Geometry Problem

MathStudent1999

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Mar 18, 2012
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17. Define T to be a triangle with sides 3,4 & 5. Define D(T,p) to be the sum of the squares of the distances from the point p to the vertices of triangle T. What is the least value of D(,T,p)? Express your answer as a common fraction.

Can someone give me a few hint and tips on this question?
 
You can define your task well with an analytic approach. Orient your triangle so the vertex at the right angle is at the Origin. Did you observe it was a right triangle. Put the other two vertices at (4,0) and (0,3).

Now, it's easy to check the total distance from any vertex:

(0,0) ==> 3^2 + 4^2 = 25
(4,0) ==> 4^2 + 5^2 = 41
(0,3) ==> 3^2 + 5^2 = 34

(0,0) wins the vertex competition.

Okay, it's edge time.

Anywhere on the x-axis, say (a,0) we have 4^2 + (a^2 + 3^2) = 25 + a^2 < 25 + 16 = 41
Minimizing this, we get a = 0, which is back at the Origin.

Anywhere on the y-axis, say (0,b) we have 3^2 + (b^2 + 4^2) = 25 + b^2 < 25 + 9 = 34
Minimizing this, we get b = 0, which is back at the Origin.

Anywhere on the hypotenuse, say (r,s) we have 5^2 + (r^2 + s^2) = 25 + r^2 + s^2 < 25 + 16 + 9 = 50
Minimizing this, we get r = 0 and s = 3, which gives 34, but we already know that.

Lesson learned. If we wander along the edges, we get greater values than at the vertices!! Wow! I smell a theorem!!!

Okay, what happens if we wander into the interior??? Really big hint: In may be similar to the hypotenuse except that we are not constrained to that line.

Note: This is called EXPLORATION. One needn't know how to proceed in order to get started. :)
 
The end: What is the least value of D(,T,p)? Express your answer as a common fraction. How do I express as a common fraction when it is cordinates? Please give examples.
 
The "inside job"!:
Code:
B(0,3)
 



         P(u,v)


C(0,0)                    A(4,0)
CP = sqrt(u^2 + v^2)
AP = sqrt[v^2 + (4 - u)^2]
BP = sqrt[u^2 + (3 - v)^2]

Wonder why question is to find minimum sum of AP^2 + BP^2 + CP^2...
I make point P at (4/3,1), so sum = 16 2/3
Agree, TK?

The "well known" one is minimum sum of AP + BP + CP; see here (7 solutions!):
http://www.cut-the-knot.org/Generalization/fermat_point.shtml

Instead of getting (4/3, 1) i got (1, 4/3)! I figured that the where the three medians of the triangle intersect would be the point where the squares pf the distance to the vertices were the smallest. So I got the equation of two of the three lines, and found the intersection point was (1, 4/3). The equations for the two lines I used were y=4x/3 and y=-8x/3 + 4. The I calculated the distance from the point to the three vertices. I then squared the three answers, added them up and got 44/3! Not 50/3. Is my answer correct or yours? Can you check?

Can you also help me on these questions?
http://www.freemathhelp.com/forum/threads/75058-Probability-Problems
 
By the way, sum of squares of shortest distances is not necessarily the minimum sum:
take 2,6,8: sum = 16, sum squares = 104
take 4,6,7: sum = 17, sum squares = 101
...get my drift?

Triangle sides being a,b,c and inside P coordinates being (u,v), then sum squares
= u^2+v^2 + u^2+a^2-2av+v^2 + v^2+b^2-2bu+u^2
= 3u^2 + 3v^2 - 2bu - 2av + a^2 + b^2

Setting to zero; 1st derivative:
6u + 6v - 2b - 2a = 0
3(u + v) = a + b

I had u = 4/3 and v = 1;
3(4/3 + 1) = 3 + 4 ?
7 = 7

edit: I got 50/3. Thanks! :)
 
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