# V.I. Arnold says Russian students can't solve this problem, but American students can — why?

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?

• 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 5:46
• 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 5:51
• @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 5:59
• 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 6:03
• 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

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.

• 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 5:55
• What about non euclidean geometry (I may be overanalyzing, but overanalyzing is fun)? – PyRulez Dec 31 '15 at 20:56
• @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 1 '16 at 21:32
• @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 4 '16 at 20:48
• 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 5 '16 at 0:38

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

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.

• how did you generate such a nice diagram? – Fattie Dec 31 '15 at 15:17
• @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
• 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
• @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
• 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 1 '16 at 16:46

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)

• 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 4 '16 at 6:15
• @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 4 '16 at 6:23
• @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 4 '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 4 '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 4 '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.

• "A right triangle with hypotenuse 100 & an altitude of 6" I guess there's a typo and should say 10. Great explanation though. – Masclins Dec 31 '15 at 7:50
• An easier proof would be to draw a Thales' circle around the hypotenuse and note that it has no point 6 inches away from the hypotenuse. – John Dvorak Dec 31 '15 at 10:20
• An easier proof would be to note that $a = \sqrt{x^2+y^2}$ and height/hypotenuse $= \frac{xy}{x^2+y^2} \le \frac{1}{2}$ by AM GM inequality. – Ishan Banerjee Dec 31 '15 at 12:23
• Calculus is a bit overkill for such a simple proof, see @Mick answer – mattecapu Dec 31 '15 at 13:53
• Calculus is used just to show the result analytically. – Harish Chandra Rajpoot Jan 1 '16 at 19:33

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:

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

$\blacksquare$

• This has got to be the correct answer. Arnold's finest joke. – PatrickT Aug 27 '16 at 7:21
• i have edited a little thanks – Bhaskara-III Sep 21 '16 at 10:33
• It is one of several different possible non-Euclidian answers. – Klaws Jan 26 '18 at 16:07

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.

• 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
• 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
• The formulation says “The altitude dropped onto it” (i.e. the hypotenuse) “ is”. – PJTraill Dec 31 '15 at 16:31
• @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
• 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 1 '16 at 0:49

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,

$$\frac{z}{y}=\frac{6}{x}=\frac{y}{10}$$

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.

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.

• nice answer..... +1 – user808376 Jul 24 '20 at 4:37

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!

• It would be better to not use abbreviations so that your answer is clearer. Also MathJax. – 6005 Dec 31 '15 at 6:55
• There are many right triangles with hypotenuse $10$ but the legs not $6$, $8$. – user236182 Dec 31 '15 at 8:57
• @user but none of them have integer-sized legs – John Dvorak Dec 31 '15 at 12:01
• @JanDvorak the question doesn't mention integer-sized legs. – IanF1 Dec 31 '15 at 15:56
• @ian indeed, but integer-sized legs are what most students expect. Not valid reasoning, but some students might have followed it anyways. – John Dvorak Dec 31 '15 at 15:59