Difference between revisions of "1973 IMO Problems/Problem 4"

Revision as of 18:36, 29 June 2025

Problem

A soldier needs to check on the presence of mines in a region having the shape of an equilateral triangle. The radius of action of his detector is equal to half the altitude of the triangle. The soldier leaves from one vertex of the triangle. What path shouid he follow in order to travel the least possible distance and still accomplish his mission?

Solution

Let our triangle be $\triangle ABC$ , let the midpoint of $AB$ be $D$ , and let the midpoint of $CD$ be $E$ . Let the height of the triangle be $h$ . Draw circles around points $B$ and $C$ with radius $\frac{h}{2}$ , and label them $c_1$ and $c_2$ . Let the intersection of $BE$ and $c_1$ be $F$ .

The path that is the solution to this problem must go from $A$ to a point on $c_2$ to a point on $c_1$ . Let us first find the shortest possible path. We will then prove that this path fits the requirements.

The shortest path is in fact the path from $A$ to $E$ to $F$ . We will prove this as follows:

Suppose that a different point on $c_2$ is the optimal point to go to. Let this point be $P$ . Then, the optimal point on $c_1$ would be the intersection of $BP$ and $c_1$ (let this point be $R$ ). Let the line through $E$ parallel to $AB$ be $l$ , and the intersection of $AP$ and $l$ be $Q$ . Then, we have

$AE + BE < AQ + BQ \leq AQ + QP + BP = AP + BP.\textbf{ (1)}$

The second inequality is true due to the triangle inequality, and we can prove the first to be true as follows:

Suppose we have a point on $l$ , $X$ . Reflect $B$ across $l$ to get $B'$ . Then, $AX + BX = AX + B'X$ , which is obviously minimized at the intersection of the two lines, $E$ .

By $\textbf{(1)}$ , we have

$AE + EF = AE + BE - \frac{h}{2} < AP + BP - \frac{h}{2} = AP + PR,$

and we have proved the claim.

This path easily covers the whole triangle. This is because if you draw a line perpendicular to $AB$ from any point in the triangle, this line will hit the path in a distance less than or equal to $\frac{h}{2}$ . We can compute the length of the path to be

$\left(\sqrt{\frac{7}{3}} - \frac{1}{2}\right)h$ $= \left(\frac{\sqrt{7}}{2} - \frac{\sqrt{3}}{4}\right)\cdot\textrm{the side length of the triangle. }\square$

~mathboy100

Remarks (added by pf02, June 2025)

The solution given above is so incomplete (and incorrect in a few details) that it can not be called a solution.

Below I will first point out the shortcomings of the solution. Then I will attempt to give a solution. In fact, I don't have a solution, but I will argue (along the lines of the solution above, but filling in many details, and correcting others) that the path described above is most likely minimal. I will highlight the points where my attempted solution fails to be rigorous.

1. The author states: "The path that is the solution to this problem must go from $A$ to a point on $\mathcal{C}_2$ to a point on $\mathcal{C}_1$ ." This needs a proof. I believe this to be true (up to a symmetry), but it is the hardest part of the solution.

2. In the solution above, the author makes the following construction: take $P$ on the arc of $\mathcal{C}_2$ inside the triangle, and take the point $R$ where $PB$ intersects $\mathcal{C}_1$ ( $PR$ is the normal from $P$ to $\mathcal{C}_1$ ). He shows that the length of the curve $APR$ is minimal when $P = E$ , where $E$ is the midpoint of the arch (also, the intersection of the arc with the height from $C$ , and also, the midpoint of the parallel $D_aD_b$ to $AB$ , going through the midpoints $D_a, D_b$ of $BC, AC$ ). (Denote by $F$ the point $R$ when $P = E$ .)

Then, the author argues that the union of circles of radius $\frac{h}{2}$ centered at all the points along $AEF$ covers the triangle. (In fact, the argument is incorrect, but the fact is true.)

This is all good, but the two statements above do not prove that $AEF$ is the shortest curve such that the union of circles of radius $\frac{h}{2}$ centered at all the points along the curve covers the triangle. In fact, they don't even prove the much easier problem we obtain if we accept that "the path that is the solution to this problem must go from $A$ to a point on $\mathcal{C}_2$ to a point on $\mathcal{C}_1$ ."

3. The author says

"This path [ $AEF$ ] easily covers the whole triangle. This is because if you draw a line perpendicular to $AB$ from any point in the triangle, this line will hit the path in a distance less than or equal to $\frac{h}{2}$ ."

In fact it is not true that "if you draw a line perpendicular to $AB$ from any point in the triangle, this line will hit the path in a distance less than or equal to $\frac{h}{2}$ ." Clearly, there are points $T$ such that a perpendicular from $T$ to $AB$ will not hit the path $AEF$ at all. Besides, this is not what we need in order to conclude that the union of circles of radius $\frac{h}{2}$ centered at all the points along $AEF$ covers the triangle.

Solution (attempted)

Let us say that a curve has the property $\mathcal{P}$ if the union of circles of radius $\frac{h}{2}$ centered at all the points along the curve covers the triangle. The problem asks us to find the shortest curve with one end at $A$ satisfying $\mathcal{P}$ .

In this attempted solution when we say that a point is in a region, we mean that the point is in the interior, or on the boundary of the region. Also, note that we will use the words curve and path interchangeably.

TO BE CONTINUED.

Template:Solutions

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	== See Also == {{IMO box\|year=1973\|num-b=3\|num-a=5}}		== See Also == {{IMO box\|year=1973\|num-b=3\|num-a=5}}

1973 IMO (Problems) • Resources
Preceded by Problem 3	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 5
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