Dr.Math -- the minimum possible value -- problems of the week
바로 콕 찍어서 - 진너자하/詩처럼아름다운數學 2007/05/27 17:52A triangle has sides 2, 3, and 4 feet.
Find the minimum possible value
for the sum of the distances from any point
to the three vertices of the triangle.
[SECRET]
Q:거리를 뜻하는 영어단어는?
A:distance
[solution]
The answer is ((29+9(5^(1/2)))/2)^(1/2) = 4.9560 feet approximately.
Here we use ^(1/2) for square root.
The details are difficult, especially in HTML.
Let the distances be a, b, c as shown, with S their sum.
Consider the sum of the unit vectors from the middle point to the three vertices.
For minimum S
this sum must be zero
since otherwise we can decrease S by moving opposite to the sum of the unit vectors.
An alternative explanation can be given by considering the gradient of S,
which is precisely the opposite of this sum of unit vectors
since the gradient of (e.g.) a is a unit vector away from the corresponding vertex.
The only way three unit vectors can add to zero is
for them to be sides of an equilateral triangle.
As a result of this the unit vectors
(and the distances a, b, c) must make angles of 120 degrees with each other.
Now the law of cosines gives the equations 4=b^2+c^2+bc (where ^2 means squaring),
9=a^2+b^2+ab, and 16=a^2+c^2+ac.
Adding these equations gives 29=2S^2-3(ab+ac+bc).
Also the differences between the three equations
give three new equations 5=a^2-c^2+(a-c)b=(a-c)S, 7=c^2-b^2+(c-b)a=(c-b)S,
and 12=a^2-b^2+(a-b)c=(a-b)S.
Squaring these three and adding gives 218=S^2(2S^2-6(ab+ac+bc)).
Eliminating ab+ac+bc with the earlier equation gives S^4-29S^2+109=0.
This may be solved by the quadratic formula.
[/SECRET]
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