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:


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?
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.
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!