Problems for Children from 5 to 15

 Author: Vladimir Igorevich Arnold
Translated from Russian to English by Vitor Goryunov and Sabir Gusein-Zade

V. I. Arnold is arguably the best Russian mathematician of the 20th century.  One of his famous quotes is the following.

A person, who had not mastered the art of the proofs in high school, is as a rule unable to distinguish correct reasoning from that which is misleading. Such people can be easily manipulated by the irresponsible politicians.

The author has written this book when Russian Parisians asked him to help their young children gain “the thinking culture” traditional to Russia.  He claimed that five year old kids solve similar problems—problems similar to problems in this book—better than pupils spoiled by coaching (Hint: tuition classes).  He also claimed that the worst in solving these simple problems are Nobel and Field price winners.

The following is the sixth problem of this book.

The hypotenuse of a right-angled triangle is 10 inches, the altitude dropped onto it is 6 inches.  Find the area of the triangle.

Well … any 8th grader should know that the area of a triangle is half of the product of the base and the height.  So the area is \(\frac{1}{2}\cdot 10\cdot 6=30\) square inches.  Right?

Analysis: Method 1

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I learned the following theorem when I was in the 8th grade.

Theorem: If the roots of a quadratic equation \(x^2+bx+c=0\) are \(r_1\) and \(r_2\), then \(r_1+r_2=-b\) and \(r_1r_2=c\), and vice versa.

I later learned that this theorem is known as the "Viete's theorem".

(I was fascinated by this result and what it can do as a child.  There were many applications of this theorem in grades 8 - 12.)

The following is a quick proof of the Viete's theorem.

\(r_1\) and \(r_2\) are roots of the quadratic equation

\(\Longleftrightarrow (x-r_1)(x-r_2)=0.\)

\(\Longleftrightarrow x^2-r_2x-r_1x+r_1r_2=0.\)

The above result follows from the distributive law.

\(\Longleftrightarrow x^2-(r_1+r_2)x+r_1r_2=0.\)

\(\Longleftrightarrow (r_1+r_2)=-b\) and \(r_1r_2=c.\)

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Now let us revisit the above problem.

Let the lengths of the segments of the hypotenuse partitioned by the foot of the altitude be \(r_1\) and \(r_2\) as shown in the following figure.

Then \(r_1+r_2=10\) and \(r_1r_2=36\).  (The explanation of the second result is given below.)*

Then by the Viete's theorem, \(r_1\) and \(r_2\) are the roots of the quadratic equation \(x^2-10x+36=0\).

The discriminant** of this equation is \(10^2-4\cdot 36\), and the discriminant is negative.  That means that this equation cannot have real solutions.  

That is, there is no such triangle.

In other words, if the hypotenuse of a right triangle is 10 units, then the hight cannot be 6 units.  This is a frivolous question.

The height of a right triangle with hypotenuse 10 units must have a height at most \(5\) units.  (If the height is \(5\) then the right triangle is an isosceles triangle.)

Footnote * 

The two right triangles \(\triangle ACD\) and \(\triangle CBD\) are similar triangles, by the \(AA\) criterion.  [By using the cyclic order, \(\angle A\) of the triangle \(\triangle ACD\) is congruent to \(\angle C\) of the triangle \(\triangle CBD\)].  

Then \(\frac{CD}{DB}=\frac{AD}{CD}\), by the similar triangles theorem.  That is, \(\frac{6}{r_2}=\frac{r_1}{6}\).  Then, by the cross-multiplication algorithm, \(r_1r_2=36\).

Footnote ** The discriminant \(\Delta\) of a quadratic equation \(ax^2+bx+c=0\) is \(b^2-4ac\).

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Analysis: Method 2

The hypotenuse is a diameter of the circumcircle of the triangle.  (This is a theorem I learned in the 7th grade geometry.)  That is, the radius of the circumcircle is 5 units.  See the following figure.

You can clearly see that the altitude must be less than or equal to the radius.  (There are several theorems that you can use here.  One such theorem is the Pythagorean theorem.)

So, the altitude of the right triangle must be at most 5 units and stating that the altitude is 6 is nonsensical.