A truck has had new brake linings fitted to all four of its wheels. A lining is considered completely worn out if it has traveled $15000 \mathrm{~km}$ on a rear wheel, or $25000 \mathrm{~km}$ on a front wheel. How far can the truck travel until all four linings are completely worn out, if at the appropriate time the front lining pair is swapped with the rear pair?
I. solution. Consider the state of the tires after, for example, $1000 \mathrm{~km}$ of travel. The rear tires have worn out by $11/5$ of their part, and the front tires by $1/25$ of their part. (We assume that the car travels on an average quality road, and the wear is directly proportional to the distance traveled.) If we now swap the tires, then after another $1000 \mathrm{~km}$ of travel, each of the four tires will have traveled an equal distance and can be considered worn out by $1/15 + 1/25 = 8/75$ of their part. The tires can travel $2000 \mathrm{~km}$ more times the ratio of the total wear to the wear that has already occurred, which is $1$ to the $8/75$ wear ratio. Therefore, the car can travel $2000 \cdot 75 / 8 = 18750 \mathrm{~km}$ before the tires are completely worn out, with half of this distance covered in the original arrangement and the other half with the swapped tires.

László Jereb (Sopron, Széchenyi I. Gymnasium II. grade)

Note. Similarly, it can be shown that if a tire wears out after $a \mathrm{~km}$ on one wheel and $b \mathrm{~km}$ on the other, then with the above swap, the distance that can be traveled with four tires is

$$
\frac{1}{\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)}
$$

This value is known as the harmonic mean of the two numbers.

II. solution. The ratio of the allowed kilometers is $15000: 25000 = 3: 5$, so if a tire is moved from the rear wheel to the front wheel, the distance it can still travel increases by a factor of $5/3$. If it is moved from the front to the rear, it decreases by a factor of $3/5$.

If the car has already traveled $x$ km with the tires, then the front tires could still travel $25000 - x$ km, but after the swap, they can only travel $\frac{3}{5}(25000 - x)$ km, while the rear tires, instead of the remaining $15000 - x$ km, can still be used for $\frac{5}{3}(15000 - x)$ km. We need to determine $x$ such that the distance the two types of tires can still travel is equal:

$$
\frac{3}{5}(25000 - x) = \frac{5}{3}(15000 - x)
$$

From this, $x = 9375 \mathrm{~km}$, and the distance traveled after the swap is $3 \cdot 15625 / 5 = 9375 \mathrm{~km}$, so the car can travel $18750 \mathrm{~km}$.

Béla Pénzes (Esztergom, Temesvári Pelbárt Gymnasium II. grade)

Note. We can also use the fact that the distance traveled before and after the swap is the same, otherwise, by the time the rear tire is completely worn out, the other tire would still be usable. This leads to the somewhat simpler equation $3(25000 - x) / 5 = x$.

![](https://cdn.mathpix.com/cropped/2024_05_02_c7e6978c848383369babg-1.jpg?height=686&width=900&top_left_y=1473&top_left_x=602)

III. solution. Viewing a new tire as $100 \%$ and a completely worn-out tire as $0 \%$, we will illustrate how the value state of the tires changes with the distance traveled. The value graph for the two types of tires from the initial setup is obtained by connecting the point $A$ on the value axis marked $100 \%$ with a straight line to the point on the distance axis marked $15000 \mathrm{~km}$ or $25000 \mathrm{~km}$. When the tires are swapped, the slopes of their graphs also swap, and the four graph segments corresponding to the two distance segments form a parallelogram, which we want to construct.

The two types of tires reach the $0 \%$ value state after traveling the same distance, so the two graphs intersect on the distance axis, which will be the point $C$ opposite $A$ in the parallelogram. The other two vertices $B$ and $D$ correspond to the tire swap, so the two vertices are above the same point on the distance axis, meaning we know the direction of the $BD$ diagonal: it is parallel to the value axis. From this, we can determine the position of the other diagonal. The segments of the lines parallel to the $BD$ diagonal that lie in the $BAD$ angle region are bisected by the $AC$ diagonal. Thus, taking any line parallel to the value axis, the segment $B^*D^*$ between the two graphs that is bisected by $F^*$ is on the $AC$ diagonal, so $AF^*$ intersects the distance axis at $C$, and the corresponding abscissa value is the sought distance.

Notes. 1. To answer the question of the problem, it is not necessary to construct $B$ and $D$, but to carry out the tire swap, it is precisely this that needs to be known, at which position of the odometer the tire swap should take place.

2. This solution is essentially a graphical version of the first solution. The similarity would be even more evident if we completed the $D^*AB^*$ triangle to a parallelogram and drew the diagonal through the point $C^*$, which would correspond to the diagonal.

3. The two diagonals bisect each other, so their intersection point is on a line parallel to the distance axis at the height corresponding to $50 \%$. If we reflect one wheel's value graph over this line, the reflection intersects the other graph at the point corresponding to the tire swap.

We can also arrive at this directly by reasoning, since after the swap, one tire will wear out as much as the other has already worn out, so its usage is equal to the other's value, and the reflected graph illustrates the wear (value decrease) of the wheel (2nd figure).

(Using an unnamed, textless graph)

![](https://cdn.mathpix.com/cropped/2024_05_02_c7e6978c848383369babg-2.jpg?height=631&width=681&top_left_y=693&top_left_x=711)