In a book of word problems by V.I Arnold, the following appears:

  1. The hypotenuse of a right-angled triangle (in a standard American examination) is 10 inches, the altitude dropped onto it is 6 inches. Find the area of the triangle.

American school students had been coping successfully with this problem for over a decade. But then Russian school students arrived from Moscow, and none of them was able to solve it as had their American peers (giving 30 square inches as the answer). Why?

Here's the book. I assume the answer is some joke at the expense of the Americans, but I don't get it. Possibly a joke about inches? Anyone?

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Eli Rose
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    The joke is at the Americans expence (naively applying formulas without thinking). Hint: Try to compute what the other two sides of the triangle must be given this information. – Winther Dec 31 '15 at 05:46
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    Is it $24$ the area – Archis Welankar Dec 31 '15 at 05:48
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    @Winther: this isn't clear to me. I can't tell from the way it's written whether "giving 30 square inches as the answer" was supposed to apply to the Russians or the Americans. – Qiaochu Yuan Dec 31 '15 at 05:49
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    @Archis Welankar The altitude is not one side of the triangle. – Henricus V. Dec 31 '15 at 05:50
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    Oh, I figured it had something to do with American teenage boys always lying about things that are 6 inches being 10 inches. – fleablood Dec 31 '15 at 05:51
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    I guess he was right. I am an American student and stared at this for a solid minute... – pancini Dec 31 '15 at 05:56
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    @ElliotG: I think this can be spun whatever way you want -- perhaps Americans do not assume that their teacher is constantly trying to punk them. – Eli Rose Dec 31 '15 at 05:59
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    I'd quibble that a triangle with a side (hypotenuse or otherwise) of 10 and an altitude to that side 6 then if we are *given* such a triangle the area is 30. That such a triangle is impossible isn't *my* fault. We weren't asked can such a triangle exist; we were asked given such a triangle what *would* they area be. And it *would* be 30. – fleablood Dec 31 '15 at 06:03
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    might be too that most ( or all ) eurasian countries work in metrical, the question wasn't graded and they were too lazy to multiply everything by 2,54 and divide it back, being school students and all. – CptEric Dec 31 '15 at 08:23
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    Am I the only one who had to look up what an "altitude" meant in this context? I've only ever seen it referred to as the "height" of a triangle, as in the usual "area = base * height / 2" formula... is this US-specific terminology or something? – Thomas Dec 31 '15 at 11:52
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    In a country where people study enough logic, I would expect an answer like: "any triangle with these properties has area 42" (or 0, 3.14, 2016, whatever number you fancy). In other words, all flying pigs have a green tail. – Marc Glisse Dec 31 '15 at 14:27
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    @fleablood the area would *also* be -18 or any number you want. – djechlin Dec 31 '15 at 14:39
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    @fleablood unless you are trying to say that when confronted with a word problem you are supposed to directly apply the nearest formula available and write down the answer, in which case, ah yes, we have arrived at American pedagogy and it may take the Russians a while to get used to that. – djechlin Dec 31 '15 at 14:41
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    @Thomas Nope, I'm from the US and I've never heard of it. I assumed it was a funny way of saying height and the mistake was using the hypotenuse's length in the standard formula until I looked at the answers. – Ixrec Dec 31 '15 at 14:41
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    *Arnold says Russian students can't solve this problem* - I wouldn't take too seriously anything Schwarzenegger says about math or Russians. – Lucian Dec 31 '15 at 15:59
  • Would that be Vladimir Ilyich Arnold? That was meant to be a joke, but I see he really was Vladimir Igorevich! – PJTraill Dec 31 '15 at 16:34
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    You need to analytically continue the properties of the triangle into a larger domain where the triangle no longer exists. – Count Iblis Dec 31 '15 at 17:23
  • @Win There is no ambiguity in the original Russian version - your initial reading is correct. – A.S. Dec 31 '15 at 17:47
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    @djechlin - While it is vacuously true that "the area would also be -18 or any number", the answer of 30 square inches is in some sense distinguished from others in that it's the answer that you would arrive at if you ignored the information stated in the question that the triangle is right-angled. Ignoring this information is not unreasonable; if we start off with the assumption that the question is being posed in good faith by a competent examiner, then we might as well discard that information without examining it any further! It is only because the (meta-)question here is from a (contd) – Hammerite Dec 31 '15 at 18:41
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    (contd) book of word games that we are clued in to further examine the superfluous information about the triangle contained in the (reported) question. As to "applying the nearest formula available", you mischaracterise an entirely reasonable application of a basic fact about geometry (again, charitably granting that the student may assume competence and good faith on the part of the examiner). Your insult to the American education system - no matter what its faults might be - is unwarranted and makes you appear childish. It is cynicism without intelligence or insight. – Hammerite Dec 31 '15 at 18:46
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    @Hammerite I'm sorry the joke didn't meet your standards of rigor. I do feel American pedagogy sets its students up as less likely to catch a glitched question like this for exactly the reasons I stated, and students fail at questions that are correct but for which the kneejerk formula's hypotheses don't actually apply. Sorry if I appeared childish to you, you appear like you're wearing a monocle to me, as long as we're talking about what people appear like. – djechlin Dec 31 '15 at 19:11
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    @fleablood Well, yes, but then by the principle of explosion it would also be 70. – Jack M Dec 31 '15 at 20:30
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    What this joke illustrates is that in math, the most important thing to do is to check whether your answer makes sense. If you don't, the answers become 'just numbers' and you yourself 'just a calculator'. And then you get things like [this](https://xkcd.com/612/). "That such a triangle is impossible isn't my fault." But that you have shut off your brain and are acting like a calculator *is* your fault. "perhaps Americans do not assume that their teacher is constantly trying to punk them". Good teachers *should* try to punk their students. They should learn to *think for themselves*. – Stijn de Witt Jan 01 '16 at 01:47
  • Yes, this book is great. Thanks for sharing this question. I shall check the book out. My knowledge of non-Euclidean geometry is limited. Can someone tell me if such a triangle is possible in non Euclidean geometry where the sum of angles in a triangle is not 180 degrees ? – Saikat Jan 01 '16 at 04:02
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    I am not sure if this relates to my speaking of BrE and not AmE or some kind of translation issue from the original language of the problem but the phrase "the altitude dropped onto it" has no obvious meaning to me. Indeed I had to read the answers to make a guess at what the problem meant at all. – Vality Jan 01 '16 at 09:28
  • @djechlin, I was wearing a monocle but it fell off when I read your post, which is what put me in such a bad mood. – Hammerite Jan 01 '16 at 17:09
  • I do not like this question but to vote to close it as primarily opinion based is strange. Arnold's intent is completely transparent (at least in context) and correctly given in the very first comment. That some, with varied degrees of seriousness, come up with alternative interpretations does not change this. – quid Jan 02 '16 at 15:44
  • @fleablood: Undefined is undefined. The answer is _not_ 30 because there is no answer. – Lightness Races in Orbit Jan 03 '16 at 16:17
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    It's not impossible. Consider the the triangle with radius 5 centered at (0,0). Consider point (x,y); y > 0. The points (-5,0) (0,-5) and (x,y) will form a right triangle with hypotenuse 10 and altitude y and area 5*y. So simply select the point $(\sqrt{11}i, 6)$. That will form a right triangle with hypotenuse 10 and altitude 6 and area 30. The fact that $(\sqrt{11}i, 6)$ does not exist should *not* be a hindrance to any theoretical mathematician. – fleablood Jan 03 '16 at 16:50
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    Here's my train of thought when first reading the question: "Hmm, if the hypotenuse is 10 inches and the height is 6, that means the base is 8, so the area is 6*8/2 = 24 square inches. Wait—why is everyone saying that the area is not 30? Oh, altitude is not the same as height? So *that's* why the Russian students couldn't figure it out..." – ETHproductions Jan 03 '16 at 19:31
  • Assuming fleablood's claim is a typo, the "corrected" statement that the points $(-5, 0)$, $(5, 0)$, and $(x,y)$ always form a right triangle with hypotenuse $10$, altitude $y$, and area $5y$ is still mistaken; they will do this iff $x^{2}+y^{2} = 25$ (this statement made without any consideration at all of complex $x,y$). Extending in the obvious way to complex $x$, we can easily observe that $(\sqrt{11}i)^{2} + 6^{2} = -121 + 36 = -85 \neq 25$. – collapsinghrung Jan 03 '16 at 19:42
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    *"V.I. Arnold says Russian students can't solve this problem, but American students can"* is great clickbait material – cat Jan 04 '16 at 01:38
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    @cat: Haha, yes. I do feel slightly guilty about the success of this question. Essentially I took existing content, gave it a clickbait-y title, and profited -- am I so different from BuzzFeed? – Eli Rose Jan 04 '16 at 01:58
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    I'm with @ETHproductions on thos one, why does everyone have to assume that the hypotenuse is at the base? I have no problem understanding "altitude" but perhaps missed the significance of the phrase "dropped onto ***it***". – slebetman Jan 04 '16 at 02:39
  • @Thomas: the (length of the) altitude *is* the height of the triangle, if you regard the hypotenuse as its base. – Qiaochu Yuan Jan 04 '16 at 06:33
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    @slebetman: What other than the hypotenuse do you think "it" can reasonably refer to? – hmakholm left over Monica Jan 04 '16 at 15:09
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    @HenningMakholm slebetman, ETHproductions, and I presumably all assumed that "it" referred to the triangle itself. – Kyle Strand Jan 04 '16 at 22:18
  • @KyleStrand: But what does it mean to drop an altitude onto _a triangle_ as a whole rather than onto one of the sides? – hmakholm left over Monica Jan 05 '16 at 00:30
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    @HenningMakholm I (and presumably the other two) am not familiar with the phrase "drop (an altitude) onto". I interpreted it simply to mean that the height in (at least) one direction was 6. If I had been in an exam situation or some other context where I needed to answer the question definitively, I'd have recognized my uncertainty and sought clarification. – Kyle Strand Jan 05 '16 at 03:10
  • i think Arnold here focuses on different language conventions used in mathematics in various areas rather than whether a specific right-angled triangle exists or not (which pertains of course to the convention used by one to interpret the question in the first place, as one answer already noted) – Nikos M. Jan 05 '16 at 23:30
  • @Thomas, every triangle has three altitudes, the distance of each corner from the opposite side. It's a standard term. – Anton Sherwood Jun 07 '16 at 18:35
  • This questions does clearly demonstrate one thing: either Americans or Russians can't answer math problems well, but certainly Russians can't pose math problems well. – tparker Jan 14 '18 at 14:10
  • I do not think that this has something to do with word problems. It should be noted that the introduction to the book has (obviously intentionally) a wrong English translation. In the original text it claims that the Russian "thinking culture" is far superior over the western one. And this should of course include the mathematical skills as well. The problem under consideration had obviously to demonstrate this alleged superiority but I doubt that every Russian student will check whether this problem has solution before giving the answer. – user Jan 27 '22 at 13:27

9 Answers9


There is no such right triangle. The maximum possible altitude is half the hypotenuse (inscribe the triangle into a circle to see this), which here is $5$ inches. You would only get $30$ square inches if you tried to compute the area without checking whether the triangle actually exists.

Qiaochu Yuan
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    Ah. Yep, he loves that kind of thing. Thanks for telling me how to see the maximum possible altitude thing, I didn't know that. – Eli Rose Dec 31 '15 at 05:55
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    What about non euclidean geometry (I may be overanalyzing, but [overanalyzing is fun](http://math.stackexchange.com/questions/1435312/are-there-real-numbers-that-are-neither-rational-nor-irrational#comment2924329_1435318))? – PyRulez Dec 31 '15 at 20:56
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    How does it answer the question about russians and americans? – Peter Franek Jan 01 '16 at 20:40
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    Aren't you American, Qiaochu? I smell a paradox here ... – Georges Elencwajg Jan 01 '16 at 21:14
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    @Peter: one of the Americans and the Russians are mindlessly applying formulas and getting $30$ square inches, while the other is checking whether the triangle exists and getting confused. But the way the joke is written I can't tell which is which. – Qiaochu Yuan Jan 01 '16 at 21:32
  • @QiaochuYuan Thanks, seems like a language nuance, but I'm starting to understand :) – Peter Franek Jan 01 '16 at 21:47
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    @PeterFranek (and QiaochuYuan): To clarify the original English: firstly, in “_But then Russian school students arrived from Moscow, and none of them was able to solve it_” the phrase “none of them” refers to “Russian school students”, so it means that “_none of the Russian students was able to solve it_”. Secondly, “_as had their American peers_” means “_as the American students had been able to solve it_”, as the previous sentence said. (“Peers” = “equals”; “successfully” is ironic.) Thus the **Americans** were _mindless_ while the **Russians** were _checking_ whether the triangle existed. – PJTraill Jan 04 '16 at 20:48
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    Even through Arnold probably meant to denounce the calculation as "mindless", I don't think that is a fair characterization. Real life is full of problems where you have an _abundance_ of data about a situation already and need to figure out the _simplest way_ to read out some missing information from them, identifying the _subset_ of the data you need for finding what you seek. It is unproductive and wasteful to insist on approaching all such situations by verifying from first principles that the givens are internally consistent before you even start to consider the actual problem. – hmakholm left over Monica Jan 05 '16 at 00:38
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    @HenningMakholm Until you encounter an issue and have to deal with the consequences: http://www5.in.tum.de/~huckle/bugse.html – James Jan 25 '17 at 12:03

There are many ways to prove that such triangle does not exist. I am using a different approach.

enter image description here

Suppose that the said right angled triangle can be formed. Then, we are interested in where should the foot of the said altitude (CD) be? [That is, how far is D (on AB) from A (or from B)?]

We assume that D is $\alpha$ and $\beta$ units from A and B respectively.

Clearly, we have $\alpha + \beta = 10$ …… (1)

Also, by a fact on right angled triangles, we have $\alpha \beta= 6^2$ ……… (2)

EDIT : That fact is "Power of a point".

To find $\alpha$ and $\beta$ is equivalent to solving the quadratic equation $x^2 – 10x + 36 = 0$.

Since the discriminant $(= [-10]^2 - 4 \times 36)$ is negative , we can conclude that such roots ($\alpha$ and $\beta$) are not real.

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    how did you generate such a nice diagram? – Fattie Dec 31 '15 at 15:17
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    @JoeBlow Using Geogebra. Image is then sent to WORD via the clipboard. After re-sizing, the adjusted image is then sent to PC Paintbrush for the final touch. It (*.png) is then uploaded. – Mick Dec 31 '15 at 18:05
  • Geogebra. Amazing, you mean you did it right there [online](http://www.geogebra.org) Brilliant, thanks for that – Fattie Dec 31 '15 at 18:10
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    i am really confused, it is clear thta $alpha+beta=10$ but how did yyou find $\alpha \beta=6^2=36$ i mean what geometry did you used to get eq(2)? – Bhaskara-III Dec 31 '15 at 18:12
  • @JoeBlow It can be downloaded as a standalone s/w. Then, you have ample of time to edit your picture. – Mick Dec 31 '15 at 18:13
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    @Bhaskara-III Any right-angled triangle with an altitude so formed is being cut into a total of 3 triangles. These 3 triangles are similar to each other. Setup the corresponding ratios to find out the said relation. – Mick Dec 31 '15 at 18:18
  • i really don't know that. please can you show me how to find that? – Bhaskara-III Dec 31 '15 at 18:21
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    i know similar triangles but how to get $\alpha\beta=6^2$ from your picture? – Bhaskara-III Dec 31 '15 at 18:23
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    @Bhaskara-III Try to read http://www.mathwarehouse.com/geometry/similar/triangles/geometric-mean.php – Mick Dec 31 '15 at 18:43
  • in your picture, diameter of circle is $AB=10$ then why did show that altitude is $CD=6$? – Bhaskara-III Jan 01 '16 at 14:26
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    @Mick: it could be better if you represent altitude $CD$ by some variable say $x$. If you take directly $CD=6$ in a semi-circle of diameter $10$ then the point $C$ should not lie on the periphery it should lie outside the periphery. See the answer by Michael Chirico. He has nicely represented all the outside points by a red line in the picture. – Bhaskara-III Jan 01 '16 at 14:34
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    @Mick: In you picture, point $C$ must lie outside the periphery because you take $CD=6$. point 'C' must not lie on the periphery of semicircle. I think picture representation by Michael Chirico is better than yours. – Bhaskara-III Jan 01 '16 at 14:50
  • @Bhaskara-III I think you are not quite familiar with “proof by contradiction”. It goes like this:- 1) We ‘assume’ the construction is possible first. 2) Then, we try to show such assumption contradicts some known facts. 3) From that, we conclude that such construction is not valid. – Mick Jan 01 '16 at 16:11
  • @Bhaskara-III As I have mentioned in the beginning that there are many ways. It depends on whether a quick result is needed or a formal vigorous proof is required. There is no need to discuss this any further. – Mick Jan 01 '16 at 16:42
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    Picture by Michael Chirico clearly shows that such a right triangle not exits while your picture is showing that such a triangle is constructed. it is also a contradiction between these two pictures here. – Bhaskara-III Jan 01 '16 at 16:46
  • FWIW, the two legs of this impossible right triangle have length $\sqrt{10(5\pm i\sqrt{11})}$ – PM 2Ring Jan 02 '16 at 13:08
  • @PM2Ring Yes. I know that the quadratic equation is solvable into complex roots. But can those values for lengths be expressible as ***real*** units with respect to our everyday physical measuring instruments? – Mick Jan 02 '16 at 14:38
  • @Mick: Of course not! :) FWIW, I wasn't implying any criticism of your excellent answer (which I've up-voted, BTW), I just wanted to add that info for the benefit of other readers, and your answer seemed an appropriate place to put it, since you mention the relevant quadratic. – PM 2Ring Jan 02 '16 at 14:48
  • Actually, I just noticed that it's possible to simplify those roots. :oops: $\sqrt{55} \pm i\sqrt{5} \approx 7.4161984871 \pm 2.2360679775i$ – PM 2Ring Jan 02 '16 at 15:05
  • @PM2Ring Thanks for the encouragement. – Mick Jan 02 '16 at 15:15
  • "by a fact on right angled triangles" ... might be more useful to someone if you make it "by the [Power of a Point Theorem](https://en.wikipedia.org/wiki/Power_of_a_point)"? – quintopia Jan 03 '16 at 07:38
  • I think it would be just easier to see that the maximum that the altitude can be is the radius of this circle, that is 5 in this case. – Aritra Das Jan 06 '16 at 08:01
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    @AritraDas As I have mentioned in the beginning, there are many ways to prove.... I think this prove (by contradiction) is equally convincing. – Mick Jan 06 '16 at 10:21

enter image description here

The red line represents all possible third vertices for triangles with base 10 and height 6;

The blue curve represents all possible third vertices for right triangles with hypotenuse 10.

The two sets have null intersection.

(in fact, the maximum possible third angle 6 units away is $\arccos(\frac{11}{61})\approx$ 79.6°)

(and yes, technically we should include the corresponding points below the segment as well)

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    I think your answer (along with all other answers, with the exception of the one by @AlanCampbell) has badly misinterpreted the problem. The problem statement says that the altitude "dropped onto the hypotenuse" -- *not* the one from dropped from the vertex with the right angle -- is 6 inches. Thus, the triangle has sides 6, 8, and 10 inches -- there's no problem at all with such a triangle existing... – Mico Jan 04 '16 at 06:15
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    @Mico though I agree it's apparent the author should have been more careful with verbiage, I disagree with you for hair-splitting reasons. the problem states "_the_ altitude"; in the 3-4-5 case it becomes immediately ambiguous as there are in fact _three_ altitudes with an end on the hypotenuse. thinking this way, there is no way the author could get away with using the definite article. the only way we can take the problem as given is to deduce the author in fact meant the altitude corresponding to the right angle. – MichaelChirico Jan 04 '16 at 06:23
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    @Mico: what do you think the 'the altitude "dropped onto the hypotenuse"' is intended to mean? As far as I am concerned it is the exact same thing as 'the one from dropped from the vertex with the right angle.' In any triangle there are three altitudes. Each can be specified by either the side on which they are dropped or by the vertex from which the are dropped (or still differently). – quid Jan 04 '16 at 20:03
  • I'll add that this could well be a quibble that is simply something lost in translation (I take it the original quote was in Russian?)... we need a native Russian speaker to comment on whether, in the original Russian, this ambiguity existed in the first place. Until the ambiguity is confirmed in the original Russian, this is simply a poor translation. – MichaelChirico Jan 04 '16 at 20:13
  • @MichaelChirico you're too kind; I followed exactly the same train of thought as you, except I attributed the odd grammar to either a poor translation or a non-native speaker doing the best they could. This subconscious hypothesis was reinforced by the odd syntax/ambiguity in the "as had their Russican peers" statement. – pjz Jan 04 '16 at 21:17

By mistake, one can fairly easily calculate the area of given right triangle as $\frac{1}{2}(10)(6)=30$ but this is incorrect. Why? Perhaps, this is the intuition behind the question that one should first check the existence of such a right triangle with given data before calculating area.

A right triangle with hypotenuse $10$ & an altitude of $6$ drawn to it doesn't exist because the maximum possible length of altitude drawn to the hypotenuse is $5$ i.e. half the length of hypotenuse. Here is an analytic proof to check existence of such a right triangle.

Statement: The maximum length of altitude, drawn from right angled vertex to the hypotenuse of length $a$ in a right triangle, is $a/2$ i.e. half the length of hypotenuse.

Proof: Let $x$ & $y$ be the legs (of variable length) of the right triangle having hypotenuse $a$ (known value) then using Pythagorean theorem, one should have $$x^2+y^2=10^2$$ $$y^2=a^2-x^2\tag 1$$ Now, the length of altitude say $p$ drawn to the hypotenuse in right triangle is given as $$=\color{blue}{\frac{(\text{leg}_1)\times (\text{leg}_2)}{(\text{hypotenuse})}}=\frac{xy}{a}$$ $$\implies p=\frac{xy}{a}$$$$\iff a^2p^2=x^2y^2\tag 2$$ let $a^2p^2=P$ (some other variable ), now setting value of $y^2$ from (1), $$P=x^2(a^2-x^2)=a^2x^2-x^4$$ $$\frac{dP}{dx}=2a^2x-4x^3$$ $$\frac{d^2P}{dx^2}=2a^2-12x^2\tag 3$$ For maxima or minima, setting $\frac{dP}{dx}=0$, $$2a^2x-4x^3=0\implies x=0,\frac{a}{\sqrt 2}, -\frac{a}{\sqrt 2}$$, But $x>0$, hence $x=\frac{a}{\sqrt 2}$. Now, setting this value of $x$ in (3), $$\frac{d^2P}{dx^2}=2a^2-12\left(\frac{a}{\sqrt 2}\right)^2=-4a^2<0$$ hence, $P$ i.e. $a^2p^2$ is maximum at $x=\frac{a}{\sqrt 2}$ hence, from (1), the corresponding value of $y$, $$y=\sqrt{a^2-\frac{a^2}{2}}=\frac{a}{\sqrt 2}$$

hence, the maximum possible length of altitude drawn (from right angled vertex ) to the hypotenuse, $$\color{red}{p}=\frac{xy}{a}=\frac{\frac{a}{\sqrt 2}\frac{a}{\sqrt 2}}{a}=\color{red}{\frac{a}{2}}$$ So if the length of altitude $p$ is greater than $\frac{a}{2}$ (half the length of hypotenuse) then such a right triangle doesn't exist.

Harish Chandra Rajpoot
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Non-Euclidian triangle on a sphere

Here a non-Euclidian answer to the question

The depicted sphere has a circumference of length 40, and thus a radius of $\frac{20}{\pi}$. The points $A$ and $B$ are located on the equator, and $C$ is located on a pole.

Now $\triangle ABC$ is a right triangle (it even has two right angles), since all meridians intersect the equator perpendicularly.

Define $AC$ to be the hypothenusa, with length $10$. The height, being the shortest distance from $B$ to its hypothenusa $AC$ is then indeed $6$. As $\triangle ABC$ covers $\frac6{40}$ of the upper hemisphere, its area is:

$$A_{\text{triangle}} = \frac{6}{40}\times\frac{1}{2}\times 4\pi r^2 = \frac{120}{\pi}$$


Job Bouwman
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Interesting - I had forgotten what an altitude was. Wikipedia says:

In geometry, an altitude of a triangle is a line segment through a vertex and perpendicular to (i.e. forming a right angle with) a line containing the base (the opposite side of the triangle). This line containing the opposite side is called the extended base of the altitude.

Who said the hypotenuse was the base? Why can't the altitude be equal to one side of the triangle?

This is a 3-4-5 right angled triangle. Or, to be precise: a 6-8-10 triangle. Hypotenuse is 10 inches, "altitude" (or height) is 6 inches, so the base is 8 inches.

Area is: 1/2 * (6) * (8) = 24 square inches.

If you insist on defining altitude as the distance from right-angled vertex to the hypotenuse, you get the problems others have already discussed.

Edit: An altitude is at right-angle to a side, and connects a side to a vertex. In this case we have an (unspecified) base, an altitude of 6 inches, and a 10 inch hypotenuse.

If the base is horizontal, and the altitude "drops" from the end, making a right-angle, then it contacts the hypotenuse line at the very end. The right-angle required of an altitude is formed at the base end, not the hypotenuse end.

Alan Campbell
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    I'm with Alan in that: *"I had no clue that 'altitude' is apparently used in such a well-agreed way, indeed, I rather disagree that it is such a well-agreed term".* However Alan, I'd point out that it is indeed meant to be a joke. The joke only "makes sense" if one interprets it in the obvious way. (One could say, it's not for example an engineering description where we'd all assert "don't use such vague terminology".) Also if I'm not mistaken it comes from an earlier historical period (and indeed from Russia?) so it's reasonable to have to tease-out that meaning, I think. – Fattie Dec 31 '15 at 15:23
  • Indeed, for me, I would translate it differently: in translating such a let us say archaic/foreign term, one would often have to spell-it-out more. – Fattie Dec 31 '15 at 15:24
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    The question doesn't just mention "an (or the) altitude", it mentions "the hypotenuse ... and the altitude dropped onto it". There is only one altitude dropped onto the hypotenuse, and that is the one from the right-angle. – IanF1 Dec 31 '15 at 15:55
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    The formulation says “The altitude dropped onto it” (i.e. the hypotenuse) “ is”. – PJTraill Dec 31 '15 at 16:31
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    @Alan: every triangle has three altitudes, one perpendicular to each side. For a right triangle, the two legs are each other's altitudes, but the problem clearly refers to the third altitude, the one dropped onto (and hence perpendicular to) the hypotenuse, which is neither of the legs. This is completely standard terminology as far as I know. – Qiaochu Yuan Dec 31 '15 at 19:19
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    This was my interpretation, too. I had never heard the phrase 'altitude dropped onto' the hypotenuse of a right triangle and figured this was a side and it was a 6-8-10 right triangle with area 24. I certainly believe Qiaochu that this is standard terminology, but I was an American student and it's not something I ever remember hearing. – Adam Acosta Jan 01 '16 at 00:49
  • @Alan Campbell: How did you get area $=1/2(6)(8)=24=?$ The hypotenuse is clearly given $10$ in the question. Where did you get $8$ from? Why did you asssume 3-4-5 right triangle ? – Bhaskara-III Jan 01 '16 at 15:02
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    @Bhaskara-III The altitude is $6$, and in a right-angled triangle, two of the three altitudes correspond to the sides. If you know you have a right-angled triangle with hypothenuse $10$, one side $6$, the other side must be $8$. – hvd Jan 02 '16 at 08:05
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    @hvd: but it is clearly given in the question that the altitude of $6$ is drawn to the hypotenuse then why did you assume it to be a leg?? – Bhaskara-III Jan 03 '16 at 04:56
  • @Bhaskara-III There are three altitudes to every triangle. If you have a right-angle triangle, two of the altitudes are equal to the sides. Given a hypotenuse and a side, Pythagoras gives us the other side. – Alan Campbell Jan 03 '16 at 06:32
  • I'd say your answer is the only one given so far that interprets the problem statement correctly. – Mico Jan 04 '16 at 06:17
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    To reiterate it more directly: This answer is **false,** or at least has nothing to do with the author's intent. There is no ambiguity whatsoever regarding which altitude/height is meant. (It is merely slightly obfuscated, maybe by intent but this might also be happenstance.) It is the one dropped from the vertex with the right angle onto the hypotenuse. The "it" must refer to hypotenuse and the from the vertex with the right angle is redundant. (Moreover the entire story-line would make no sense were it like in this answer, but this is tangential.) – quid Jan 04 '16 at 20:13
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    @IanF1 I interpreted "it" to refer to the entire triangle, i.e., "one of the altitudes of the triangle is 6." I realize that the original phrasing is probably *supposed* to indicate that the altitude *perpendicular to the hypotenuse* is 6, but like several other people commenting, I don't remember ever hearing the "dropped onto" terminology before, so my guess was the one that made it possible for the triangle to exist. If I were a student and this were an exam, I'd simply have asked for clarification. – Kyle Strand Jan 04 '16 at 22:24
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    "In this case we have an (unspecified) base" No, it **is specified** to be the hypotenuse as it says "the altitude dropped onto it is 6 inches" and "it" cannot refer to anything but the earlier mentioned hypotenuse. – quid Jan 12 '16 at 13:37

Let $ABC$ be a right angled triangle with hypotenuse $AC=10$, base $BA=y$(say) and perpendicular $CB=x$ (say). Drop perpendicular $BD=6$ (if it exists) on $AB$. It is easy to see that $\Delta ABC\sim \Delta ADB$ then using ratios,


From last two ratios, $xy=60$

Also $x^2+y^2=10^2$ as $ABC$ is right angled at $B$.

This gives $$x^2+\left(\frac{60}{x}\right)^2=100$$ or $$x^4-100x^2+3600=0$$ which has no real roots.

Harish Chandra Rajpoot
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Nitin Uniyal
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Let $\Delta ABC$ be our triangle, $\measuredangle ACB=90^{\circ}$ and $CD$ be an altitude of the triangle.

Thus, by AM-GM $$6=CD=\sqrt{AD\cdot BD}\leq\frac{AD+BD}{2}=\frac{10}{2}=5,$$ which is a contradiction, which says this triangle does not exist.

Michael Rozenberg
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This is from a paper full of trick questions, so likely, is criticizing the American students. The crux of the problem is whether the discrepancy is noticed, and how it is handled.

In middle school, kids are taught a simplified formula, Triangle Area $\Delta = \dfrac{1}{2}\times b\times h$, or Area equals Base times Height.

In Geometry, the full version is taught,

$$\Delta = \dfrac{1}{2}\times b\times a$$ $$\implies \Delta = \dfrac{1}{2}\times \text{base (any side of triangle)} \times \text{altitude (line perpendicular to base and going to opposite vertex)}$$

The definitions of base and altitude are critical and repeatedly taught. They have been standard since the time of Euclid, but teaching the simplified version causes confusion.

Using this formula, the $30 \,\mathrm{in^2}$ area would be true for a triangle with hypotenuse of $10\,\mathrm{in^2}$ and corresponding altitude of $6\,\mathrm{in^2}$.

Kids are taught that the sides of a right triangle with hypotenuse length $10 \,\mathrm{in}$ and a side of $6 \,\mathrm{in}$ is a Pythagorian Triple, $6, 8, 10$. For this $6:8:10$ triangle, $6$ and $8$ are perpendicular and thus altitudes of each other, the Area $\Delta = \dfrac{1}{2}\times 6\times 8 = 24 \,\mathrm{in^2}$. The altitude for the hypotenuse can be found by $\Delta = 24 = \dfrac{1}{2}\times 10\times \text{altitude}$, and $\text{altitude} = 4.8 \,\mathrm{in}$.

Thus, using simple tools taught to the students, the "altitude $6$" triangle cannot be a right triangle, since the right triangle with side $6$ and Hypotenuse $10$ has a side of $6$, not the altitude to the hypotenuse. Expecting complicated proofs from secondary students is unreasonable, but applying the Pythagorean Theorem and area formulae are standard.

The typical student probably sees the $6$ and $10$ as parts of a Pythagorean Triple and "knows" it is a right triangle, then finds the obvious area without further thought. Thus, answering the question as 30 could imply sloppy work or lack of knowledge or understanding.

Alternately, it could be that the American students saw the discrepancy, and made a judgement call on how to answer a test question, assuming a typographical error.

PS. Yuan's inscribing the triangle in a circle to determine the maximum possible altitude to the hypotenuse is simple, brilliant, and uses concepts taught in Geometry!

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D. Fraser
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