    – Eric Duminil
    Oct 18, 2017 at 6:32
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    $\begingroup$ Interesting. What are the other ways? Wouldn't it make sense to use all the provided numbers? BTW, I've updated my answer to show what a correct 3D diagram would look like. $\endgroup$
    – Eric Duminil
    Oct 18, 2017 at 12:57
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    $\begingroup$ Have you taken note that most other answers seem to use the numbers quite naturally? In my opinion it is because of the core reason that they are the only set of information presented that isn't unambiguous. Your answer is helpful in that it points out the rest, but that doesn't mean that the numbers are necessarily wrong. As @EricDuminil has said " It isn't to scale probably to prevent the reader from simply guessing or measuring the height". $\endgroup$
    – Lamar Latrell
    Oct 20, 2017 at 6:20
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Visualising the model is a bit tricky. The eye cannot see the tiny differences. One of the great examples that eyes can be trapped is the missing square puzzle. Therefore, we cannot rely on the visual presentation of the model to calculate the height.

Here, I will try to present the problem mathematically:

enter image description here

We know the length of $BC=6m$ which represents the horizontal shadow, $CD=4m$ which represents the vertical shadow, $BE=4m$, $AE=x$

height of the red bar is $AE + BE = 4m + x$

$\angle{AED}=90^\circ$ and $\angle{ADE}=60^\circ$ and $\angle{ADE}=30^\circ$

So, $AD=2x$ and $ED=x\sqrt{3}=6m \Rightarrow x = \frac{6}{\sqrt{3}}$

$AB = 4m + \frac{6}{\sqrt{3}}m \approx 7.464m$

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