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

Question

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

6. 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?

Answer 1

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.

Answer 2

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.

Answer 3

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)

Answer 4

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.

Answer 5

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:

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

$\blacksquare$

Answer 6

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.

Answer 7

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.

Answer 8

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.

Answer 9

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.

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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!