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Difference between revisions of "1973 IMO Problems/Problem 4"
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The shortest path is in fact the path from <math>A</math> to <math>E</math> to <math>F</math>. We will prove this as follows:
 
The shortest path is in fact the path from <math>A</math> to <math>E</math> to <math>F</math>. We will prove this as follows:
  
−
Suppose that a different point on <math>c_2</math> is the optimal point to go to. Let this point be <math>P</math>. Then, the optimal point on <math>c_1</math> would be the intersection of <math>BP</math> and <math>c_1</math>. Let the line through <math>E</math> parallel to <math>AB</math> be <math>l</math>, and the intersection of <math>AP</math> and <math>l</math> be <math>Q</math>. Then, we have
+
Suppose that a different point on <math>c_2</math> is the optimal point to go to. Let this point be <math>P</math>. Then, the optimal point on <math>c_1</math> would be the intersection of <math>BP</math> and <math>c_1</math> (let this point be <math>R</math>). Let the line through <math>E</math> parallel to <math>AB</math> be <math>l</math>, and the intersection of <math>AP</math> and <math>l</math> be <math>Q</math>. Then, we have
  
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<cmath>AE + BE \leq AQ + BQ \leq AQ + QP + BP = AP + BP.\label{(1)}</cmath>
+
<cmath>AE + BE < AQ + BQ \leq AQ + QP + BP = AP + BP.\textbf{ (1)}</cmath>
  
 
The second inequality is true due to the triangle inequality, and we can prove the first to be true as follows:
 
The second inequality is true due to the triangle inequality, and we can prove the first to be true as follows:
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Suppose we have a point on <math>l</math>, <math>X</math>. Reflect <math>B</math> across <math>l</math> to get <math>B'</math>. Then, <math>AX + BX = AX + B'X</math>, which is obviously minimized at the intersection of the two lines, <math>E</math>.  
 
Suppose we have a point on <math>l</math>, <math>X</math>. Reflect <math>B</math> across <math>l</math> to get <math>B'</math>. Then, <math>AX + BX = AX + B'X</math>, which is obviously minimized at the intersection of the two lines, <math>E</math>.  
  
 +
By <math>\textbf{(1)}</math>, we have
  
 +
<cmath>AE + EF = AE + BE - \frac{h}{2} < AP + BP - \frac{h}{2} = AP + PR, </cmath>
 +
 +
and we have proved the claim.
 +
 +
This path easily covers the whole triangle. This is because if you draw a line perpendicular to <math>AB</math> from any point in the triangle, this line will hit the path in a distance less than or equal to <math>\frac{h}{2}</math>. We can compute the length of the path to be
 +
 +
<cmath>\left(\sqrt{\frac{7}{3}} - \frac{1}{2}\right)h</cmath>
 +
<cmath> = \left(\frac{\sqrt{7}}{2} - \frac{\sqrt{3}}{4}\right)\cdot\textrm{the side length of the triangle. }\square</cmath>
 +
 +
~mathboy100
  
 
== See Also == {{IMO box|year=1973|num-b=3|num-a=5}}
 
== See Also == {{IMO box|year=1973|num-b=3|num-a=5}}

Latest revision as of 22:48, 12 December 2022

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

See Also

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