A soldier needs to check if there are any mines

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A soldier needs to check if there are any mines

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Source: IMO ShortList 1973, Yugoslavia 2, IMO 1973, Day 2, Problem 1

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orl

#1 h Oct 29, 2005, 10:17 PM • 1 Y

Y by Adventure10

A soldier needs to check if there are any mines in the interior or on the sides of an equilateral triangle $ABC.$ His detector can detect a mine at a maximum distance equal to half the height of the triangle. The soldier leaves from one of the vertices of the triangle. Which is the minimum distance that he needs to traverse so that at the end of it he is sure that he completed successfully his mission?

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#2 h Oct 30, 2005, 5:02 AM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ the triangle, with side $a$ . Let $h=a\frac{\sqrt{3}}{4}$ the half altitude

The soldier starts moving from $A$ .

Let $D,E,F$ the midpoints of $AB,BC,CA$ respectively.

If he moves on the segment $DE$ (after $A\to D$ ), then he can reach the most points of the area but he will lose the vertice $C$ .

The problem is of course at the vertices.

We construct the triangle $DEF$ and let $K,L$ the midpoints of $DE,EF$ .

From $K$ , he can reach the point $B$ since $BK=h$ .
Also, from $L$ he can reach the point $C$ .

If he could start from $K$ , then the better way would be $K\to L\to M$ where $M$ is the midpoint of $DF$ .
Actually, each of these points covers a large percentage of the triangle, for example the circle $(K,h)$ includes the triangle $BDE,DEF$ and much more area.

By the way, if could "jump" from $K$ to $L$ and then to $M$ , the area would be covered perfectly.

I think that, since he must start from $A$ , the most convinience way is to get first to the point $K$ and then go to the point $L$

The vertical distance from $A$ to $K$ is $2h-\frac{h}{2}=\frac{3h}{2}$

The horizontal distance is $\frac{a}{8}$

$AK^2 = (\frac{3h}{2})^2+(\frac{a}{8})^2 =$

$(\frac{3a\sqrt{3}}{8})^2 + (\frac{a}{8})^2 =$

$= a^2 \frac{9 \cdot 3 +1}{8^2}=$

$= a^2 \frac{7}{16}\Rightarrow$

$AK = a\frac{\sqrt{7}}{4}$ and $KL = \frac{a}{4}$

$AK+KL = a\frac{\sqrt{7}+1}{4}$

I don't know if this is the minimum or if this is the better way to face the problem. At least I tried

Maybe it should be easier using calculus?

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Flakky

#3 h May 31, 2008, 6:36 PM • 2 Y

Y by Adventure10, Mango247

This solution is wrong!
Check http://www.mat.itu.edu.tr/gungor/IMO/http://www.kalva.demon.co.uk/imo/isoln/isoln734.html
The answer is $a\frac {2\sqrt {7} - \sqrt {3}}{4}$ which is less then $a\frac {\sqrt {7} + 1}{4}$

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77ant

#4 h May 31, 2008, 7:43 PM • 2 Y

Y by Adventure10, Mango247

Geometric Problems on MAXIMA and MINIMA (by Author "Titu Andreescu")

"Let h be the length of the altitude of the given equilateral triangle ABC. Assume that the soldier's path starts at the point A. Consider the circles k1 and k2 with centers B and C, respectively, both with radius h/2. In order to check the points B and C, the soldier's path must have common points with both k1 and k2. Assume that the total length of the path is t and it has a common point M with k2 first and then a common point N with k1. Denote by D the common point of k2 and the altitude through C in triangle ABC and by L the line through D parallel to AB. Adding the constant h/2 to t and using the triangle inequality, one gets t+(h/2)≥AM+MN+NB=AM+MP+PN+NB≥AP+PB, where P is the intersection point MN and L. On the other hand, Heron's problem shows that AP+PB≥AD+DB, where equality occurs precisely when P=D. This implies t+(h/2)≥AD+DB,
i.e. , t≥AD+DE, where E is the point of intersection of DB and k1."

⇒shortest path length=a×{2×(√7/4) - (√3/4)}, where a is the length of equilateral triangle ABC.

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Soski

#5 h Dec 20, 2021, 11:52 PM

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Let d be the distance from any of the vetices to the mid point of the triangle's altitude. Let h be the altitude of the equilateral triangle. In order to do his task, the soldier must travel the distance d plus the diference (d - h/2). Hence:

displacement = d + d - h/2 ~ 0.8899 units of length,

as 77ant suggested

This post has been edited 1 time. Last edited by Soski, Dec 20, 2021, 11:53 PM
Reason: typing mistake

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#6 h Jan 2, 2022, 9:04 AM

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Let h be the altitude of triangle ABC and t be the radius of the detector, $r=\frac{h}{2}$

Let the soldier start from A, so the critical points to detect are B and C --> the paths to be taken are $A\to D$ and $D\to E$ , where D is a point with the distance $r=\frac{h}{2}$ from B and E is a point with the distance $r=\frac{h}{2}$ from C.

The length of the paths is $s=\|A\to D\|+\|D\to E\|$

Let's imagine the soldier is already on D (although we don't know where). To take the shortest path from D to E, the soldier must walk in the direction to C
--> E is on the line DC. So $\|D\to E\|=\|D\to C\| - \|E\to C\|=\|D\to C\| - \frac{h}{2}$

--> $s=\|A\to D\|+\|D\to C\|-r$

s is minimal, if and only if $\|A\to D\|+\|D\to C\|$ is minimal. $\|A\to D\|+\|D\to C\|$ is minimal, if and only if $\|A\to D\|=\|D\to C\|$ .
--> D must be on the line from B to F, where F is the midpoint of line AC.

Because BF is an altitude of triangle ABC, so D must be the midpoint of line BF.

We can calculate s as a function of h, but it's not the answer from the question. The correct answer is:
The soldier must walk from A to D and then to E. D is the midpoint of line BF (the altitude of the triangle ABC from B). E is on line DC with $EC=\frac{h}{2}$

This post has been edited 2 times. Last edited by andib2n, Jan 2, 2022, 9:06 AM
Reason: typo

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